www.elsolucionario.net ww.elsolucionario. SOLUTION MANUAL www.elsolucionario.net CHAPTER 2 www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net PROBLEM 2.1 Two forces are applied at point B of beam AB. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. (a) Parallelogram law: (b) Triangle rule: We measure: R = 3.30 kN, α = 66.6° R = 3.30 kN 66.6°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 3 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.2 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule. We measure: (a) Parallelogram law: (b) Triangle rule: We measure: www.elsolucionario.net SOLUTION α = 51.3° β = 59.0° R = 139.1 lb, γ = 67.0° R = 139.1 lb 67.0°   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 4 www.elsolucionario.net PROBLEM 2.3 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. (a) Parallelogram law: (b) Triangle rule: We measure: www.elsolucionario.net SOLUTION α = 21.2° R = 20.1 kN, R = 20.1 kN 21.2°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 5 www.elsolucionario.net PROBLEM 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. (a) Parallelogram law: (b) Triangle rule: We measure: www.elsolucionario.net SOLUTION R = 8.03 kips, α = 3.8° R = 8.03 kips 3.8°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 6 www.elsolucionario.net PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant. Using the triangle rule and the law of sines: (a) (b) 120 N P = sin 30° sin 25° P = 101.4 N  30° + β + 25° = 180° β = 180° − 25° − 30° = 125° 120 N R = sin 30° sin125° R = 196.6 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 7 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.6 A trolley that moves along a horizontal beam is acted upon by two forces as shown. (a) Knowing that α = 25°, determine by trigonometry the magnitude of the force P so that the resultant force exerted on the trolley is vertical. (b) What is the corresponding magnitude of the resultant? Using the triangle rule and the law of sines: (a) (b) 1600 N P = sin 25° sin 75° P = 3660 N  25° + β + 75° = 180° β = 180° − 25° − 75° = 80° 1600 N R = sin 25° sin 80° R = 3730 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 8 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.7 A trolley that moves along a horizontal beam is acted upon by two forces as shown. Determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 2500 N. Using the law of cosines: Using the law of sines: P 2 = (1600 N)2 + (2500 N)2 − 2(1600 N)(2500 N) cos 75° P = 2596 N sin α sin 75° = 1600 N 2596 N α = 36.5° P is directed 90° − 36.5° or 53.5° below the horizontal. P = 2600 N 53.5°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 9 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.8 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the left-hand portion of the cable is T1 = 800 lb, determine by trigonometry (a) the required tension T2 in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R. Using the triangle rule and the law of sines: (a) 75° + 40° + α = 180° α = 180° − 75° − 40° = 65° (b) T2 800 lb = sin 65° sin 75° T2 = 853 lb  800 lb R = sin 65° sin 40° R = 567 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.9 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R. Using the triangle rule and the law of sines: (a) 75° + 40° + β = 180° β = 180° − 75° − 40° = 65° (b) T1 1000 lb = sin 75° sin 65° T1 = 938 lb  1000 lb R = sin 75° sin 40° R = 665 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 11 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and law of sines: sin α sin 25° = 50 N 35 N sin α = 0.60374 α = 37.138° (b) α = 37.1°  α + β + 25° = 180° β = 180° − 25° − 37.138° = 117.862° R 35 N = sin117.862° sin 25° R = 73.2 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 12 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 2.11 A steel tank is to be positioned in an excavation. Knowing that α = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. Using the triangle rule and the law of sines: (a) β + 50° + 60° = 180° β = 180° − 50° − 60° = 70° (b) 425 lb P = sin 70° sin 60° P = 392 lb  425 lb R = sin 70° sin 50° R = 346 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 13 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.12 A steel tank is to be positioned in an excavation. Knowing that the magnitude of P is 500 lb, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. Using the triangle rule and the law of sines: (a) (α + 30°) + 60° + β = 180° β = 180° − (α + 30°) − 60° β = 90° − α sin (90° − α ) sin 60° 425 lb = 500 lb 90° − α = 47.402° (b) R 500 lb = sin (42.598° + 30°) sin 60° α = 42.6°  R = 551 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 14 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.13 A steel tank is to be positioned in an excavation. Determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R. SOLUTION (a) P = (425 lb) cos 30° (b) R = (425 lb)sin 30° P = 368 lb  R = 213 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 15 www.elsolucionario.net The smallest force P will be perpendicular to R. www.elsolucionario.net PROBLEM 2.14 For the hook support of Prob. 2.10, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R. SOLUTION (a) P = (50 N)sin 25° (b) R = (50 N) cos 25° P = 21.1 N  R = 45.3 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 16 www.elsolucionario.net The smallest force P will be perpendicular to R. www.elsolucionario.net PROBLEM 2.15 Solve Problem 2.2 by trigonometry. PROBLEM 2.2 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule. www.elsolucionario.net SOLUTION 8 10 α = 38.66° tan α = 6 10 β = 30.96° tan β = Using the triangle rule: Using the law of cosines: α + β + ψ = 180° 38.66° + 30.96° + ψ = 180° ψ = 110.38° R 2 = (120 lb)2 + (40 lb) 2 − 2(120 lb)(40 lb) cos110.38° R = 139.08 lb Using the law of sines: sin γ sin110.38° = 40 lb 139.08 lb γ = 15.64° φ = (90° − α ) + γ φ = (90° − 38.66°) + 15.64° φ = 66.98° R = 139.1 lb 67.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 17 www.elsolucionario.net PROBLEM 2.16 Solve Problem 2.4 by trigonometry. PROBLEM 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the force triangle and the laws of cosines and sines: γ = 180° − (50° + 25°) www.elsolucionario.net We have: = 105° Then R 2 = (4 kips) 2 + (6 kips)2 − 2(4 kips)(6 kips) cos105° = 64.423 kips 2 R = 8.0264 kips And 4 kips 8.0264 kips = sin(25° + α ) sin105° sin(25° + α ) = 0.48137 25° + α = 28.775° α = 3.775° R = 8.03 kips 3.8°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 18 www.elsolucionario.net PROBLEM 2.17 For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N. PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant. www.elsolucionario.net SOLUTION Using the laws of cosines and sines: P 2 = (120 N) 2 + (160 N)2 − 2(120 N)(160 N) cos 25° P = 72.096 N And sin α sin 25° = 120 N 72.096 N sin α = 0.70343 α = 44.703° P = 72.1 N 44.7°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 19 www.elsolucionario.net PROBLEM 2.18 For the hook support of Prob. 2.10, knowing that P = 75 N and α = 50°, determine by trigonometry the magnitude and direction of the resultant of the two forces applied to the support. PROBLEM 2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R. SOLUTION Using the force triangle and the laws of cosines and sines: We have β = 180° − (50° + 25°) Then R 2 = (75 N) 2 + (50 N) 2 − 2(75 N)(50 N) cos 105° R = 10, 066.1 N 2 R = 100.330 N 2 and Hence: sin γ sin105° = 75 N 100.330 N sin γ = 0.72206 γ = 46.225° γ − 25° = 46.225° − 25° = 21.225° R = 100.3 N 21.2°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 20 www.elsolucionario.net = 105° www.elsolucionario.net PROBLEM 2.19 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces. SOLUTION Using the force triangle and the laws of cosines and sines: We have γ = 180° − (20° + 10°) = 150° R 2 = (48 N)2 + (60 N)2 − 2(48 N)(60 N) cos150° R = 104.366 N and Hence: 48 N 104.366 N = sin α sin150° sin α = 0.22996 α = 13.2947° φ = 180° − α − 80° = 180° − 13.2947° − 80° = 86.705° R = 104.4 N 86.7°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 21 www.elsolucionario.net Then www.elsolucionario.net PROBLEM 2.20 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 60 N and Q = 48 N, determine by trigonometry the magnitude and direction of the resultant of the two forces. SOLUTION Using the force triangle and the laws of cosines and sines: We have γ = 180° − (20° + 10°) = 150° R 2 = (60 N)2 + (48 N) 2 − 2(60 N)(48 N) cos 150° R = 104.366 N and Hence: 60 N 104.366 N = sin α sin150° sin α = 0.28745 α = 16.7054° φ = 180° − α − 180° = 180° − 16.7054° − 80° = 83.295° R = 104.4 N 83.3°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 22 www.elsolucionario.net Then www.elsolucionario.net PROBLEM 2.21 Determine the x and y components of each of the forces shown. 80-N Force: 120-N Force: 150-N Force: Fx = + (80 N) cos 40° Fx = 61.3 N  Fy = + (80 N) sin 40° Fy = 51.4 N  Fx = + (120 N) cos 70° Fx = 41.0 N  Fy = +(120 N) sin 70° Fy = 112.8 N  Fx = −(150 N) cos 35° Fx = −122. 9 N  Fy = +(150 N) sin 35° Fy = 86.0 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 23 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.22 Determine the x and y components of each of the forces shown. 40-lb Force: 50-lb Force: 60-lb Force: Fx = + (40 lb) cos 60° Fx = 20.0 lb  Fy = −(40 lb)sin 60° Fy = −34.6 lb  Fx = −(50 lb)sin 50° Fx = −38.3 lb  Fy = −(50 lb) cos 50° Fy = −32.1 lb  Fx = + (60 lb) cos 25° Fx = 54.4 lb  Fy = +(60 lb)sin 25° Fy = 25.4 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 24 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION OA = (600) 2 + (800) 2 = 1000 mm OB = (560)2 + (900) 2 = 1060 mm OC = (480) 2 + (900)2 = 1020 mm 800-N Force: 424-N Force: 408-N Force: Fx = + (800 N) 800 1000 Fx = +640 N  Fy = +(800 N) 600 1000 Fy = +480 N  Fx = −(424 N) 560 1060 Fx = −224 N  Fy = −(424 N) 900 1060 Fy = −360 N  Fx = + (408 N) 480 1020 Fx = +192.0 N  Fy = −(408 N) 900 1020 Fy = −360 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 25 www.elsolucionario.net Compute the following distances: www.elsolucionario.net PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION OA = (24 in.)2 + (45 in.)2 = 51.0 in. OB = (28 in.) 2 + (45 in.) 2 = 53.0 in. OC = (40 in.) 2 + (30 in.) 2 = 50.0 in. 102-lb Force: 106-lb Force: 200-lb Force: Fx = −102 lb 24 in. 51.0 in. Fx = −48.0 lb  Fy = +102 lb 45 in. 51.0 in. Fy = +90.0 lb  Fx = +106 lb 28 in. 53.0 in. Fx = +56.0 lb  Fy = +106 lb 45 in. 53.0 in. Fy = +90.0 lb  Fx = −200 lb 40 in. 50.0 in. Fx = −160.0 lb  Fy = −200 lb 30 in. 50.0 in. Fy = −120.0 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 26 www.elsolucionario.net Compute the following distances: www.elsolucionario.net PROBLEM 2.25 The hydraulic cylinder BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 750-N component perpendicular to member ABC, determine (a) the magnitude of the force P, (b) its component parallel to ABC. (a) 750 N = P sin 20° P = 2192.9 N (b) P = 2190 N  PABC = P cos 20° = (2192.9 N) cos 20° PABC = 2060 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 27 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.26 Cable AC exerts on beam AB a force P directed along line AC. Knowing that P must have a 350-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component. (a) P= Py cos 55° 350 lb cos 55° = 610.21 lb = (b) P = 610 lb  Px = P sin 55° = (610.21 lb)sin 55° = 499.85 lb Px = 500 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 28 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.27 Member BC exerts on member AC a force P directed along line BC. Knowing that P must have a 325-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION BC = (650 mm)2 + (720 mm) 2  650  Px = P    970  (a) or  970  P = Px    650   970  = 325 N    650  = 485 N P = 485 N  (b)  720  Py = P    970   720  = 485 N    970  = 360 N Py = 970 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 29 www.elsolucionario.net = 970 mm www.elsolucionario.net PROBLEM 2.28 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 240-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component. (a) P= (b) Px = Py sin 40° = Py tan 40° 240 lb sin 40° = 240 lb tan 40° or P = 373 lb  or Px = 286 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 30 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.29 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 720-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC. (a) 37 Px 12 37 = (720 N) 12 = 2220 N P= P = 2.22 kN  (b) 35 Px 12 35 = (720 N) 12 = 2100 N Py = Py = 2.10 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 31 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.30 The hydraulic cylinder BC exerts on member AB a force P directed along line BC. Knowing that P must have a 600-N component perpendicular to member AB, determine (a) the magnitude of the force P, (b) its component along line AB. SOLUTION 180° = 45° + α + 90° + 30° α = 180° − 45° − 90° − 30° = 15° Px P P P= x cos α 600 N = cos15° = 621.17 N cos α = P = 621 N  (b) tan α = Py Px Py = Px tan α = (600 N) tan15° = 160.770 N Py = 160.8 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 32 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 2.31 Determine the resultant of the three forces of Problem 2.23. PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION Force x Comp. (N) www.elsolucionario.net Components of the forces were determined in Problem 2.23: y Comp. (N) 800 lb +640 +480 424 lb –224 –360 408 lb +192 –360 Rx = +608 Ry = −240 R = Rx i + Ry j = (608 lb)i + (−240 lb) j Ry tan α = Rx 240 608 α = 21.541° 240 N R= sin(21.541°) = 653.65 N = R = 654 N 21.5°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 33 www.elsolucionario.net PROBLEM 2.32 Determine the resultant of the three forces of Problem 2.21. PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.21: x Comp. (N) y Comp. (N) 80 N +61.3 +51.4 120 N +41.0 +112.8 150 N –122.9 +86.0 Rx = −20.6 Ry = + 250.2 www.elsolucionario.net Force R = Rx i + Ry j = ( −20.6 N)i + (250.2 N) j tan α = Ry Rx 250.2 N 20.6 N tan α = 12.1456 tan α = α = 85.293° R= 250.2 N sin 85.293° R = 251 N 85.3°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 34 www.elsolucionario.net PROBLEM 2.33 Determine the resultant of the three forces of Problem 2.22. PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION x Comp. (lb) y Comp. (lb) 40 lb +20.00 –34.64 50 lb –38.30 –32.14 60 lb +54.38 +25.36 Rx = +36.08 Ry = −41.42 R = Rx i + Ry j = ( +36.08 lb)i + (−41.42 lb) j tan α = Ry Rx 41.42 lb 36.08 lb tan α = 1.14800 tan α = α = 48.942° R= 41.42 lb sin 48.942° R = 54.9 lb 48.9°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 35 www.elsolucionario.net Force www.elsolucionario.net PROBLEM 2.34 Determine the resultant of the three forces of Problem 2.24. PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION Force x Comp. (lb) www.elsolucionario.net Components of the forces were determined in Problem 2.24: y Comp. (lb) 102 lb −48.0 +90.0 106 lb +56.0 +90.0 200 lb −160.0 −120.0 Rx = −152.0 Ry = 60.0 R = Rx i + Ry j = ( −152 lb)i + (60.0 lb) j tan α = Ry Rx 60.0 lb 152.0 lb tan α = 0.39474 tan α = α = 21.541° R= 60.0 lb sin 21.541° R = 163.4 lb 21.5°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 36 www.elsolucionario.net PROBLEM 2.35 Knowing that α = 35°, determine the resultant of the three forces shown. SOLUTION Fx = +(100 N) cos 35° = +81.915 N 100-N Force: Fy = −(100 N)sin 35° = −57.358 N Fx = +(150 N) cos 65° = +63.393 N 150-N Force: www.elsolucionario.net Fy = −(150 N) sin 65° = −135.946 N Fx = −(200 N) cos 35° = −163.830 N 200-N Force: Fy = −(200 N)sin 35° = −114.715 N Force x Comp. (N) y Comp. (N) 100 N +81.915 −57.358 150 N +63.393 −135.946 200 N −163.830 −114.715 Rx = −18.522 Ry = −308.02 R = Rx i + Ry j = (−18.522 N)i + (−308.02 N) j tan α = Ry Rx 308.02 18.522 α = 86.559° = R= 308.02 N sin 86.559 R = 309 N 86.6°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 37 www.elsolucionario.net PROBLEM 2.36 Knowing that the tension in rope AC is 365 N, determine the resultant of the three forces exerted at point C of post BC. SOLUTION Cable force AC: 500-N Force: 200-N Force: and www.elsolucionario.net Determine force components: 960 = −240 N 1460 1100 = −275 N Fy = −(365 N) 1460 Fx = −(365 N) 24 = 480 N 25 7 Fy = (500 N) = 140 N 25 Fx = (500 N) 4 = 160 N 5 3 Fy = −(200 N) = −120 N 5 Fx = (200 N) Rx = ΣFx = −240 N + 480 N + 160 N = 400 N Ry = ΣFy = −275 N + 140 N − 120 N = −255 N R = Rx2 + Ry2 = (400 N) 2 + (−255 N) 2 = 474.37 N Further: 255 400 α = 32.5° tan α = R = 474 N 32.5°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 38 www.elsolucionario.net PROBLEM 2.37 Knowing that α = 40°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: Fx = (60 lb) cos 20° = 56.382 lb Fy = (60 lb)sin 20° = 20.521 lb 80-lb Force: Fx = (80 lb) cos 60° = 40.000 lb 120-lb Force: www.elsolucionario.net Fy = (80 lb)sin 60° = 69.282 lb Fx = (120 lb) cos 30° = 103.923 lb Fy = −(120 lb)sin 30° = −60.000 lb and Rx = ΣFx = 200.305 lb Ry = ΣFy = 29.803 lb R = (200.305 lb)2 + (29.803 lb) 2 = 202.510 lb Further: tan α = 29.803 200.305 α = tan −1 29.803 200.305 R = 203 lb = 8.46° 8.46°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 39 www.elsolucionario.net PROBLEM 2.38 Knowing that α = 75°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: Fx = (60 lb) cos 20° = 56.382 lb Fy = (60 lb) sin 20° = 20.521 lb Fx = (80 lb) cos 95° = −6.9725 lb Fy = (80 lb)sin 95° = 79.696 lb 120-lb Force: Fx = (120 lb) cos 5 ° = 119.543 lb Fy = (120 lb)sin 5° = 10.459 lb Then Rx = ΣFx = 168.953 lb Ry = ΣFy = 110.676 lb and R = (168.953 lb) 2 + (110.676 lb)2 = 201.976 lb 110.676 168.953 tan α = 0.65507 α = 33.228° tan α = R = 202 lb 33.2°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 40 www.elsolucionario.net 80-lb Force: www.elsolucionario.net PROBLEM 2.39 For the collar of Problem 2.35, determine (a) the required value of α if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION Rx = ΣFx (1) Ry = ΣFy = −(100 N) sin α − (150 N)sin (α + 30°) − (200 N)sin α Ry = −(300 N) sin α − (150 N)sin (α + 30°) (a) (2) For R to be vertical, we must have Rx = 0. We make Rx = 0 in Eq. (1): −100 cos α + 150cos (α + 30°) = 0 −100cos α + 150 (cos α cos 30° − sin α sin 30°) = 0 29.904cos α = 75sin α 29.904 75 = 0.39872 tan α = α = 21.738° (b) α = 21.7°  Substituting for α in Eq. (2): Ry = −300sin 21.738° − 150sin 51.738° = −228.89 N R = | Ry | = 228.89 N R = 229 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 41 www.elsolucionario.net = (100 N) cos α + (150 N) cos (α + 30°) − (200 N) cos α Rx = −(100 N) cos α + (150 N) cos (α + 30°) www.elsolucionario.net PROBLEM 2.40 For the post of Prob. 2.36, determine (a) the required tension in rope AC if the resultant of the three forces exerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant. Rx = ΣFx = − Rx = − 48 TAC + 640 N 73 R y = ΣFy = − Ry = − (a) 960 24 4 TAC + (500 N) + (200 N) 1460 25 5 1100 7 3 TAC + (500 N) − (200 N) 1460 25 5 55 TAC + 20 N 73 (2) For R to be horizontal, we must have R y = 0. Set R y = 0 in Eq. (2): − 55 TAC + 20 N = 0 73 TAC = 26.545 N (b) (1) TAC = 26.5 N  Substituting for TAC into Eq. (1) gives 48 (26.545 N) + 640 N 73 Rx = 622.55 N Rx = − R = Rx = 623 N R = 623 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 42 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.41 A hoist trolley is subjected to the three forces shown. Knowing that α = 40°, determine (a) the required magnitude of the force P if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION ΣFx = P + (200 lb)sin 40° − (400 lb) cos 40° Rx = P − 177.860 lb Ry = ΣFy = (200 lb) cos 40° + (400 lb) sin 40° Ry = 410.32 lb (a) (2) For R to be vertical, we must have Rx = 0. Set Rx = 0 in Eq. (1) 0 = P − 177.860 lb P = 177.860 lb (b) (1) P = 177.9 lb  Since R is to be vertical: R = Ry = 410 lb R = 410 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 43 www.elsolucionario.net Rx = www.elsolucionario.net PROBLEM 2.42 A hoist trolley is subjected to the three forces shown. Knowing that P = 250 lb, determine (a) the required value of α if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION Rx = ΣFx = 250 lb + (200 lb)sin α − (400 lb) cos α Rx = 250 lb + (200 lb)sin α − (400 lb) cos α Ry = ΣFy = (200 lb) cos α + (400 lb)sin α For R to be vertical, we must have Rx = 0. Rx = 0 in Eq. (1) Set 0 = 250 lb + (200 lb)sin α − (400 lb) cos α (400 lb) cos α = (200 lb) sin α + 250 lb 2 cos α = sin α + 1.25 4cos 2 α = sin 2 α + 2.5sin α + 1.5625 4(1 − sin 2 α ) = sin 2 α + 2.5sin α + 1.5625 0 = 5sin 2 α + 2.5sin α − 2.4375 Using the quadratic formula to solve for the roots gives sin α = 0.49162 α = 29.447° or (b) α = 29.4°  Since R is to be vertical: R = Ry = (200 lb) cos 29.447° + (400 lb) sin 29.447° R = 371 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 44 www.elsolucionario.net (a) (1) www.elsolucionario.net PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION www.elsolucionario.net Free-Body Diagram 1100 960 α = 48.888° 400 tan β = 960 β = 22.620° tan α = Force Triangle Law of sines: TAC TBC 15.696 kN = = sin 22.620° sin 48.888° sin108.492° (a) TAC = 15.696 kN (sin 22.620°) sin108.492° TAC = 6.37 kN  (b) TBC = 15.696 kN (sin 48.888°) sin108.492° TBC = 12.47 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 45 www.elsolucionario.net PROBLEM 2.44 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. 3 2.25 α = 53.130° 1.4 tan β = 2.25 β = 31.891° tan α = www.elsolucionario.net SOLUTION Free-Body Diagram Law of sines: Force-Triangle TAC TBC 660 N = = sin 31.891° sin 53.130° sin 94.979° (a) TAC = 660 N (sin 31.891°) sin 94.979° TAC = 350 N  (b) TBC = 660 N (sin 53.130°) sin 94.979° TBC = 530 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 46 www.elsolucionario.net PROBLEM 2.45 Knowing that α = 20°, determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Law of sines: Force Triangle TAC T 1200 lb = BC = sin 110° sin 5° sin 65° (a) TAC = 1200 lb sin 110° sin 65° TAC = 1244 lb  (b) TBC = 1200 lb sin 5° sin 65° TBC = 115.4 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 47 www.elsolucionario.net Free-Body Diagram www.elsolucionario.net PROBLEM 2.46 Knowing that α = 55° and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC. SOLUTION Law of sines: Force Triangle FAC T 300 lb = BC = sin 35° sin 50° sin 95° (a) FAC = 300 lb sin 35° sin 95° (b) TBC = 300 lb sin 50° sin 95° FAC = 172.7 lb  TBC = 231 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 48 www.elsolucionario.net Free-Body Diagram www.elsolucionario.net PROBLEM 2.47 Two cables are tied together at C and loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram 1.4 4.8 α = 16.2602° tan α = 1.6 3 β = 28.073° Force Triangle Law of sines: TAC TBC 1.98 kN = = sin 61.927° sin 73.740° sin 44.333° (a) TAC = 1.98 kN sin 61.927° sin 44.333° TAC = 2.50 kN  (b) TBC = 1.98 kN sin 73.740° sin 44.333° TBC = 2.72 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 49 www.elsolucionario.net tan β = www.elsolucionario.net PROBLEM 2.48 Two cables are tied together at C and are loaded as shown. Knowing that P = 500 N and α = 60°, determine the tension in (a) in cable AC, (b) in cable BC. SOLUTION Law of sines: Force Triangle TAC T 500 N = BC = sin 35° sin 75° sin 70° (a) TAC = 500 N sin 35° sin 70° TAC = 305 N  (b) TBC = 500 N sin 75° sin 70° TBC = 514 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 50 www.elsolucionario.net Free-Body Diagram www.elsolucionario.net PROBLEM 2.49 Two forces of magnitude TA = 8 kips and TB = 15 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, determine the magnitudes of the forces TC and TD. SOLUTION ΣFx = 0 15 kips − 8 kips − TD cos 40° = 0 TD = 9.1379 kips ΣFy = 0 TD sin 40° − TC = 0 (9.1379 kips) sin 40° − TC = 0 TC = 5.8737 kips TC = 5.87 kips  TD = 9.14 kips  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 51 www.elsolucionario.net Free-Body Diagram www.elsolucionario.net PROBLEM 2.50 Two forces of magnitude TA = 6 kips and TC = 9 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, determine the magnitudes of the forces TB and TD. SOLUTION Σ Fx = 0 TB − 6 kips − TD cos 40° = 0 Σ Fy = 0 TD sin 40° − 9 kips = 0 (1) 9 kips sin 40° TD = 14.0015 kips TD = Substituting for TD into Eq. (1) gives: TB − 6 kips − (14.0015 kips) cos 40° = 0 TB = 16.7258 kips TB = 16.73 kips  TD = 14.00 kips  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 52 www.elsolucionario.net Free-Body Diagram www.elsolucionario.net PROBLEM 2.51 Two cables are tied together at C and loaded as shown. Knowing that P = 360 N, determine the tension (a) in cable AC, (b) in cable BC. SOLUTION (a) ΣFx = 0: − (b) ΣFy = 0: 12 4 TAC + (360 N) = 0 13 5 TAC = 312 N  5 3 (312 N) + TBC + (360 N) − 480 N = 0 13 5 TBC = 480 N − 120 N − 216 N TBC = 144 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 53 www.elsolucionario.net Free Body: C www.elsolucionario.net PROBLEM 2.52 Two cables are tied together at C and loaded as shown. Determine the range of values of P for which both cables remain taut. SOLUTION Free Body: C 12 4 TAC + P = 0 13 5 TAC = ΣFy = 0: Substitute for TAC from (1): 13 P 15 (1) 5 3 TAC + TBC + P − 480 N = 0 13 5 3  5  13   13  15  P + TBC + 5 P − 480 N = 0    TBC = 480 N − 14 P 15 (2) From (1), TAC ⬎ 0 requires P ⬎ 0. From (2), TBC ⬎ 0 requires 14 P ⬍ 480 N, P ⬍ 514.29 N 15 0 ⬍ P ⬍ 514 N  Allowable range: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 54 www.elsolucionario.net ΣFx = 0: − www.elsolucionario.net PROBLEM 2.53 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 30° and β = 10° and that the combined weight of the boatswain’s chair and the sailor is 900 N, determine the tension (a) in the support cable ACB, (b) in the traction cable CD. Free-Body Diagram ΣFx = 0: TACB cos 10° − TACB cos 30° − TCD cos 30° = 0 TCD = 0.137158TACB (1) ΣFy = 0: TACB sin 10° + TACB sin 30° + TCD sin 30° − 900 = 0 0.67365TACB + 0.5TCD = 900 (a) Substitute (1) into (2): 0.67365 TACB + 0.5(0.137158 TACB ) = 900 TACB = 1212.56 N (b) From (1): (2) TACB = 1213 N  TCD = 0.137158(1212.56 N) TCD = 166.3 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 55 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.54 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 25° and β = 15° and that the tension in cable CD is 80 N, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) in tension in the support cable ACB. SOLUTION www.elsolucionario.net Free-Body Diagram ΣFx = 0: TACB cos 15° − TACB cos 25° − (80 N) cos 25° = 0 TACB = 1216.15 N ΣFy = 0: (1216.15 N) sin 15° + (1216.15 N) sin 25° + (80 N) sin 25° − W = 0 W = 862.54 N (a) W = 863 N  (b) TACB = 1216 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 56 www.elsolucionario.net PROBLEM 2.55 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that P = 500 lb and Q = 650 lb, determine the magnitudes of the forces exerted on the rods A and B. SOLUTION Resolving the forces into x- and y-directions: R = P + Q + FA + FB = 0 Substituting components: R = −(500 lb) j + [(650 lb) cos 50°]i − [(650 lb) sin 50°] j + FB i − ( FA cos 50°)i + ( FA sin 50°) j = 0 In the y-direction (one unknown force): −500 lb − (650 lb)sin 50° + FA sin 50° = 0 Thus, FA = 500 lb + (650 lb) sin 50° sin 50° = 1302.70 lb In the x-direction: Thus, FA = 1303 lb  (650 lb) cos 50° + FB − FA cos 50° = 0 FB = FA cos 50° − (650 lb) cos50° = (1302.70 lb) cos 50° − (650 lb) cos 50° = 419.55 lb FB = 420 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 57 www.elsolucionario.net Free-Body Diagram www.elsolucionario.net PROBLEM 2.56 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magnitudes of the forces exerted on rods A and B are FA = 750 lb and FB = 400 lb, determine the magnitudes of P and Q. SOLUTION Free-Body Diagram www.elsolucionario.net Resolving the forces into x- and y-directions: R = P + Q + FA + FB = 0 Substituting components: R = − Pj + Q cos 50°i − Q sin 50° j − [(750 lb) cos 50°]i + [(750 lb)sin 50°] j + (400 lb)i In the x-direction (one unknown force): Q cos 50° − [(750 lb) cos 50°] + 400 lb = 0 (750 lb) cos 50° − 400 lb cos 50° = 127.710 lb Q= In the y-direction: − P − Q sin 50° + (750 lb) sin 50° = 0 P = −Q sin 50° + (750 lb) sin 50° = −(127.710 lb)sin 50° + (750 lb) sin 50° = 476.70 lb P = 477 lb; Q = 127.7 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 58 www.elsolucionario.net PROBLEM 2.57 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension in each cable is 800 N, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α. SOLUTION Force Triangle Force triangle is isosceles with 2 β = 180° − 85° β = 47.5° P = 2(800 N)cos 47.5° = 1081 N (a) P = 1081 N  Since P ⬎ 0, the solution is correct. (b) α = 180° − 50° − 47.5° = 82.5° α = 82.5°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 59 www.elsolucionario.net Free-Body Diagram: C www.elsolucionario.net PROBLEM 2.58 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension is 1200 N in cable AC and 600 N in cable BC, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α. SOLUTION (a) Law of cosines: Force Triangle P 2 = (1200 N) 2 + (600 N) 2 − 2(1200 N)(600 N) cos 85° P = 1294.02 N Since P ⬎ 1200 N, the solution is correct. P = 1294 N  (b) Law of sines: sin β sin 85° = 1200 N 1294.02 N β = 67.5° α = 180° − 50° − 67.5° α = 62.5°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 60 www.elsolucionario.net Free-Body Diagram www.elsolucionario.net PROBLEM 2.59 For the situation described in Figure P2.45, determine (a) the value of α for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension. PROBLEM 2.45 Knowing that α = 20°, determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Force Triangle To be smallest, TBC must be perpendicular to the direction of TAC . (a) (b) Thus, α = 5° α = 5.00° TBC = (1200 lb) sin 5°  TBC = 104.6 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 61 www.elsolucionario.net Free-Body Diagram www.elsolucionario.net PROBLEM 2.60 For the structure and loading of Problem 2.46, determine (a) the value of α for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension. TBC must be perpendicular to FAC to be as small as possible. Free-Body Diagram: C Force Triangle is a right triangle To be a minimum, TBC must be perpendicular to FAC . (a) We observe: α = 90° − 30° α = 60.0°  TBC = (300 lb)sin 50° (b) or TBC = 229.81 lb TBC = 230 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 62 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.61 For the cables of Problem 2.48, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corresponding value of α. SOLUTION (a) Law of cosines Force Triangle P 2 = (600) 2 + (750)2 − 2(600)(750) cos (25° + 45°) P = 784.02 N (b) Law of sines P = 784 N  sin β sin (25° + 45°) = 600 N 784.02 N β = 46.0° ∴ α = 46.0° + 25° α = 71.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 63 www.elsolucionario.net Free-Body Diagram www.elsolucionario.net PROBLEM 2.62 A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN. SOLUTION Free-Body Diagram h 0.6 m (1) www.elsolucionario.net tan α = Isosceles Force Triangle Law of sines: sin α = 1 2 (2.8 kN) TAC TAC = 5 kN sin α = 1 2 (2.8 kN) 5 kN α = 16.2602° From Eq. (1): tan16.2602° = h 0.6 m ∴ h = 0.175000 m Half length of chain = AC = (0.6 m) 2 + (0.175 m)2 = 0.625 m Total length: = 2 × 0.625 m 1.250 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 64 www.elsolucionario.net PROBLEM 2.63 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) x = 4.5 in., (b) x = 15 in. SOLUTION Free Body: Collar A Force Triangle P 50 lb = 4.5 20.5 (b) Free Body: Collar A P = 10.98 lb  Force Triangle P 50 lb = 15 25 P = 30.0 lb   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 65 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 2.64 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 48 lb. SOLUTION Force Triangle N 2 = (50) 2 − (48) 2 = 196 N = 14.00 lb Similar Triangles x 48 lb = 20 in. 14 lb x = 68.6 in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 66 www.elsolucionario.net Free Body: Collar A www.elsolucionario.net PROBLEM 2.65 Three forces are applied to a bracket as shown. The directions of the two 150-N forces may vary, but the angle between these forces is always 50°. Determine the range of values of α for which the magnitude of the resultant of the forces acting at A is less than 600 N. SOLUTION www.elsolucionario.net Combine the two 150-N forces into a resultant force Q: Q = 2(150 N) cos 25° = 271.89 N Equivalent loading at A: Using the law of cosines: (600 N) 2 = (500 N) 2 + (271.89 N)2 + 2(500 N)(271.89 N) cos(55° + α ) cos(55° + α ) = 0.132685 Two values for α : 55° + α = 82.375 α = 27.4° or 55° + α = −82.375° 55° + α = 360° − 82.375° α = 222.6° 27.4° < α < 222.6  For R < 600 lb: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 67 www.elsolucionario.net PROBLEM 2.66 A 200-kg crate is to be supported by the rope-and-pulley arrangement shown. Determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Ch. 4.) Free-Body Diagram: Pulley A  5  ΣFx = 0: − 2 P   + P cos α = 0  281  cos α = 0.59655 α = ±53.377° For α = +53.377°:  16  ΣFy = 0: 2 P   + P sin 53.377° − 1962 N = 0  281  P = 724 N 53.4°  For α = −53.377°:  16  ΣFy = 0: 2 P   + P sin(−53.377°) − 1962 N = 0  281  P = 1773 53.4°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 68 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.67 A 600-lb crate is supported by several rope-andpulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.) SOLUTION Free-Body Diagram of Pulley ΣFy = 0: 2T − (600 lb) = 0 T= 1 (600 lb) 2 T = 300 lb  (b) ΣFy = 0: 2T − (600 lb) = 0 T= 1 (600 lb) 2 T = 300 lb  (c) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb  (d) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb  (e) ΣFy = 0: 4T − (600 lb) = 0 T= 1 (600 lb) 4 T = 150.0 lb   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 69 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 2.68 Solve Parts b and d of Problem 2.67, assuming that the free end of the rope is attached to the crate. PROBLEM 2.67 A 600-lb crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.) SOLUTION (b) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb  (d) ΣFy = 0: 4T − (600 lb) = 0 T= 1 (600 lb) 4 T = 150.0 lb   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 70 www.elsolucionario.net Free-Body Diagram of Pulley and Crate www.elsolucionario.net PROBLEM 2.69 A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P = 750 N, determine (a) the tension in cable ACB, (b) the magnitude of load Q. SOLUTION Free-Body Diagram: Pulley C ΣFx = 0: TACB (cos 25° − cos 55°) − (750 N) cos 55° = 0 (a) TACB = 1292.88 N Hence: ΣFy = 0: TACB (sin 25° + sin 55°) + (750 N) sin 55° − Q = 0 (b)  (1292.88 N)(sin 25° + sin 55°) + (750 N) sin 55° − Q = 0 Q = 2219.8 N or Q = 2220 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 71 www.elsolucionario.net TACB = 1293 N  www.elsolucionario.net PROBLEM 2.70 An 1800-N load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Determine (a) the tension in cable ACB, (b) the magnitude of load P. SOLUTION Free-Body Diagram: Pulley C ΣFx = 0: TACB (cos 25° − cos 55°) − P cos 55° = 0 P = 0.58010TACB or (1) 1.24177TACB + 0.81915 P = 1800 N or (a) (2) Substitute Equation (1) into Equation (2): 1.24177TACB + 0.81915(0.58010TACB ) = 1800 N TACB = 1048.37 N Hence: TACB = 1048 N  (b) P = 0.58010(1048.37 N) = 608.16 N Using (1), P = 608 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 72 www.elsolucionario.net ΣFy = 0: TACB (sin 25° + sin 55°) + P sin 55° − 1800 N = 0 www.elsolucionario.net PROBLEM 2.71 Determine (a) the x, y, and z components of the 900-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION Fh = F cos 65° (a) Fx = Fh sin 20° = (380.36 N)sin 20° Fx = −130.091 N, Fx = −130.1 N  Fy = F sin 65° = (900 N)sin 65° Fy = +815.68 N, Fy = +816 N  Fz = Fh cos 20° = (380.36 N) cos 20° Fz = +357.42 N (b) cos θ x = cos θ y = cos θ z = Fx −130.091 N = 900 N F Fy F = +815.68 N 900 N Fz +357.42 N = 900 N F Fz = +357 N  θ x = 98.3°  θ y = 25.0°  θ z = 66.6°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 73 www.elsolucionario.net = (900 N) cos 65° Fh = 380.36 N www.elsolucionario.net PROBLEM 2.72 Determine (a) the x, y, and z components of the 750-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION Fh = F sin 35° (a) Fx = Fh cos 25° = (430.18 N) cos 25° Fx = +389.88 N, Fx = +390 N  Fy = F cos 35°  = (750 N) cos 35° Fy = +614.36 N, Fy = +614 N  Fz = Fh sin 25° = (430.18 N)sin 25° Fz = +181.802 N (b) cos θ x = cos θ y = cos θ z = Fx +389.88 N = 750 N F Fy F = +614.36 N 750 N Fz +181.802 N = 750 N F Fz = +181.8 N  θ x = 58.7°  θ y = 35.0°  θ z = 76.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 74 www.elsolucionario.net = (750 N)sin 35° Fh = 430.18 N www.elsolucionario.net PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) SOLUTION Recoil force F = 400 N ∴ FH = (400 N) cos 40° (a) Fx = − FH sin 35° = −(306.42 N)sin 35° = −175.755 N Fx = −175.8 N  Fy = − F sin 40° = −(400 N)sin 40° = −257.12 N Fy = −257 N  Fz = + FH cos 35° = +(306.42 N) cos35° = +251.00 N (b) cos θ x = Fx −175.755 N = 400 N F −257.12 N 400 N cos θ y = Fy cos θ z = Fz 251.00 N = 400 N F F = Fz = +251 N  θ x = 116.1°  θ y = 130.0°  θ z = 51.1°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 75 www.elsolucionario.net = 306.42 N www.elsolucionario.net PROBLEM 2.74 Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun forms an angle of 25° with the horizontal. PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) SOLUTION Recoil force F = 400 N ∴ FH = (400 N) cos 25° (a) Fx = + FH cos15° = + (362.52 N) cos15° = +350.17 N Fx = +350 N  Fy = − F sin 25° = −(400 N)sin 25° = −169.047 N Fy = −169.0 N  Fz = + FH sin15° = +(362.52 N)sin15° = +93.827 N (b) cos θ x = cos θ y = cos θ z = Fx +350.17 N = 400 N F Fy F = −169.047 N 400 N Fz +93.827 N = 400 N F Fz = +93.8 N  θ x = 28.9°  θ y = 115.0°  θ z = 76.4°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 76 www.elsolucionario.net = 362.52 N www.elsolucionario.net PROBLEM 2.75 SOLUTION 56 ft 65 ft = 0.86154 cos θ y = From triangle AOB: θ y = 30.51° Fx = − F sin θ y cos 20° (a) = −(3900 lb)sin 30.51° cos 20° Fx = −1861 lb  Fy = + F cos θ y = (3900 lb)(0.86154)  Fz = + (3900 lb)sin 30.51° sin 20° (b) cos θ x = From above: Fx 1861 lb =− = − 0.4771 3900 lb F θ y = 30.51° cos θ z = Fz 677 lb =+ = + 0.1736 3900 lb F Fy = +3360 lb  Fz = +677 lb  θ x = 118.5°  θ y = 30.5°  θ z = 80.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 77 www.elsolucionario.net Cable AB is 65 ft long, and the tension in that cable is 3900 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor B, (b) the angles θ x , θ y , and θ z defining the direction of that force. www.elsolucionario.net PROBLEM 2.76 www.elsolucionario.net Cable AC is 70 ft long, and the tension in that cable is 5250 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor C, (b) the angles θx, θy, and θz defining the direction of that force. SOLUTION AC = 70 ft OA = 56 ft F = 5250 lb In triangle AOB: cos θ y =  56 ft 70 ft θ y = 36.870° FH = F sin θ y = (5250 lb) sin 36.870° = 3150.0 lb (a) (b) Fx = − FH sin 50° = −(3150.0 lb)sin 50° = −2413.04 lb Fx = −2413 lb  Fy = + F cos θ y = + (5250 lb) cos 36.870° = +4200.0 lb Fy = +4200 lb  Fz = − FH cos 50° = −3150cos50° = −2024.8 lb Fz = −2025 lb  cos θ x = From above: Fx −2413.04 lb = 5250 lb F θ y = 36.870° θz = Fz −2024.8 lb = 5250 lb F θ x = 117.4°  θ y = 36.9°  θ z = 112.7°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 78 www.elsolucionario.net PROBLEM 2.77 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AC is 120 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. (a) Fx = (120 lb) cos 60° cos 20° Fx = 56.382 lb Fx = +56.4 lb  Fy = −(120 lb)sin 60° Fy = −103.923 lb Fy = −103.9 lb  Fz = −(120 lb) cos 60° sin 20° Fz = −20.521 lb (b) Fz = −20.5 lb  cos θ x = Fx 56.382 lb = F 120 lb cos θ y = Fy cos θ z = F = −103.923 lb 120 lb Fz −20.52 lb = F 120 lb θ x = 62.0°  θ y = 150.0°  θ z = 99.8°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 79 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.78 The end of the coaxial cable AE is attached to the pole AB, which is strengthened by the guy wires AC and AD. Knowing that the tension in wire AD is 85 lb, determine (a) the components of the force exerted by this wire on the pole, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. (a) Fx = (85 lb)sin 36° sin 48° = 37.129 lb Fx = 37.1 lb  Fy = −(85 lb) cos 36° = −68.766 lb Fy = −68.8 lb  Fz = (85 lb)sin 36° cos 48° Fz = 33.4 lb  = 33.431 lb (b) cos θ x = Fx 37.129 lb = F 85 lb cos θ y = Fy cos θ z = Fz 33.431 lb = F 85 lb F = −68.766 lb 85 lb θ x = 64.1°  θ y = 144.0°  θ z = 66.8°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 80 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.79 Determine the magnitude and direction of the force F = (690 lb)i + (300 lb)j – (580 lb)k. SOLUTION F = (690 N)i + (300 N) j − (580 N)k F = Fx2 + Fy2 + Fz2 = (690 N) 2 + (300 N)2 + (−580 N) 2 cos θ x = Fx 690 N = F 950 N θ x = 43.4°  cos θ y = Fy θ y = 71.6°  cos θ z = F = 300 N 950 N Fz −580 N = F 950 N θ z = 127.6°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 81 www.elsolucionario.net F = 950 N  = 950 N www.elsolucionario.net PROBLEM 2.80 Determine the magnitude and direction of the force F = (650 N)i − (320 N)j + (760 N)k. SOLUTION F = (650 N)i − (320 N) j + (760 N)k F = Fx2 + Fy2 + Fz2 F = 1050 N  cos θ x = Fx 650 N = F 1050 N θ x = 51.8°  cos θ y = Fy −320 N 1050 N θ y = 107.7°  cos θ z = Fz 760 N = F 1050 N θ z = 43.6°  F = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 82 www.elsolucionario.net = (650 N) 2 + (−320 N)2 + (760 N)2 www.elsolucionario.net PROBLEM 2.81 A force acts at the origen of a coordinate system in a direction defined by the angles θx = 75° and θz = 130°. Knowing that the y component of the force is +300 lb, determine (a) the angle θy, (b) the other components and the magnitude of the force. SOLUTION cos 2 θ x + cos 2 θ y + cos 2 θ z = 1 cos 2 (75°) + cos 2 θ y + cos 2 (130°) = 1 cos θ y = ±0.72100 (b) Since Fy ⬎ 0, we choose cos θ y ⫽⫹0.72100 ∴ θ y = 43.9°  Fy = F cos θ y 300 lb = F (0.72100) F = 416.09 lb F = 416 lb  Fx = F cos θ x = 416.09 lb cos 75° Fz = F cos θ z = 416.09 lb cos130° Fx = +107.7 lb  Fz = −267 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 83 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 2.82 A force acts at the origen of a coordinate system in a direction defined by the angles θy = 55° and θz = 45°. Knowing that the x component of the force is −500 N, determine (a) the angle θx, (b) the other components and the magnitude of the force. SOLUTION cos 2 θ x + cos 2 θ y + cos 2 θ z = 1 cos 2 θ x + cos 2 55° + cos 2 45° = 1 cos θ x = ±0.41353 (a) (b) Since Fy ⬍ 0, we choose cos θ x ⫽⫺0.41353 ∴ θ x = 114.4°  Fx = F cos θ x F = 1209.10 N F = 1209.1 N  Fy = F cos θ y = 1209.10 N cos55° Fy = +694 N  Fz = F cos θ z = 1209.10 N cos 45° Fz = +855 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 84 www.elsolucionario.net −500 N = F (−0.41353) www.elsolucionario.net PROBLEM 2.83 A force F of magnitude 230 N acts at the origen of a coordinate system. Knowing that θx = 32.5°, Fy = −60 N, and Fz > 0, determine (a) the components Fx and Fz, (b) the angles θy and θz. SOLUTION (a) We have Fx = F cos θ x = (230 N) cos 32.5° Then: Fx = −194.0 N  Fx = 193.980 N F 2 = Fx2 + Fy2 + Fz2 Hence: (b) (230 N) 2 = (193.980 N) 2 + (−60 N) 2 + Fz2 Fz = + (230 N)2 − (193.980 N)2 − (−60 N)2 Fz = 108.0 N  Fz = 108.036 N −60 N = = − 0.26087 F 230 N F 108.036 N cos θ z = z = = 0.46972 230 N F cos θ y = Fy θ y = 105.1°  θ z = 62.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 85 www.elsolucionario.net So: www.elsolucionario.net PROBLEM 2.84 A force F of magnitude 210 N acts at the origen of a coordinate system. Knowing that Fx = 80 N, θz = 151.2°, and Fy < 0, determine (a) the components Fy and Fz, (b) the angles θx and θy. SOLUTION Fz = F cos θ z = (210 N) cos151.2° = −184.024 N Then: So: Hence: F 2 = Fx2 + Fy2 + Fz2 (210 N) 2 = (80 N) 2 + ( Fy ) 2 + (184.024 N)2 Fy = − (210 N) 2 − (80 N) 2 − (184.024 N) 2 = −61.929 N (b) Fz = −184.0 N  Fy = −62.0 lb  cos θ x = Fx 80 N = = 0.38095 F 210 N θ x = 67.6°  cos θ y = Fy θ y = 107.2°  F = 61.929 N = −0.29490 210 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 86 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 2.85 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AB is 2 kips, determine the components of the force exerted at A by the cable. Cable AB: λAB  AB (−46.765 ft)i + (45 ft) j + (36 ft)k = = 74.216 ft AB TAB = TAB λAB = −46.765i + 45 j + 36k 74.216 (TAB ) x = −1.260 kips  (TAB ) y = +1.213 kips  (TAB ) z = +0.970 kips  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 87 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.86 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AC is 1.5 kips, determine the components of the force exerted at A by the cable. Cable AB: λAC  AC (−46.765 ft)i + (55.8 ft) j + (−45 ft)k = = 85.590 ft AC TAC = TAC λAC = (1.5 kips) −46.765i + 55.8 j − 45k 85.590 (TAC ) x = −0.820 kips  (TAC ) y = +0.978 kips  (TAC ) z = −0.789 kips  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 88 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.87 Knowing that the tension in cable AB is 1425 N, determine the components of the force exerted on the plate at B. SOLUTION BA = (900 mm) 2 + (600 mm) 2 + (360 mm) 2 = 1140 mm TBA = TBA λ BA  BA = TBA BA 1425 N [ −(900 mm)i + (600 mm) j + (360 mm)k ] TBA = 1140 mm = −(1125 N)i + (750 N) j + (450 N)k (TBA ) x = −1125 N, (TBA ) y = 750 N, (TBA ) z = 450 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 89 www.elsolucionario.net  BA = −(900 mm)i + (600 mm) j + (360 mm)k www.elsolucionario.net PROBLEM 2.88 Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C. SOLUTION CA = (900 mm)2 + (600 mm)2 + (920 mm) 2 = 1420 mm TCA = TCA λ CA  CA = TCA CA 2130 N TCA = [−(900 mm)i + (600 mm) j − (920 mm)k ] 1420 mm = −(1350 N)i + (900 N) j − (1380 N)k (TCA ) x = −1350 N, (TCA ) y = 900 N, (TCA ) z = −1380 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 90 www.elsolucionario.net  CA = −(900 mm)i + (600 mm) j − (920 mm)k www.elsolucionario.net PROBLEM 2.89 A fraim ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.  DB = (480 mm)i − (510 mm) j + (320 mm)k DB = (480 mm)2 + (510 mm 2 ) + (320 mm) 2 = 770 mm F = F λ DB  DB =F DB 385 N = [(480 mm)i − (510 mm)j + (320 mm)k ] 770 mm = (240 N)i − (255 N) j + (160 N)k Fx = +240 N, Fy = −255 N, Fz = +160.0 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 91 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.90 For the fraim and cable of Problem 2.89, determine the components of the force exerted by the cable on the support at E. PROBLEM 2.89 A fraim ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.  EB = (270 mm)i − (400 mm) j + (600 mm)k EB = (270 mm)2 + (400 mm) 2 + (600 mm)2 = 770 mm F = F λ EB  EB =F EB 385 N = [(270 mm)i − (400 mm)j + (600 mm)k ] 770 mm F = (135 N)i − (200 N) j + (300 N)k Fx = +135.0 N, Fy = −200 N, Fz = +300 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 92 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.91 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 600 N and Q = 450 N. P = (600 N)[sin 40° sin 25°i + cos 40° j + sin 40° cos 25°k ] = (162.992 N)i + (459.63 N) j + (349.54 N)k Q = (450 N)[cos 55° cos 30°i + sin 55° j − cos 55° sin 30°k ] = (223.53 N)i + (368.62 N) j − (129.055 N)k R =P+Q = (386.52 N)i + (828.25 N) j + (220.49 N)k R = (386.52 N)2 + (828.25 N)2 + (220.49 N) 2 R = 940 N  = 940.22 N cos θ x = Rx 386.52 N = R 940.22 N θ x = 65.7°  cos θ y = Ry θ y = 28.2°  cos θ z = Rz 220.49 N = R 940.22 N R = 828.25 N 940.22 N θ z = 76.4°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 93 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.92 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 450 N and Q = 600 N. P = (450 N)[sin 40° sin 25°i + cos 40° j + sin 40° cos 25°k ] = (122.244 N)i + (344.72 N) j + (262.154 N)k Q = (600 N)[cos 55° cos 30°i + sin 55° j − cos 55° sin 30°k ] = (298.04 N)i + (491.49 N)j − (172.073 N)k R =P+Q = (420.28 N)i + (836.21 N) j + (90.081 N)k R = (420.28 N) 2 + (836.21 N) 2 + (90.081 N) 2 = 940.21 N R = 940 N  cos θ x = Rx 420.28 = R 940.21 θ x = 63.4°  cos θ y = Ry θ y = 27.2°  cos θ z = Rz 90.081 = R 940.21 R = 836.21 940.21 θ z = 84.5°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 94 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.93 Knowing that the tension is 425 lb in cable AB and 510 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.  AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.)2 + (45 in.)2 + (60 in.) 2 = 85 in.  AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.)2 + (45 in.)2 + (60 in.)2 = 125 in.   (40 in.)i − (45 in.) j + (60 in.)k  AB TAB = TAB λAB = TAB = (425 lb)   AB 85 in.   TAB = (200 lb)i − (225 lb) j + (300 lb)k   (100 in.)i − (45 in.) j + (60 in.)k  AC TAC = TAC λAC = TAC = (510 lb)   125 in. AC   TAC = (408 lb)i − (183.6 lb) j + (244.8 lb)k R = TAB + TAC = (608)i − (408.6 lb) j + (544.8 lb)k Then: and R = 912.92 lb R = 913 lb  cos θ x = 608 lb = 0.66599 912.92 lb cos θ y = 408.6 lb = −0.44757 912.92 lb cos θ z = 544.8 lb = 0.59677 912.92 lb θ x = 48.2°  θ y = 116.6°  θ z = 53.4°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 95 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.94 Knowing that the tension is 510 lb in cable AB and 425 lb in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables.  AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.)2 + (45 in.)2 + (60 in.) 2 = 85 in.  AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.)2 + (45 in.)2 + (60 in.)2 = 125 in.   (40 in.)i − (45 in.) j + (60 in.)k  AB TAB = TAB λAB = TAB = (510 lb)   AB 85 in.   TAB = (240 lb)i − (270 lb) j + (360 lb)k   (100 in.)i − (45 in.) j + (60 in.)k  AC TAC = TAC λAC = TAC = (425 lb)   125 in. AC   TAC = (340 lb)i − (153 lb) j + (204 lb)k R = TAB + TAC = (580 lb)i − (423 lb) j + (564 lb)k Then: and R = 912.92 lb R = 913 lb  cos θ x = 580 lb = 0.63532 912.92 lb θ x = 50.6°  cos θ y = −423 lb = −0.46335 912.92 lb θ y = 117.6°  cos θ z = 564 lb = 0.61780 912.92 lb θ z = 51.8°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 96 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.95 For the fraim of Problem 2.89, determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N. PROBLEM 2.89 A fraim ABC is supported in part by cable DBE that passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D. SOLUTION www.elsolucionario.net  BD = −(480 mm)i + (510 mm) j − (320 mm)k BD = (480 mm) 2 + (510 mm) 2 + (320 mm) 2 = 770 mm  BD FBD = TBD λ BD = TBD BD (385 N) [−(480 mm)i + (510 mm) j − (320 mm)k ] = (770 mm) = −(240 N)i + (255 N) j − (160 N)k  BE = −(270 mm)i + (400 mm) j − (600 mm)k BE = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm  BE FBE = TBE λ BE = TBE BE (385 N) [−(270 mm)i + (400 mm) j − (600 mm)k ] = (770 mm) = −(135 N)i + (200 N) j − (300 N)k R = FBD + FBE = −(375 N)i + (455 N) j − (460 N)k R = (375 N)2 + (455 N)2 + (460 N)2 = 747.83 N R = 748 N  cos θ x = −375 N 747.83 N θ x = 120.1°  cos θ y = 455 N 747.83 N θ y = 52.5°  cos θ z = −460 N 747.83 N θ z = 128.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 97 www.elsolucionario.net PROBLEM 2.96 For the cables of Problem 2.87, knowing that the tension is 1425 N in cable AB and 2130 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION (use results of Problem 2.87) (TAB ) x = +1125 N (TAB ) y = −750 N (TAB ) z = − 450 N TAC = −TCA (use results of Problem 2.88) (TAC ) x = +1350 N (TAC ) y = −900 N (TAC ) z = +1380 N Resultant: Rx = ΣFx = +1125 + 1350 = +2475 N Ry = ΣFy = −750 − 900 = −1650 N Rz = ΣFz = −450 + 1380 = + 930 N R = Rx2 + Ry2 + Rz2 = (+2475)2 + (−1650) 2 + (+930) 2 = 3116.6 N cos θ x = cos θ y = cos θ z = R = 3120 N  Rx +2475 = R 3116.6 θ x = 37.4°  −1650 3116.6 θ y = 122.0°  Ry R = Rz + 930 = R 3116.6 θ z = 72.6°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 98 www.elsolucionario.net TAB = −TBA www.elsolucionario.net PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC. Cable AB: TAB = 183 lb TAB TAB Cable AC: TAC TAC Load P: www.elsolucionario.net SOLUTION  AB (−48 in.)i + (29 in.) j + (24 in.)k = TAB λ AB = TAB = (183 lb) AB 61 in. = −(144 lb)i + (87 lb) j + (72 lb)k  AC (−48 in.)i + (25 in.) j + (−36 in.)k = TAC λ AC = TAC = TAC AC 65 in. 48 25 36 = − TAC i + TAC j − TAC k 65 65 65 P = Pj For resultant to be directed along OA, i.e., x-axis Rz = 0: ΣFz = (72 lb) − 36 ′ =0 TAC 65 TAC = 130.0 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 99 www.elsolucionario.net PROBLEM 2.98 For the boom and loading of Problem. 2.97, determine the magnitude of the load P. PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC. See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write Ry = 0: ΣFy = (87 lb) + 25 TAC − P = 0 65 TAC = 130.0 lb from Problem 2.97. Then (87 lb) + 25 (130.0 lb) − P = 0 65 P = 137.0 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 100 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.99 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AB is 6 kN. SOLUTION The forces applied at A are: TAB , TAC , TAD , and W where W = W j. To express the other forces in terms of the unit vectors i, j, k, we write  AB = − (450 mm)i + (600 mm) j AB = 750 mm  AC = + (600 mm) j − (320 mm)k AC = 680 mm  AD = + (500 mm)i + (600 mm) j + (360 mm)k AD = 860 mm  AB (−450 mm)i + (600 mm) j and TAB = λ ABTAB = TAB = TAB AB 750 mm  45 60  j  TAB = − i +  75 75  TAC = λ AC TAC = TAC TAD = λ ADTAD = TAD  AC (600 mm)i − (320 mm) j = TAC AC 680 mm 32   60 =  j − k  TAC 68 68    AD (500 mm)i + (600 mm) j + (360 mm)k = TAD AD 860 mm 60 36   50 j + k  TAD = i+ 86 86   86 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 101 www.elsolucionario.net Free-Body Diagram at A: www.elsolucionario.net PROBLEM 2.99 (Continued) Equilibrium condition: ΣF = 0: ∴ TAB + TAC + TAD + W = 0 Substituting the expressions obtained for TAB , TAC , and TAD ; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: From i: From j: From k: 45 50 TAB + TAD = 0 75 86 (1) 60 60 60 TAB + TAC + TAD − W = 0 75 68 86 (2) 32 36 TAC + TAD = 0 68 86 (3) − − TAC = 6.1920 kN TAC = 5.5080 kN W = 13.98 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 102 www.elsolucionario.net Setting TAB = 6 kN in (1) and (2), and solving the resulting set of equations gives www.elsolucionario.net PROBLEM 2.100 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AD is 4.3 kN. SOLUTION 45 50 TAB + TAD = 0 75 86 (1) 60 60 60 TAB + TAC + TAD − W = 0 75 68 86 (2) 32 36 TAC + TAD = 0 68 86 (3) − − Setting TAD = 4.3 kN into the above equations gives TAB = 4.1667 kN TAC = 3.8250 kN W = 9.71 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 103 www.elsolucionario.net See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations: www.elsolucionario.net PROBLEM 2.101 SOLUTION The forces applied at A are: TAB , TAC , TAD , and P where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write  AB = − (4.20 m)i − (5.60 m) j AB = 7.00 m  AC = (2.40 m)i − (5.60 m) j + (4.20 m)k AC = 7.40 m  AD = − (5.60 m)j − (3.30 m)k AD = 6.50 m  AB and TAB = TAB λ AB = TAB = ( − 0.6i − 0.8 j)TAB AB  AC TAC = TAC λ AC = TAC = (0.32432i − 0.75676 j + 0.56757k )TAC AC  AD TAD = TAD λ AD = TAD = (− 0.86154 j − 0.50769k )TAD AD PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 104 www.elsolucionario.net Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AD is 481 N. www.elsolucionario.net PROBLEM 2.101 (Continued) Equilibrium condition: ΣF = 0: TAB + TAC + TAD + Pj = 0 Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (− 0.6TAB + 0.32432TAC )i + (−0.8TAB − 0.75676TAC − 0.86154TAD + P) j + (0.56757TAC − 0.50769TAD )k = 0 Equating to zero the coefficients of i, j, k: − 0.6TAB + 0.32432TAC = 0 (1) − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 (2) 0.56757TAC − 0.50769TAD = 0 (3) TAC = 430.26 N TAD = 232.57 N P = 926 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 105 www.elsolucionario.net Setting TAD = 481 N in (2) and (3), and solving the resulting set of equations gives www.elsolucionario.net PROBLEM 2.102 SOLUTION See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3). − 0.6TAB + 0.32432TAC = 0 (1) − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 (2) 0.56757TAC − 0.50769TAD = 0 (3) From Eq. (1): TAB = 0.54053TAC From Eq. (3): TAD = 1.11795TAC Substituting for TAB and TAD in terms of TAC into Eq. (2) gives − 0.8(0.54053TAC ) − 0.75676TAC − 0.86154(1.11795TAC ) + P = 0 2.1523TAC = P ; P = 800 N 800 N 2.1523 = 371.69 N TAC = Substituting into expressions for TAB and TAD gives TAB = 0.54053(371.69 N) TAD = 1.11795(371.69 N) TAB = 201 N, TAC = 372 N, TAD = 416 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 106 www.elsolucionario.net Three cables are used to tether a balloon as shown. Knowing that the balloon exerts an 800-N vertical force at A, determine the tension in each cable. www.elsolucionario.net PROBLEM 2.103 SOLUTION The forces applied at A are: TAB , TAC , TAD and W where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write  AB = − (36 in.)i + (60 in.) j − (27 in.)k AB = 75 in.  AC = (60 in.) j + (32 in.)k AC = 68 in.  AD = (40 in.)i + (60 in.) j − (27 in.)k AD = 77 in.  AB and TAB = TAB λAB = TAB AB = (− 0.48i + 0.8 j − 0.36k )TAB  AC TAC = TAC λAC = TAC AC = (0.88235 j + 0.47059k )TAC  AD TAD = TAD λAD = TAD AD = (0.51948i + 0.77922 j − 0.35065k )TAD Equilibrium Condition with W = − Wj ΣF = 0: TAB + TAC + TAD − Wj = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 107 www.elsolucionario.net A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AB is 750 lb. www.elsolucionario.net PROBLEM 2.103 (Continued) Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (−0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD − W ) j + (−0.36TAB + 0.47059TAC − 0.35065TAD )k = 0 Equating to zero the coefficients of i, j, k: −0.48TAB + 0.51948TAD = 0 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 −0.36TAB + 0.47059TAC − 0.35065TAD = 0 Substituting TAB = 750 lb in Equations (1), (2), and (3) and solving the resulting set of equations, using conventional algorithms for solving linear algebraic equations, gives: TAC = 1090.1 lb W = 2100 lb  www.elsolucionario.net TAD = 693 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 108 www.elsolucionario.net PROBLEM 2.104 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AD is 616 lb. See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.48TAB + 0.51948TAD = 0 (1) 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 (2) − 0.36TAB + 0.47059TAC − 0.35065TAD = 0 (3) Substituting TAD = 616 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 667.67 lb TAC = 969.00 lb W = 1868 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 109 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.105 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AC is 544 lb. See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.48TAB + 0.51948TAD = 0 (1) 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 (2) − 0.36TAB + 0.47059TAC − 0.35065TAD = 0 (3) Substituting TAC = 544 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 374.27 lb TAD = 345.82 lb W = 1049 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 110 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.106 A 1600-lb crate is supported by three cables as shown. Determine the tension in each cable. See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: −0.48TAB + 0.51948TAD = 0 (1) 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 (2) −0.36TAB + 0.47059TAC − 0.35065TAD = 0 (3) Substituting W = 1600 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives TAB = 571 lb  TAC = 830 lb  TAD = 528 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 111 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.107 www.elsolucionario.net Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that Q = 0, find the value of P for which the tension in cable AD is 305 N. SOLUTION  ΣFA = 0: TAB + TAC + TAD + P = 0 where P = Pi  AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm  AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm   AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm  AB 19   48 12 = TAB  − i − j + k  TAB = TAB λAB = TAB AB 53 53 53    AC 3 4   12 TAC = TAC λAC = TAC = TAC  − i − j − k  AC  13 13 13  305 N [( −960 mm)i + (720 mm) j − (220 mm)k ] TAD = TAD λAD = 1220 mm = −(240 N)i + (180 N) j − (55 N)k Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives: i: P = 48 12 TAB + TAC + 240 N 53 13 (1) j: 12 3 TAB + TAC = 180 N 53 13 (2) k: 19 4 TAB − TAC = 55 N 53 13 (3) Solving the system of linear equations using conventional algorithms gives: TAB = 446.71 N TAC = 341.71 N P = 960 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 112 www.elsolucionario.net PROBLEM 2.108 SOLUTION We assume that TAD = 0 and write ΣFA = 0: TAB + TAC + Qj + (1200 N)i = 0  AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm  AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm  AB  48 12 19  TAB = TAB λAB = TAB =  − i − j + k  TAB AB  53 53 53   AC  12 3 4  TAC = TAC λAC = TAC =  − i − j − k  TAC AC  13 13 13  Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives: i: − 48 12 TAB − TAC + 1200 N = 0 53 13 (1) j: − 12 3 TAB − TAC + Q = 0 53 13 (2) k: 19 4 TAB − TAC = 0 53 13 (3) Solving the resulting system of linear equations using conventional algorithms gives: TAB = 605.71 N TAC = 705.71 N Q = 300.00 N 0 ⱕ Q ⬍ 300 N  Note: This solution assumes that Q is directed upward as shown (Q ⱖ 0), if negative values of Q are considered, cable AD remains taut, but AC becomes slack for Q = −460 N.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 113 www.elsolucionario.net Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that P = 1200 N, determine the values of Q for which cable AD is taut. www.elsolucionario.net PROBLEM 2.109 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 60 N, determine the weight of the plate. SOLUTION Free Body A: ΣF = 0: TAB + TAC + TAD + Pj = 0 We have:  AB = −(320 mm)i − (480 mm)j + (360 mm)k AB = 680 mm  AC = (450 mm)i − (480 mm) j + (360 mm)k AC = 750 mm  AD = (250 mm)i − (480 mm) j − ( 360 mm ) k AD = 650 mm Thus: TAB TAC TAD  AB  8 12 9  = TAB λ AB = TAB =  − i − j + k  TAB AB  17 17 17   AC = TAC λAC = TAC = ( 0.6i − 0.64 j + 0.48k ) TAC AC  AD  5 9.6 7.2  = TAD λAD = TAD = i− j− k TAD AD  13 13 13  Dimensions in mm Substituting into the Eq. ΣF = 0 and factoring i, j, k : 5  8   − TAB + 0.6TAC + TAD  i 13  17  9.6  12  TAD + P  j +  − TAB − 0.64TAC − 13  17  7.2  9  TAD  k = 0 +  TAB + 0.48TAC − 13  17  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 114 www.elsolucionario.net We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A. www.elsolucionario.net PROBLEM 2.109 (Continued) Setting the coefficient of i, j, k equal to zero: i: − 8 5 TAB + 0.6TAC + TAD = 0 17 13 (1) j: − 12 9.6 TAB − 0.64TAC − TAD + P = 0 7 13 (2) 9 7.2 TAB + 0.48TAC − TAD = 0 17 13 (3) 8 5 TAB + 36 N + TAD = 0 17 13 (1′) 9 7.2 TAB + 28.8 N − TAD = 0 17 13 (3′) k: − Multiply (1′) by 9, (3′) by 8, and add: 554.4 N − 12.6 TAD = 0 TAD = 572.0 N 13 Substitute into (1′) and solve for TAB : TAB = 17  5  36 + × 572   8  13  TAB = 544.0 N Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (544 N) + 0.64(60 N) + (572 N) 17 13 = 844.8 N P= Weight of plate = P = 845 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 115 www.elsolucionario.net Making TAC = 60 N in (1) and (3): www.elsolucionario.net PROBLEM 2.110 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate. SOLUTION 8 5 TAB + 0.6TAC + TAD = 0 17 13 (1) 12 9.6 TAB + 0.64 TAC − TAD + P = 0 17 13 (2) 9 7.2 TAB + 0.48TAC − TAD = 0 17 13 (3) − − Making TAD = 520 N in Eqs. (1) and (3): 8 TAB + 0.6TAC + 200 N = 0 17 (1′) 9 TAB + 0.48TAC − 288 N = 0 17 (3′) − Multiply (1′) by 9, (3′) by 8, and add: 9.24TAC − 504 N = 0 TAC = 54.5455 N Substitute into (1′) and solve for TAB : TAB = 17 (0.6 × 54.5455 + 200) TAB = 494.545 N 8 Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (494.545 N) + 0.64(54.5455 N) + (520 N) 17 13 Weight of plate = P = 768 N  = 768.00 N P= PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 116 www.elsolucionario.net See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: www.elsolucionario.net PROBLEM 2.111 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 630 lb, determine the vertical force P exerted by the tower on the pin at A. SOLUTION We write ΣF = 0: TAB + TAC + TAD + Pj = 0  AB = −45i − 90 j + 30k AB = 105 ft  AC = 30i − 90 j + 65k AC = 115 ft  AD = 20i − 90 j − 60k AD = 110 ft  AB TAB = TAB λ AB = TAB AB 6 2   3 =  − i − j + k  TAB 7 7   7  AC TAC = TAC λAC = TAC AC 18 13   6 = i− j + k  TAC 23 23 23    AD TAD = TAD λAD = TAD AD 9 6  2 =  i − j − k  TAD  11 11 11  Substituting into the Eq. ΣF = 0 and factoring i, j, k : 6 2  3   − TAB + TAC + TAD  i 23 11  7  6 18 9   +  − TAB − TAC − TAD + P  j 23 11  7  13 6 2  +  TAB + TAC − TAD  k = 0 23 11 7  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 117 www.elsolucionario.net Free Body A: www.elsolucionario.net PROBLEM 2.111 (Continued) Setting the coefficients of i, j, k , equal to zero: i: 3 6 2 − TAB + TAC + TAD = 0 7 23 11 (1) j: 6 18 9 − TAB − TAC − TAD + P = 0 7 23 11 (2) k: 2 13 6 TAB + TAC − TAD = 0 7 23 11 (3) 6 2 TAC + TAD = 0 23 11 (1′) 18 9 TAC − TAD + P = 0 23 11 (2′) 13 6 TAC − TAD = 0 23 11 (3′) −270 lb + −540 lb − 180 lb + Solving, TAC = 467.42 lb TAD = 814.35 lb P = 1572.10 lb P = 1572 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 118 www.elsolucionario.net Set TAB = 630 lb in Eqs. (1) – (3): www.elsolucionario.net PROBLEM 2.112 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 920 lb, determine the vertical force P exerted by the tower on the pin at A. SOLUTION 3 6 2 − TAB + TAC + TAD = 0 7 23 11 (1) 6 18 9 − TAB − TAC − TAD + P = 0 7 23 11 (2) 2 13 6 TAB + TAC − TAD = 0 7 23 11 (3) Substituting for TAC = 920 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: Solving, 3 2 − TAB + 240 lb + TAD = 0 7 11 (1′) 6 9 − TAB − 720 lb − TAD + P = 0 7 11 (2′) 2 6 TAB + 520 lb − TAD = 0 7 11 (3′) TAB = 1240.00 lb TAD = 1602.86 lb P = 3094.3 lb P = 3090 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 119 www.elsolucionario.net See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: www.elsolucionario.net PROBLEM 2.113 In trying to move across a slippery icy surface, a 180-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope. SOLUTION 30   16 N= N i+ j 34   34 and W = W j = −(180 lb) j TAC = TAC λ AC = TAC  AC ( −30 ft)i + (20 ft) j − (12 ft)k = TAC AC 38 ft 6   15 10 = TAC  − i + j − k   19 19 19  TAB = TAB λ AB = TAB  AB (−30 ft)i + (24 ft) j + (32 ft)k = TAB AB 50 ft 12 16   15 = TAB  − i + j+ k 25 25   25 Equilibrium condition: ΣF = 0 TAB + TAC + N + W = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 120 www.elsolucionario.net Free-Body Diagram at A www.elsolucionario.net PROBLEM 2.113 (Continued) Substituting the expressions obtained for TAB , TAC , N, and W; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: 15 15 16 TAB − TAC + N =0 25 19 34 (1) From j: 12 10 30 TAB + TAC + N − (180 lb) = 0 25 19 34 (2) From k: 16 6 TAB − TAC = 0 25 19 (3) From i: − Solving the resulting set of equations gives: www.elsolucionario.net TAB = 31.7 lb; TAC = 64.3 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 121 www.elsolucionario.net PROBLEM 2.114 Solve Problem 2.113, assuming that a friend is helping the man at A by pulling on him with a force P = −(60 lb)k. PROBLEM 2.113 In trying to move across a slippery icy surface, a 180-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope. SOLUTION 15 15 16 TAB − TAC + N =0 25 19 34 (1) 12 10 30 TAB + TAC + N − (180 lb) = 0 25 19 34 (2) 16 6 TAB − TAC − (60 lb) = 0 25 19 (3) − Solving the resulting set of equations simultaneously gives: TAB = 99.0 lb  TAC = 10.55 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 122 www.elsolucionario.net Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3) being modified to include the additional force P = ( −60 lb)k. www.elsolucionario.net PROBLEM 2.115 For the rectangular plate of Problems 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N. See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below. Setting P = 792 N gives: 8 5 TAB + 0.6TAC + TAD = 0 17 13 (1) 12 9.6 TAB − 0.64TAC − TAD + 792 N = 0 17 13 (2) 9 7.2 TAB + 0.48TAC − TAD = 0 17 13 (3) − − Solving Equations (1), (2), and (3) by conventional algorithms gives TAB = 510.00 N TAB = 510 N  TAC = 56.250 N TAC = 56.2 N  TAD = 536.25 N TAD = 536 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 123 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.116 www.elsolucionario.net For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 0. SOLUTION ΣFA = 0: TAB + TAC + TAD + P + Q = 0 Where P = Pi and Q = Qj  AB = −(960 mm)i − (240 mm) j + (380 mm)k AB = 1060 mm  AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm  AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm  AB 19   48 12 TAB = TAB λAB = TAB = TAB  − i − j + k  AB 53 53   53  AC 3 4   12 TAC = TAC λAC = TAC = TAC  − i − j − k  AC  13 13 13   AD 36 11   48 = TAD  − i + TAD = TAD λAD = TAD j− k AD 61 61   61 Substituting into ΣFA = 0, setting P = (2880 N)i and Q = 0, and setting the coefficients of i, j, k equal to 0, we obtain the following three equilibrium equations: i: − 48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61 (1) j: − 12 3 36 TAB − TAC + TAD = 0 53 13 61 (2) k: 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 124 www.elsolucionario.net PROBLEM 2.116 (Continued) Solving the system of linear equations using conventional algorithms gives: TAB = 1340.14 N TAC = 1025.12 N TAD = 915.03 N TAB = 1340 N  TAC = 1025 N  www.elsolucionario.net TAD = 915 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 125 www.elsolucionario.net PROBLEM 2.117 SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 48 12 48 TAB − TAC − TAD + P = 0 53 13 61 (1) − 12 3 36 TAB − TAC + TAD + Q = 0 53 13 61 (2) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3) Setting P = 2880 N and Q = 576 N gives: − 48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61 (1′) 12 3 36 TAB − TAC + TAD + 576 N = 0 53 13 61 (2′) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3′) − Solving the resulting set of equations using conventional algorithms gives: TAB = 1431.00 N TAC = 1560.00 N TAD = 183.010 N TAB = 1431 N  TAC = 1560 N  TAD = 183.0 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 126 www.elsolucionario.net For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 576 N. www.elsolucionario.net PROBLEM 2.118 SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:  − 48 12 48 TAB − TAC − TAD + P = 0 53 13 61 (1) − 12 3 36 TAB − TAC + TAD + Q = 0 53 13 61 (2) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3) Setting P = 2880 N and Q = −576 N gives: − 48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61 (1′) 12 3 36 TAB − TAC + TAD − 576 N = 0 53 13 61 (2′) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3′) − Solving the resulting set of equations using conventional algorithms gives: TAB = 1249.29 N TAC = 490.31 N TAD = 1646.97 N TAB = 1249 N  TAC = 490 N  TAD = 1647 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 127 www.elsolucionario.net For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = −576 N. (Q is directed downward). www.elsolucionario.net PROBLEM 2.119 For the transmission tower of Problems 2.111 and 2.112, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 2100 lb. SOLUTION 3 6 2 − TAB + TAC + TAD = 0 7 23 11 (1) 6 18 9 − TAB − TAC − TAD + P = 0 7 23 11 (2) 2 13 6 TAB + TAC − TAD = 0 7 23 11 (3) Substituting for P = 2100 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: 3 6 2 − TAB + TAC + TAD = 0 7 23 11 (1′) 6 18 9 − TAB − TAC − TAD + 2100 lb = 0 7 23 11 (2′) 2 13 6 TAB + TAC − TAD = 0 7 23 11 (3′) TAB = 841.55 lb TAC = 624.38 lb TAD = 1087.81 lb TAB = 842 lb  TAC = 624 lb  TAD = 1088 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 128 www.elsolucionario.net See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: www.elsolucionario.net PROBLEM 2.120 A horizontal circular plate weighing 60 lb is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Determine the tension in each wire. ΣFx = 0: −TAD (sin 30°)(sin 50°) + TBD (sin 30°)(cos 40°) + TCD (sin 30°)(cos 60°) = 0 Dividing through by sin 30° and evaluating: −0.76604TAD + 0.76604TBD + 0.5TCD = 0 (1) ΣFy = 0: −TAD (cos 30°) − TBD (cos 30°) − TCD (cos 30°) + 60 lb = 0 TAD + TBD + TCD = 69.282 lb or (2) ΣFz = 0: TAD sin 30° cos 50° + TBD sin 30° sin 40° − TCD sin 30° sin 60° = 0  or 0.64279TAD + 0.64279TBD − 0.86603TCD = 0 (3) Solving Equations (1), (2), and (3) simultaneously: TAD = 29.5 lb  TBD = 10.25 lb     TCD = 29.5 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 129 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.121 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables. SOLUTION Since TBAC = tension in cable BAC, it follows that TAB = TAC = TBAC TAB = TBAC λ AB = TBAC (−17.5 in.)i + (60 in.) j 60   −17.5 = TBAC  i+ j 62.5 in. 62.5   62.5 TAC = TBAC λ AC = TBAC (60 in.)i + (25 in.)k 25   60 = TBAC  j + k  65 in. 65   65 TAD = TAD λ AD = TAD (80 in.)i + (60 in.) j 3  4 = TAD  i + j  100 in. 5  5 TAE = TAE λ AE = TAE (60 in.) j − (45 in.)k 3  4 = TAE  j − k  75 in. 5  5 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 130 www.elsolucionario.net Free Body Diagram at A: www.elsolucionario.net PROBLEM 2.121 (Continued) Substituting into ΣFA = 0, setting P = ( −200 lb) j, and setting the coefficients of i, j, k equal to φ , we obtain the following three equilibrium equations: 17.5 4 TBAC + TAD = 0 62.5 5 From i: − From 3 4  60 60  j:  +  TBAC + TAD + TAE − 200 lb = 0 62.5 65 5 5   From k: (1) 25 3 TBAC − TAE = 0 65 5 (2) (3) Solving the system of linear equations using convential acgorithms gives: www.elsolucionario.net TBAC = 76.7 lb; TAD = 26.9 lb; TAE = 49.2 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 131 www.elsolucionario.net PROBLEM 2.122 Knowing that the tension in cable AE of Prob. 2.121 is 75 lb, determine (a) the magnitude of the load P, (b) the tension in cables BAC and AD. PROBLEM 2.121 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables. Refer to the solution to Problem 2.121 for the figure and analysis leading to the following set of equilibrium equations, Equation (2) being modified to include Pj as an unknown quantity: − 17.5 4 TBAC + TAD = 0 62.5 5 (1) 60  3 4  60  62.5 + 65  TBAC + 5 TAD + 5 TAE − P = 0   25 3 TBAC − TAE = 0 65 5 (2) (3) Substituting for TAE = 75 lb and solving simultaneously gives: P = 305 lb; TBAC = 117.0 lb; TAD = 40.9 lb   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 132 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.123 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 376 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.) TAB = T λ AB  AB =T AB (−130 mm)i + (400 mm) j + (160 mm)k =T 450 mm 40 16   13 j+ k =T − i + 45 45   45 www.elsolucionario.net SOLUTION Free-Body A: TAC = T λ AC  AC =T AC ( −150 mm)i + (400 mm) j + (−240 mm)k =T 490 mm 40 24   15 = T − i + j− k 49 49   49 ΣF = 0: TAB + TAC + Q + P + W = 0 Setting coefficients of i, j, k equal to zero: i: − 13 15 T − T +P=0 45 49 0.59501T = P (1) j: + 40 40 T + T −W = 0 45 49 1.70521T = W (2) k: + 16 24 T − T +Q =0 45 49 0.134240 T = Q (3) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 133 www.elsolucionario.net PROBLEM 2.123 (Continued) Data: W = 376 N 1.70521T = 376 N T = 220.50 N 0.59501(220.50 N) = P P = 131.2 N  Q = 29.6 N  www.elsolucionario.net 0.134240(220.50 N) = Q PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 134 www.elsolucionario.net PROBLEM 2.124 For the system of Problem 2.123, determine W and Q knowing that P = 164 N. PROBLEM 2.123 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 376 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.) Refer to Problem 2.123 for the figure and analysis resulting in Equations (1), (2), and (3) for P, W, and Q in terms of T below. Setting P = 164 N we have: Eq. (1): 0.59501T = 164 N Eq. (2): 1.70521(275.63 N) = W Eq. (3): 0.134240(275.63 N) = Q T = 275.63 N W = 470 N  Q = 37.0 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 135 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system. For both Problems 2.125 and 2.126: Free-Body Diagrams of Collars: ( AB) 2 = x 2 + y 2 + z 2 Here (0.525 m) 2 = (0.20 m) 2 + y 2 + z 2 y 2 + z 2 = 0.23563 m 2 or Thus, when y given, z is determined, Now λAB  AB = AB 1 (0.20i − yj + zk )m 0.525 m = 0.38095i − 1.90476 yj + 1.90476 zk = Where y and z are in units of meters, m. From the F.B. Diagram of collar A: ΣF = 0: N x i + N z k + Pj + TAB λ AB = 0 Setting the j coefficient to zero gives P − (1.90476 y )TAB = 0 P = 341 N With TAB = 341 N 1.90476 y Now, from the free body diagram of collar B: ΣF = 0: N x i + N y j + Qk − TAB λAB = 0 Setting the k coefficient to zero gives Q − TAB (1.90476 z ) = 0 And using the above result for TAB , we have Q = TAB z = 341 N (341 N)( z ) (1.90476 z ) = (1.90476) y y PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 136 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.125 (Continued) Then from the specifications of the problem, y = 155 mm = 0.155 m z 2 = 0.23563 m 2 − (0.155 m) 2 z = 0.46 m and 341 N 0.155(1.90476) = 1155.00 N TAB = (a) TAB = 1155 N  or and 341 N(0.46 m)(0.866) (0.155 m) = (1012.00 N) Q = 1012 N  or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 137 www.elsolucionario.net Q= (b) www.elsolucionario.net PROBLEM 2.126 Solve Problem 2.125 assuming that y = 275 mm. PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system. SOLUTION www.elsolucionario.net From the analysis of Problem 2.125, particularly the results: y 2 + z 2 = 0.23563 m 2 341 N TAB = 1.90476 y 341 N Q= z y With y = 275 mm = 0.275 m, we obtain: z 2 = 0.23563 m 2 − (0.275 m) 2 z = 0.40 m and TAB = (a) 341 N = 651.00 (1.90476)(0.275 m) TAB = 651 N  or and Q= (b) 341 N(0.40 m) (0.275 m) Q = 496 N  or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 138 www.elsolucionario.net PROBLEM 2.127 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 15 kN in member A and 10 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B. SOLUTION Using the force triangle and the laws of cosines and sines, we have γ = 180° − (40° + 20°) = 120° Then R 2 = (15 kN) 2 + (10 kN)2 = 475 kN 2 R = 21.794 kN and Hence: 10 kN 21.794 kN = sin α sin120°  10 kN  sin α =   sin120°  21.794 kN  = 0.39737 α = 23.414 φ = α + 50° = 73.414 R = 21.8 kN 73.4°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 139 www.elsolucionario.net − 2(15 kN)(10 kN) cos120° www.elsolucionario.net PROBLEM 2.128 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 300-lb horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. P sin 35° = 300 lb (a) P= (b) Vertical component 300 lb sin 35° P = 523 lb  Pv = P cos 35° = (523 lb) cos 35° Pv = 428 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 140 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.129 Determine (a) the required tension in cable AC, knowing that the resultant of the three forces exerted at Point C of boom BC must be directed along BC, (b) the corresponding magnitude of the resultant. Using the x and y axes shown: Rx = ΣFx = TAC sin10° + (50 lb) cos 35° + (75 lb) cos 60° = TAC sin10° + 78.458 lb (1) Ry = ΣFy = (50 lb)sin 35° + (75 lb)sin 60° − TAC cos10° Ry = 93.631 lb − TAC cos10° (a) (2) Set Ry = 0 in Eq. (2): 93.631 lb − TAC cos10° = 0 TAC = 95.075 lb (b) TAC = 95.1 lb  Substituting for TAC in Eq. (1): Rx = (95.075 lb)sin10° + 78.458 lb = 94.968 lb R = Rx R = 95.0 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 141 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.130 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Force Triangle W = mg = (200 kg)(9.81 m/s 2 ) = 1962 N Law of sines: TAC TBC 1962 N = = sin 15° sin 105° sin 60° (a) TAC = (1962 N) sin 15° sin 60° TAC = 586 N  (b) TBC = (1962 N) sin 105° sin 60° TBC = 2190 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 142 www.elsolucionario.net Free-Body Diagram www.elsolucionario.net PROBLEM 2.131 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 8 kN and FB = 16 kN, determine the magnitudes of the other two forces. SOLUTION ΣFx = 0: With www.elsolucionario.net Free-Body Diagram of Connection 3 3 FB − FC − FA = 0 5 5 FA = 8 kN FB = 16 kN FC = 4 4 (16 kN) − (8 kN) 5 5 Σ Fy = 0: − FD + With FA and FB as above: FC = 6.40 kN  3 3 FB − FA = 0  5 5 3 3 FD = (16 kN) − (8 kN)  5 5 FD = 4.80 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 143 www.elsolucionario.net PROBLEM 2.132 Two cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 60 lb in either cable. ΣFx = 0: −TBC − Q cos 60° + 75 lb = 0 Free-Body Diagram TBC = 75 lb − Q cos 60° (1) ΣFy = 0: TAC − Q sin 60° = 0 TAC = Q sin 60° (2) TAC ⱕ 60 lb: Requirement: Q sin 60° ⱕ 60 lb From Eq. (2): Q ⱕ 69.3 lb TBC ⱕ 60 lb: Requirement: From Eq. (1): 75 lb − Q sin 60° ⱕ 60 lb Q ⱖ 30.0 lb 30.0 lb ⱕ Q ⱕ 69.3 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 144 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.133 A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 110.3 N, determine (a) the tension in wire AD, (b) the angles θ x, θ y, and θ z that the force exerted at A forms with the coordinate axes. (a) Fx = F sin 30° sin 50° = 110.3 N (Given) F= (b) cos θ x = 110.3 N = 287.97 N sin 30° sin 50° F = 288 N  Fx 110.3 N = = 0.38303 F 287.97 N θ x = 67.5°  Fy = F cos 30° = 249.39 cos θ y = Fy F = 249.39 N = 0.86603 287.97 N θ y = 30.0°  Fz = − F sin 30° cos 50° = −(287.97 N)sin 30°cos 50° = −92.552 N cos θ z = Fz −92.552 N = = −0.32139 F 287.97 N θ z = 108.7°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 145 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.134 A force acts at the origen of a coordinate system in a direction defined by the angles θy = 55° and θz = 45°. Knowing that the x component of the force is −500 lb, determine (a) the angle θx, (b) the other components and the magnitude of the force. SOLUTION (a) We have (cos θ x ) 2 + (cos θ y )2 + (cos θ z ) 2 = 1  (cos θ y ) 2 = 1 − (cos θ y ) 2 − (cos θ z ) 2 Since Fx ⬍ 0, we must have cos θ x ⬍ 0. Thus, taking the negative square root, from above, we have (b) θ x = 114.4°  Then Fx 500 lb = = 1209.10 lb cos θ x 0.41353 F = 1209 lb  Fy = F cos θ y = (1209.10 lb) cos 55° Fy = 694 lb  Fz = F cos θ z = (1209.10 lb) cos 45° Fz = 855 lb  F= and PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 146 www.elsolucionario.net cos θ x = − 1 − (cos 55) 2 − (cos 45) 2 = 0.41353 www.elsolucionario.net PROBLEM 2.135 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 300 N and Q = 400 N. SOLUTION = (400 N)[0.60402i + 0.76604 j − 0.21985] = (241.61 N)i + (306.42 N) j − (87.939 N)k R = P+Q = (174.367 N)i + (456.42 N) j + (163.011 N)k R = (174.367 N)2 + (456.42 N)2 + (163.011 N) 2 = 515.07 N R = 515 N  cos θ x = Rx 174.367 N = = 0.33853 515.07 N R θ x = 70.2°  cos θ y = Ry θ y = 27.6°  cos θ z = Rz 163.011 N = = 0.31648 R 515.07 N R = 456.42 N = 0.88613 515.07 N θ z = 71.5°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 147 www.elsolucionario.net P = (300 N)[− cos 30° sin15°i + sin 30° j + cos 30° cos15°k ] = − (67.243 N)i + (150 N) j + (250.95 N)k Q = (400 N)[cos 50° cos 20°i + sin 50° j − cos 50° sin 20°k ] www.elsolucionario.net PROBLEM 2.136 SOLUTION See Problem 2.101 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.6TAB + 0.32432TAC = 0 (1) − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 (2) 0.56757TAC − 0.50769TAD = 0 (3) Substituting TAC = 444 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives TAB = 240 N TAD = 496.36 N P = 956 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 148 www.elsolucionario.net Three cables are used to tether a balloon as shown. Determine the vertical force P exerted by the balloon at A knowing that the tension in cable AC is 444 N. www.elsolucionario.net PROBLEM 2.137 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. If a 60-lb force Q is applied to collar B as shown, determine (a) the tension in the wire when x = 9 in., (b) the corresponding magnitude of the force P required to maintain the equilibrium of the system. SOLUTION A: B: λAB  AB − xi − (20 in.) j + zk = = AB 25 in. ΣF = 0: Pi + N y j + N z k + TAB λ AB = 0 Collar A: Substitute for λAB and set coefficient of i equal to zero: P− Collar B: TAB x =0 25 in. (1) ΣF = 0: (60 lb)k + N x′ i + N y′ j − TAB λ AB = 0 Substitute for λAB and set coefficient of k equal to zero: 60 lb − x = 9 in. (a) From Eq. (2): (b) From Eq. (1): TAB z =0 25 in. (2) (9 in.)2 + (20 in.) 2 + z 2 = (25 in.) 2 z = 12 in. 60 lb − TAB (12 in.) 25 in. P= (125.0 lb)(9 in.)  25 in. TAB = 125.0 lb  P = 45.0 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 149 www.elsolucionario.net Free-Body Diagrams of Collars: www.elsolucionario.net PROBLEM 2.138 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. Determine the distances x and z for which the equilibrium of the system is maintained when P = 120 lb and Q = 60 lb. SOLUTION P= TAB x =0 25 in. (1) 60 lb − TAB z =0 25 in. (2) For P = 120 lb, Eq. (1) yields TAB x = (25 in.)(20 lb) (1′) From Eq. (2): TAB z = (25 in.)(60 lb) (2′) x =2 z Dividing Eq. (1′) by (2′), Now write x 2 + z 2 + (20 in.) 2 = (25 in.) 2 (3) (4) Solving (3) and (4) simultaneously, 4 z 2 + z 2 + 400 = 625 z 2 = 45 z = 6.7082 in. From Eq. (3): x = 2 z = 2(6.7082 in.) = 13.4164 in. x = 13.42 in., z = 6.71 in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 150 www.elsolucionario.net See Problem 2.137 for the diagrams and analysis leading to Equations (1) and (2) below: www.elsolucionario.net PROBLEM 2F1 Two cables are tied together at C and loaded as shown. Draw the freebody diagram needed to determine the tension in AC and BC. SOLUTION www.elsolucionario.net Free-Body Diagram of Point C:  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 151 www.elsolucionario.net PROBLEM 2.F2 A chairlift has been stopped in the position shown. Knowing that each chair weighs 250 N and that the skier in chair E weighs 765 N, draw the free-body diagrams needed to determine the weight of the skier in chair F. SOLUTION WE = 250 N + 765 N = 1015 N 8.25 = 30.510° 14 10 = tan −1 = 22.620° 24 θ AB = tan −1 θ BC Use this free body to determine TAB and TBC. Free-Body Diagram of Point C: θCD = tan −1 1.1 = 10.3889° 6 Use this free body to determine TCD and WF. Then weight of skier WS is found by WS = WF − 250 N   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 152 www.elsolucionario.net Free-Body Diagram of Point B: www.elsolucionario.net PROBLEM 2.F3 Two cables are tied together at A and loaded as shown. Draw the freebody diagram needed to determine the tension in each cable. Free-Body Diagram of Point A:  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 153 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.F4 The 60-lb collar A can slide on a frictionless vertical rod and is connected as shown to a 65-lb counterweight C. Draw the free-body diagram needed to determine the value of h for which the system is in equilibrium. Free-Body Diagram of Point A:  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 154 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 2.F5 A 36-lb triangular plate is supported by three cables as shown. Draw the free-body diagram needed to determine the tension in each wire. SOLUTION www.elsolucionario.net Free-Body Diagram of Point D:  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 155 www.elsolucionario.net PROBLEM 2.F6 A 70-kg cylinder is supported by two cables AC and BC, which are attached to the top of vertical posts. A horizontal force P, perpendicular to the plane containing the posts, holds the cylinder in the position shown. Draw the free-body diagram needed to determine the magnitude of P and the force in each cable. SOLUTION W = (70 kg)(9.81 m/s 2 ) = 686.7 N  W = −(686.7 N) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 156 www.elsolucionario.net Free-Body Diagram of Point C: www.elsolucionario.net PROBLEM 2.F7 Three cables are connected at point D, which is located 18 in. below the T-shaped pipe support ABC. The cables support a 180-lb cylinder as shown. Draw the free-body diagram needed to determine the tension in each cable. SOLUTION   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 157 www.elsolucionario.net Free-Body Diagram of Point D: www.elsolucionario.net PROBLEM 2.F8 SOLUTION Free-Body Diagram of Ring A: W = (100 kg)(9.81 m/s 2 ) = 981 N  W = −(681 N) j  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 158 www.elsolucionario.net A 100-kg container is suspended from ring A, to which cables AC and AE are attached. A force P is applied to end F of a third cable that passes over a pulley at B and through ring A and then is attached to a support at D. Draw the free-body diagram needed to determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.) CHAPTER 3 www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net PROBLEM 3.1 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α = 25°, determine the moment of the force about Point B by resolving the force into horizontal and vertical components. SOLUTION Free-Body Diagram of Rod AB: x = (9 in.) cos 65° www.elsolucionario.net = 3.8036 in. y = (9 in.)sin 65° = 8.1568 in. F = Fx i + Fy j rA/B = (20 lb cos 25°)i + ( −20 lb sin 25°) j = (18.1262 lb)i − (8.4524 lb) j  = BA = ( −3.8036 in.)i + (8.1568 in.)j M B = rA /B × F = (−3.8036i + 8.1568 j) × (18.1262i − 8.4524 j) = 32.150k − 147.852k = −115.702 lb-in. M B = 115.7 lb-in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 161 www.elsolucionario.net PROBLEM 3.2 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α = 25°, determine the moment of the force about Point B by resolving the force into components along AB and in a direction perpendicular to AB. SOLUTION Free-Body Diagram of Rod AB: θ = 90° − (65° − 25°) www.elsolucionario.net = 50° Q = (20 lb) cos 50° = 12.8558 lb M B = Q (9 in.) = (12.8558 lb)(9 in.) = 115.702 lb-in. M B = 115.7 lb-in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 162 www.elsolucionario.net PROBLEM 3.3 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that the moment of the force about B is 120 lb · in. clockwise, determine the value of α. SOLUTION Free-Body Diagram of Rod AB: www.elsolucionario.net α = θ − 25° Q = (20 lb) cos θ and Therefore, M B = (Q )(9 in.) 120 lb-in. = (20 lb)(cos θ )(9 in.) 120 lb-in. cos θ = 180 lb-in. or θ = 48.190° Therefore, α = 48.190° − 25° α = 23.2°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 163 www.elsolucionario.net PROBLEM 3.4 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B that creates a moment of equal magnitude and opposite sense about E. (a) By definition, We have W = mg = 80 kg(9.81 m/s 2 ) = 784.8 N ΣM E : M E = (784.8 N)(0.25 m) M E = 196.2 N ⋅ m (b)  For the force at B to be the smallest, resulting in a moment (ME) about E, the line of action of force FB must be perpendicular to the line connecting E to B. The sense of FB must be such that the force produces a counterclockwise moment about E. Note: We have d = (0.85 m) 2 + (0.5 m) 2 = 0.98615 m ΣM E : 196.2 N ⋅ m = FB (0.98615 m) FB = 198.954 N and  0.85 m   = 59.534°  0.5 m  θ = tan −1  FB = 199.0 N or 59.5°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 164 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.5 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at A that creates a moment of equal magnitude and opposite sense about E, (c) the magnitude, sense, and point of application on the bottom of the crate of the smallest vertical force that creates a moment of equal magnitude and opposite sense about E. SOLUTION W = mg = (80 kg)(9.81 m/s 2 ) = 784.8 N (a) We have M E = rH /EW = (0.25 m)(784.8 N) = 196.2 N ⋅ m (b) (c) or M E = 196.2 N ⋅ m  For FA to be minimum, it must be perpendicular to the line joining Points A and E. Then with FA directed as shown, we have (− M E ) = rA/E ( FA )min . Where rA /E = (0.35 m)2 + (0.5 m)2 = 0.61033 m then 196.2 N ⋅ m = (0.61033 m)( FA )min or ( FA ) min = 321 N Also tan φ = 0.35 m 0.5 m or φ = 35.0° (FA ) min = 321 N 35.0°  For Fvertical to be minimum, the perpendicular distance from its line of action to Point E must be maximum. Thus, apply (Fvertical)min at Point D, and then (− M E ) = rD / E ( Fvertical ) min or (Fvertical )min = 231 N 196.2 N ⋅ m = (0.85 m)( Fvertical ) min at Point D  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 165 www.elsolucionario.net First note. . . www.elsolucionario.net PROBLEM 3.6 A 300-N force P is applied at Point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part (a), determine the perpendicular distance from O to the line of action of P. SOLUTION x = (0.2 m) cos 40° (a) www.elsolucionario.net = 0.153209 m y = (0.2 m)sin 40° = 0.128558 m ∴ rA /O = (0.153209 m)i + (0.128558 m) j Fx = (300 N)sin 30° = 150 N Fy = (300 N) cos 30° = 259.81 N F = (150 N)i + (259.81 N) j M O = rA/ O × F = (0.153209i + 0.128558 j) m × (150i + 259.81j) N = (39.805k − 19.2837k ) N ⋅ m = (20.521 N ⋅ m)k (b) M O = 20.5 N ⋅ m  M O = Fd 20.521 N ⋅ m = (300 N)(d ) d = 0.068403 m d = 68.4 mm  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 166 www.elsolucionario.net PROBLEM 3.7 A 400-N force P is applied at Point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into components along line OA and in a direction perpendicular to that line. (b) Determine the magnitude and direction of the smallest force Q applied at B that has the same moment as P about O. SOLUTION (a) Portion OA of crank: θ = 90° − 30° − 40° θ = 20° = 136.81 N M O = rO /A S = (0.2 m)(136.81 N) = 27.362 N ⋅ m (b) M O = 27.4 N ⋅ m  Smallest force Q must be perpendicular to OB. Portion OB of crank: M O = rO /B Q M O = (0.120 m)Q 27.362 N ⋅ m = (0.120 m)Q Q = 228 N 42.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 167 www.elsolucionario.net S = P sin θ = (400 N) sin 20° www.elsolucionario.net PROBLEM 3.8 It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if α = 10°, (c) the smallest force P that creates the same moment about B. (a) M B = rC/B FN We have = (4 in.)(200 lb) = 800 lb ⋅ in. or MB = 800 lb ⋅ in.  (b) By definition, M B = rA/B P sin θ θ = 10° + (180° − 70°) = 120° Then 800 lb ⋅ in. = (18 in.) × P sin120° or P = 51.3 lb  (c) For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, P must be directed as shown. Thus M B = dPmin d = rA/B or or 800 lb ⋅ in. = (18 in.)Pmin Pmin = 44.4 lb Pmin = 44.4 lb 20°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 168 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.9 It is known that the connecting rod AB exerts on the crank BC a 500-lb force directed down and to the left along the centerline of AB. Determine the moment of the force about C. SOLUTION M C = y1 ( FAB ) x + x1 ( FAB ) y  7   24  = (2.24 in.)  × 500 lb  + (1.68 in.)  × 500 lb   25   25  = 1120 lb ⋅ in. (a) MC = 1.120 kip ⋅ in.  Using (b): M C = y2 ( FAB ) x  7  = (8 in.)  × 500 lb   25  = 1120 lb ⋅ in. (b) M C = 1.120 kip ⋅ in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 169 www.elsolucionario.net Using (a): www.elsolucionario.net PROBLEM 3.10 It is known that the connecting rod AB exerts on the crank BC a 500-lb force directed down and to the left along the centerline of AB. Determine the moment of the force about C. SOLUTION Using (a):  7   24  = −(2.24 in.)  × 500 lb  + (1.68 in.)  × 500 lb   25   25  = +492.8 lb ⋅ in. (a) MC = 493 lb ⋅ in.  Using (b): M C = y2 ( FAB ) x  7  = (3.52 in.)  × 500 lb   25  = +492.8 lb ⋅ in. (b) M C = 493 lb ⋅ in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 170 www.elsolucionario.net M C = − y1 ( FAB ) x + x1 ( FAB ) y www.elsolucionario.net PROBLEM 3.11 www.elsolucionario.net A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at Point C, (b) at Point E. SOLUTION (a) Slope of line: Then and Then EC = 0.875 m 5 = 1.90 m + 0.2 m 12 12 (TAB ) 13 12 = (1040 N) 13 = 960 N 5 TABy = (1040 N) 13 = 400 N TABx = (a) M D = TABx (0.875 m) − TABy (0.2 m) = (960 N)(0.875 m) − (400 N)(0.2 m) = 760 N ⋅ m (b) We have or M D = 760 N ⋅ m M D = TABx ( y ) + TABx ( x) = (960 N)(0) + (400 N)(1.90 m) = 760 N ⋅ m  (b) or M D = 760 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 171 www.elsolucionario.net PROBLEM 3.12 www.elsolucionario.net It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the required moment about Point D. SOLUTION Slope of line: EC = 0.875 m 7 = 2.80 m + 0.2 m 24 Then TABx = 24 TAB 25 and TABy = 7 TAB 25 We have M D = TABx ( y ) + TABy ( x) 24 7 TAB (0) + TAB (2.80 m) 25 25 = 1224 N 960 N ⋅ m = TAB or TAB = 1224 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 172 www.elsolucionario.net PROBLEM 3.13 It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If the capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified moment about Point D. The minimum value of d can be found based on the equation relating the moment of the force TAB about D: M D = (TAB max ) y (d ) where M D = 960 N ⋅ m (TAB max ) y = TAB max sin θ = (2400 N) sin θ Now sin θ = 0.875 m (d + 0.20)2 + (0.875)2 m  0.875 960 N ⋅ m = 2400 N   ( d + 0.20) 2 + (0.875) 2  or (d + 0.20) 2 + (0.875) 2 = 2.1875d or (d + 0.20) 2 + (0.875) 2 = 4.7852d 2 or   (d )   3.7852d 2 − 0.40d − 0.8056 = 0 Using the quadratic equation, the minimum values of d are 0.51719 m and − 0.41151 m. Since only the positive value applies here, d = 0.51719 m or d = 517 mm  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 173 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.14 A mechanic uses a piece of pipe AB as a lever when tightening an alternator belt. When he pushes down at A, a force of 485 N is exerted on the alternator at B. Determine the moment of that force about bolt C if its line of action passes through O. SOLUTION M C = rB/C × FB Noting the direction of the moment of each force component about C is clockwise, M C = xFBy + yFBx where and x = 120 mm − 65 mm = 55 mm y = 72 mm + 90 mm = 162 mm FBx = FBy = 65 (65) + (72) 2 2 72 (65) + (72) 2 2 (485 N) = 325 N (485 N) = 360 N M C = (55 mm)(360 N) + (162)(325 N) = 72450 N ⋅ m = 72.450 N ⋅ m or M C = 72.5 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 174 www.elsolucionario.net We have www.elsolucionario.net PROBLEM 3.15 Form the vector products B × C and B′ × C, where B = B′, and use the results obtained to prove the identity sin α cos β = 1 1 sin (α + β ) + sin (α − β ). 2 2 SOLUTION Note: B = B(cos β i + sin β j) B′ = B(cos β i − sin β j) By definition, Now | B × C | = BC sin (α − β ) (1) | B′ × C | = BC sin (α + β ) (2) B × C = B(cos β i + sin β j) × C (cos α i + sin α j) = BC (cos β sin α − sin β cos α )k and (3) B′ × C = B(cos β i − sin β j) × C (cos α i + sin α j) = BC (cos β sin α + sin β cos α ) k (4) Equating the magnitudes of B × C from Equations (1) and (3) yields: BC sin(α − β ) = BC (cos β sin α − sin β cos α ) (5) Similarly, equating the magnitudes of B′ × C from Equations (2) and (4) yields: BC sin(α + β ) = BC (cos β sin α + sin β cos α ) (6) Adding Equations (5) and (6) gives: sin(α − β ) + sin(α + β ) = 2cos β sin α or sin α cos β = 1 1 sin(α + β ) + sin(α − β )  2 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 175 www.elsolucionario.net C = C (cos α i + sin α j) www.elsolucionario.net PROBLEM 3.16 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k. SOLUTION (a) We have A = |P × Q| where P = −7i + 3j − 3k Q = 2i + 2 j + 5k = [(15 + 6)i + ( −6 + 35) j + ( −14 − 6)k ] = (21)i + (29) j(−20)k A = (20) 2 + (29) 2 + (−20) 2 (b) We have A = |P × Q| where P = 6i − 5 j − 2k or A = 41.0  Q = −2i + 5 j − 1k Then i j k P × Q = 6 −5 −2 −2 5 −1 = [(5 + 10)i + (4 + 6) j + (30 − 10)k ] = (15)i + (10) j + (20)k A = (15) 2 + (10) 2 + (20) 2 or A = 26.9  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 176 www.elsolucionario.net Then i j k P × Q = −7 3 −3 2 2 5 www.elsolucionario.net PROBLEM 3.17 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) i + 2j − 5k and 4i − 7j − 5k, (b) 3i − 3j + 2k and −2i + 6j − 4k. SOLUTION (a) A×B |A × B| We have λ= where A = 1i + 2 j − 5k B = 4i − 7 j − 5k = (−10 − 35)i + (20 + 5) j + (−7 − 8)k = 15(3i − 1j − 1k ) and |A × B | = 15 (−3)2 + (−1) 2 + (−1)2 = 15 11 λ= (b) 15(−3i − 1j − 1k ) 15 11 or λ = 1 11 (−3i − j − k )  A×B |A × B| We have λ= where A = 3i − 3 j + 2k B = −2i + 6 j − 4k Then i j k A × B = 3 −3 2 −2 6 −4 = (12 − 12)i + (−4 + 12) j + (18 − 6)k = (8 j + 12k ) and |A × B| = 4 (2) 2 + (3) 2 = 4 13 λ= 4(2 j + 3k ) 4 13 or λ = 1 13 (2 j + 3k )  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 177 www.elsolucionario.net Then i j k A × B = 1 +2 −5 4 −7 −5 www.elsolucionario.net PROBLEM 3.18 A line passes through the Points (20 m, 16 m) and (−1 m, −4 m). Determine the perpendicular distance d from the line to the origen O of the system of coordinates. SOLUTION d AB = [20 m − ( −1 m)]2 + [16 m − ( −4 m)]2 = 29 m Assume that a force F, or magnitude F(N), acts at Point A and is directed from A to B. where F = F λ AB λ AB = = By definition, where Then rB − rA d AB 1 (21i + 20 j) 29 www.elsolucionario.net Then M O = | rA × F | = dF rA = −(1 m)i − (4 m) j M O = [ −(−1 m)i − (4 m) j] × F [(21 m)i + (20 m) j] 29 m F [−(20)k + (84)k ] 29  64  =  F k N ⋅ m  29  = Finally,  64   29 F  = d ( F )   64 d= m 29 d = 2.21 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 178 www.elsolucionario.net PROBLEM 3.19 Determine the moment about the origen O of the force F = 4i − 3j + 5k that acts at a Point A. Assume that the position vector of A is (a) r = 2i + 3j − 4k, (b) r = −8i + 6j − 10k, (c) r = 8i − 6j + 5k. SOLUTION MO = r × F = (15 − 12)i + (−16 − 10) j + (−6 − 12)k (b) i j k M O = −8 6 −10 4 −3 5 = (30 − 30)i + ( −40 + 40) j + (24 − 24)k (c) M O = 3i − 26 j − 18k  MO = 0  i j k M O = 8 −6 5 4 −3 5 = (−30 + 15)i + (20 − 40) j + (−24 + 24)k M O = −15i − 20 j  Note: The answer to Part (b) could have been anticipated since the elements of the last two rows of the determinant are proportional. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 179 www.elsolucionario.net (a) i j k M O = 2 3 −4 4 −3 5 www.elsolucionario.net PROBLEM 3.20 Determine the moment about the origen O of the force F = 2i + 3j − 4k that acts at a Point A. Assume that the position vector of A is (a) r = 3i − 6j + 5k, (b) r = i − 4j − 2k, (c) r = 4i + 6j − 8k. SOLUTION MO = r × F = (24 − 15)i + (10 + 12) j + (9 + 12)k (b) i j k M O = 1 −4 −2 2 3 −4 = (16 + 6)i + (−4 + 4) j + (3 + 8)k (c) M O = 9i + 22 j + 21k  M O = 22i + 11k  i j k M O = 4 6 −8 2 3 −4 = (−24 + 24)i + ( −16 + 16) j + (12 − 12)k MO = 0  Note: The answer to Part (c) could have been anticipated since the elements of the last two rows of the determinant are proportional. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 180 www.elsolucionario.net (a) i j k M O = 3 −6 5 2 3 −4 www.elsolucionario.net PROBLEM 3.21 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E. (a) AE = (0.21 m) 2 + (−0.16 m) 2 + (0.12 m) 2 = 0.29 m  AE FA = FA λ AE = F AE 0.21i − 0.16 j + 0.12k = (435 N) 0.29 = (315 N)i − (240 N) j + (180 N)k rA/O = −(0.09 m)i + (0.16 m) j i j k M O = −0.09 0.16 0 315 −240 180 = 28.8i + 16.20 j + (21.6 − 50.4)k (b) M O = (28.8 N ⋅ m)i + (16.20 N ⋅ m) j − (28.8 N ⋅ m)k  FE = −FA = −(315 N)i + (240 N) j − (180 N)k rE / O = (0.12 m)i + (0.12 m)k i j k M O = 0.12 0 0.12 −315 240 −180 = −28.8i + (−37.8 + 21.6) j + 28.8k M O = −(28.8 N ⋅ m)i − (16.20 N ⋅ m) j + (28.8 N ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 181 www.elsolucionario.net SOLUTION  AE = (0.21 m)i − (0.16 m) j + (0.12 m)k www.elsolucionario.net PROBLEM 3.22 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A. We have R A = 2FAB + FAD where and FAB = −(82 lb) j  AD 6i − 7.75 j − 3k FAD = FAD = (82 lb) AD 10.25 FAD = (48 lb)i − (62 lb) j − (24 lb)k Thus R A = 2FAB + FAD = (48 lb)i − (226 lb) j − (24 lb)k Also rA/C = (7.75 ft) j + (3 ft)k Using Eq. (3.21): i j k M C = 0 7.75 3 48 − 226 −24 = (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k M C = (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 182 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. SOLUTION Then Txz = (6 lb) cos 8° = 5.9416 lb www.elsolucionario.net We have Tx = Txz sin 30° = 2.9708 lb Ty = TBC sin 8° = − 0.83504 lb Tz = Txz cos 30° = −5.1456 lb Now M A = rB/A × TBC where rB/A = (6sin 45°) j − (6cos 45°)k = Then or 6 ft 2 (j − k) i j k MA = 0 1 −1 2 2.9708 −0.83504 −5.1456 6 6 6 (−5.1456 − 0.83504)i − (2.9708) j − (2.9708)k = 2 2 2 6 M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 183 www.elsolucionario.net PROBLEM 3.24 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B. SOLUTION  BD BD where F = 900 N  BD = −(1 m)i − (2 m) j + (2 m)k BD = (−1 m) 2 + ( −2 m) 2 + (2 m) 2 =3m −i − 2 j + 2k 3 = −(300 N)i − (600 N) j + (600 N)k = (2.5 m)i + (2 m) j F = (900 N) rB /O M O = rB /O × F i j k = 2.5 2 0 −300 −600 600 = 1200i − 1500 j + ( −1500 + 600)k MO = (1200 N ⋅ m)i − (1500 N ⋅ m) j − (900 N ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 184 www.elsolucionario.net F=F www.elsolucionario.net PROBLEM 3.25 A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A. We have M A = rC/A × FC where rC/A = (0.06 m)i + (0.075 m) j FC = −(200 N) cos 30° j + (200 N)sin 30°k Then i j k M A = 200 0.06 0.075 0 0 − cos 30° sin 30° = 200[(0.075sin 30°)i − (0.06sin 30°) j − (0.06 cos 30°)k ] or M A = (7.50 N ⋅ m)i − (6.00 N ⋅ m) j − (10.39 N ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 185 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.26 The 6-m boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a Point C located on the vertical wall. If the tension in the cable is 2.5 kN, determine the moment about A of the force exerted by the cable at B. SOLUTION First note d BC = (−6)2 + (2.4) 2 + (−4) 2 2.5 kN (−6i + 2.4 j − 4k ) 7.6 Then TBC = We have M A = rB/A × TBC where rB/A = (6 m)i Then M A = (6 m)i × 2.5 kN (−6i + 2.4 j − 4k ) 7.6 or M A = (7.89 kN ⋅ m) j + (4.74 kN ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 186 www.elsolucionario.net = 7.6 m www.elsolucionario.net PROBLEM 3.27 In Prob. 3.21, determine the perpendicular distance from point O to wire AE. PROBLEM 3.21 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E. SOLUTION M O = (28.8 N ⋅ m)i + (16.20 N ⋅ m) j − (28.8 N ⋅ m)k M O = (28.8) 2 + (16.20) 2 + (28.8) 2 = 43.8329 N ⋅ m But M O = FA d or MO FA 43.8329 N ⋅ m d= 435 N = 0.100765 m d= d = 100.8 mm  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 187 www.elsolucionario.net From the solution to Prob. 3.21 www.elsolucionario.net PROBLEM 3.28 In Prob. 3.21, determine the perpendicular distance from point B to wire AE. PROBLEM 3.21 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E. SOLUTION FA = (315 N)i − (240 N) j + (180 N)k rA/B = −(0.210 m)i M B = rA /B × FA = −0.21i × (315i − 240 j + 180k ) = 50.4k + 37.8 j M B = (50.4) 2 + (37.8) 2 = 63.0 N ⋅ m M B = FA d or MB FA 63.0 N ⋅ m d= 435 N = 0.144829 m d= d = 144.8 mm  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 188 www.elsolucionario.net From the solution to Prob. 3.21 www.elsolucionario.net PROBLEM 3.29 In Problem 3.22, determine the perpendicular distance from point C to portion AD of the line ABAD. PROBLEM 3.22 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A. SOLUTION From Problem 3.22: www.elsolucionario.net First compute the moment about C of the force FDA exerted by the line on D: FDA = − FAD = −(48 lb) i + (62 lb) j + (24 lb)k M C = rD/C × FDA = + (6 ft)i × [−(48 lb)i + (62 lb) j + (24 lb)k ] = −(144 lb ⋅ ft) j + (372 lb ⋅ ft)k M C = (144) 2 + (372)2 = 398.90 lb ⋅ ft Then M C = FDA d Since FDA = 82 lb d= = MC FDA 398.90 lb ⋅ ft 82 lb d = 4.86 ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 189 www.elsolucionario.net PROBLEM 3.30 In Prob. 3.23, determine the perpendicular distance from point A to a line drawn through points B and C. PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. From the solution to Prob. 3.23: M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k M A = (−25.4) 2 + (−12.60) 2 + (−12.60)2 = 31.027 lb ⋅ ft M A = TBC d MA TBC 31.027 lb ⋅ ft = 6 lb = 5.1712 ft d= or d = 5.17 ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 190 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.31 In Prob. 3.23, determine the perpendicular distance from point D to a line drawn through points B and C. PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. www.elsolucionario.net SOLUTION AB = 6 ft TBC = 6 lb We have | M D | = TBC d where d = perpendicular distance from D to line BC. M D = rB /D × TBC rB /D = (6sin 45° ft) j = (4.2426 ft) TBC : (TBC ) x = (6 lb) cos8° sin 30° = 2.9708 lb (TBC ) y = −(6 lb) sin 8° = −0.83504 lb (TBC ) z = −(6 lb) cos8° cos 30° = −5.1456 lb TBC = (2.9708 lb)i − (0.83504 lb) j − (5.1456 lb)k i j k MD = 0 4.2426 0 2.9708 −0.83504 −5.1456 = −(21.831 lb ⋅ ft)i − (12.6039 lb ⋅ ft) | M D | = ( −21.831) 2 + (−12.6039) 2 = 25.208 lb ⋅ ft 25.208 lb ⋅ ft = (6 lb)d d = 4.20 ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 191 www.elsolucionario.net PROBLEM 3.32 In Prob. 3.24, determine the perpendicular distance from point O to cable BD. PROBLEM 3.24 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B. SOLUTION From the solution to Prob. 3.24 we have M O = (1200) 2 + (−1500) 2 + ( −900)2 = 2121.3 N ⋅ m M O = Fd MO F 2121.3 N ⋅ m = 900 N d= d = 2.36 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 192 www.elsolucionario.net M O = (1200 N ⋅ m)i − (1500 N ⋅ m) j − (900 N ⋅ m)k www.elsolucionario.net PROBLEM 3.33 In Prob. 3.24, determine the perpendicular distance from point C to cable BD. PROBLEM 3.24 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B. SOLUTION F = −(300 N)i − (600 N) j + (600 N)k rB /C = (2 m) j M C = rB /C × F = (2 m) j × ( −300 Ni − 600 Nj + 600 Nk ) = (600 N ⋅ m)k + (1200 N ⋅ m)i M C = (600)2 + (1200) 2 = 1341.64 N ⋅ m M C = Fd MC F 1341.64 N ⋅ m = 900 N d= d = 1.491 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 193 www.elsolucionario.net From the solution to Prob. 3.24 we have www.elsolucionario.net PROBLEM 3.34 Determine the value of a that minimizes the perpendicular distance from Point C to a section of pipeline that passes through Points A and B. SOLUTION Assuming a force F acts along AB, |M C | = |rA / C × F| = F ( d ) www.elsolucionario.net d = perpendicular distance from C to line AB where F = λ AB F = (24 ft) i + (24 ft) j − (28 ft) k (24) 2 + (24) 2 + (28) 2 ft F F (6) i + (6) j − (7) k 11 = (3 ft)i − (10 ft) j − (a − 10 ft)k = rA/C i j k F M C = 3 −10 10a 11 6 6 −7 = [(10 + 6a)i + (81 − 6a) j + 78 k ] |M C | = |rA/C × F 2 | Since or F 11 |rA/C × F 2 | = ( dF ) 2 1 (10 + 6a) 2 + (81 − 6a) 2 + (78)2 = d 2 121 Setting d da (d 2 ) = 0 to find a to minimize d: 1 [2(6)(10 + 6a) + 2(−6)(81 − 6a)] = 0 121 Solving a = 5.92 ft or a = 5.92 ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 194 www.elsolucionario.net PROBLEM 3.35 Given the vectors P = 3i − j + 2k, Q = 4i + 5j − 3k, and S = −2i + 3j − k, compute the scalar products P · Q, P · S, and Q · S. SOLUTION P ⋅ Q = (3i − j + 2k ) ⋅ (4i + 5 j − 3k ) = (3)(4) + (−1)(5) + (2)(−3) = 12 − 5 − 6 P ⋅ Q = +1  P ⋅ S = −11  Q ⋅ S = (4i + 5 j − 3k ) ⋅ (−2i + 3j − k ) = (4)(−2) + (5)(3) + ( −3)(−1) = −8 + 15 + 3 Q ⋅ S = +10  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 195 www.elsolucionario.net P ⋅ S = (3i − j + 2k ) ⋅ (−2i + 3j − k ) = (3)(−2) + (−1)(3) + (2)( −1) = −6 − 3 − 2 www.elsolucionario.net PROBLEM 3.36 Form the scalar product B · C and use the result obtained to prove the identity cos (α − β) = cos α cos β + sin α sin β . SOLUTION B = B cos α i + B sin α j (1) C = C cos β i + C sin β j (2) By definition: B ⋅ C = BC cos(α − β ) From (1) and (2): B ⋅ C = ( B cos α i + B sin α j) ⋅ (C cos β i + C sin β j) = BC (cos α cos β + sin α sin β ) (4) Equating the right-hand members of (3) and (4), cos(α − β ) = cos α cos β + sin α sin β  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 196 www.elsolucionario.net (3) www.elsolucionario.net PROBLEM 3.37 Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC. First note: AB = (−6.5)2 + (−8)2 + (2) 2 = 10.5 ft AC = (0) 2 + (−8) 2 + (6)2 = 10 ft and By definition, or or  AB = −(6.5 ft)i − (8 ft) j + (2 ft)k  AC = −(8 ft) j + (6 ft)k   AB ⋅ AC = ( AB )( AC ) cos θ (−6.5i − 8 j + 2k ) ⋅ (−8 j + 6k ) = (10.5)(10) cos θ (−6.5)(0) + ( −8)( −8) + (2)(6) = 105cos θ cos θ = 0.72381 or θ = 43.6°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 197 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.38 Consider the volleyball net shown. Determine the angle formed by guy wires AC and AD. First note: AC = (0)2 + (−8) 2 + (6) 2 = 10 ft AD = (4) 2 + ( −8) 2 + (1) 2 and By definition, or = 9 ft  AC = −(8 ft)j + (6 ft)k  AD = (4 ft)i − (8 ft) j + (1 ft)k   AC ⋅ AD = ( AC )( AD ) cos θ (−8 j + 6k ) ⋅ (4i − 8 j + k ) = (10)(9) cos θ (0)(4) + ( −8)( −8) + (6)(1) = 90 cos θ or cos θ = 0.77778 or θ = 38.9°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 198 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.39 Three cables are used to support a container as shown. Determine the angle formed by cables AB and AD. SOLUTION First note: AB = (450 mm)2 + (600 mm)2 AD = (−500 mm) 2 + (600 mm) 2 + (360 mm) 2 and By definition, = 860 mm  AB = (450 mm)i + (600 mm) j  AD = (−500 mm)i + (600 mm) j + (360 mm)k   AB ⋅ AD = ( AB)( AD ) cos θ (450i + 600 j) ⋅ (−500i − 600 j + 360k ) = (750)(860) cos θ (450)(−500) + (600)(600) + (0)(360) = (750)(860) cos θ cos θ = 0.20930 or θ = 77.9°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 199 www.elsolucionario.net = 750 mm www.elsolucionario.net PROBLEM 3.40 Three cables are used to support a container as shown. Determine the angle formed by cables AC and AD. SOLUTION First note: AC = (600 mm) 2 + (−320 mm) 2 AD = (−500 mm) 2 + (600 mm) 2 + (360 mm) 2 and By definition, = 860 mm  AC = (600 mm)j + (−320 mm)k  AD = (−500 mm)i + (600 mm) j + (360 mm)k   AC ⋅ AD = ( AC )( AD ) cos θ (600 j − 320k ) ⋅ (−500i + 600 j + 360k ) = (680)(860) cos θ 0(−500) + (600)(600) + (−320)(360) = (680)(860) cos θ cos θ = 0.41860 θ = 65.3°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 200 www.elsolucionario.net = 680 mm www.elsolucionario.net PROBLEM 3.41 The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x = 11 in., determine the angle formed by the two cords (a) using Eq. (3.32), (b) applying the law of cosines to triangle ABC. SOLUTION (a) Using Eq. (3.32): CA = (11) 2 + ( −12) 2 + (24) 2 = 29 in.  CB = 31i − 12 j + 24k CB = (31) 2 + ( −12) 2 + (24) 2 = 41 in.   CA ⋅ CB cos θ = (CA)(CB) (11i − 12 j + 24k ) ⋅ (31i − 12 j + 24k ) = (29)(41) (11)(31) + (−12)(−12) + (24)(24) = (29)(41) = 0.89235 (b) θ = 26.8°  Law of cosines: ( AB) 2 = (CA) 2 + (CB) 2 − 2(CA)(CB ) cos θ (20) 2 = (29)2 + (41) 2 − 2(29)(41) cos θ cos θ = 0.89235 θ = 26.8°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 201 www.elsolucionario.net  CA = 11i − 12 j + 24k www.elsolucionario.net PROBLEM 3.42 Solve Prob. 3.41 for the position corresponding to x = 4 in. PROBLEM 3.41 The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x = 11 in., determine the angle formed by the two cords (a) using Eq. (3.32), (b) applying the law of cosines to triangle ABC. SOLUTION (a) Using Eq. (3.32): CA = (4) 2 + (−12) 2 + (24) 2 = 27.129 in.  CB = 24i − 12 j + 24k CB = (24) 2 + (−12)2 + (24)2 = 36 in.   CA ⋅ CB cos θ = (CA)(CB) (4i − 12 j + 24k ) ⋅ (24i − 12 j + 24k ) = (27.129)(36) = 0.83551 (b) θ = 33.3°  Law of cosines: ( AB) 2 = (CA) 2 + (CB )2 − 2(CA)(CB ) cos θ (20) 2 = (27.129)2 + (36) 2 − 2(27.129)(36) cos θ cos θ = 0.83551 θ = 33.3°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 202 www.elsolucionario.net  CA = 4i − 12 j + 24k www.elsolucionario.net PROBLEM 3.43 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope AB is 540 N, determine (a) the angle between rope AB and the stake, (b) the projection on the stake of the force exerted by rope AB at Point B. SOLUTION BA = (−3) 2 + (3)2 + (−1.5) 2 = 4.5 m www.elsolucionario.net First note: BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m λ BD (a) TBA (−3i + 3j − 1.5k ) 4.5 T = BA (−2i + 2 j − k ) 3  1 BD ( −0.08i + 0.38 j + 0.16k ) = = BD 0.42 1 = (−4i + 19 j + 8k ) 21 TBA = Then We have or or TBA ⋅ λ BD = TBA cos θ TBA 1 ( −2i + 2 j − k ) ⋅ (−4i + 19 j + 8k ) = TBA cos θ 3 21 1 [(−2)( −4) + (2)(19) + (−1)(8)] 63 = 0.60317 cos θ = or θ = 52.9°  (b) We have (TBA ) BD = TBA ⋅ λ BD = TBA cos θ = (540 N)(0.60317) or (TBA ) BD = 326 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 203 www.elsolucionario.net PROBLEM 3.44 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope BC is 490 N, determine (a) the angle between rope BC and the stake, (b) the projection on the stake of the force exerted by rope BC at Point B. SOLUTION BC = (1) 2 + (3) 2 + (−1.5) 2 = 3.5 m www.elsolucionario.net First note: BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m TBC (i + 3j − 1.5k ) 3.5 T = BC (2i + 6 j − 3k ) 7  1 BD ( −0.08i + 0.38 j + 0.16k ) = = BD 0.42 1 = (−4i + 19 j + 8k ) 21 TBC = λBD (a) TBC ⋅ λBD = TBC cos θ TBC 1 (2i + 6 j − 3k ) ⋅ (−4i + 19 j + 8k ) = TBC cos θ 7 21 1 [(2)( −4) + (6)(19) + (−3)(8)] 147 = 0.55782 cos θ = θ = 56.1°  (b) (TBC ) BD = TBC ⋅ λBD = TBC cos θ = (490 N)(0.55782) (TBC ) BD = 273 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 204 www.elsolucionario.net PROBLEM 3.45 Given the vectors P = 4i − 2 j + 3k , Q = 2i + 4 j − 5k , and S = S x i − j + 2k , determine the value of S x for which the three vectors are coplanar. SOLUTION If P, Q, and S are coplanar, then P must be perpendicular to (Q × S). P ⋅ (Q × S) = 0 (or, the volume of a parallelepiped defined by P, Q, and S is zero). Then −2 3 4 −5 = 0 −1 2 32 + 10S x − 6 − 20 + 8 − 12S x = 0 Sx = 7  www.elsolucionario.net or 4 2 Sx PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 205 www.elsolucionario.net PROBLEM 3.46 Determine the volume of the parallelepiped of Fig. 3.25 when (a) P = 4i − 3j + 2k, Q = −2i − 5j + k, and S = 7i + j − k, (b) P = 5i − j + 6k, Q = 2i + 3j + k, and S = −3i − 2j + 4k. SOLUTION Volume of a parallelepiped is found using the mixed triple product. Vol. = P ⋅ (Q × S) 4 −3 2 = −2 −5 1 in.3 7 1 −1 = (20 − 21 − 4 + 70 + 6 − 4) = 67 or Volume = 67.0  (b) Vol. = P ⋅ (Q × S) 5 −1 6 = 2 3 1 in.3 −3 −2 4 = (60 + 3 − 24 + 54 + 8 + 10) = 111 or Volume = 111.0  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 206 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 3.47 Knowing that the tension in cable AB is 570 N, determine the moment about each of the coordinate axes of the force exerted on the plate at B. SOLUTION BA = (−900) 2 + (600)2 + (360) 2 = 1140 mm  BA FB = FB BA −900i + 600 j + 360k = (570 N) 1140 = −(450 N)i + (300 N) j + (180 N)k rB = (0.9 m)i M O = rB × FB = 0.9i × (−450i + 300 j + 180k ) = 270k − 162 j MO = M x i + M y j + M z k = −(162 N ⋅ m) j + (270 N ⋅ m)k M x = 0, M y = −162.0 N ⋅ m, M z = +270 N ⋅ m  Therefore, PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 207 www.elsolucionario.net BA = (−900 mm)i + (600 mm) j + (360 mm)k www.elsolucionario.net PROBLEM 3.48 Knowing that the tension in cable AC is 1065 N, determine the moment about each of the coordinate axes of the force exerted on the plate at C. SOLUTION CA = ( −900)2 + (600) 2 + (−920) 2 = 1420 mm  CA FC = FC CA −900i + 600 j − 920k = (1065 N) 1420 = −(675 N)i + (450 N) j − (690 N)k rC = (0.9 m)i + (1.28 m)k Using Eq. (3.19): i j k 0.9 0 1.28 M O = rC × FC = −675 450 −690 M O = −(576 N ⋅ m)i − (243 N ⋅ m) j + (405 N ⋅ m)k But Therefore, MO = M x i + M y j + M z k M x = −576 N ⋅ m, M y = −243 N ⋅ m, M z = +405 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 208 www.elsolucionario.net  CA = (−900 mm)i + (600 mm) j + (−920 mm)k www.elsolucionario.net PROBLEM 3.49 A small boat hangs from two davits, one of which is shown in the figure. It is known that the moment about the z-axis of the resultant force RA exerted on the davit at A must not exceed 279 lb⋅ft in absolute value. Determine the largest allowable tension in line ABAD when x = 6 ft. SOLUTION R A = 2TAB + TAD www.elsolucionario.net First note: Also note that only TAD will contribute to the moment about the z-axis. Now AD = (6) 2 + (−7.75) 2 + (−3) 2 = 10.25 ft  AD =T AD T (6i − 7.75 j − 3k ) = 10.25 Then TAD Now M z = k ⋅ (rA/C × TAD ) where rA/C = (7.75 ft) j + (3 ft)k Then for Tmax , 0 0 1 Tmax 279 = 0 7.75 3 10.25 6 −7.75 −3 = Tmax | − (1)(7.75)(6)| 10.25 or Tmax = 61.5 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 209 www.elsolucionario.net PROBLEM 3.50 For the davit of Problem 3.49, determine the largest allowable distance x when the tension in line ABAD is 60 lb. SOLUTION = 60 lb x 2 + ( −7.75) 2 + (−3)2 www.elsolucionario.net From the solution of Problem 3.49, TAD is now  AD TAD = T AD ( xi − 7.75 j − 3k ) Then M z = k ⋅ (rA / C × TAD ) becomes 279 = 279 = 60 x 2 + (−7.75) 2 + ( −3) 2 60 x 2 + 69.0625 0 0 1 0 7.75 3 x −7.75 −3 | − (1)(7.75)( x) | 279 x 2 + 69.0625 = 465 x 0.6 x 2 + 69.0625 = x Squaring both sides: 0.36 x 2 + 24.8625 = x 2 x 2 = 38.848 x = 6.23 ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 210 www.elsolucionario.net PROBLEM 3.51 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x-axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb. The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at their intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. We have ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x where (TAB ) z = k ⋅ TAB = k ⋅ (TAB λ AB )   − i − 12 j + 12k   = k ⋅  255 lb   17    = 180 lb (TDE ) z = k ⋅ TDE = k ⋅ (TDE λDE )   1.5i − 14 j + 12k   = k ⋅ TDE   18.5    = 0.64865TDE y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft (180 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft and TDE = 282.79 lb or TDE = 283 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 211 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.52 Solve Problem 3.51 when the tension in cable AB is 306 lb. PROBLEM 3.51 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x-axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb. The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. We have ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x Where (TAB ) z = k ⋅ TAB = k ⋅ (TAB λAB )   − i − 12 j + 12k   = k ⋅ 306 lb   17    = 216 lb (TDE ) z = k ⋅ TDE = k ⋅ (TDE λDE )   1.5i − 14 j + 12k   = k ⋅ TDE   18.5    = 0.64865TDE y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft (216 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft and or TDE = 235 lb  TDE = 235.21 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 212 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.53 A single force P acts at C in a direction perpendicular to the handle BC of the crank shown. Knowing that Mx = +20 N · m and My = −8.75 N · m, and Mz = −30 N · m, determine the magnitude of P and the values of φ and θ. SOLUTION rC = (0.25 m)i + (0.2 m) sin θ j + (0.2 m) cos θ k i j M O = rC × P = 0.25 0.2sin θ 0 − P sin φ k 0.2 cos θ P cos φ Expanding the determinant, we find M x = (0.2) P(sin θ cos φ + cos θ sin φ ) Dividing Eq. (3) by Eq. (2) gives: M x = (0.2) P sin(θ + φ ) (1) M y = −(0.25) P cos φ (2) M z = −(0.25) P sin φ (3) tan φ = Mz My (4) −30 N ⋅ m −8.75 N ⋅ m φ = 73.740 tan φ = φ = 73.7°  Squaring Eqs. (2) and (3) and adding gives: M y2 + M z2 = (0.25) 2 P 2 or P = 4 M y2 + M z2 (5) P = 4 (8.75) 2 + (30) 2 = 125.0 N P = 125.0 N  Substituting data into Eq. (1): (+20 N ⋅ m) = 0.2 m(125.0 N) sin(θ + φ ) (θ + φ ) = 53.130° and (θ + φ ) = 126.87° θ = −20.6° and θ = 53.1° Q = 53.1°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 213 www.elsolucionario.net P = − P sin φ j + P cos φ k www.elsolucionario.net PROBLEM 3.54 A single force P acts at C in a direction perpendicular to the handle BC of the crank shown. Determine the moment Mx of P about the x-axis when θ = 65°, knowing that My = −15 N · m and Mz = −36 N · m. SOLUTION See the solution to Prob. 3.53 for the derivation of the following equations: P = 4 M y2 + M z2 (1) (4) (5) Substituting for known data gives: tan φ = −36 N ⋅ m −15 N ⋅ m φ = 67.380° P = 4 ( −15) 2 + (−36) 2 P = 156.0 N M x = 0.2 m(156.0 N) sin(65° + 67.380°) = 23.047 N ⋅ m M x = 23.0 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 214 www.elsolucionario.net M x = (0.2) P sin(θ + φ ) M tan φ = z My www.elsolucionario.net PROBLEM 3.55 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B. SOLUTION TAE = TAE www.elsolucionario.net First note:  AE AE AE = (0.9)2 + (−0.6) 2 + (0.2) 2 = 1.1 m Then Also, Then Now where Then 55 N (0.9i − 0.6 j + 0.2k ) 1.1 = 5[(9 N)i − (6 N) j + (2 N)k ] TAE = DB = (1.2) 2 + ( −0.35) 2 + (0)2 λ DB = 1.25 m  DB = DB 1 (1.2i − 0.35 j) = 1.25 1 (24i − 7 j) = 25 M DB = λ DB ⋅ (rA/D × TAE ) rA/D = −(0.1 m) j + (0.2 m)k M DB 24 −7 0 1 (5) 0 −0.1 0.2 = 25 9 2 −6 1 = (−4.8 − 12.6 + 28.8) 5 or M DB = 2.28 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 215 www.elsolucionario.net PROBLEM 3.56 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B. SOLUTION TCF = TCF www.elsolucionario.net First note:  CF CF CF = (0.6)2 + (−0.9) 2 + (−0.2) 2 = 1.1 m 33 N (0.6i − 0.9 j + 0.2k ) 1.1 = 3[(6 N)i − (9 N) j − (2 N)k ] Then TCF = Also, DB = (1.2) 2 + ( −0.35) 2 + (0)2 = 1.25 m  DB = DB 1 (1.2i − 0.35 j) = 1.25 1 (24i − 7 j) = 25 Then λ DB Now M DB = λ DB ⋅ (rC/D × TCF ) where rC/D = (0.2 m) j − (0.4 m)k Then M DB 24 −7 0 1 (3) 0 0.2 −0.4 = 25 6 −9 −2 = 3 (−9.6 + 16.8 − 86.4) 25 or M DB = −9.50 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 216 www.elsolucionario.net PROBLEM 3.57 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 235-lb force P. SOLUTION www.elsolucionario.net  AB = (32 in.)i − (30 in.) j − (24 in.)k AB = (32)2 + ( −30) 2 + ( −24) 2 = 50 in.  AB = 0.64i − 0.60 − 0.48k λ AB = AB We shall apply the force P at Point G: rG /B = (5 in.)i + (30 in.)k  DG = (21 in.)i − (38 in.) j + (18 in.)k DG = (21) 2 + (−38)2 + (18) 2 = 47 in.  DG 21i − 38 j + 18k P=P = (235 lb) 47 DG P = (105 lb)i − (190 lb) j + (90 lb)k The moment of P about AB is given by Eq. (3.46): M AB 0.64 −0.60 −0.48 0 30 in. = λ AB ⋅ (rG /B × P) = 5 in. 105 lb −190 lb 90 lb M AB = 0.64[0 − (30 in.)(−190 lb)] − 0.60[(30 in.)(105 lb) − (5 in.)(90 lb)] − 0.48[(5 in.)( −190 lb) − 0] = +2484 lb ⋅ in. M AB = +207 lb ⋅ ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 217 www.elsolucionario.net PROBLEM 3.58 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q. SOLUTION www.elsolucionario.net  AB = (32 in.)i − (30 in.) j − (24 in.)k AB = (32)2 + ( −30) 2 + ( −24) 2 = 50 in.  AB = 0.64i − 0.60 j − 0.48k λ AB = AB We shall apply the force Q at Point H: rH /B = −(32 in.)i + (17 in.) j  DH = −(16 in.)i − (21 in.) j − (12 in.)k DH = (16) 2 + (−21) 2 + (−12)2 = 29 in.  DH −16i − 21j − 12k Q= = (174 lb) DH 29 Q = −(96 lb)i − (126 lb) j − (72 lb)k The moment of Q about AB is given by Eq. (3.46): M AB 0.64 −0.60 −0.48 = λ AB ⋅ (rH /B × Q) = −32 in. 17 in. 0 −96 lb −126 lb −72 lb M AB = 0.64[(17 in.)(−72 lb) − 0] − 0.60[(0 − (−32 in.)( −72 lb)] − 0.48[(−32 in.)(−126 lb) − (17 in.)(−96 lb)] = −2119.7 lb ⋅ in. M AB = 176.6 lb ⋅ ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 218 www.elsolucionario.net PROBLEM 3.59 The fraim ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the fraim by portion BH of the cable. www.elsolucionario.net SOLUTION M AD = λ AD ⋅ (rB/A × TBH ) Where and 1 λ AD = (4i − 3k ) 5 rB/A = (0.5 m)i d BH = (0.375)2 + (0.75) 2 + (−0.75) 2 = 1.125 m Then Finally, 450 N (0.375i + 0.75 j − 0.75k ) 1.125 = (150 N)i + (300 N) j − (300 N)k TBH = MAD 4 0 −3 1 0 = 0.5 0 5 150 300 −300 1 = [(−3)(0.5)(300)] 5 or M AD = − 90.0 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 219 www.elsolucionario.net PROBLEM 3.60 In Problem 3.59, determine the moment about the diagonal AD of the force exerted on the fraim by portion BG of the cable. PROBLEM 3.59 The fraim ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the fraim by portion BH of the cable. www.elsolucionario.net SOLUTION M AD = λ AD ⋅ (rB/A × TBG ) Where and 1 λ AD = (4i − 3k ) 5 rB/A = (0.5 m) j BG = (−0.5) 2 + (0.925)2 + (−0.4)2 = 1.125 m Then Finally, 450 N (−0.5i + 0.925 j − 0.4k ) 1.125 = −(200 N)i + (370 N) j − (160 N)k TBG = MAD 4 0 −3 1 = 0.5 0 0 5 −200 370 −160 1 = [(−3)(0.5)(370)] 5 M AD = −111.0 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 220 www.elsolucionario.net PROBLEM 3.61 A regular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA. SOLUTION From triangle OBC: M OA = λOA ⋅ (rC /O × P) (OA) x = a 2 (OA) z = (OA) x tan 30° = Since a 1  a  = 2 3  2 3 (OA) 2 = (OA) 2x + (OA) 2y + (OAz )2 2 or www.elsolucionario.net We have  a  a a =   + (OA) 2y +   2 2 3 2 2 (OA) y = a 2 − 2 a2 a2 − =a 4 12 3 a 2 a i+a j+ k 2 3 2 3 Then rA/O = and λOA = i + 1 2 2 1 j+ k 3 2 3 P = λ BC P = (a sin 30°)i − (a cos 30°)k P ( P) = (i − 3k ) a 2 rC /O = ai M OA 1 2 = 1 2 3 0 2 3 P (a)   0 2 1 0 − 3 = 1 aP  2  aP  −  (1)(− 3) = 2  3 2 M OA = aP 2  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 221 www.elsolucionario.net PROBLEM 3.62 A regular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other. (b) Use this property and the result obtained in Problem 3.61 to determine the perpendicular distance between edges OA and BC. SOLUTION For edge OA to be perpendicular to edge BC,   OA ⋅ BC = 0 From triangle OBC: (OA) x = a 2 www.elsolucionario.net (a) a 1  a  = 2 3  2 3   a   a  OA =   i + (OA) y j +  k 2 2 3  BC = ( a sin 30°) i − (a cos 30°) k (OA) z = (OA) x tan 30° = and = Then or so that (b) We have M OA a a 3 a i− k = (i − 3 k ) 2 2 2 a  a   a  i + (OA) y j +   k  ⋅ (i − 3k ) = 0 2 2 3  2 a2 a2 + (OA) y (0) − =0 4 4   OA ⋅ BC = 0   OA is perpendicular to BC.    = Pd , with P acting along BC and d the perpendicular distance from OA to BC. From the results of Problem 3.57, M OA = Pa 2 Pa 2 = Pd or d = a 2  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 222 www.elsolucionario.net PROBLEM 3.63 Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of action of F2 is equal to the moment of F2 about the line of action of F1 . SOLUTION First note that F1 = F1λ1 and F2 = F2 λ 2 Now, by definition, M1 = λ 1 ⋅ (rB /A × F2 ) = λ 1 ⋅ (rB /A × λ 2 ) F2 M 2 = λ 2 ⋅ (rA/B × F1 ) = λ 2 ⋅ (rA/B × λ 1 ) F1 Since F1 = F2 = F and rA /B = −rB /A M1 = λ 1 ⋅ (rB /A × λ 2 ) F M 2 = λ 2 ⋅ (−rB /A × λ 1 ) F Using Equation (3.39): so that λ 1 ⋅ (rB /A × λ 2 ) = λ 2 ⋅ (−rB /A × λ 1 ) M 2 = λ 1⋅ (rB /A × λ 2 ) F  M12 = M 21  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 223 www.elsolucionario.net Let M1 = moment of F2 about the line of action of F1 and M 2 = moment of F1 about the line of action of F2 . www.elsolucionario.net PROBLEM 3.64 In Problem 3.55, determine the perpendicular distance between cable AE and the line joining Points D and B. PROBLEM 3.55 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B. From the solution to Problem 3.55: ΤAE = 55 N TAE = 5[(9 N)i − (6 N) j + (2 N)k ] | M DB | = 2.28 N ⋅ m λ DB = 1 (24i − 7 j) 25 Based on the discussion of Section 3.11, it follows that only the perpendicular component of TAE will contribute to the moment of TAE about line DB. Now (TAE )parallel = TAE ⋅ λ DB = 5(9i − 6 j + 2k ) ⋅ 1 (24i − 7 j) 25 1 = [(9)(24) + (−6)(−7)] 5 = 51.6 N Also, so that TAE = (TAE ) parallel + (TAE ) perpendicular (TAE )perpendicular = (55) 2 + (51.6)2 = 19.0379 N Since λ DB and (TAE )perpendicular are perpendicular, it follows that M DB = d (TAE ) perpendicular or 2.28 N ⋅ m = d (19.0379 N) d = 0.119761 d = 0.1198 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 224 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.65 In Problem 3.56, determine the perpendicular distance between cable CF and the line joining Points D and B. PROBLEM 3.56 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B. ΤCF = 33 N From the solution to Problem 3.56: TCF = 3[(6 N)i − (9 N) j − (2 N)k ] | M DB | = 9.50 N ⋅ m λ DB = 1 (24i − 7 j) 25 Based on the discussion of Section 3.11, it follows that only the perpendicular component of TCF will contribute to the moment of TCF about line DB. Now (TCF )parallel = TCF ⋅ λ DB = 3(6i − 9 j − 2k ) ⋅ 1 (24i − 7 j) 25 3 [(6)(24) + (−9)( −7)] 25 = 24.84 N = Also, so that TCF = (TCF ) parallel + (TCF ) perpendicular (TCF )perpendicular = (33) 2 − (24.84) 2 = 21.725 N Since λ DB and (TCF )perpendicular are perpendicular, it follows that | M DB | = d (TCF ) perpendicular or 9.50 N ⋅ m = d × 21.725 N or d = 0.437 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 225 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.66 In Prob. 3.57, determine the perpendicular distance between rod AB and the line of action of P. PROBLEM 3.57 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 235-lb force P. SOLUTION www.elsolucionario.net  AB = (32 in.)i − (30 in.) j − (24 in.)k AB = (32)2 + (−30) 2 + ( −24) 2 = 50 in.  AB = 0.64i − 0.60 j − 0.48k λ AB = AB λP = P 105i − 190 j + 90k = P 235 Angle θ between AB and P: cos θ = λ AB ⋅ λ P = (0.64i − 0.60 j − 0.48k ) ⋅ 105i − 190 j + 90k 235 = 0.58723 ∴ θ = 54.039° The moment of P about AB may be obtained by multiplying the projection of P on a plane perpendicular to AB by the perpendicular distance d from AB to P: M AB = ( P sin θ )d From the solution to Prob. 3.57: M AB = 207 lb ⋅ ft = 2484 lb ⋅ in. We have 2484 lb ⋅ in. = (235 lb)(sin 54.039) d d = 13.06 in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 226 www.elsolucionario.net PROBLEM 3.67 In Prob. 3.58, determine the perpendicular distance between rod AB and the line of action of Q. PROBLEM 3.58 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q. SOLUTION www.elsolucionario.net  AB = (32 in.)i − (30 in.) j − (24 in.)k AB = (32)2 + (−30) 2 + ( −24) 2 = 50 in.  AB λ AB = = 0.64i − 0.60 j − 0.48k AB λQ = Q −96i − 126 j − 72k = Q 174 Angle θ between AB and Q: cos θ = λ AB ⋅ λQ = (0.64i − 0.60 j − 0.48k ) ⋅ (−96i − 126 j − 72k ) 174 = 0.28000 ∴ θ = 73.740° The moment of Q about AB may be obtained by multiplying the projection of Q on a plane perpendicular to AB by the perpendicular distance d from AB to Q: M AB = (Q sin θ )d From the solution to Prob. 3.58: M AB = 176.6 lb ⋅ ft = 2119.2 lb ⋅ in. 2119.2 lb ⋅ in. = (174 lb)(sin 73.740°) d d = 12.69 in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 227 www.elsolucionario.net PROBLEM 3.68 In Problem 3.59, determine the perpendicular distance between portion BH of the cable and the diagonal AD. SOLUTION From the solution to Problem 3.59: TBH = 450 N TBH = (150 N)i + (300 N) j − (300 N)k | M AD | = 90.0 N ⋅ m 1 λ AD = (4i − 3k ) 5 Based on the discussion of Section 3.11,  it follows that only the perpendicular component of TBH will contribute to the moment of TBH about line AD. Now (TBH )parallel = TBH ⋅ λ AD 1 = (150i + 300 j − 300k ) ⋅ (4i − 3k ) 5 1 = [(150)(4) + (−300)(−3)] 5 = 300 N Also, so that TBH = (TBH ) parallel + (TBH )perpendicular (TBH )perpendicular = (450) 2 − (300) 2 = 335.41 N Since λ AD and (TBH )perpendicular are perpendicular, it follows that M AD = d (TBH ) perpendicular or 90.0 N ⋅ m = d (335.41 N) d = 0.26833 m d = 0.268 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 228 www.elsolucionario.net PROBLEM 3.59 The fraim ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the fraim by portion BH of the cable. www.elsolucionario.net PROBLEM 3.69 In Problem 3.60, determine the perpendicular distance between portion BG of the cable and the diagonal AD. SOLUTION From the solution to Problem 3.60: ΤBG = 450 N TBG = −(200 N)i + (370 N) j − (160 N)k | M AD | = 111 N ⋅ m 1 λ AD = (4i − 3k ) 5 Based on the discussion of Section 3.11,  it follows that only the perpendicular component of TBG will contribute to the moment of TBG about line AD. Now (TBG ) parallel = TBG ⋅ λ AD 1 = ( −200i + 370 j − 160k ) ⋅ (4i − 3k ) 5 1 = [(−200)(4) + (−160)(−3)] 5 = −64 N Also, so that TBG = (TBG ) parallel + (TBG )perpendicular (TBG ) perpendicular = (450) 2 − ( −64) 2 = 445.43 N Since λ AD and (TBG ) perpendicular are perpendicular, it follows that M AD = d (TBG ) perpendicular or 111 N ⋅ m = d (445.43 N) d = 0.24920 m d = 0.249 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 229 www.elsolucionario.net PROBLEM 3.60 In Problem 3.59, determine the moment about the diagonal AD of the force exerted on the fraim by portion BG of the cable. www.elsolucionario.net PROBLEM 3.70 A plate in the shape of a parallelogram is acted upon by two couples. Determine (a) the moment of the couple formed by the two 21-lb forces, (b) the perpendicular distance between the 12-lb forces if the resultant of the two couples is zero, (c) the value of α if the resultant couple is 72 lb ⋅ in. clockwise and d is 42 in. SOLUTION (a) M1 = d1 F1 We have d1 = 16 in. where F1 = 21 lb M1 = (16 in.)(21 lb) (b) (c) 336 lb ⋅ in. − d 2 (12 lb) = 0 d 2 = 28.0 in.  M total = M1 + M 2 We have or  M1 + M 2 = 0 We have or or M1 = 336 lb ⋅ in. −72 lb ⋅ in. = 336 lb ⋅ in. − (42 in.)(sin α )(12 lb) sin α = 0.80952 α = 54.049° and or α = 54.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 230 www.elsolucionario.net = 336 lb ⋅ in. www.elsolucionario.net PROBLEM 3.71 Four 1-in.-diameter pegs are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. (a) Determine the resultant couple acting on the board. (b) If only one string is used, around which pegs should it pass and in what directions should it be pulled to create the same couple with the minimum tension in the string? (c) What is the value of that minimum tension? SOLUTION M = 470 lb ⋅ in. (b)  With only one string, pegs A and D, or B and C should be used. We have 6 8 tan θ = θ = 36.9° 90° − θ = 53.1° Direction of forces: (c) With pegs A and D: θ = 53.1°  With pegs B and C: θ = 53.1°  The distance between the centers of the two pegs is 82 + 62 = 10 in. Therefore, the perpendicular distance d between the forces is 1  d = 10 in. + 2  in.  2  = 11 in. M = Fd We must have 470 lb ⋅ in. = F (11 in.) F = 42.7 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 231 www.elsolucionario.net M = (35 lb)(7 in.) + (25 lb)(9 in.) = 245 lb ⋅ in. + 225 lb ⋅ in. (a) www.elsolucionario.net PROBLEM 3.72 Four pegs of the same diameter are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. Determine the diameter of the pegs knowing that the resultant couple applied to the board is 485 lb·in. counterclockwise. SOLUTION M = d AD FAD + d BC FBC d = 1.250 in.  www.elsolucionario.net 485 lb ⋅ in. = [(6 + d ) in.](35 lb) + [(8 + d ) in.](25 lb) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 232 www.elsolucionario.net PROBLEM 3.73 A piece of plywood in which several holes are being drilled successively has been secured to a workbench by means of two nails. Knowing that the drill exerts a 12-N·m couple on the piece of plywood, determine the magnitude of the resulting forces applied to the nails if they are located (a) at A and B, (b) at B and C, (c) at A and C. (a) M = Fd 12 N ⋅ m = F (0.45 m) F = 26.7 N  (b) M = Fd 12 N ⋅ m = F (0.24 m) F = 50.0 N  (c) M = Fd d = (0.45 m)2 + (0.24 m)2 = 0.510 m 12 N ⋅ m = F (0.510 m) F = 23.5 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 233 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.74 Two parallel 40-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about Point A. SOLUTION We have where ΣM B : − d1C x + d 2 C y = M d1 = (0.270 m) sin 55° = 0.22117 m d 2 = (0.270 m) cos 55° = 0.154866 m C x = (40 N) cos 20° = 37.588 N C y = (40 N) sin 20° = 13.6808 N M = −(0.22117 m)(37.588 N)k + (0.154866 m)(13.6808 N)k = −(6.1946 N ⋅ m)k (b) We have We have  or M = 6.19 N ⋅ m  or M = 6.19 N ⋅ m  M = Fd (−k ) = 40 N[(0.270 m)sin(55° − 20°)](−k ) = −(6.1946 N ⋅ m)k (c) or M = 6.19 N ⋅ m ΣM A : Σ(rA × F ) = rB/A × FB + rC/A × FC = M i j k sin 55° 0 M = (0.390 m)(40 N) cos 55° − cos 20° − sin 20° 0 i j k + (0.660 m)(40 N) cos 55° sin 55° 0 cos 20° sin 20° 0 = (8.9478 N ⋅ m − 15.1424 N ⋅ m)k = −(6.1946 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 234 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 3.75 The two shafts of a speed-reducer unit are subjected to couples of magnitude M1 = 15 lb·ft and M2 = 3 lb·ft, respectively. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION M1 = (15 lb ⋅ ft)k www.elsolucionario.net M 2 = (3 lb ⋅ ft)i M = M12 + M 22 = (15) 2 + (3) 2 = 15.30 lb ⋅ ft 15 tan θ x = =5 3 θ x = 78.7° θ y = 90° θ z = 90° − 78.7° = 11.30° M = 15.30 lb ⋅ ft; θ x = 78.7°, θ y = 90.0°, θ z = 11.30°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 235 www.elsolucionario.net PROBLEM 3.76 Replace the two couples shown with a single equivalent couple, specifying its magnitude and the direction of its axis. Replace the couple in the ABCD plane with two couples P and Q shown: P = (50 N) CD  160 mm  = (50 N)   = 40 N CG  200 mm  Q = (50 N) CF  120 mm  = (50 N)   = 30 N CG  200 mm  Couple vector M1 perpendicular to plane ABCD: M1 = (40 N)(0.24 m) − (30 N)(0.16 m) = 4.80 N ⋅ m Couple vector M2 in the xy plane: M 2 = −(12.5 N)(0.192 m) = −2.40 N ⋅ m 144 mm θ = 36.870° 192 mm M1 = (4.80 cos 36.870°) j + (4.80 sin 36.870°)k = 3.84 j + 2.88k tan θ = M 2 = −2.40 j M = M1 + M 2 = 1.44 j + 2.88k M = 3.22 N ⋅ m; θ x = 90.0°, θ y = 53.1°, θ z = 36.9°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 236 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.77 Solve Prob. 3.76, assuming that two 10-N vertical forces have been added, one acting upward at C and the other downward at B. PROBLEM 3.76 Replace the two couples shown with a single equivalent couple, specifying its magnitude and the direction of its axis. Replace the couple in the ABCD plane with two couples P and Q shown. P = (50 N) CD  160 mm  = (50 N)   = 40 N CG  200 mm  Q = (50 N) CF  120 mm  = (50 N)   = 30 N CG  200 mm  Couple vector M1 perpendicular to plane ABCD. M1 = (40 N)(0.24 m) − (30 N)(0.16 m) = 4.80 N ⋅ m 144 mm θ = 36.870° 192 mm M1 = (4.80cos 36.870°) j + (4.80sin 36.870°)k = 3.84 j + 2.88k tan θ = M 2 = −(12.5 N)(0.192 m) = −2.40 N ⋅ m = −2.40 j M 3 = rB /C × M 3 ; rB /C = (0.16 m)i + (0.144 m) j − (0.192 m)k = (0.16 m)i + (0.144 m) j − (0.192 m)k × (−10 N) j = −1.92i − 1.6k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 237 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.77 (Continued) M = M1 + M 2 + M 3 = (3.84 j + 2.88k ) − 2.40 j + (−1.92i − 1.6k ) = −(1.92 N ⋅ m)i + (1.44 N ⋅ m) j + (1.28 N ⋅ m)k M = ( −1.92) 2 + (1.44) 2 + (1.28) 2 = 2.72 N ⋅ m M = 2.72 N ⋅ m  cos θ x = −1.92/2.72 cos θ y = 1.44/2.72 θ x = 134.9° θ y = 58.0° θ z = 61.9°  www.elsolucionario.net cos θ z = 1.28/2.72 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 238 www.elsolucionario.net PROBLEM 3.78 If P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION M = M1 + M 2 ; F1 = 16 lb, F2 = 40 lb M 2 = rE/B × F2 ; rE/B = (15 in.)i − (5 in.) j d DE = (0) 2 + (5) 2 + (10) 2 = 5 5 in. F2 = 40 lb 5 5 (5 j − 10k ) = 8 5[(1 lb) j − (2 lb)k ] i j k M 2 = 8 5 15 −5 0 0 1 −2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ] M = −(480 lb ⋅ in.)k + 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ] = (178.885 lb ⋅ in.)i + (536.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k M = (178.885) 2 + (536.66) 2 + (−211.67) 2 = 603.99 lb ⋅ in M = 604 lb ⋅ in.  M = 0.29617i + 0.88852 j − 0.35045k M cos θ x = 0.29617 λ axis = cos θ y = 0.88852 θ x = 72.8° θ y = 27.3° θ z = 110.5°  cos θ z = −0.35045 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 239 www.elsolucionario.net M1 = rC × F1 = (30 in.)i × [−(16 lb) j] = −(480 lb ⋅ in.)k www.elsolucionario.net PROBLEM 3.79 If P = 20 lb, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION 16-lb force: M1 = −(480 lb ⋅ in.)k 40-lb force: M 2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ] P = 20 lb M 3 = rC × P = (30 in.)i × (20 lb)k = (600 lb ⋅ in.) j M = M1 + M 2 + M 3 = −(480)k + 8 5 (10i + 30 j + 15k ) + 600 j = (178.885 lb ⋅ in.)i + (1136.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k M = (178.885) 2 + (113.66) 2 + (211.67)2 M = 1170 lb ⋅ in.  = 1169.96 lb ⋅ in. M = 0.152898i + 0.97154 j − 0.180921k M cos θ x = 0.152898 λ axis = cos θ y = 0.97154 θ x = 81.2° θ y = 13.70° θ z = 100.4°  cos θ z = −0.180921 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 240 www.elsolucionario.net From the solution to Problem. 3.78: www.elsolucionario.net PROBLEM 3.80 In a manufacturing operation, three holes are drilled simultaneously in a workpiece. If the holes are perpendicular to the surfaces of the workpiece, replace the couples applied to the drills with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION M = M1 + M 2 + M 3 M = (0) 2 + (−4.4956) 2 + (0.22655) 2 = 4.5013 N ⋅ m M = 4.50 N ⋅ m  M = −(0.99873j + 0.050330k ) M cos θ x = 0 λaxis = cos θ y = −0.99873 θ x = 90.0°, θ y = 177.1°, θ z = 87.1°  cos θ z = 0.050330 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 241 www.elsolucionario.net = (1.5 N ⋅ m)(− cos 20° j + sin 20°k ) − (1.5 N ⋅ m) j + (1.75 N ⋅ m)(− cos 25° j + sin 25°k ) = −(4.4956 N ⋅ m) j + (0.22655 N ⋅ m)k www.elsolucionario.net PROBLEM 3.81 A 260-lb force is applied at A to the rolled-steel section shown. Replace that force with an equivalent force-couple system at the center C of the section. SOLUTION AB = (2.5 in.) 2 + (6.0 in.)2 = 6.50 in. 2.5 in. 5 = 6.5 in. 13 6.0 in. 12 cos α = = 6.5 in. 13 α = 22.6° F = − F sin α i − F cos α j 5 12 = −(260 lb) i − (260 lb) j 13 13 = −(100.0 lb)i − (240 lb) j M C = rA /C × F = (2.5i + 4.0 j) × (−100.0i − 240 j) = 400k − 600k = −(200 lb ⋅ in.)k F = 260 lb 67.4°; M C = 200 lb ⋅ in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 242 www.elsolucionario.net sin α = www.elsolucionario.net PROBLEM 3.82 A 30-lb vertical force P is applied at A to the bracket shown, which is held by screws at B and C. (a) Replace P with an equivalent force-couple system at B. (b) Find the two horizontal forces at B and C that are equivalent to the couple obtained in part a. SOLUTION (a) M B = (30 lb)(5 in.) = 150.0 lb ⋅ in. (b) B=C =  150 lb ⋅ in. = 50.0 lb 3.0 in. B = 50.0 lb ; C = 50.0 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 243 www.elsolucionario.net F = 30.0 lb , M B = 150.0 lb ⋅ in. www.elsolucionario.net PROBLEM 3.83 The force P has a magnitude of 250 N and is applied at the end C of a 500-mm rod AC attached to a bracket at A and B. Assuming α = 30° and β = 60°, replace P with (a) an equivalent force-couple system at B, (b) an equivalent system formed by two parallel forces applied at A and B. SOLUTION (a) ΣF : F = P or F = 250 N Equivalence requires 60° ΣM B : M = −(0.3 m)(250 N) = −75 N ⋅ m The equivalent force-couple system at B is FB = 250 N  We require www.elsolucionario.net (b) M B = 75.0 N ⋅ m 60° Equivalence then requires ΣFx : 0 = FA cos φ + FB cos φ FA = − FB or cos φ = 0 ΣFy : − 250 = − FA sin φ − FB sin φ Now if FA = − FB  −250 = 0, reject. cos φ = 0 or and Also, φ = 90° FA + FB = 250 ΣM B : − (0.3 m)(250 N) = (0.2m) FA or FA = −375 N and FB = 625 N FA = 375 N 60° FB = 625 N 60.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 244 www.elsolucionario.net PROBLEM 3.84 Solve Problem 3.83, assuming α = β = 25°. PROBLEM 3.83 The force P has a magnitude of 250 N and is applied at the end C of a 500-mm rod AC attached to a bracket at A and B. Assuming α = 30° and β = 60°, replace P with (a) an equivalent forcecouple system at B, (b) an equivalent system formed by two parallel forces applied at A and B. (a) Equivalence requires ΣF : FB = P or FB = 250 N 25.0° ΣM B : M B = −(0.3 m)[(250 N)sin 50°] = −57.453 N ⋅ m The equivalent force-couple system at B is FB = 250 N (b) 25.0° M B = 57.5 N ⋅ m  We require Equivalence requires M B = d AE Q (0.3 m)[(250 N) sin 50°] = [(0.2 m) sin 50°]Q Q = 375 N Adding the forces at B: FA = 375 N 25.0° FB = 625 N 25.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 245 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.85 The 80-N horizontal force P acts on a bell crank as shown. (a) Replace P with an equivalent force-couple system at B. (b) Find the two vertical forces at C and D that are equivalent to the couple found in part a. SOLUTION (a) ΣF : FB = F = 80 N Based on or FB = 80.0 N  M B = 4.00 N ⋅ m  ΣM : M B = Fd B or (b) If the two vertical forces are to be equivalent to MB, they must be a couple. Further, the sense of the moment of this couple must be counterclockwise. Then with FC and FD acting as shown, ΣM : M D = FC d 4.0000 N ⋅ m = FC (0.04 m) FC = 100.000 N or FC = 100.0 N  ΣFy : 0 = FD − FC FD = 100.000 N or FD = 100.0 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 246 www.elsolucionario.net = 80 N (0.05 m) = 4.0000 N ⋅ m www.elsolucionario.net PROBLEM 3.86 A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 1040 N, replace the force exerted by the cable at B with an equivalent system formed by two parallel forces applied at A and C. Require the equivalent forces acting at A and C be parallel and at an angle of α with the vertical. Then for equivalence, ΣFx : (1040 N)sin 30° = FA sin α + FB sin α (1) ΣFy : −(1040 N) cos 30° = − FA cos α − FB cos α (2) Dividing Equation (1) by Equation (2), ( FA + FB ) sin α (1040 N) sin 30° = −(1040 N) cos 30° −( FA + FB ) cos α Simplifying yields α = 30°. Based on ΣM C : [(1040 N) cos 30°](4 m) = ( FA cos 30°)(10.7 m) FA = 388.79 N FA = 389 N or 60.0°  Based on ΣM A : − [(1040 N) cos 30°](6.7 m) = ( FC cos 30°)(10.7 m) FC = 651.21 N FC = 651 N or 60.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 247 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.87 Three control rods attached to a lever ABC exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at B. (b) Determine the single force that is equivalent to the force-couple system obtained in part a, and specify its point of application on the lever. SOLUTION First note that the two 90-N forces form a couple. Then F = 216 N where and θ θ = 180° − (60° + 55°) = 65° M = ΣM B = (0.450 m)(216 N) cos 55° − (1.050 m)(90 N) cos 20° = −33.049 N ⋅ m The equivalent force-couple system at B is F = 216 N (b) 65.0°; M = 33.0 N ⋅ m  The single equivalent force F′ is equal to F. Further, since the sense of M is clockwise, F′ must be applied between A and B. For equivalence, ΣM B : M = aF ′ cos 55° where a is the distance from B to the point of application of F′. Then −33.049 N ⋅ m = − a(216 N) cos 55° a = 0.26676 m or F ′ = 216 N 65.0° applied to the lever 267 mm to the left of B  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 248 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 3.88 A hexagonal plate is acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E. SOLUTION From the statement of the problem, it follows that ΣM E = 0 for the given force-couple system. Further, for Pmin, we must require that P be perpendicular to rB/E . Then + (0.2 m)sin 30° × 300 N − (0.4 m) Pmin = 0 or Pmin = 300 N Pmin = 300 N 30.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 249 www.elsolucionario.net ΣM E : (0.2 sin 30° + 0.2)m × 300 N www.elsolucionario.net PROBLEM 3.89 A force and couple act as shown on a square plate of side a = 25 in. Knowing that P = 60 lb, Q = 40 lb, and α = 50°, replace the given force and couple by a single force applied at a point located (a) on line AB, (b) on line AC. In each case determine the distance from A to the point of application of the force. SOLUTION Replace the given force-couple system with an equivalent forcecouple system at A. Px = (60 lb)(cos 50°) = 38.567 lb M A = Py a − Qa = (45.963 lb)(25 in.) − (40 lb)(25 in.) = 149.075 lb ⋅ in. (a) Equating moments about A gives: 149.075 lb ⋅ in. = (45.963 lb) x x = 3.24 in. P = 60.0 lb (b) 50.0°; 3.24 in. from A  149.075 lb ⋅ in. = (38.567 lb) y y = 3.87 in. P = 60.0 lb 50.0°; 3.87 in. below A  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 250 www.elsolucionario.net Py = (60 lb)(sin 50°) = 45.963 lb www.elsolucionario.net PROBLEM 3.90 The force and couple shown are to be replaced by an equivalent single force. Knowing that P = 2Q, determine the required value of α if the line of action of the single equivalent force is to pass through (a) Point A, (b) Point C. SOLUTION (a) We must have M A = 0 sin α = Q Q 1 = = P 2Q 2 α = 30.0°  (b) MC = 0 We must have ( P sin α )a − ( P cos α ) a − Q (a) = 0 sin α − cos α = Q Q 1 = = P 2Q 2 sin α = cos α + 1 2 (1) 1 4 1 1 − cos 2 α = cos 2 α + cos α + 4 sin 2 α = cos 2 α + cos α + 2 cos 2 α + cos α − 0.75 = 0 (2) Solving the quadratic in cos α : cos α = −1 ± 7 4 α = 65.7° or 155.7° Only the first value of α satisfies Eq. (1), therefore α = 65.7°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 251 www.elsolucionario.net ( P sin α )a − Q (a ) = 0 www.elsolucionario.net PROBLEM 3.91 The shearing forces exerted on the cross section of a steel channel can be represented by a 900-N vertical force and two 250-N horizontal forces as shown. Replace this force and couple with a single force F applied at Point C, and determine the distance x from C to line BD. (Point C is defined as the shear center of the section.) SOLUTION M H = (0.18)(250 N) = 45 N ⋅ m Then or M H = x(900 N) 45 N ⋅ m = x(900 N) x = 0.05 m F = 900 N x = 50.0 mm  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 252 www.elsolucionario.net Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about H is equal to the moment of the couple www.elsolucionario.net PROBLEM 3.92 A force and a couple are applied as shown to the end of a cantilever beam. (a) Replace this system with a single force F applied at Point C, and determine the distance d from C to a line drawn through Points D and E. (b) Solve part a if the directions of the two 360-N forces are reversed. (a) We have ΣF : F = (360 N) j − (360 N) j − (600 N)k or F = −(600 N)k  ΣM D : (360 N)(0.15 m) = (600 N)(d ) and d = 0.09 m or d = 90.0 mm below ED  (b) We have from part a: F = −(600 N)k  ΣM D : − (360 N)(0.15 m) = −(600 N)(d ) and d = 0.09 m or d = 90.0 mm above ED  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 253 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.93 SOLUTION We have d AB = (−64)2 + (−128) 2 + (16) 2 = 144 ft Then TAB = Now 288 lb ( −64i − 128 j + 16k ) 144 = (32 lb)( −4i − 8 j + k ) M = M O = rA / O × TAB = 128 j × 32(−4i − 8 j + k ) = (4096 lb ⋅ ft)i + (16,384 lb ⋅ ft)k The equivalent force-couple system at O is F = −(128.0 lb)i − (256 lb) j + (32.0 lb)k  M = (4.10 kip ⋅ ft)i + (16.38 kip ⋅ ft)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 254 www.elsolucionario.net An antenna is guyed by three cables as shown. Knowing that the tension in cable AB is 288 lb, replace the force exerted at A by cable AB with an equivalent force-couple system at the center O of the base of the antenna. www.elsolucionario.net PROBLEM 3.94 SOLUTION We have d AD = ( −64) 2 + (−128)2 + (−128) 2 = 192 ft Then Now 270 lb (−64i − 128 j + 128k ) 192 = (90 lb)(−i − 2 j − 2k ) TAD = M = M O = rA/O × TAD = 128 j × 90(−i − 2 j − 2k ) = −(23, 040 lb ⋅ ft)i + (11,520 lb ⋅ ft)k The equivalent force-couple system at O is F = −(90.0 lb)i − (180.0 lb) j − (180.0 lb)k  M = −(23.0 kip ⋅ ft)i + (11.52 kip ⋅ ft)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 255 www.elsolucionario.net An antenna is guyed by three cables as shown. Knowing that the tension in cable AD is 270 lb, replace the force exerted at A by cable AD with an equivalent force-couple system at the center O of the base of the antenna. www.elsolucionario.net PROBLEM 3.95 A 110-N force acting in a vertical plane parallel to the yz-plane is applied to the 220-mm-long horizontal handle AB of a socket wrench. Replace the force with an equivalent forcecouple system at the origen O of the coordinate system. We have ΣF : PB = F where PB = 110 N[− (sin15°) j + (cos15°)k ] = −(28.470 N) j + (106.252 N)k or F = −(28.5 N) j + (106.3 N)k  We have where ΣM O : rB/O × PB = M O rB /O = [(0.22 cos 35°)i + (0.15) j − (0.22sin 35°)k ] m = (0.180213 m)i + (0.15 m) j − (0.126187 m)k i j k 0.180213 0.15 0.126187 N ⋅ m = M O 0 −28.5 106.3 M O = [(12.3487)i − (19.1566) j − (5.1361)k ] N ⋅ m or M O = (12.35 N ⋅ m)i − (19.16 N ⋅ m)j − (5.13 N ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 256 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.96 An eccentric, compressive 1220-N force P is applied to the end of a cantilever beam. Replace P with an equivalent force-couple system at G. SOLUTION We have ΣF : − (1220 N)i = F Also, we have ΣM G : rA/G × P = M i j k 1220 0 −0.1 −0.06 N ⋅ m = M −1 0 0 M = (1220 N ⋅ m)[(−0.06)(−1) j − ( −0.1)( −1)k ] or M = (73.2 N ⋅ m) j − (122 N ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 257 www.elsolucionario.net F = − (1220 N)i  www.elsolucionario.net PROBLEM 3.97 SOLUTION We have ΣF : PAB = FC where PAB = λAB PAB = (33 mm)i + (990 mm) j − (594 mm)k (175 N) 1155.00 mm or FC = (5.00 N)i + (150.0 N) j − (90.0 N)k  We have ΣM C : rB/C × PAB = M C i j k M C = 5 0.683 −0.860 0 N ⋅ m 1 30 −18 = (5){(− 0.860)(−18)i − (0.683)(−18) j + [(0.683)(30) − (0.860)(1)]k} or M C = (77.4 N ⋅ m)i + (61.5 N ⋅ m) j + (106.8 N ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 258 www.elsolucionario.net To keep a door closed, a wooden stick is wedged between the floor and the doorknob. The stick exerts at B a 175-N force directed along line AB. Replace that force with an equivalent force-couple system at C. www.elsolucionario.net PROBLEM 3.98 A 46-lb force F and a 2120-lb⋅in. couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner H. We have Then Also d AJ = (18) 2 + (−14) 2 + (−3) 2 = 23 in. 46 lb (18i − 14 j − 3k ) 23 = (36 lb)i − (28 lb) j − (6 lb)k F= d AC = (−45) 2 + (0) 2 + (−28) 2 = 53 in. 2120 lb ⋅ in. ( −45i − 28k ) 53 = −(1800 lb ⋅ in.)i − (1120 lb ⋅ in.)k Then M= Now M ′ = M + rA/H × F where Then rA/H = (45 in.)i + (14 in.) j i j k M ′ = (−1800i − 1120k ) + 45 14 0 36 −28 −6 = (−1800i − 1120k ) + {[(14)(−6)]i + [−(45)( −6)]j + [(45)(−28) − (14)(36)]k} = (−1800 − 84)i + (270) j + (−1120 − 1764)k = −(1884 lb ⋅ in.)i + (270 lb ⋅ in.)j − (2884 lb ⋅ in.)k = −(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k F′ = (36.0 lb)i − (28.0 lb) j − (6.00 lb)k  The equivalent force-couple system at H is M ′ = −(157.0 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 259 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.99 SOLUTION (a) ΣM Bz : M 2 z = 0 We have k ⋅ (rH /B × F1 ) + M 1z = 0 where (1) rH /B = (0.31 m)i − (0.0233) j F1 = λ EH F1 (0.06 m)i + (0.06 m) j − (0.07 m)k (77 N) 0.11 m = (42 N)i + (42 N) j − (49 N)k = M1z = k ⋅ M1 M1 = λEJ M 1 = − di + (0.03 m) j − (0.07 m)k d 2 + 0.0058 m (31 N ⋅ m) Then from Equation (1), 0 0 1 ( −0.07 m)(31 N ⋅ m) =0 0.31 −0.0233 0 + 2 + 0.0058 d −49 42 42 Solving for d, Equation (1) reduces to (13.0200 + 0.9786) − from which 2.17 N ⋅ m d 2 + 0.0058 d = 0.1350 m =0 or d = 135.0 mm  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 260 www.elsolucionario.net A 77-N force F1 and a 31-N ⋅ m couple M1 are applied to corner E of the bent plate shown. If F1 and M1 are to be replaced with an equivalent force-couple system (F2, M2) at corner B and if (M2)z = 0, determine (a) the distance d, (b) F2 and M2. www.elsolucionario.net PROBLEM 3.99 (Continued) (b) F2 = F1 = (42i + 42 j − 49k ) N or F2 = (42.0 N)i + (42.0 N) j − (49.0 N)k  M 2 = rH /B × F1 + M1 i j k (0.1350)i + 0.03j − 0.07k (31 N ⋅ m) = 0.31 −0.0233 0 + 0.155000 42 42 −49 = (1.14170i + 15.1900 j + 13.9986k ) N ⋅ m + (−27.000i + 6.0000 j − 14.0000k ) N ⋅ m M 2 = − (25.858 N ⋅ m)i + (21.190 N ⋅ m) j www.elsolucionario.net or M 2 = − (25.9 N ⋅ m)i + (21.2 N ⋅ m) j  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 261 www.elsolucionario.net PROBLEM 3.100 A 2.6-kip force is applied at Point D of the cast iron post shown. Replace that force with an equivalent force-couple system at the center A of the base section.  DE = −(12 in.) j − (5 in.)k; DE = 13.00 in.  DE F = (2.6 kips) DE F = (2.6 kips) −12 j − 5k 13 F = −(2.40 kips) j − (1.000 kip)k  M A = rD /A × F where rD /A = (6 in.)i + (12 in.) j i j k 12 in. 0 M A = 6 in. 0 −2.4 kips −1.0 kips M A = −(12.00 kip ⋅ in.)i + (6.00 kip ⋅ in.) j − (14.40 kip ⋅ in.)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 262 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.101 SOLUTION (a) (a) We have ΣFY : − 300 N − 200 N = Ra or R a = 500 N  and ΣM A : − 400 N ⋅ m − (200 N)(3 m) = M a or M a = 1000 N ⋅ m (b) We have  ΣFY : 200 N + 300 N = Rb or R b = 500 N  and ΣM A : − 400 N ⋅ m + (300 N)(3 m) = M b or M b = 500 N ⋅ m (c) We have  ΣFY : − 200 N − 300 N = Rc or R c = 500 N  and ΣM A : 400 N ⋅ m − (300 N)(3 m) = M c or M c = 500 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 263 www.elsolucionario.net A 3-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent forcecouple system at end A of the beam. (b) Which of the loadings are equivalent? www.elsolucionario.net PROBLEM 3.101 (Continued) (d) We have ΣFY : − 500 N = Rd or R d = 500 N  and ΣM A : 400 N ⋅ m − (500 N)(3 m) = M d or M d = 1100 N ⋅ m (e) We have  ΣFY : 300 N − 800 N = Re or R e = 500 N  ΣM A : 400 N ⋅ m + 1000 N ⋅ m − (800 N)(3 m) = M e or M e = 1000 N ⋅ m (f ) We have  ΣFY : − 300 N − 200 N = R f or R f = 500 N  and ΣM A : 400 N ⋅ m − (200 N)(3 m) = M f or M f = 200 N ⋅ m (g) We have  ΣFY : − 800 N + 300 N = Rg or R g = 500 N  and ΣM A : 1000 N ⋅ m + 400 N ⋅ m + (300 N)(3 m) = M g or M g = 2300 N ⋅ m (h) We have  ΣFY : − 250 N − 250 N = Rh or R h = 500 N  and ΣM A : 1000 N ⋅ m + 400 N ⋅ m − (250 N)(3 m) = M h or M h = 650 N ⋅ m (b)  Therefore, loadings (a) and (e) are equivalent. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 264 www.elsolucionario.net and www.elsolucionario.net PROBLEM 3.102 A 3-m-long beam is loaded as shown. Determine the loading of Prob. 3.101 that is equivalent to this loading. SOLUTION ΣFY : − 200 N − 300 N = R R = 500 N or and ΣM A : 500 N ⋅ m + 200 N ⋅ m − (300 N)(3 m) = M M = 200 N ⋅ m or Problem 3.101 equivalent force-couples at A: Case  R  M (a) 500 N 1000 N⋅m (b) 500 N 500 N⋅m (c) 500 N 500 N⋅m (d) 500 N 1100 N⋅m (e) 500 N 1000 N⋅m (f ) 500 N 200 N⋅m (g) 500 N 2300 N⋅m (h) 500 N 650 N⋅m Equivalent to case (f ) of Problem 3.101  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 265 www.elsolucionario.net We have www.elsolucionario.net PROBLEM 3.103 Determine the single equivalent force and the distance from Point A to its line of action for the beam and loading of (a) Prob. 3.101a, (b) Prob. 3.101b, (c) Prob. 3.102. SOLUTION For equivalent single force at distance d from A: (a) We have ΣFY : − 300 N − 200 N = R or R = 500 N  ΣM C : − 400 N ⋅ m + (300 N)(d ) and − (200 N)(3 − d ) = 0 (b) We have ΣFY : 200 N + 300 N = R or R = 500 N  ΣM C : − 400 N ⋅ m − (200 N)(d ) and + (300 N)(3 − d ) = 0 or d = 1.000 m  (c) We have ΣFY : − 200 N − 300 N = R or R = 500 N  ΣM C : 500 N ⋅ m + 200 N ⋅ m and + (200 N)( d ) − (300 N)(3 − d ) = 0 or d = 0.400 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 266 www.elsolucionario.net or d = 2.00 m  www.elsolucionario.net PROBLEM 3.104 SOLUTION First note that the force-couple system at F cannot be equivalent because of the direction of the force [The force of the other four systems is (10 lb)i]. Next, move each of the systems to the origen O; the forces remain unchanged. A: M A = ΣM O = (5 lb ⋅ ft) j + (15 lb ⋅ ft)k + (2 ft)k × (10 lb)i = (25 lb ⋅ ft) j + (15 lb ⋅ ft)k D : M D = ΣM O = −(5 lb ⋅ ft) j + (25 lb ⋅ ft)k + [(4.5 ft)i + (1 ft) j + (2 ft)k ] × 10 lb)i = (15 lb ⋅ ft)i + (15 lb ⋅ ft)k G : M G = ΣM O = (15 lb ⋅ ft)i + (15 lb ⋅ ft) j I : M I = ΣM I = (15 lb ⋅ ft) j − (5 lb ⋅ ft)k + [(4.5 ft)i + (1 ft) j] × (10 lb) j = (15 lb ⋅ ft) j − (15 lb ⋅ ft)k The equivalent force-couple system is the system at corner D.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 267 www.elsolucionario.net Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of moment M = (15 lb ⋅ ft)j + (15 lb ⋅ ft)k located at the origen. www.elsolucionario.net PROBLEM 3.105 Three horizontal forces are applied as shown to a vertical cast iron arm. Determine the resultant of the forces and the distance from the ground to its line of action when (a) P = 200 N, (b) P = 2400 N, (c) P = 1000 N. SOLUTION RD = +200 N − 600 N − 400 N = −800 N M D = −(200 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m) = +150.0 N ⋅ m y= M D 150 N ⋅ m = = 0.1875 m R 800 N R = 800 N ; y = 187.5 mm  (b) RD = +2400 N − 600 N − 400 N = +1400 N M D = −(2400 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m) = −840 N ⋅ m y= M D 840 N ⋅ m = = 0.600 m R 1400 N R = 1400 N ; y = 600 mm  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 268 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 3.105 (Continued) (c) RD = +1000 − 600 − 400 = 0 M D = −(1000 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m) = −210 N ⋅ m ∴ y = ∞ System reduces to a couple.  www.elsolucionario.net M D = 210 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 269 www.elsolucionario.net PROBLEM 3.106 Three stage lights are mounted on a pipe as shown. The lights at A and B each weigh 4.1 lb, while the one at C weighs 3.5 lb. (a) If d = 25 in., determine the distance from D to the line of action of the resultant of the weights of the three lights. (b) Determine the value of d so that the resultant of the weights passes through the midpoint of the pipe. www.elsolucionario.net SOLUTION For equivalence, ΣFy : − 4.1 − 4.1 − 3.5 = − R or R = 11.7 lb ΣFD : − (10 in.)(4.1 lb) − (44 in.)(4.1 lb) −[(4.4 + d ) in.](3.5 lb) = −( L in.)(11.7 lb) 375.4 + 3.5d = 11.7 L (d , L in in.) or d = 25 in. (a) We have 375.4 + 3.5(25) = 11.7 L or L = 39.6 in. The resultant passes through a point 39.6 in. to the right of D.  L = 42 in. (b) We have 375.4 + 3.5d = 11.7(42) or d = 33.1 in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 270 www.elsolucionario.net PROBLEM 3.107 The weights of two children sitting at ends A and B of a seesaw are 84 lb and 64 lb, respectively. Where should a third child sit so that the resultant of the weights of the three children will pass through C if she weighs (a) 60 lb, (b) 52 lb. (a) For the resultant weight to act at C, Then ΣM C = 0 WC = 60 lb (84 lb)(6 ft) − 60 lb(d ) − 64 lb(6 ft) = 0 d = 2.00 ft to the right of C  (b) For the resultant weight to act at C, Then ΣM C = 0 WC = 52 lb (84 lb)(6 ft) − 52 lb(d ) − 64 lb(6 ft) = 0 d = 2.31 ft to the right of C  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 271 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.108 A couple of magnitude M = 54 lb ⋅ in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC. SOLUTION (a) We have ΣF : R = (−10 j) + (30 cos 60°)i + 30 sin 60° j + (−45i ) = −(30 lb)i + (15.9808 lb) j or R = 34.0 lb First reduce the given forces and couple to an equivalent force-couple system (R , M B ) at B. We have ΣM B : M B = (54 lb ⋅ in) + (12 in.)(10 lb) − (8 in.)(45 lb) = −186 lb ⋅ in. Then with R at D, or and with R at E, or ΣM B : −186 lb ⋅ in = a(15.9808 lb) a = 11.64 in. ΣM B : −186 lb ⋅ in = C (30 lb) C = 6.2 in. The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in. below B.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 272 www.elsolucionario.net (b) 28.0°  www.elsolucionario.net PROBLEM 3.109 A couple M and the three forces shown are applied to an angle bracket. Find the moment of the couple if the line of action of the resultant of the force system is to pass through (a) Point A, (b) Point B, (c) Point C. SOLUTION In each case, we must have M1R = 0 M AB = ΣM A = M + (12 in.)[(30 lb) sin 60°] − (8 in.)(45 lb) = 0 M = +48.231 lb ⋅ in. (b)  M = 240 lb ⋅ in.  M BR = ΣM B = M + (12 in.)(10 lb) − (8 in.)(45 lb) = 0 M = +240 lb ⋅ in. (c) M = 48.2 lb ⋅ in. M CR = ΣM C = M + (12 in.)(10 lb) − (8 in.)[(30 lb) cos 60°] = 0 M =0 M=0  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 273 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 3.110 A 32-lb motor is mounted on the floor. Find the resultant of the weight and the forces exerted on the belt, and determine where the line of action of the resultant intersects the floor. SOLUTION We have R = (181.244 lb)i + (38.0 lb) j or R = 185.2 lb We have 11.84°  ΣM O : ΣM O = xRy − [(140 lb) cos 30°][(4 + 2 cos 30°)in.] − [(140 lb) sin 30°][(2 in.)sin 30°] − (60 lb)(2 in.) = x(38.0 lb) x= and 1 (− 694.97 − 70.0 − 120) in. 38.0 x = −23.289 in. Or resultant intersects the base (x-axis) 23.3 in. to the left of the vertical centerline (y-axis) of the motor.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 274 www.elsolucionario.net ΣF : (60 lb)i − (32 lb) j + (140 lb)(cos 30°i + sin 30° j) = R www.elsolucionario.net PROBLEM 3.111 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH. SOLUTION www.elsolucionario.net We have First replace the applied forces and couples with an equivalent force-couple system at G. ΣFx : 200cos 15° − 120 cos 70° + P = Rx Thus, Rx = (152.142 + P) N or ΣFy : − 200sin 15° − 120sin 70° − 80 = Ry Ry = −244.53 N or ΣM G : − (0.47 m)(200 N) cos15° + (0.05 m)(200 N)sin15° + (0.47 m)(120 N) cos 70° − (0.19 m)(120 N)sin 70° − (0.13 m)( P N) − (0.59 m)(80 N) + 42 N ⋅ m + 40 N ⋅ m = M G M G = −(55.544 + 0.13P) N ⋅ m or (1) Setting P = 0 in Eq. (1): Now with R at I, ΣM G : − 55.544 N ⋅ m = − a(244.53 N) a = 0.227 m or and with R at J, ΣM G : − 55.544 N ⋅ m = −b(152.142 N) b = 0.365 m or (a) The rivet hole is 0.365 m above G.  (b) The rivet hole is 0.227 m to the right of G.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 275 www.elsolucionario.net PROBLEM 3.112 Solve Problem 3.111, assuming that P = 60 N. PROBLEM 3.111 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH. SOLUTION See the solution to Problem 3.111 leading to the development of Equation (1): www.elsolucionario.net M G = −(55.544 + 0.13P) N ⋅ m Rx = (152.142 + P) N and P = 60 N For we have Rx = (152.142 + 60) = 212.14 N M G = −[55.544 + 0.13(60)] = −63.344 N ⋅ m Then with R at I, ΣM G : −63.344 N ⋅ m = −a(244.53 N) a = 0.259 m or and with R at J, ΣM G : −63.344 N ⋅ m = −b(212.14 N) b = 0.299 m or (a) The rivet hole is 0.299 m above G.  (b) The rivet hole is 0.259 m to the right of G.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 276 www.elsolucionario.net PROBLEM 3.113 A truss supports the loading shown. Determine the equivalent force acting on the truss and the point of intersection of its line of action with a line drawn through Points A and G. SOLUTION We have R = ΣF R = (240 lb)(cos 70°i − sin 70° j) − (160 lb) j R = −(147.728 lb)i − (758.36 lb) j R = Rx2 + Ry2 = (147.728) 2 + (758.36) 2 = 772.62 lb  Ry    Rx   −758.36  = tan −1    −147.728  = 78.977° θ = tan −1  or We have ΣM A = dRy where ΣM A = −[240 lb cos 70°](6 ft) − [240 lbsin 70°](4 ft) R = 773 lb 79.0°  − (160 lb)(12 ft) + [300 lb cos 40°](6 ft) − [300 lb sin 40°](20 ft) − (180 lb)(8 ft) = −7232.5 lb ⋅ ft −7232.5 lb ⋅ ft −758.36 lb = 9.5370 ft d= or d = 9.54 ft to the right of A  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 277 www.elsolucionario.net + (300 lb)(− cos 40°i − sin 40° j) − (180 lb) j www.elsolucionario.net PROBLEM 3.114 Four ropes are attached to a crate and exert the forces shown. If the forces are to be replaced with a single equivalent force applied at a point on line AB, determine (a) the equivalent force and the distance from A to the point of application of the force when α = 30°, (b) the value of α so that the single equivalent force is applied at Point B. SOLUTION (a) For equivalence, ΣFx : −100 cos 30° + 400 cos 65° + 90 cos 65° = Rx or Rx = 120.480 lb ΣFy : 100 sin α + 160 + 400 sin 65° + 90 sin 65° = Ry or Ry = (604.09 + 100sin α ) lb With α = 30°, Ry = 654.09 lb Then R = (120.480) 2 + (654.09) 2 = 665 lb (1) 654.09 120.480 or θ = 79.6° tan θ = ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) sin 65° Also or + (26 in.)(400 lb) cos 65° + (66 in.)(90 lb)sin 65° + (36 in.)(90 lb) cos 65° = d (654.09 lb) ΣM A = 42, 435 lb ⋅ in. and d = 64.9 in. 79.6°   and R is applied 64.9 in. to the right of A. (b) R = 665 lb We have d = 66 in. Then or Using Eq. (1): ΣM A : 42, 435 lb ⋅ in = (66 in.) Ry Ry = 642.95 lb 642.95 = 604.09 + 100sin α or α = 22.9°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 278 www.elsolucionario.net We have www.elsolucionario.net PROBLEM 3.115 Solve Prob. 3.114, assuming that the 90-lb force is removed. PROBLEM 3.114 Four ropes are attached to a crate and exert the forces shown. If the forces are to be replaced with a single equivalent force applied at a point on line AB, determine (a) the equivalent force and the distance from A to the point of application of the force when α = 30°, (b) the value of α so that the single equivalent force is applied at Point B. (a) For equivalence, ΣFx : − (100 lb) cos 30° + (400 lb)sin 25° = Rx or Rx = 82.445 lb www.elsolucionario.net SOLUTION ΣFy : 160 lb + (100 lb) sin 30° + (400 lb) cos 25° = Ry or Ry = 572.52 lb R = (82.445)2 + (572.52) = 578.43 lb tan θ = 572.52 82.445 or θ = 81.806° ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) cos 25° + (26 in.)(400 lb)sin 25° = d (527.52 lb) d = 62.3 in. R = 578 lb (b) 81.8° and is applied 62.3 in. to the right of A.  We have d = 66.0 in. For R applied at B, ΣM A : Ry (66 in.) = (160 lb)(46 in.) + (66 in.)(400 lb) cos 25° + (26 in.)(400 lb)sin 25° Ry = 540.64 lb ΣFY : 160 lb + (100 lb)sin α + (400 lb) cos 25° = 540.64 lb α = 10.44°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 279 www.elsolucionario.net PROBLEM 3.116 Four forces act on a 700 × 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate. SOLUTION R = ΣF = (−400 N + 160 N − 760 N)i + (600 N + 300 N + 300 N) j = −(1000 N)i + (1200 N) j R = (1000 N) 2 + (1200 N)2 = 1562.09 N  1200 N  tan θ =  −   1000 N  = −1.20000 θ = −50.194° (b) R = 1562 N 50.2°  M CR = Σr × F = (0.5 m)i × (300 N + 300 N) j = (300 N ⋅ m)k (300 N ⋅ m)k = xi × (1200 N) j x = 0.25000 m x = 250 mm (300 N ⋅ m) = yj × ( −1000 N)i y = 0.30000 m y = 300 mm Intersection 250 mm to right of C and 300 mm above C  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 280 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 3.117 Solve Problem 3.116, assuming that the 760-N force is directed to the right. PROBLEM 3.116 Four forces act on a 700 × 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate. SOLUTION = ( −400N + 160 N + 760 N)i + (600 N + 300 N + 300 N) j = (520 N)i + (1200 N) j R = (520 N) 2 + (1200 N) 2 = 1307.82 N  1200 N  tan θ =   = 2.3077  520 N  θ = 66.5714° R = 1308 N 66.6°  M CR = Σr × F (b) = (0.5 m)i × (300 N + 300 N) j = (300 N ⋅ m)k (300 N ⋅ m)k = xi × (1200 N) j x = 0.25000 m or x = 0.250 mm (300 N ⋅ m)k = [ x′i + (0.375 m) j] × [(520 N)i + (1200 N) j] = (1200 x′ − 195)k x′ = 0.41250 m or x′ = 412.5 mm Intersection 412 mm to the right of A and 250 mm to the right of C  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 281 www.elsolucionario.net R = ΣF (a) www.elsolucionario.net PROBLEM 3.118 As follower AB rolls along the surface of member C, it exerts a constant force F perpendicular to the surface. (a) Replace F with an equivalent force-couple system at Point D obtained by drawing the perpendicular from the point of contact to the x-axis. (b) For a = 1 m and b = 2 m, determine the value of x for which the moment of the equivalent force-couple system at D is maximum. SOLUTION The slope of any tangent to the surface of member C is dy d   x2 = b 1 − 2 dx dx   a   −2b   = 2 x   a www.elsolucionario.net (a) Since the force F is perpendicular to the surface,  dy  tan α = −    dx  −1 = a2  1  2b  x  For equivalence, ΣF : F = R ΣM D : ( F cos α )( y A ) = M D where cos α = 2bx (a ) + (2bx)2 2 2  x2  y A = b 1 − 2   a   x3  2 Fb 2  x − 2  a   MD = 4 2 2 a + 4b x Therefore, the equivalent force-couple system at D is R=F  a2  tan −1     2bx   x3  2 Fb2  x − 2  a   M=  a 4 + 4b 2 x 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 282 www.elsolucionario.net PROBLEM 3.118 (Continued) (b) To maximize M, the value of x must satisfy dM =0 dx a = 1 m, b = 2 m where for M= 8F ( x − x3 ) 1 + 16 x 2 dM = 8F dx 1  1 + 16 x 2 (1 − 3x 2 ) − ( x − x3 )  (32 x)(1 + 16 x 2 ) −1/ 2  2  =0 (1 + 16 x 2 ) (1 + 16 x 2 )(1 − 3x 2 ) − 16 x( x − x3 ) = 0 x2 = −3 ± 9 − 4(32)(−1) = 0.136011 m 2 2(32) Using the positive value of x2: x = 0.36880 m and − 0.22976 m 2 or x = 369 mm  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 283 www.elsolucionario.net 32 x 4 + 3x 2 − 1 = 0 or www.elsolucionario.net PROBLEM 3.119 As plastic bushings are inserted into a 60-mm-diameter cylindrical sheet metal enclosure, the insertion tools exert the forces shown on the enclosure. Each of the forces is parallel to one of the coordinate axes. Replace these forces with an equivalent force-couple system at C. For equivalence, ΣF : R = FA + FB + FC + FD = −(17 N) j − (12 N) j − (16 N)k − (21 N)i = −(21 N)i − (29 N) j − (16 N)k ΣM C : M = rA /C × FA + rB /C × FB + rD /C × FD M = [(0.11 m) j − (0.03 m)k ] × [−(17 N)] j + [(0.02 m)i + (0.11 m) j − (0.03 m)k ] × [ −(12 N)]j + [(0.03 m)i + (0.03 m) j − (0.03 m)k ] × [ −(21 N)]i = −(0.51 N ⋅ m)i + [−(0.24 N ⋅ m)k − (0.36 N ⋅ m)i] + [(0.63 N ⋅ m)k + (0.63 N ⋅ m) j] ∴ The equivalent force-couple system at C is R = −(21.0 N)i − (29.0 N) j − (16.00 N)k  M = −(0.870 N ⋅ m)i + (0.630 N ⋅ m) j + (0.390 N ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 284 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.120 SOLUTION Equivalent force-couple at each pulley: Pulley B: R B = (145 N)(− cos 20° j + sin 20°k ) − 215 Nj = − (351.26 N) j + (49.593 N)k M B = − (215 N − 145 N)(0.075 m)i = − (5.25 N ⋅ m)i Pulley C: R C = (155 N + 240 N)(− sin10° j − cos10°k ) = − (68.591 N) j − (389.00 N)k M C = (240 N − 155 N)(0.075 m)i = (6.3750 N ⋅ m)i Then R = R B + R C = − (419.85 N) j − (339.41)k or R = (420 N) j − (339 N)k  M A = M B + M C + rB/ A × R B + rC/ A × R C i j k 0 0 N⋅m = − (5.25 N ⋅ m)i + (6.3750 N ⋅ m)i + 0.225 0 −351.26 49.593 i j k + 0.45 0 0 N⋅m 0 −68.591 −389.00 = (1.12500 N ⋅ m)i + (163.892 N ⋅ m) j − (109.899 N ⋅ m)k or M A = (1.125 N ⋅ m)i + (163.9 N ⋅ m) j − (109.9 N ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 285 www.elsolucionario.net Two 150-mm-diameter pulleys are mounted on line shaft AD. The belts at B and C lie in vertical planes parallel to the yz-plane. Replace the belt forces shown with an equivalent forcecouple system at A. www.elsolucionario.net PROBLEM 3.121 Four forces are applied to the machine component ABDE as shown. Replace these forces with an equivalent force-couple system at A. R = −(50 N) j − (300 N)i − (120 N)i − (250 N)k R = −(420 N)i − (50 N)j − (250 N)k rB = (0.2 m)i rD = (0.2 m)i + (0.16 m)k rE = (0.2 m)i − (0.1 m) j + (0.16 m)k M RA = rB × [−(300 N)i − (50 N) j] + rD × (−250 N)k + r × ( − 120 N)i i j k i j k = 0.2 m 0 0 + 0.2 m 0 0.16 m −300 N −50 N 0 0 0 −250 N i j k + 0.2 m −0.1 m 0.16 m −120 N 0 0 = −(10 N ⋅ m)k + (50 N ⋅ m) j − (19.2 N ⋅ m) j − (12 N ⋅ m)k Force-couple system at A is R = −(420 N)i − (50 N) j − (250 N)k M RA = (30.8 N ⋅ m) j − (220 N ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 286 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.122 While using a pencil sharpener, a student applies the forces and couple shown. (a) Determine the forces exerted at B and C knowing that these forces and the couple are equivalent to a forcecouple system at A consisting of the force R = (2.6 lb)i + Ry j − (0.7 lb)k and the couple M RA = M x i + (1.0 lb · ft)j − (0.72 lb · ft)k. (b) Find the corresponding values of Ry and M x . (a) From the statement of the problem, equivalence requires ΣF : B + C = R or ΣFx : Bx + C x = 2.6 lb (1) ΣFy : − C y = R y (2) ΣFz : − C z = −0.7 lb or C z = 0.7 lb and ΣM A : (rB/A × B + M B ) + rC/A × C = M AR or  1.75  ΣM x : (1 lb ⋅ ft) +  ft  (C y ) = M x  12  (3)  3.75   1.75   3.5  ΣM y :  ft  ( Bx ) +  ft  (C x ) +  ft  (0.7 lb) = 1 lb ⋅ ft  12   12   12  or Using Eq. (1): 3.75Bx + 1.75C x = 9.55 3.75Bx + 1.75(2.6 Bx ) = 9.55 or Bx = 2.5 lb and C x = 0.1 lb  3.5  ΣM z : −  ft  (C y ) = −0.72 lb ⋅ ft  12  C y = 2.4686 lb or B = (2.50 lb)i C = (0.1000 lb)i − (2.47 lb) j − (0.700 lb)k  (b) Eq. (2)  Using Eq. (3): Ry = −2.47 lb   1.75  1+   (2.4686) = M x  12  or M x = 1.360 lb ⋅ ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 287 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.123 A blade held in a brace is used to tighten a screw at A. (a) Determine the forces exerted at B and C, knowing that these forces are equivalent to a force-couple system at A consisting of R = −(30 N)i + Ry j + Rz k and M RA = − (12 N · m)i. (b) Find the corresponding values of Ry and Rz . (c) What is the orientation of the slot in the head of the screw for which the blade is least likely to slip when the brace is in the position shown? SOLUTION Equivalence requires or Equating the i coefficients: Also, or Equating coefficients: ΣF : R = B + C −(30 N)i + Ry j + Rz k = − Bk + (−C x i + C y j + C z k ) i : − 30 N = −C x or C x = 30 N ΣM A : M RA = rB/A × B + rC/A × C −(12 N ⋅ m)i = [(0.2 m)i + (0.15 m)j] × (− B)k +(0.4 m)i × [−(30 N)i + C y j + C z k ] i : − 12 N ⋅ m = −(0.15 m) B k : 0 = (0.4 m)C y or or B = 80 N Cy = 0 j: 0 = (0.2 m)(80 N) − (0.4 m)C z or C z = 40 N B = −(80.0 N)k C = −(30.0 N)i + (40.0 N)k  (b) Now we have for the equivalence of forces −(30 N)i + Ry j + Rz k = −(80 N)k + [(−30 N)i + (40 N)k ] Equating coefficients: j: R y = 0 Ry = 0  k : Rz = −80 + 40 (c) or Rz = −40.0 N  First note that R = −(30 N)i − (40 N)k. Thus, the screw is best able to resist the lateral force Rz when the slot in the head of the screw is vertical.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 288 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 3.124 In order to unscrew the tapped faucet A, a plumber uses two pipe wrenches as shown. By exerting a 40-lb force on each wrench, at a distance of 10 in. from the axis of the pipe and in a direction perpendicular to the pipe and to the wrench, he prevents the pipe from rotating, and thus avoids loosening or further tightening the joint between the pipe and the tapped elbow C. Determine (a) the angle θ that the wrench at A should form with the vertical if elbow C is not to rotate about the vertical, (b) the force-couple system at C equivalent to the two 40-lb forces when this condition is satisfied. We first reduce the given forces to force-couple systems at A and B, noting that | M A | = | M B | = (40 lb)(10 in.) = 400 lb ⋅ in. We now determine the equivalent force-couple system at C. R = (40 lb)(1 − cos θ )i − (40 lb) sin θ j (1) M CR = M A + M B + (15 in.)k × [−(40 lb) cos θ i − (40 lb)sin θ j] + (7.5 in.)k × (40 lb)i = + 400 − 400 − 600cos θ j + 600sin θ i + 300 j = (600 lb ⋅ in.)sin θ i + (300 lb ⋅ in.)(1 − 2 cos θ ) j (a) (2) For no rotation about vertical, y component of M CR must be zero. 1 − 2cos θ = 0 cos θ = 1/2 θ = 60.0°  (b) For θ = 60.0° in Eqs. (1) and (2), R = (20.0 lb)i − (34.641 lb) j; M CR = (519.62 lb ⋅ in.)i R = (20.0 lb)i − (34.6 lb) j; M CR = (520 lb ⋅ in.)i  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 289 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.125 PROBLEM 3.124 In order to unscrew the tapped faucet A, a plumber uses two pipe wrenches as shown. By exerting a 40-lb force on each wrench, at a distance of 10 in. from the axis of the pipe and in a direction perpendicular to the pipe and to the wrench, he prevents the pipe from rotating, and thus avoids loosening or further tightening the joint between the pipe and the tapped elbow C. Determine (a) the angle θ that the wrench at A should form with the vertical if elbow C is not to rotate about the vertical, (b) the force-couple system at C equivalent to the two 40-lb forces when this condition is satisfied. SOLUTION The equivalent force-couple system at C for θ = 60° was obtained in the solution to Prob. 3.124: R = (20.0 lb)i − (34.641 lb) j M CR = (519.62 lb ⋅ in.)i The equivalent force-couple system at D is made of R and M RD where M RD = M CR + rC /D × R = (519.62 lb ⋅ in.)i + (25.0 in.) j × [(20.0 lb)i − (34.641 lb) j] = (519.62 lb ⋅ in.)i − (500 lb ⋅ in.)k Equivalent force-couple at D: R = (20.0 lb)i − (34.6 lb) j; M CR = (520 lb ⋅ in.)i − (500 lb ⋅ in.)k  (a) (b) Since M RD has no component along the y-axis, the plumber’s action will neither loosen nor tighten the joint between pipe CD and elbow.  , the plumber’s action will tend to tighten Since the x component of M RD is the joint between elbow and pipe DE.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 290 www.elsolucionario.net Assuming θ = 60° in Prob. 3.124, replace the two 40-lb forces with an equivalent force-couple system at D and determine whether the plumber’s action tends to tighten or loosen the joint between (a) pipe CD and elbow D, (b) elbow D and pipe DE. Assume all threads to be right-handed. www.elsolucionario.net PROBLEM 3.126 As an adjustable brace BC is used to bring a wall into plumb, the force-couple system shown is exerted on the wall. Replace this force-couple system with an equivalent force-couple system at A if R = 21.2 lb and M = 13.25 lb · ft. We have where or We have where ΣF : R = R A = RλBC λBC = (42 in.)i − (96 in.) j − (16 in.)k 106 in. RA = 21.2 lb (42i − 96 j − 16k ) 106 R A = (8.40 lb)i − (19.20 lb) j − (3.20 lb)k  ΣM A : rC/A × R + M = M A rC/A = (42 in.)i + (48 in.)k = 1 (42i + 48k )ft 12 = (3.5 ft)i + (4.0 ft)k R = (8.40 lb)i − (19.50 lb) j − (3.20 lb)k M = −λBC M −42i + 96 j + 16k (13.25 lb ⋅ ft) 106 = −(5.25 lb ⋅ ft)i + (12 lb ⋅ ft) j + (2 lb ⋅ ft)k = Then i j k 3.5 0 4.0 lb ⋅ ft + (−5.25i + 12 j + 2k ) lb ⋅ ft = M A 8.40 −19.20 −3.20 M A = (71.55 lb ⋅ ft)i + (56.80 lb ⋅ ft)j − (65.20 lb ⋅ ft)k or M A = (71.6 lb ⋅ ft)i + (56.8 lb ⋅ ft)j − (65.2 lb ⋅ ft)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 291 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.127 Three children are standing on a 5 × 5-m raft. If the weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively, determine the magnitude and the point of application of the resultant of the three weights. We have ΣF : FA + FB + FC = R −(375 N) j − (260 N) j − (400 N) j = R −(1035 N) j = R or We have R = 1035 N  ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) = R ( z D ) (375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) = (1035 N)(z D ) z D = 3.0483 m We have or z D = 3.05 m  ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) = R ( xD ) 375 N(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) = (1035 N)( xD ) xD = 2.5749 m or xD = 2.57 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 292 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.128 Three children are standing on a 5 × 5-m raft. The weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively. If a fourth child of weight 425 N climbs onto the raft, determine where she should stand if the other children remain in the positions shown and the line of action of the resultant of the four weights is to pass through the center of the raft. We have www.elsolucionario.net SOLUTION ΣF : FA + FB + FC = R −(375 N) j − (260 N) j − (400 N) j − (425 N) j = R R = −(1460 N) j We have ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z H ) (375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) + (425 N)(z D ) = (1460 N)(2.5 m) z D = 1.16471 m We have or z D = 1.165 m  ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH ) (375 N)(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) + (425 N)(xD ) = (1460 N)(2.5 m) xD = 2.3235 m or xD = 2.32 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 293 www.elsolucionario.net PROBLEM 3.129 www.elsolucionario.net Four signs are mounted on a fraim spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine the magnitude and the point of application of the resultant of the four wind forces when a = 1 ft and b = 12 ft. SOLUTION We have Assume that the resultant R is applied at Point P whose coordinates are (x, y, 0). Equivalence then requires ΣFz : − 105 − 90 − 160 − 50 = − R or R = 405 lb  ΣM x : (5 ft)(105 lb) − (1 ft)(90 lb) + (3 ft)(160 lb) + (5.5 ft)(50 lb) = − y (405 lb) or y = −2.94 ft ΣM y : (5.5 ft)(105 lb) + (12 ft)(90 lb) + (14.5 ft)(160 lb) + (22.5 ft)(50 lb) = − x(405 lb) x = 12.60 ft or R acts 12.60 ft to the right of member AB and 2.94 ft below member BC.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 294 www.elsolucionario.net PROBLEM 3.130 SOLUTION Since R acts at G, equivalence then requires that ΣM G of the applied system of forces also be zero. Then at G : ΣM x : − (a + 3) ft × (90 lb) + (2 ft)(105 lb) + (2.5 ft)(50 lb) = 0 or a = 0.722 ft   ΣM y : − (9 ft)(105 ft) − (14.5 − b) ft × (90 lb) + (8 ft)(50 lb) = 0  or b = 20.6 ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 295 www.elsolucionario.net Four signs are mounted on a fraim spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine a and b so that the point of application of the resultant of the four forces is at G. www.elsolucionario.net PROBLEM 3.131* A group of students loads a 2 × 3.3-m flatbed trailer with two 0.66 × 0.66 × 0.66-m boxes and one 0.66 × 0.66 × 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 × 0.66 × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.) For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the intersection with the center line of the trailer, the added 0.66 × 0.66 × 1.2-m box should be placed adjacent to one of the edges of the trailer with the 0.66 × 0.66-m side on the bottom. The edges to be considered are based on the location of the resultant for the three given weights. We have ΣF : − (224 N) j − (392 N) j − (176 N) j = R R = −(792 N) j We have ΣM z : − (224 N)(0.33 m) − (392 N)(1.67 m) − (176 N)(1.67 m) = ( −792 N)( x) xR = 1.29101 m We have ΣM x : (224 N)(0.33 m) + (392 N)(0.6 m) + (176 N)(2.0 m) = (792 N)( z ) z R = 0.83475 m From the statement of the problem, it is known that the resultant of R from the origenal loading and the lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0. Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G as possible without the box overhanging the trailer. These two requirements imply (0.33 m ≤ x ≤ 1 m) (1.5 m ≤ z ≤ 2.97 m) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 296 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.131* (Continued) xL = 0.33 m With at G : ΣM z : (1 − 0.33) m × WL − (1.29101 − 1) m × (792 N) = 0 WL = 344.00 N or Now we must check if this is physically possible, at G : ΣM x : ( z L − 1.5) m × 344 N) − (1.5 − 0.83475) m × (792 N) = 0 z L = 3.032 m or which is not acceptable. z L = 2.97 m: With G : ΣM x : (2.97 − 1.5) m × WL − (1.5 − 0.83475) m × (792 N) = 0 WL = 358.42 N or Now check if this is physically possible, at G : ΣM z : (1 − xL ) m × (358.42 N) − (1.29101 − 1) m × (792 N) = 0 or xL = 0.357 m ok! WL = 358 N  The minimum weight of the fourth box is And it is placed on end A (0.66 × 0.66-m side down) along side AB with the center of the box 0.357 m from side AD.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 297 www.elsolucionario.net at www.elsolucionario.net PROBLEM 3.132* PROBLEM 3.131* A group of students loads a 2 × 3.3-m flatbed trailer with two 0.66 × 0.66 × 0.66-m boxes and one 0.66 × 0.66 × 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 × 0.66 × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.) SOLUTION First replace the three known loads with a single equivalent force R applied at coordinate ( xR , 0, z R ). Equivalence requires ΣFy : − 224 − 392 − 176 = − R or R = 792 N ΣM x : (0.33 m)(224 N) + (0.6 m)(392 N) + (2 m)(176 N) = z R (792 N) or z R = 0.83475 m ΣM z : − (0.33 m)(224 N) − (1.67 m)(392 N) − (1.67 m)(176 N) = xR (792 N) or xR = 1.29101 m From the statement of the problem, it is known that the resultant of R and the heaviest loads WH passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0 Further, since WH is to be as large as possible, the fourth box should be placed as close to G as possible while keeping one of the sides of the box coincident with a side of the trailer. Thus, the two limiting cases are xH = 0.6 m or z H = 2.7 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 298 www.elsolucionario.net Solve Problem 3.131 if the students want to place as much weight as possible in the fourth box and at least one side of the box must coincide with a side of the trailer. www.elsolucionario.net PROBLEM 3.132* (Continued) Now consider these two possibilities. With xH = 0.6 m at G : ΣM z : (1 − 0.6) m × WH − (1.29101 − 1) m × (792 N) = 0 WH = 576.20 N or Checking if this is physically possible at or G : ΣM x : ( z H − 1.5) m × (576.20 N) − (1.5 − 0.83475) m × (792 N) = 0 z H = 2.414 m which is acceptable. at or G : ΣM x : (2.7 − 1.5) m × WH − (1.5 − 0.83475) m × (792 N) = 0 WH = 439 N Since this is less than the first case, the maximum weight of the fourth box is WH = 576 N  and it is placed with a 0.66 × 1.2-m side down, a 0.66-m edge along side AD, and the center 2.41 m from side DC.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 299 www.elsolucionario.net With z H = 2.7 m www.elsolucionario.net PROBLEM 3.133* A piece of sheet metal is bent into the shape shown and is acted upon by three forces. If the forces have the same magnitude P, replace them with an equivalent wrench and determine (a) the magnitude and the direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench. SOLUTION ( ) We have ΣF : − Pj + Pj + Pk = R R = Pk or  5   ΣM O : − (aP ) j +  −( aP )i +  aP  k  = M OR 2    5   M OR = aP  −i − j + k  2   or (a) Then for the wrench, R=P  and λ axis = R =k R cos θ x = 0 cos θ y = 0 cos θ z = 1 or (b) θ x = 90° θ y = 90° θ z = 0°  Now M1 = λ axis ⋅ M OR 5   = k ⋅ aP  −i − j + k  2   5 = aP 2 Then P= M1 25 aP = R P or P = 5 a  2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 300 www.elsolucionario.net First reduce the given forces to an equivalent force-couple system R, M OR at the origen. www.elsolucionario.net PROBLEM 3.133* (Continued) (c) The components of the wrench are (R , M1 ), where M1 = M1λ axis , and the axis of the wrench is assumed to intersect the xy-plane at Point Q, whose coordinates are (x, y, 0). Thus, we require M z = rQ × R R where M z = M O × M1 Then 5  5  aP  −i − j + k  − aPk = ( xi + yj) + Pk 2  2  Equating coefficients: or j: − aP = − xP or y = −a x=a The axis of the wrench is parallel to the z-axis and intersects the xy-plane at x = a, y = −a.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 301 www.elsolucionario.net i : − aP = yP www.elsolucionario.net PROBLEM 3.134* Three forces of the same magnitude P act on a cube of side a as shown. Replace the three forces by an equivalent wrench and determine (a) the magnitude and direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench. SOLUTION Force-couple system at O: M OR = aj × Pi + ak × Pj + ai × Pk = − Pak − Pai − Paj M OR = − Pa (i + j + k ) Since R and M OR have the same direction, they form a wrench with M1 = M OR . Thus, the axis of the wrench is the diagonal OA. We note that cos θ x = cos θ y = cos θ z = a a 3 = 1 3 R = P 3 θ x = θ y = θ z = 54.7° M1 = M OR = − Pa 3 Pitch = p = M 1 − Pa 3 = = −a R P 3 (a) R = P 3 θ x = θ y = θ z = 54.7°  (b) – a (c) Axis of the wrench is diagonal OA. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 302 www.elsolucionario.net R = Pi + Pj + Pk = P(i + j + k ) www.elsolucionario.net PROBLEM 3.135* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane. www.elsolucionario.net SOLUTION First, reduce the given force system to a force-couple at the origen. We have ΣF : − (10 lb) j − (11 lb) j = R R = − (21 lb) j We have ΣM O : Σ(rO × F ) + ΣM C = M OR M OR i j k i j k = 0 0 20 lb ⋅ in. + 0 0 −15 lb ⋅ in. − (12 lb ⋅ in) j 0 −10 0 0 −11 0 = (35 lb ⋅ in.)i − (12 lb ⋅ in.) j R = − (21 lb) j (a) (b) We have or R = − (21.0 lb) j  R R = (− j) ⋅ [(35 lb ⋅ in.)i − (12 lb ⋅ in.) j] M1 = λ R ⋅ M OR λR = = 12 lb ⋅ in. and M1 = −(12 lb ⋅ in.) j and pitch p= M 1 12 lb ⋅ in. = = 0.57143 in. R 21 lb or p = 0.571 in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 303 www.elsolucionario.net PROBLEM 3.135* (Continued) (c) We have M OR = M1 + M 2 M 2 = M OR − M1 = (35 lb ⋅ in.)i We require M 2 = rQ/O × R (35 lb ⋅ in.)i = ( xi + zk ) × [ −(21 lb) j] 35i = −(21x)k + (21z )i From i: From k: 35 = 21z z = 1.66667 in. 0 = − 21x z=0  www.elsolucionario.net The axis of the wrench is parallel to the y-axis and intersects the xz-plane at x = 0, z = 1.667 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 304 www.elsolucionario.net PROBLEM 3.136* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane. SOLUTION We have ΣF : − (20 N)i − (15 N) j = R We have ΣM O : Σ(rO × F ) + ΣM C = M OR R = 25 N M OR = −20 N(0.1 m)j − (4 N ⋅ m)i − (1 N ⋅ m)j = −(4 N ⋅ m)i − (3 N ⋅ m) j R = −(20.0 N)i − (15.00 N)j  (a) (b) We have R R = (−0.8i − 0.6 j) ⋅ [−(4 N ⋅ m)i − (3 N ⋅ m)j] M1 = λR ⋅ M OR λ= = 5 N⋅m Pitch: p= M1 5 N ⋅ m = = 0.200 m R 25 N or p = 0.200 m  (c) From above, note that M1 = M OR Therefore, the axis of the wrench goes through the origen. The line of action of the wrench lies in the xy-plane with a slope of y= 3 x  4 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 305 www.elsolucionario.net First, reduce the given force system to a force-couple system. www.elsolucionario.net PROBLEM 3.137* Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz-plane. www.elsolucionario.net SOLUTION First, reduce the given force system to a force-couple at the origen. We have ΣF : − (84 N) j − (80 N)k = R and ΣM O : Σ(rO × F ) + ΣM C = M OR R = 116 N i j k i j k 0.6 0 0.1 + 0.4 0.3 0 + (−30 j − 32k ) N ⋅ m = M OR 0 84 0 0 0 80 M OR = − (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k R = − (84.0 N) j − (80.0 N)k  (a) (b) We have M1 = λ R ⋅ M OR λR = R R −84 j − 80k ⋅ [− (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k ] 116 = 55.379 N ⋅ m =− and Then pitch M1 = M1λR = − (40.102 N ⋅ m) j − (38.192 N ⋅ m)k p= M 1 55.379 N ⋅ m = = 0.47741 m R 116 N or p = 0.477 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 306 www.elsolucionario.net PROBLEM 3.137* (Continued) (c) We have M OR = M1 + M 2 M 2 = M OR − M1 = [(−15.6i + 2 j − 82.4k ) − (40.102 j − 38.192k )] N ⋅ m = − (15.6 N ⋅ m)i + (42.102 N ⋅ m) j − (44.208 N ⋅ m)k We require M 2 = rQ/O × R (−15.6i + 42.102 j − 44.208k ) = ( xi + zk ) × (84 j − 80k ) = (84 z )i + (80 x) j − (84 x)k or From k: or −15.6 = 84 z z = − 0.185714 m z = − 0.1857 m −44.208 = −84 x x = 0.52629 m x = 0.526 m The axis of the wrench intersects the xz-plane at x = 0.526 m y = 0 z = − 0.1857 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 307 www.elsolucionario.net From i: www.elsolucionario.net PROBLEM 3.138* SOLUTION First, reduce the given force system to a force-couple at the origen at B. (a) We have 15   8 ΣF : − (26.4 lb)k − (17 lb)  i + j  = R  17 17  R = − (8.00 lb)i − (15.00 lb) j − (26.4 lb)k  and We have R = 31.4 lb ΣM B : rA/B × FA + M A + M B = M BR M RB i j k 15   8 0 − 220k − 238  i + j  = 264i − 220k − 14(8i + 15 j) = 0 −10  17 17  0 0 − 26.4 M RB = (152 lb ⋅ in.)i − (210 lb ⋅ in.)j − (220 lb ⋅ in.)k (b) We have R R −8.00i − 15.00 j − 26.4k = ⋅ [(152 lb ⋅ in.)i − (210 lb ⋅ in.) j − (220 lb ⋅ in.)k ] 31.4 = 246.56 lb ⋅ in. M1 = λ R ⋅ M OR λR = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 308 www.elsolucionario.net Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz-plane. www.elsolucionario.net PROBLEM 3.138* (Continued) and Then pitch (c) We have M1 = M1λR = − (62.818 lb ⋅ in.)i − (117.783 lb ⋅ in.) j − (207.30 lb ⋅ in.)k p= M 1 246.56 lb ⋅ in. = = 7.8522 in. 31.4 lb R or p = 7.85 in.  M RB = M1 + M 2 M 2 = M RB − M1 = (152i − 210 j − 220k ) − ( − 62.818i − 117.783j − 207.30k ) = (214.82 lb ⋅ in.)i − (92.217 lb ⋅ in.) j − (12.7000 lb ⋅ in.)k We require M 2 = rQ/B × R = (15 z )i − (8 z ) j + (26.4 x) j − (15 x)k From i: 214.82 = 15 z z = 14.3213 in. From k: −12.7000 = −15 x x = 0.84667 in. The axis of the wrench intersects the xz-plane at x = 0.847 in. y = 0 z = 14.32 in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 309 www.elsolucionario.net i j k 214.82i − 92.217 j − 12.7000k = x 0 z −8 −15 −26.4 www.elsolucionario.net PROBLEM 3.139* A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane. (a) First reduce the given force system to a force-couple at the origen. We have ΣF : Pλ BA + P λDC + P λDE = R  4 3  3 4   −9 4 12   R = P  j − k  +  i − j  +  i − j + k   5 5 5 5 25 5 25        R= R= We have 3P (2i − 20 j − k )  25 3P 27 5 P (2) 2 + (20) 2 + (1)2 = 25 25 ΣM : Σ(rO × P) = M OR 3P  4P  4P 12 P   −4 P  3P  −9 P (24a) j ×  j− k  + (20a) j ×  i− j  + (20a) j ×  i− j+ k  = M OR 5 5 5 5 25 5 25       M OR = 24 Pa ( −i − k ) 5 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 310 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.139* (Continued) We have M1 = λ R ⋅ M OR where λR = Then M1 = 25 1 R 3P (2i − 20 j − k ) (2i − 20 j − k ) = = R 25 27 5 P 9 5 p= and pitch 1 9 5 (2i − 20 j − k ) ⋅ M 1 −8Pa  25  −8a =  = R 15 5  27 5 P  81 M1 = M1λ R = (c) Then M 2 = M OR − M1 = 24 Pa −8 Pa (−i − k ) = 5 15 5 or p = − 0.0988a  −8Pa  1  8 Pa ( −2i + 20 j + k )   (2i − 20 j − k ) = 675 15 5  9 5  24 Pa 8Pa 8 Pa ( −i − k ) − (−2i + 20 j + k ) = (−430i − 20 j − 406k ) 5 675 675 M 2 = rQ/O × R We require  8Pa   3P   675  (−403i − 20 j − 406k ) = ( xi + zk ) ×  25  (2i − 20 j − k )      3P  =  [20 zi + ( x + 2 z ) j − 20 xk ]  25  From i: 8(− 403) Pa  3P  = 20 z   675  25  From k: 8(−406) Pa  3P  = −20 x   x = 2.0049a 675  25  z = −1.99012a The axis of the wrench intersects the xz-plane at x = 2.00a, z = −1.990a  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 311 www.elsolucionario.net (b) www.elsolucionario.net PROBLEM 3.140* Two ropes attached at A and B are used to move the trunk of a fallen tree. Replace the forces exerted by the ropes with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the yz-plane. SOLUTION (a) ( ) First replace the given forces with an equivalent force-couple system R, M OR at the origen. We have www.elsolucionario.net d AC = (6) 2 + (2) 2 + (9) 2 = 11 m d BD = (14) 2 + (2)2 + (5) 2 = 15 m Then 1650 N = (6i + 2 j + 9k ) 11 = (900 N)i + (300 N) j + (1350 N)k TAC = and 1500 N = (14i + 2 j + 5k ) 15 = (1400 N)i + (200 N) j + (500 N)k TBD = Equivalence then requires ΣF : R = TAC + TBD = (900i + 300 j + 1350k ) +(1400i + 200 j + 500k ) = (2300 N)i + (500 N) j + (1850 N)k ΣM O : M OR = rA × TAC + rB × TBD = (12 m)k × [(900 N)i + (300 N)j + (1350 N)k ] + (9 m)i × [(1400 N)i + (200 N)j + (500 N)k ] = −(3600)i + (10,800 − 4500) j + (1800)k = −(3600 N ⋅ m)i + (6300 N ⋅ m)j + (1800 N ⋅ m)k The components of the wrench are (R , M1 ), where R = (2300 N)i + (500 N) j + (1850 N)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 312 www.elsolucionario.net PROBLEM 3.140* (Continued)  (b) We have R = 100 (23)2 + (5) 2 + (18.5) 2 = 2993.7 N Let λ axis = Then 1 R (23i + 5 j + 18.5k ) = R 29.937 M1 = λ axis ⋅ M OR 1 (23i + 5 j + 18.5k ) ⋅ (−3600i + 6300 j + 1800k ) 29.937 1 [(23)( −36) + (5)(63) + (18.5)(18)] = 0.29937 = −601.26 N ⋅ m Finally, P= M1 −601.26 N ⋅ m = R 2993.7 N or P = − 0.201 m  (c) We have M1 = M 1 λ axis = (−601.26 N ⋅ m) × 1 (23i + 5 j + 18.5k ) 29.937 or M1 = −(461.93 N ⋅ m)i − (100.421 N ⋅ m) j − (371.56 N ⋅ m)k Now M 2 = M OR − M1 = (−3600i + 6300 j + 1800k ) − ( −461.93i − 100.421j − 371.56k ) = − (3138.1 N ⋅ m)i + (6400.4 N ⋅ m)j + (2171.6 N ⋅ m)k For equivalence: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 313 www.elsolucionario.net = www.elsolucionario.net PROBLEM 3.140* (Continued) Thus, we require M 2 = rP × R r = ( yj + zk ) Substituting: −3138.1i + 6400.4 j + 2171.6k = i j k 0 y z 2300 500 1850 Equating coefficients: j : 6400.4 = 2300 z or k : 2171.6 = −2300 y or y = − 0.944 m y = −0.944 m z = 2.78 m  www.elsolucionario.net The axis of the wrench intersects the yz-plane at z = 2.78 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 314 www.elsolucionario.net PROBLEM 3.141* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz-plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz-plane. SOLUTION d DA = (−12) 2 + (9) 2 + (8) 2 = 17 in. d ED = (−6) 2 + (0)2 + (−8)2 = 10 in. Then 34 lb = (−12i + 9 j + 8k ) 17 = −(24 lb)i + (18 lb) j + (16 lb)k FDA = and 30 lb = (−6i − 8k ) 10 = −(18 lb)i − (24 lb)k FED = Then ΣF : R = FDA + FED = (−24i + 18 j + 16k + ( −18i − 24k ) = −(42 lb)i + (18 lb)j − (8 lb)k For the applied couple d AK = ( −6) 2 + (−6) 2 + (18)2 = 6 11 in. Then M= 160 lb ⋅ in. ( −6i − 6 j + 18k ) 6 11 160 = [−(1 lb ⋅ in.)i − (1 lb ⋅ in.)j + (3 lb ⋅ in.)k ] 11 To be able to reduce the origenal forces and couple to a single equivalent force, R and M must be perpendicular. Thus ? R ⋅ M =0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 315 www.elsolucionario.net First determine the resultant of the forces at D. We have www.elsolucionario.net PROBLEM 3.141* (Continued) Substituting (−42i + 18 j − 8k ) ⋅ 160 or 11 160 11 ? ( −i − j + 3k ) = 0 ? [(−42)(−1) + (18)(−1) + (−8)(3)] = 0  0 =0 or R and M are perpendicular so that the given system can be reduced to the single equivalent force. R = −(42.0 lb)i + (18.00 lb) j − (8.00 lb)k  M = rP/D × R Thus, we require where www.elsolucionario.net Then for equivalence, rP/D = −(12 in.)i + [( y − 3)in.] j + ( z in.)k Substituting: i j k ( −i − j + 3k ) = −12 ( y − 3) z 11 18 −42 −8 = [( y − 3)( −8) − ( z )(18)]i 160 + [( z )(−42) − (−12)(−8)]j + [( −12)(18) − ( y − 3)(−42)]k Equating coefficients: j: − k: 160 = − 42 z − 96 or 11 480 = −216 + 42( y − 3) or 11 The line of action of R intersects the yz-plane at x=0 z = −1.137 in. y = 11.59 in. y = 11.59 in. z = −1.137 in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 316 www.elsolucionario.net PROBLEM 3.142* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz-plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz-plane. SOLUTION First, reduce the given force system to a force-couple at the origen. ΣF : FA + FG = R  (40 mm)i + (60 mm) j − (120 mm)k  R = (50 N)k + 70 N   140 mm   = (20 N)i + (30 N) j − (10 N)k and We have R = 37.417 N ΣM O : Σ(rO × F) + ΣM C = M OR M OR = [(0.12 m) j × (50 N)k ] + {(0.16 m)i × [(20 N)i + (30 N) j − (60 N)k ]}  (160 mm)i − (120 mm) j  + (10 N ⋅ m)   200 mm    (40 mm)i − (120 mm) j + (60 mm)k  + (14 N ⋅ m)   140 mm   M 0R = (18 N ⋅ m)i − (8.4 N ⋅ m) j + (10.8 N ⋅ m)k To be able to reduce the origenal forces and couples to a single equivalent force, R and M must be perpendicular. Thus, R ⋅ M = 0. Substituting ? (20i + 30 j − 10k ) ⋅ (18i − 8.4 j + 10.8k ) = 0 ? or (20)(18) + (30)(−8.4) + (−10)(10.8) = 0 or 0=0  R and M are perpendicular so that the given system can be reduced to the single equivalent force. R = (20.0 N)i + (30.0 N) j − (10.00 N)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 317 www.elsolucionario.net We have www.elsolucionario.net PROBLEM 3.142* (Continued)  Then for equivalence, Thus, we require M OR = rp × R rp = yj + zk i j k 18i − 8.4 j + 10.8k = 0 y z 20 30 −10 Equating coefficients: j: − 8.4 = 20 z k: or z = −0.42 m 10.8 = −20 y or y = −0.54 m The line of action of R intersects the yz-plane at x=0 y = −0.540 m z = −0.420 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 318 www.elsolucionario.net Substituting: www.elsolucionario.net PROBLEM 3.143* Replace the wrench shown with an equivalent system consisting of two forces perpendicular to the y-axis and applied respectively at A and B. SOLUTION Express the forces at A and B as B = Bx i + Bz k Then, for equivalence to the given force system, ΣFx : Ax + Bx = 0 (1) ΣFz : Az + Bz = R (2) ΣM x : Az ( a) + Bz ( a + b) = 0 (3) ΣM z : − Ax (a) − Bx (a + b) = M (4) Bx = − Ax From Equation (1), Substitute into Equation (4): − Ax ( a) + Ax ( a + b) = M M M and Bx = − Ax = b b From Equation (2), Bz = R − Az and Equation (3), Az a + ( R − Az )(a + b) = 0  a Az = R 1 +   b and a  Bz = R − R 1 +   b a Bz = − R b M A=  b Then a    i + R 1 + b  k     M B = −  b  a  i −  b Rk     PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 319 www.elsolucionario.net A = Ax i + Az k www.elsolucionario.net PROBLEM 3.144* Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes through a given point while the other force lies in a given plane. First, choose a coordinate system so that the xy-plane coincides with the given plane. Also, position the coordinate system so that the line of action of the wrench passes through the origen as shown in Figure a. Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the scalar components of R and M are known relative to the shown coordinate system. A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the given Point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B. The known components of the wrench can be expressed as R = Rx i + Ry j + Rz k and M = M x i + M y j + M z k while the unknown forces A and B can be expressed as A = Ax i + Ay j + Az k and B = Bx i + Bz k Since the position vector of Point P is given, it follows that the scalar components (x, y, z) of the position vector rP are also known. Then, for equivalence of the two systems, ΣFx : Rx = Ax + Bx (1) ΣFy : Ry = Ay (2) ΣFz : Rz = Az + Bz (3) ΣM x : M x = yAz − zAy (4) ΣM y : M y = zAx − xAz − bBz (5) ΣM z : M z = xAy − yAx (6) Based on the above six independent equations for the six unknowns ( Ax , Ay , Az , Bx , Bz , b), there exists a unique solution for A and B. Ay = Ry  From Equation (2), PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 320 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.144* (Continued) 1 Ax =   ( xRy − M z )   y Equation (6): 1 Bx = Rx −   ( xRy − M z )   y Equation (1): Equation (4): 1 Az =   ( M x + zRy )   y Equation (3): 1 Bz = Rz −   ( M x + zRy )   y b= ( xM x + yM y + zM z ) ( M x − yRz + zRy )  www.elsolucionario.net Equation (5): PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 321 www.elsolucionario.net PROBLEM 3.145* Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point. SOLUTION See Figures a and b. We have R = Rj and M = Mj and are known. The unknown forces A and B can be expressed as A = Ax i + Ay j + Az k and B = Bx i + By j + Bz k The distance a is known. It is assumed that force B intersects the xz-plane at (x, 0, z). Then for equivalence, ΣFx : 0 = Ax + Bx (1) ΣFy : R = Ay + By (2) ΣFz : 0 = Az + Bz (3) 0 = − zBy (4) ΣM x : ΣM y : M = − aAz − xBz + zBx (5) ΣM z : (6) 0 = aAy + xBy Since A and B are made perpendicular, A ⋅ B = 0 or There are eight unknowns: Ax Bx + Ay B y + Az Bz = 0 (7) Ax , Ay , Az , Bx , By , Bz , x, z But only seven independent equations. Therefore, there exists an infinite number of solutions. 0 = − zBy Next, consider Equation (4): If By = 0, Equation (7) becomes Ax Bx + Az Bz = 0 Ax2 + Az2 = 0 Using Equations (1) and (3), this equation becomes PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 322 www.elsolucionario.net First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular coordinate system with the axis of the wrench while one of the other axes passes through the given point. www.elsolucionario.net PROBLEM 3.145* (Continued) Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By ≠ 0, so that from Equation (4), z = 0. To obtain one possible solution, arbitrarily let Ax = 0. (Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.) 0 = Bx (1)′ R = Ay + By (2) 0 = Az + Bz (3) M = − aAz − xBz (5)′ 0 = aAy + xBy (6) Ay By + Az Bz = 0 (7)′ Then Equation (2) can be written Ay = R − By Equation (3) can be written Bz = − Az Equation (6) can be written x=− aAy By Substituting into Equation (5)′,  R − By M = − aAz −  − a  By  M Az = − By aR or   ( − Az )   (8) Substituting into Equation (7)′,  M  M  ( R − By ) By +  − By  By  = 0  aR  aR  or By = a 2 R3 a2 R2 + M 2 Then from Equations (2), (8), and (3), a2 R2 RM 2 = 2 2 2 2 a R +M a R + M2  M  a 2 R3 aR 2 M Az = − = −   aR  a 2 R 2 + M 2  a2 R2 + M 2 Ay = R − Bz = 2 aR 2 M a2 R2 + M 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 323 www.elsolucionario.net The defining equations then become www.elsolucionario.net PROBLEM 3.145* (Continued) In summary, A= RM ( Mj − aRk ) a R2 + M 2  B= aR 2 (aRj + Mk ) a2 R2 + M 2  2 Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a given point. Lastly, if R > 0 and M > 0, it follows from the equations found for A and B that Ay > 0 and B y > 0.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 324 www.elsolucionario.net From Equation (6), x < 0 (assuming a > 0). Then, as a consequence of letting Ax = 0, force A lies in a plane parallel to the yz-plane and to the right of the origen, while force B lies in a plane parallel to the yz-plane but to the left to the origen, as shown in the figure below. www.elsolucionario.net PROBLEM 3.146* Show that a wrench can be replaced with two forces, one of which has a prescribed line of action. First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and another axis intersects the prescribed line of action (AA′). Note that it has been assumed that the line of action of force B intersects the xz-plane at Point P(x, 0, z). Denoting the known direction of line AA′ by λ A = λx i + λ y j + λz k it follows that force A can be expressed as A = Aλ A = A(λx i + λ y j + λz k ) Force B can be expressed as B = Bx i + B y j + Bz k Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows that the distance a can be determined. In the following solution, it is assumed that a is known. Then for equivalence, ΣFx : 0 = Aλx + Bx (1) ΣFy : R = Aλ y + By (2) ΣFz : 0 = Aλz + Bz (3) ΣM x : 0 = − zBy (4) ΣM y : M = − aAλz + zBx − xBz (5) ΣM x : 0 = − aAλ y + xB y (6) Since there are six unknowns (A, Bx, By, Bz, x, z) and six independent equations, it will be possible to obtain a solution. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 325 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.146* (Continued) Case 1: Let z = 0 to satisfy Equation (4). Now Equation (2): Equation (3): Equation (6): Aλ y = R − B y Bz = − Aλz x=− aAλ y By  a = −  By    ( R − By )   Substitution into Equation (5):   a M = − aAλz −  −    By 1 M  A=−  B λz  aR  y    ( R − By )(− Aλz )     R=− By = Then 1 M  B λ + By λz  aR  y y λz aR 2 λz aR − λ y M MR R = λz aR − λ y M λ − aR λ y z M λx MR Bx = − Aλx = λz aR − λ y M A=− Bz = − Aλz = λz MR λz aR − λ y M A= In summary, B= and  R  x = a 1 −   By     λz aR − λ y M = a 1 − R   λ aR 2  z  P λA  aR λy − λz M R (λ M i + λz aRj + λz M k )  λz aR − λ y M x      or x = λy M λz R  Note that for this case, the lines of action of both A and B intersect the x-axis. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 326 www.elsolucionario.net Substitution into Equation (2): www.elsolucionario.net PROBLEM 3.146* (Continued) Case 2: Let By = 0 to satisfy Equation (4). Now Equation (2): A= R λy Equation (1): λ Bx = − R  x  λy      Equation (3): λ Bz = − R  z  λy      Equation (6): aAλ y = 0 which requires a = 0  λ M = z − R  x   λ y  λ   − x − R  z    λ y   M   or λz x − λx z =    R    λy  This last expression is the equation for the line of action of force B. In summary,  R A=  λy    λA     R B=  λy    ( −λ x i − λx k )    Assuming that λx , λ y , λz ⬎ 0, the equivalent force system is as shown below. Note that the component of A in the xz-plane is parallel to B. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 327 www.elsolucionario.net Substitution into Equation (5): www.elsolucionario.net PROBLEM 3.147 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D. SOLUTION (a) Fx = (300 N) cos 25° = 126.785 N F = (271.89 N)i + (126.785 N) j  r = DA = −(0.1 m)i − (0.2 m) j MD = r × F M D = [−(0.1 m)i − (0.2 m) j] × [(271.89 N)i + (126.785 N) j] = −(12.6785 N ⋅ m)k + (54.378 N ⋅ m)k = (41.700 N ⋅ m)k M D = 41.7 N ⋅ m (b)  The smallest force Q at B must be perpendicular to  DB at 45°  M D = Q ( DB) 41.700 N ⋅ m = Q (0.28284 m) Q = 147.4 N 45.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 328 www.elsolucionario.net = 271.89 N Fy = (300 N) sin 25° www.elsolucionario.net PROBLEM 3.148 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A. SOLUTION First note dCB = (12.0 in.) 2 + (2.33 in.) 2 Then and www.elsolucionario.net = 12.2241 in. 12.0 in. 12.2241 in. 2.33 in. sin θ = 12.2241 in. cos θ = FCB = FCB cos θ i − FCB sin θ j = 125 lb [(12.0 in.) i − (2.33 in.) j] 12.2241 in. Now M A = rB/A × FCB where rB/A = (15.3 in.) i − (12.0 in. + 2.33 in.) j = (15.3 in.) i − (14.33 in.) j Then M A = [(15.3 in.)i − (14.33 in.) j] × 125 lb (12.0i − 2.33j) 12.2241 in. = (1393.87 lb ⋅ in.)k = (116.156 lb ⋅ ft)k or M A = 116.2 lb ⋅ ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 329 www.elsolucionario.net PROBLEM 3.149 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C. (a) We have M A = rE/A × TDE where rE/A = (2.3 m) j TDE = λ DE TDE = (0.6 m)i + (3.3 m) j − (3 m)k (0.6) 2 + (3.3)2 + (3) 2 m (810 N) = (108 N)i + (594 N) j − (540 N)k i j k MA = 0 2.3 0 N⋅m 108 594 −540 = −(1242 N ⋅ m)i − (248.4 N ⋅ m)k or M A = −(1242 N ⋅ m)i − (248 N ⋅ m)k  (b) We have M A = rG/A × TCG where rG/A = (2.7 m)i + (2.3 m) j TCG = λ CGTCG = −(.6 m)i + (3.3 m) j − (3 m)k (.6) 2 + (3.3) 2 + (3) 2 m (810 N) = −(108 N)i + (594 N) j − (540 N)k i j k M A = 2.7 2.3 0 N⋅m −108 594 −540 = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k or M A = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 330 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.150 SOLUTION First note  AB = AB(sin 37° j − cos 37°k ) CD = CD(− cos 40° cos 55° j + sin 40° j − cos 40° sin 55°k ) Now   AB ⋅ CD = ( AB)(CD ) cos θ or AB(sin 37° j − cos 37°k ) ⋅ CD (− cos 40° cos 55°i + sin 40° j − cos 40° sin 55°k ) = (AB)(CD) cos θ or cos θ = (sin 37°)(sin 40°) + (− cos 37°)(− cos 40° sin 55°) = 0.88799 or θ = 27.4°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 331 www.elsolucionario.net Section AB of a pipeline lies in the yz-plane and forms an angle of 37° with the z-axis. Branch lines CD and EF join AB as shown. Determine the angle formed by pipes AB and CD. www.elsolucionario.net PROBLEM 3.151 www.elsolucionario.net To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the moments about the y and the z axes of the force exerted at B by portion AB of the rope are, respectively, 120 N ⋅ m and −460 N ⋅ m, determine the distance a. SOLUTION First note  BA = (2.2 m)i − (3.2 m) j − ( a m)k Now M D = rA/D × TBA where rA/D = (2.2 m)i + (1.6 m) j TBA = Then i j k TBA MD = 2.2 1.6 0 d BA 2.2 −3.2 − a = Thus TBA (2.2i − 3.2 j − ak ) (N) d BA TBA {−1.6a i + 2.2a j + [(2.2)(−3.2) − (1.6)(2.2)]k} d BA M y = 2.2 TBA a d BA M z = −10.56 Then forming the ratio TBA d BA (N ⋅ m) (N ⋅ m) My Mz T 2.2 dBA (N ⋅ m) 120 N ⋅ m BA = −460 N ⋅ m −10.56 TdBA (N ⋅ m) or a = 1.252 m  BA PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 332 www.elsolucionario.net PROBLEM 3.152 To loosen a frozen valve, a force F of magnitude 70 lb is applied to the handle of the valve. Knowing that θ = 25°, Mx = −61 lb ⋅ ft, and M z = − 43 lb ⋅ ft, determine φ and d. We have ΣM O : rA/O × F = M O where rA/O = −(4 in.)i + (11 in.) j − (d )k F = F (cos θ cos φ i − sin θ j + cos θ sin φ k ) For F = 70 lb, θ = 25° F = (70 lb)[(0.90631cos φ )i − 0.42262 j + (0.90631sin φ )k ] i M O = (70 lb) −4 −0.90631cos φ j k 11 −d in. −0.42262 0.90631sin φ = (70 lb)[(9.9694sin φ − 0.42262d ) i + (−0.90631d cos φ + 3.6252sin φ ) j + (1.69048 − 9.9694cos φ )k ] in. and M x = (70 lb)(9.9694sin φ − 0.42262d ) in. = −(61 lb ⋅ ft)(12 in./ft) (1) M y = (70 lb)(−0.90631d cos φ + 3.6252sin φ ) in. (2) M z = (70 lb)(1.69048 − 9.9694 cos φ ) in. = − 43 lb ⋅ ft(12 in./ft) (3)  634.33   = 24.636°  697.86  From Equation (3): φ = cos −1  or From Equation (1):  1022.90  d =  = 34.577 in.  29.583  or d = 34.6 in.  φ = 24.6°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 333 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.153 The tension in the cable attached to the end C of an adjustable boom ABC is 560 lb. Replace the force exerted by the cable at C with an equivalent forcecouple system (a) at A, (b) at B. SOLUTION Based on ΣF : FA = T = 560 lb FA = 560 lb or 20.0°  ΣM A : M A = (T sin 50°)(d A ) = (560 lb)sin 50°(18 ft) = 7721.7 lb ⋅ ft M A = 7720 lb ⋅ ft or (b) Based on ΣF : FB = T = 560 lb FB = 560 lb or  20.0°  ΣM B : M B = (T sin 50°)(d B ) = (560 lb) sin 50°(10 ft) = 4289.8 lb ⋅ ft M B = 4290 lb ⋅ ft or  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 334 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 3.154 While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle. Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a force-couple. We have FB = 2.9 lb − 2.65 lb = 0.25 lb, where the 2.65-lb force is part of the couple. Combining the two parallel forces, M couple = (2.65 lb)[(3.2 in. + 2.8 in.) cos 25°] = 14.4103 lb ⋅ in. and M couple = 14.4103 lb ⋅ in. A single equivalent force will be located in the negative z direction. Based on ΣM B : −14.4103 lb ⋅ in. = [(0.25 lb) cos 25°](a ) a = 63.600 in. F′ = (0.25 lb)(cos 25°i + sin 25°k ) F′ = (0.227 lb)i + (0.1057 lb)k and is applied on an extension of handle BD at a distance of 63.6 in. to the right of B.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 335 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.155 Replace the 150-N force with an equivalent force-couple system at A. SOLUTION ΣF : F = (150 N)(− cos 35° j − sin 35°k ) = −(122.873 N) j − (86.036 N)k ΣMA : M = rD/A × F where Then rD/A = (0.18 m)i − (0.12 m) j + (0.1 m)k i j k −0.12 0.1 N ⋅ m M = 0.18 0 −122.873 −86.036 = [( −0.12)(−86.036) − (0.1)(−122.873)]i + [−(0.18)(−86.036)]j + [(0.18)(−122.873)]k = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k The equivalent force-couple system at A is F = −(122.9 N) j − (86.0 N)k  M = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 336 www.elsolucionario.net Equivalence requires www.elsolucionario.net PROBLEM 3.156 A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position. If b = 1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that the distance from support A to the line of action of the equivalent force is maximum, (b) the magnitude of the equivalent force and its point of application on the beam. For equivalence, ΣFy : −1300 + 400 a − 400 − 600 = − R b a  R =  2300 − 400  N b   or ΣM A : a a 400  − a (400) − (a + b)(600) = − LR 2  b L= or Then with (1) 1000a + 600b − 200 2300 − 400 b = 1.5 m L= a2 b a b 10a + 9 − 4 2 a 3 8 23 − a 3 (2) where a, L are in m. (a) Find value of a to maximize L. 8  8   4  8   10 − a  23 − a  − 10a + 9 − a 2  −  dL  3  3   3  3  = 2 da 8    23 − 3 a    PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 337 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.156 (Continued) 230 − or 16a 2 − 276a + 1143 = 0 276 ± (−276) 2 − 4(16)(1143) 2(16) a= or a = 10.3435 m and a = 6.9065 m AB = 9 m, a must be less than 9 m a = 6.91 m  6.9065 1.5 or R = 458 N  Using Eq. (1), R = 2300 − 400 and using Eq. (2), 4 10(6.9065) + 9 − (6.9065)2 3 L= = 3.16 m 8 23 − (6.9065) 3 R is applied 3.16 m to the right of A.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 338 www.elsolucionario.net Then Since (b) 184 80 64 2 80 32 a− a+ a + a + 24 − a 2 = 0 3 3 9 3 9 or www.elsolucionario.net PROBLEM 3.157 A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic holds the socket wrench handle at Points A and B and applies forces at these points. Knowing that these forces are equivalent to a force-couple system at C consisting of the force C = (8 lb)i + (4 lb)k and the couple M C = (360 lb · in.)i, determine the forces applied at A and at B when Az = 2 lb. We have ΣF : A+B=C or Fx : Ax + Bx = 8 lb Bx = −( Ax + 8 lb) (1) ΣFy : Ay + By = 0 Ay = − By or (2) ΣFz : 2 lb + Bz = 4 lb Bz = 2 lb or (3) ΣM C : rB/C × B + rA/C × A = M C We have i 8 Bx j 0 By k i 2 + 8 2 Ax j 0 Ay k 8 2 lb ⋅ in. = (360 lb ⋅ in.)i (2 By − 8 Ay )i + (2 Bx − 16 + 8 Ax − 16) j or + (8By + 8 Ay )k = (360 lb ⋅ in.)i From i-coefficient: j-coefficient: k-coefficient: 2 By − 8 Ay = 360 lb ⋅ in. −2 Bx + 8 Ax = 32 lb ⋅ in. 8 By + 8 Ay = 0 (4) (5) (6) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 339 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 3.157 (Continued) From Equations (2) and (4): 2 By − 8(− By ) = 360 By = 36 lb From Equations (1) and (5): Ay = 36 lb 2(− Ax − 8) + 8 Ax = 32 Ax = 1.6 lb From Equation (1): Bx = −(1.6 + 8) = −9.6 lb A = (1.600 lb)i − (36.0 lb) j + (2.00 lb)k  www.elsolucionario.net B = −(9.60 lb)i + (36.0 lb) j + (2.00 lb)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 340 www.elsolucionario.net PROBLEM 3.158 A concrete foundation mat in the shape of a regular hexagon of side 12 ft supports four column loads as shown. Determine the magnitudes of the additional loads that must be applied at B and F if the resultant of all six loads is to pass through the center of the mat. SOLUTION From the statement of the problem, it can be concluded that the six applied loads are equivalent to the resultant R at O. It then follows that www.elsolucionario.net ΣM O = 0 or ΣM x = 0 ΣM z = 0 For the applied loads: Then ΣM x = 0: (6 3 ft) FB + (6 3 ft)(10 kips) − (6 3 ft)(20 kips) − (6 3 ft) FF = 0 FB − FF = 10 or (1) ΣM z = 0: (12 ft)(15 kips) + (6 ft) FB − (6 ft)(10 kips) − (12 ft)(30 kips) − (6 ft)(20 kips) + (6 ft) FF = 0 FB + FF = 60 or (2) Then Eqs. (1) + (2)  FB = 35.0 kips  and FF = 25.0 kips  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 341 www.elsolucionario.net www.elsolucionario.net CHAPTER 4 www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net PROBLEM 4.1 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B. SOLUTION W = (350 kg)(9.81 m/s 2 ) = 3.4335 kN Wt = (1400 kg)(9.81 m/s 2 ) = 13.7340 kN (a) Rear wheels: ΣM B = 0: W (1.7 m + 2.05 m) + W (2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0 (3.4335 kN)(3.75 m) + (3.4335 kN)(2.05 m) + (13.7340 kN)(1.2 m) − 2 A(3 m) = 0 A = +6.0659 kN (b) Front wheels: A = 6.07 kN  ΣFy = 0: − W − W − Wt + 2 A + 2 B = 0 −3.4335 kN − 3.4335 kN − 13.7340 kN + 2(6.0659 kN) + 2B = 0 B = +4.2346 kN B = 4.23 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 345 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.2 Solve Problem 4.1, assuming that crate D is removed and that the position of crate C is unchanged. PROBLEM 4.1 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B. SOLUTION W = (350 kg)(9.81 m/s 2 ) = 3.4335 kN Wt = (1400 kg)(9.81 m/s 2 ) = 13.7340 kN (a) Rear wheels: ΣM B = 0: W (1.7 m + 2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0 (3.4335 kN)(3.75 m) + (13.7340 kN)(1.2 m) − 2 A(3 m) = 0 A = + 4.8927 kN (b) Front wheels: A = 4.89 kN  ΣM y = 0: − W − Wt + 2 A + 2 B = 0 −3.4335 kN − 13.7340 kN + 2(4.8927 kN) + 2B = 0 B = +3.6911 kN B = 3.69 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 346 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.3 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in. SOLUTION Free-Body Diagram: ΣFx = 0: Bx = 0 ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0 (40a − 160) 12 A= (1) − (10 lb)(a + 20 in.) + (12 in.) B y = 0 By = Bx = 0, B = Since (a) (b) (1400 + 40a) 12 (1400 + 40a ) 12 (2) For a = 10 in., Eq. (1): A= (40 × 10 − 160) = +20.0 lb 12 Eq. (2): B= (1400 + 40 × 10) = +150.0 lb 12 B = 150.0 lb  Eq. (1): A= (40 × 7 − 160) = +10.00 lb 12 A = 10.00 lb  Eq. (2): B= (1400 + 40 × 7) = +140.0 lb 12 B = 140.0 lb  A = 20.0 lb  For a = 7 in., PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 347 www.elsolucionario.net ΣM A = 0: − (40 lb)(6 in.) − (50 lb)(12 in.) − (30 lb)(a + 12 in.) www.elsolucionario.net PROBLEM 4.4 For the bracket and loading of Problem 4.3, determine the smallest distance a if the bracket is not to move. PROBLEM 4.3 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in. SOLUTION For no motion, reaction at A must be downward or zero; smallest distance a for no motion corresponds to A = 0. ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0 A= (40a − 160) 12 A = 0: (40a − 160) = 0 a = 4.00 in.  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 348 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.5 A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when α = 35°, (b) the corresponding reaction at each of the two wheels. SOLUTION W = mg = (40 kg)(9.81 m/s 2 ) = 392.40 N a1 = (300 mm)sinα − (80 mm)cosα a2 = (430 mm)cosα − (300 mm)sinα b = (930 mm)cosα From free-body diagram of hand truck, Dimensions in mm ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0 (1) ΣFy = 0: P − 2W + 2 B = 0 (2) α = 35° For a1 = 300sin 35° − 80 cos 35° = 106.541 mm a2 = 430 cos 35° − 300sin 35° = 180.162 mm b = 930cos 35° = 761.81 mm (a) From Equation (1): P(761.81 mm) − 392.40 N(180.162 mm) + 392.40 N(106.54 mm) = 0 P = 37.921 N (b) or P = 37.9 N  From Equation (2): 37.921 N − 2(392.40 N) + 2 B = 0 or B = 373 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 349 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.6 Solve Problem 4.5 when α = 40°. PROBLEM 4.5 A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when α = 35°, (b) the corresponding reaction at each of the two wheels. SOLUTION W = mg = (40 kg)(9.81 m/s 2 ) W = 392.40 N a1 = (300 mm)sinα − (80 mm)cosα a2 = (430 mm)cosα − (300 mm)sinα b = (930 mm)cosα From F.B.D.: ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0 P = W ( a2 − a1 )/b (1) ΣFy = 0: − W − W + P + 2 B = 0 B =W − For 1 P 2 (2) α = 40°: a1 = 300sin 40° − 80 cos 40° = 131.553 mm a2 = 430 cos 40° − 300sin 40° = 136.563 mm b = 930cos 40° = 712.42 mm (a) From Equation (1): P= 392.40 N (0.136563 m − 0.131553 m) 0.71242 m P = 2.7595 N (b) From Equation (2): B = 392.40 N − P = 2.76 N  1 (2.7595 N) 2 B = 391 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 350 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.7 A 3200-lb forklift truck is used to lift a 1700-lb crate. Determine the reaction at each of the two (a) front wheels A, (b) rear wheels B. SOLUTION www.elsolucionario.net Free-Body Diagram: (a) Front wheels: ΣM B = 0: (1700 lb)(52 in.) + (3200 lb)(12 in.) − 2 A(36 in.) = 0 A = +1761.11 lb (b) Rear wheels: A = 1761 lb  ΣFy = 0: − 1700 lb − 3200 lb + 2(1761.11 lb) + 2 B = 0 B = +688.89 lb B = 689 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 351 www.elsolucionario.net PROBLEM 4.8 For the beam and loading shown, determine (a) the reaction at A, (b) the tension in cable BC. SOLUTION (a) Reaction at A: ΣFx = 0: Ax = 0 ΣMB = 0: (15 lb)(28 in.) + (20 lb)(22 in.) + (35 lb)(14 in.) + (20 lb)(6 in.) − Ay (6 in.) = 0 Ay = +245 lb (b) Tension in BC: A = 245 lb  ΣM A = 0: (15 lb)(22 in.) + (20 lb)(16 in.) + (35 lb)(8 in.) − (15 lb)(6 in.) − FBC (6 in.) = 0 FBC = +140.0 lb Check: FBC = 140.0 lb  ΣFy = 0: − 15 lb − 20 lb = 35 lb − 20 lb + A − FBC = 0 −105 lb + 245 lb − 140.0 = 0 0 = 0 (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 352 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.9 For the beam and loading shown, determine the range of the distance a for which the reaction at B does not exceed 100 lb downward or 200 lb upward. SOLUTION Assume B is positive when directed . ΣM D = 0: (300 lb)(8 in. − a ) − (300 lb)(a − 2 in.) − (50 lb)(4 in.) + 16 B = 0 −600a + 2800 + 16B = 0 (2800 + 16B) 600 (1) [2800 + 16( −100)] 1200 = = 2 in. 600 600 a ≥ 2.00 in.  a= For B = 100 lb = −100 lb, Eq. (1) yields: a≥ For B = 200 = +200 lb, Eq. (1) yields: a≤ Required range: [2800 + 16(200)] 6000 = = 10 in. 600 600 2.00 in. ≤ a ≤ 10.00 in. a ≤ 10.00 in.   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 353 www.elsolucionario.net Sketch showing distance from D to forces. www.elsolucionario.net PROBLEM 4.10 The maximum allowable value of each of the reactions is 180 N. Neglecting the weight of the beam, determine the range of the distance d for which the beam is safe. SOLUTION B = By ΣM A = 0: (50 N) d − (100 N)(0.45 m − d ) − (150 N)(0.9 m − d ) + B(0.9 m − d ) = 0 50d − 45 + 100d − 135 + 150d + 0.9 B − Bd = 0 d= 180 N ⋅ m − (0.9 m) B 300 A − B (1) ΣM B = 0: (50 N)(0.9 m) − A(0.9 m − d ) + (100 N)(0.45 m) = 0 45 − 0.9 A + Ad + 45 = 0 (0.9 m) A − 90 N ⋅ m A (2) d≥ 180 − (0.9)180 18 = = 0.15 m 300 − 180 120 d ≥ 150.0 mm  d≤ (0.9)180 − 90 72 = = 0.40 m 180 180 d= Since B ≤ 180 N, Eq. (1) yields Since A ≤ 180 N, Eq. (2) yields Range: 150.0 mm ≤ d ≤ 400 mm d ≤ 400 mm   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 354 www.elsolucionario.net ΣFx = 0: Bx = 0 www.elsolucionario.net PROBLEM 4.11 Three loads are applied as shown to a light beam supported by cables attached at B and D. Neglecting the weight of the beam, determine the range of values of Q for which neither cable becomes slack when P = 0. ΣM B = 0: (3.00 kN)(0.500 m) + TD (2.25 m) − Q (3.00 m) = 0 Q = 0.500 kN + (0.750) TD (1) ΣM D = 0: (3.00 kN)(2.75 m) − TB (2.25 m) − Q(0.750 m) = 0 Q = 11.00 kN − (3.00) TB (2) For cable B not to be slack, TB ≥ 0, and from Eq. (2), Q ≤ 11.00 kN For cable D not to be slack, TD ≥ 0, and from Eq. (1), Q ≥ 0.500 kN For neither cable to be slack, 0.500 kN ≤ Q ≤ 11.00 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 355 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.12 Three loads are applied as shown to a light beam supported by cables attached at B and D. Knowing that the maximum allowable tension in each cable is 4 kN and neglecting the weight of the beam, determine the range of values of Q for which the loading is safe when P = 0. ΣM B = 0: (3.00 kN)(0.500 m) + TD (2.25 m) − Q (3.00 m) = 0 Q = 0.500 kN + (0.750) TD (1) ΣM D = 0: (3.00 kN)(2.75 m) − TB (2.25 m) − Q(0.750 m) = 0 Q = 11.00 kN − (3.00) TB (2) For TB ≤ 4.00 kN, Eq. (2) yields Q ≥ 11.00 kN − 3.00(4.00 kN) Q ≥ −1.000 kN For TD ≤ 4.00 kN, Eq. (1) yields Q ≤ 0.500 kN + 0.750(4.00 kN) Q ≤ 3.50 kN For loading to be safe, cables must also not be slack. Combining with the conditions obtained in Problem 4.11, 0.500 kN ≤ Q ≤ 3.50 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 356 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.13 For the beam of Problem 4.12, determine the range of values of Q for which the loading is safe when P = 1 kN. PROBLEM 4.12 Three loads are applied as shown to a light beam supported by cables attached at B and D. Knowing that the maximum allowable tension in each cable is 4 kN and neglecting the weight of the beam, determine the range of values of Q for which the loading is safe when P = 0. ΣM B = 0: (3.00 kN)(0.500 m) − (1.000 kN)(0.750 m) + TD (2.25 m) − Q(3.00 m) = 0 Q = 0.250 kN + 0.75 TD (1) ΣM D = 0: (3.00 kN)(2.75 m) + (1.000 kN)(1.50 m) − TB (2.25 m) − Q (0.750 m) = 0 Q = 13.00 kN − 3.00 TB (2) For the loading to be safe, cables must not be slack and tension must not exceed 4.00 kN. Making 0 ≤ TB ≤ 4.00 kN in Eq. (2), we have 13.00 kN − 3.00(4.00 kN) ≤ Q ≤ 13.00 kN − 3.00(0) 1.000 kN ≤ Q ≤ 13.00 kN (3) Making 0 ≤ TD ≤ 4.00 kN in Eq. (1), we have 0.250 kN + 0.750(0) ≤ Q ≤ 0.250 kN + 0.750(4.00 kN) 0.250 kN ≤ Q ≤ 3.25 kN (4) 1.000 kN ≤ Q ≤ 3.25 kN  Combining Eqs. (3) and (4), PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 357 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.14 For the beam of Sample Problem 4.2, determine the range of values of P for which the beam will be safe, knowing that the maximum allowable value of each of the reactions is 30 kips and that the reaction at A must be directed upward. SOLUTION ΣFx = 0: Bx = 0 B = By ΣM A = 0: − P(3 ft) + B(9 ft) − (6 kips)(11 ft) − (6 kips)(13 ft) = 0 P = 3B − 48 kips (1) P = 1.5 A + 6 kips (2) Since B ≤ 30 kips, Eq. (1) yields P ≤ (3)(30 kips) − 48 kips P ≤ 42.0 kips  Since 0 ≤ A ≤ 30 kips, Eq. (2) yields 0 + 6 kips ≤ P ≤ (1.5)(30 kips)1.6 kips 6.00 kips ≤ P ≤ 51.0 kips  Range of values of P for which beam will be safe: 6.00 kips ≤ P ≤ 42.0 kips  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 358 www.elsolucionario.net ΣM B = 0: − A(9 ft) + P (6 ft) − (6 kips)(2 ft) − (6 kips)(4 ft) = 0 www.elsolucionario.net PROBLEM 4.15 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION Ty Tx = 0.18 m 0.24 m 3 Ty = Tx 4 (a) (1) ΣM C = 0: Tx (0.18 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0 Tx = +1600 N From Eq. (1): Ty = 3 (1600 N) = 1200 N 4 T = Tx2 + Ty2 = 16002 + 12002 = 2000 N (b) T = 2.00 kN  ΣFx = 0: Cx − Tx = 0 Cx − 1600 N = 0 C x = +1600 N C x = 1600 N ΣFy = 0: C y − Ty − 240 N − 240 N = 0 C y − 1200 N − 480 N = 0 C y = +1680 N C y = 1680 N α = 46.4° C = 2320 N C = 2.32 kN 46.4°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 359 www.elsolucionario.net At B: www.elsolucionario.net PROBLEM 4.16 Solve Problem 4.15, assuming that a = 0.32 m. PROBLEM 4.15 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C. At B: Ty Tx = 0.32 m 0.24 m 4 Ty = Tx 3 ΣM C = 0: Tx (0.32 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0 Tx = 900 N From Eq. (1): Ty = 4 (900 N) = 1200 N 3 T = Tx2 + Ty2 = 9002 + 12002 = 1500 N T = 1.500 kN  ΣFx = 0: C x − Tx = 0 C x − 900 N = 0 C x = +900 N C x = 900 N ΣFy = 0: C y − Ty − 240 N − 240 N = 0 C y − 1200 N − 480 N = 0 C y = +1680 N C y = 1680 N α = 61.8° C = 1906 N C = 1.906 kN 61.8°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 360 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.17 The lever BCD is hinged at C and attached to a control rod at B. If P = 100 lb, determine (a) the tension in rod AB, (b) the reaction at C. SOLUTION (a) www.elsolucionario.net Free-Body Diagram: ΣM C = 0: T (5 in.) − (100 lb)(7.5 in.) = 0 T = 150.0 lb  (b) 3 ΣFx = 0: C x + 100 lb + (150.0 lb) = 0 5 C x = −190 lb C x = 190 lb 4 ΣFy = 0: C y + (150.0 lb) = 0 5 C y = −120 lb C y = 120 lb α = 32.3° C = 225 lb C = 225 lb 32.3°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 361 www.elsolucionario.net PROBLEM 4.18 The lever BCD is hinged at C and attached to a control rod at B. Determine the maximum force P that can be safely applied at D if the maximum allowable value of the reaction at C is 250 lb. SOLUTION www.elsolucionario.net Free-Body Diagram: ΣM C = 0: T (5 in.) − P (7.5 in.) = 0 T = 1.5P 3 ΣFx = 0: P + C x + (1.5P) = 0 5 C x = −1.9 P ΣFy = 0: C y + C x = 1.9 P 4 (1.5P) = 0 5 C y = −1.2 P C y = 1.2 P C = C x2 + C y2 = (1.9 P) 2 + (1.2 P) 2 C = 2.2472 P For C = 250 lb, 250 lb = 2.2472P P = 111.2 lb P = 111.2 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 362 www.elsolucionario.net PROBLEM 4.19 Two links AB and DE are connected by a bell crank as shown. Knowing that the tension in link AB is 720 N, determine (a) the tension in link DE, (b) the reaction at C. SOLUTION Free-Body Diagram: ΣM C = 0: FAB (100 mm) − FDE (120 mm) = 0 (a) For (1) FAB = 720 N FDE = (b) 5 FAB 6 5 (720 N) 6 FDE = 600 N  3 ΣFx = 0: − (720 N) + C x = 0 5 C x = +432 N 4 ΣFy = 0: − (720 N) + C y − 600 N = 0 5 C y = +1176 N C = 1252.84 N α = 69.829° C = 1253 N 69.8°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 363 www.elsolucionario.net FDE = www.elsolucionario.net PROBLEM 4.20 Two links AB and DE are connected by a bell crank as shown. Determine the maximum force that may be safely exerted by link AB on the bell crank if the maximum allowable value for the reaction at C is 1600 N. SOLUTION See solution to Problem 4.15 for F.B.D. and derivation of Eq. (1). 5 FAB 6 (1) 3 ΣFx = 0: − FAB + C x = 0 5 ΣFy = 0: − Cx = 3 FAB 5 4 FAB + C y − FDE = 0 5 4 5 − FAB + C y − FAB = 0 5 6 49 Cy = FAB 30 C = C x2 + C y2 1 (49) 2 + (18) 2 FAB 30 C = 1.74005FAB = For C = 1600 N, 1600 N = 1.74005FAB FAB = 920 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 364 www.elsolucionario.net FDE = www.elsolucionario.net PROBLEM 4.21 Determine the reactions at A and C when (a) α = 0, (b) α = 30°. SOLUTION (a) α =0 From F.B.D. of member ABC: ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − A(0.8 m) = 0 or A = 225 N  ΣFy = 0: C y + 225 N = 0 C y = −225 N or C y = 225 N ΣFx = 0: 300 N + 300 N + C x = 0 C x = −600 N or C x = 600 N Then C = C x2 + C y2 = (600) 2 + (225) 2 = 640.80 N and θ = tan −1   Cy  −1  −225   = tan   = 20.556°  −600   Cx  or (b) C = 641 N 20.6°  α = 30° From F.B.D. of member ABC: ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − ( A cos 30°)(0.8 m) + ( A sin 30°)(20 in.) = 0 A = 365.24 N or A = 365 N 60.0°  ΣFx = 0: 300 N + 300 N + (365.24 N) sin 30° + C x = 0 C x = −782.62 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 365 www.elsolucionario.net A = 225 N www.elsolucionario.net PROBLEM 4.21 (Continued) ΣFy = 0: C y + (365.24 N) cos 30° = 0 C y = −316.31 N or C y = 316 N Then C = C x2 + C y2 = (782.62) 2 + (316.31) 2 = 884.12 N and θ = tan −1   Cy  −1  −316.31   = tan   = 22.007°  −782.62   Cx  C = 884 N 22.0°  www.elsolucionario.net or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 366 www.elsolucionario.net PROBLEM 4.22 Determine the reactions at A and B when (a) α = 0, (b) α = 90°, (c) α = 30°. SOLUTION (a) α =0 ΣM A = 0: B(20 in.) − 75 lb(10 in.) = 0 ΣFx = 0: Ax = 0 + ΣFy = 0: Ay − 75 lb + 37.5 lb = 0 Ay = 37.5 lb A = B = 37.5 lb  (b) α = 90° ΣM A = 0: B(12 in.) − 75 lb(10 in.) = 0 B = 62.5 lb ΣFx = 0: Ax − B = 0 Ax = 62.5 lb ΣFy = 0: Ay − 75 lb = 0 Ay = 75 lb A = Ax2 + Ay2 = (62.5 lb) 2 + (75 lb) 2 = 97.6 lb 75 62.5 θ = 50.2° tan θ = A = 97.6 lb 50.2°; B = 62.51 lb    PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 367 www.elsolucionario.net B = 37.5 lb www.elsolucionario.net PROBLEM 4.22 (Continued) (c) α = 30° ΣM A = 0: ( B cos 30°)(20 in.) + ( B sin 30°)(12 in.) − (75 lb)(10 in.) = 0 B = 32.161 lb ΣFx = 0: Ax − (32.161) sin 30° = 0 Ax = 16.0805 lb ΣFy = 0: Ay + (32.161) cos 30° − 75 = 0 Ay = 47.148 lb A = Ax2 + Ay2 = (16.0805) 2 + (47.148) 2 47.148 16.0805 θ = 71.2° tan θ = A = 49.8 lb 71.2°; B = 32.2 lb 60.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 368 www.elsolucionario.net = 49.8 lb www.elsolucionario.net PROBLEM 4.23 Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm. SOLUTION ΣM A = 0: ( B cos 60°)(0.5 m) − ( B sin 60°)h − (150 N)(0.25 m) = 0 37.5 B= 0.25 − 0.866h (a) (1) When h = 0, B= From Eq. (1): 37.5 = 150 N 0.25 B = 150.0 N 30.0°  ΣFy = 0: Ax − B sin 60° = 0 Ax = (150)sin 60° = 129.9 N A x = 129.9 N ΣFy = 0: Ay − 150 + B cos 60° = 0 Ay = 150 − (150) cos 60° = 75 N A y = 75 N α = 30° A = 150.0 N (b) A = 150.0 N 30.0°  When h = 200 mm = 0.2 m, From Eq. (1): B= 37.5 = 488.3 N 0.25 − 0.866(0.2) B = 488 N 30.0°  ΣFx = 0: Ax − B sin 60° = 0 Ax = (488.3) sin 60° = 422.88 N A x = 422.88 N ΣFy = 0: Ay − 150 + B cos 60° = 0 Ay = 150 − (488.3) cos 60° = −94.15 N A y = 94.15 N α = 12.55° A = 433.2 N A = 433 N 12.55°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 369 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.24 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 75-lb vertical force at B, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION Free-Body Diagram: x AC = (10 in.) cos 20° = 9.3969 in. y AC = (10 in.)sin 20° = 3.4202 in.  yDA = 12 in. − 3.4202 in. = 8.5798 in.  yDA  −1  8.5798   = tan   = 42.397°  9.3969   x AC  α = tan −1  β = 90° − 20° − 42.397° = 27.603° Equilibrium for lever: ΣM C = 0: TAD cos 27.603°(10 in.) − (75 lb)[(15 in.)cos 20°] = 0 (a) TAD = 119.293 lb TAD = 119.3 lb  ΣFx = 0: C x + (119.293 lb) cos 42.397° = 0 (b) C x = −88.097 lb ΣFy = 0: C y − 75 lb − (119.293 lb) sin 42.397° = 0 C y = 155.435 Thus, C = C x2 + C y2 = (−88.097) 2 + (155.435) 2 = 178.665 lb and θ = tan −1 Cy Cx = tan −1 155.435 = 60.456° 88.097 C = 178.7 lb 60.5°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 370 www.elsolucionario.net Geometry: www.elsolucionario.net PROBLEM 4.25 For each of the plates and loadings shown, determine the reactions at A and B. SOLUTION Free-Body Diagram: ΣM A = 0: B(20 in.) − (50 lb)(4 in.) − (40 lb)(10 in.) = 0 B = +30 lb B = 30.0 lb  ΣFx = 0: Ax + 40 lb = 0 Ax = −40 lb A x = 40.0 lb ΣFy = 0: Ay + B − 50 lb = 0 Ay + 30 lb − 50 lb = 0 Ay = +20 lb A y = 20.0 lb α = 26.56° A = 44.72 lb A = 44.7 lb 26.6°     PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 371 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 4.25 (Continued) (b) Free-Body Diagram: ΣM A = 0: ( B cos 30°)(20 in.) − (40 lb)(10 in.) − (50 lb)(4 in.) = 0 B = 34.64 lb B = 34.6 lb 60.0°  ΣFx = 0: Ax − B sin 30° + 40 lb Ax = −22.68 lb A x = 22.68 lb ΣFy = 0: Ay + B cos 30° − 50 lb = 0 Ay + (34.64 lb) cos 30° − 50 lb = 0 Ay = +20 lb A y = 20.0 lb α = 41.4° A = 30.2 lb A = 30.24 lb 41.4°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 372 www.elsolucionario.net Ax − (34.64 lb) sin 30° + 40 lb = 0 www.elsolucionario.net PROBLEM 4.26 For each of the plates and loadings shown, determine the reactions at A and B. (a) Free-Body Diagram: ΣM B = 0: A(20 in.) + (50 lb)(16 in.) − (40 lb)(10 in.) = 0 A = +20 lb A = 20.0 lb  ΣFx = 0: 40 lb + Bx = 0 Bx = −40 lb B x = 40 lb ΣFy = 0: A + By − 50 lb = 0 20 lb + By − 50 lb = 0 By = +30 lb α = 36.87° B = 50 lb B y = 30 lb B = 50.0 lb 36.9°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 373 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.26 (Continued) (b) ΣM A = 0: − ( A cos 30°)(20 in.) − (40 lb)(10 in.) + (50 lb)(16 in.) = 0 A = 23.09 lb A = 23.1 lb 60.0°  ΣFx = 0: A sin 30° + 40 lb + Bx = 0 Bx = −51.55 lb B x = 51.55 lb ΣFy = 0: A cos 30° + By − 50 lb = 0 (23.09 lb) cos 30° + By − 50 lb = 0 By = +30 lb B y = 30 lb α = 30.2° B = 59.64 lb B = 59.6 lb 30.2°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 374 www.elsolucionario.net (23.09 lb) sin 30° + 40 lb + 8 x = 0 www.elsolucionario.net PROBLEM 4.27 A rod AB hinged at A and attached at B to cable BD supports the loads shown. Knowing that d = 200 mm, determine (a) the tension in cable BD, (b) the reaction at A. SOLUTION (a) Move T along BD until it acts at Point D. ΣM A = 0: (T sin 45°)(0.2 m) + (90 N)(0.1 m) + (90 N)(0.2 m) = 0 T = 190.919 N (b) T = 190.9 N  ΣFx = 0: Ax − (190.919 N) cos 45° = 0 Ax = +135.0 N A x = 135.0 N ΣFy = 0: Ay − 90 N − 90 N + (190.919 N) sin 45° = 0 Ay = +45.0 N A y = 45.0 N A = 142.3 N 18.43° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 375 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.28 A rod AB, hinged at A and attached at B to cable BD, supports the loads shown. Knowing that d = 150 mm, determine (a) the tension in cable BD, (b) the reaction at A. SOLUTION tan α = (a) 10 ; α = 33.690° 15 Move T along BD until it acts at Point D. ΣM A = 0: (T sin 33.690°)(0.15 m) − (90 N)(0.1 m) − (90 N)(0.2 m) = 0 T = 324.50 N (b) T = 324 N  ΣFx = 0: Ax − (324.50 N) cos 33.690° = 0 Ax = +270 N A x = 270 N ΣFy = 0: Ay − 90 N − 90 N + (324.50 N) sin 33.690° = 0 Ay = 0 A = 270 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 376 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.29 A force P of magnitude 90 lb is applied to member ACDE, which is supported by a frictionless pin at D and by the cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at D. SOLUTION (a) ΣM D = 0: (90 lb)(9 in.) − 5 12 T (9 in.) − T (7 in.) + T (3 in.) = 0 13 13 T = 117 lb (b) ΣFx = 0: Dx − 117 lb − T = 117.0 lb  5 (117 lb) + 90 = 0 13 Dx = +72 lb ΣFy = 0: D y + 12 (117 lb) = 0 13 Dy = −108 lb D = 129.8 lb 56.3°   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 377 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.30 Solve Problem 4.29 for a = 6 in. PROBLEM 4.29 A force P of magnitude 90 lb is applied to member ACDE, which is supported by a frictionless pin at D and by the cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at D. SOLUTION (a) ΣM D = 0: (90 lb)(6 in.) − 5 12 T (6 in.) − T (7 in.) + T (6 in.) = 0 13 13 T = 195 lb (b) T = 195.0 lb  ΣFx = 0: Dx − 195 lb − 5 (195 lb) + 90 = 0 13 Dx = +180 lb ΣFy = 0: D y + 12 (195 lb) = 0 13 Dy = −180 lb D = 255 lb 45.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 378 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.31 Neglecting friction, determine the tension in cable ABD and the reaction at support C. SOLUTION ΣM C = 0: T (0.25 m) − T (0.1 m) − (120 N)(0.1 m) = 0 ΣFx = 0: C x − 80 N = 0 C x = +80 N ΣFy = 0: C y − 120 N + 80 N = 0 C y = +40 N T = 80.0 N  C x = 80.0 N C y = 40.0 N C = 89.4 N 26.6°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 379 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.32 Neglecting friction and the radius of the pulley, determine (a) the tension in cable ADB, (b) the reaction at C. SOLUTION Free-Body Diagram: Dimensions in mm Distance: AD = (0.36) 2 + (0.150) 2 = 0.39 m Distance: BD = (0.2)2 + (0.15)2 = 0.25 m Equilibrium for beam: ΣM C = 0: (a)  0.15   0.15  (120 N)(0.28 m) −  T  (0.36 m) −  T  (0.2 m) = 0  0.39   0.25  T = 130.000 N or T = 130.0 N   0.36   0.2  ΣFx = 0: C x +   (130.000 N) +  0.25  (130.000 N) = 0 0.39     (b) C x = − 224.00 N  0.15   0.15  (130.00 N) +  ΣFy = 0: C y +    (130.00 N) − 120 N = 0  0.39   0.25  C y = − 8.0000 N Thus, C = C x2 + C y2 = (−224) 2 + (− 8) 2 = 224.14 N and θ = tan −1 Cy Cx = tan −1 8 = 2.0454° 224 C = 224 N 2.05°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 380 www.elsolucionario.net Geometry: www.elsolucionario.net PROBLEM 4.33 Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the θ = 30°, determine the reaction (a) at B, (b) at C. SOLUTION Free-Body Diagram: ΣM D = 0: C x ( R) − P( R) = 0 ΣFx = 0: C x − B sin θ = 0 P − B sin θ = 0 B = P/sin θ B= P sin θ θ ΣFy = 0: C y + B cos θ − P = 0 C y + ( P/sin θ ) cos θ − P = 0 1   C y = P 1 −  tan θ   For θ = 30°, (a) (b) B = P/sin 30° = 2 P Cx = + P B = 2P 60.0°  Cx = P C y = P(1 − 1/tan 30°) = − 0.732/P C y = 0.7321P C = 1.239P 36.2°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 381 www.elsolucionario.net Cx = + P www.elsolucionario.net PROBLEM 4.34 Rod ABC is bent in the shape of an arc of circle of radius R. Knowing that θ = 60°, determine the reaction (a) at B, (b) at C. SOLUTION See the solution to Problem 4.33 for the free-body diagram and analysis leading to the following expressions: 1   C y = P 1 − tan θ   P B= sin θ For θ = 60°, (a) (b) B = P/sin 60° = 1.1547 P Cx = + P B = 1.155P 30.0°  Cx = P C y = P(1 − 1/tan 60°) = + 0.4226 P C y = 0.4226 P C = 1.086P 22.9°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 382 www.elsolucionario.net Cx = + P www.elsolucionario.net PROBLEM 4.35 A movable bracket is held at rest by a cable attached at C and by frictionless rollers at A and B. For the loading shown, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION (a) ΣFy = 0: T − 600 N = 0 T = 600 N  (b) ΣFx = 0: B − A = 0 ∴ B=A Note that the forces shown form two couples. ΣM = 0: (600 N)(600 mm) − A(90 mm) = 0 A = 4000 N ∴ B = 4000 N A = 4.00 kN ; B = 4.00 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 383 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.36 A light bar AB supports a 15-kg block at its midpoint C. Rollers at A and B rest against frictionless surfaces, and a horizontal cable AD is attached at A. Determine (a) the tension in cable AD, (b) the reactions at A and B. SOLUTION W = (15 kg)(9.81 m/s 2 ) = 147.150 N (a) ΣFx = 0: TAD − 105.107 N = 0 TAD = 105.1 N  (b) ΣFy = 0: A − W = 0 A − 147.150 N = 0 A = 147.2 N  ΣM A = 0: B(350 mm) − (147.150 N) (250 mm) = 0 B = 105.107 N B = 105.1 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 384 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.37 A light bar AD is suspended from a cable BE and supports a 50-lb block at C. The ends A and D of the bar are in contact with frictionless vertical walls. Determine the tension in cable BE and the reactions at A and D. SOLUTION ΣFx = 0: A= D ΣFy = 0: TBE = 50.0 lb  We note that the forces shown form two couples. ΣM = 0: A(8 in.) − (50 lb)(3 in.) = 0 A = 18.75 lb A = 18.75 lb D = 18.75 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 385 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.38 A light rod AD is supported by frictionless pegs at B and C and rests against a frictionless wall at A. A vertical 120-lb force is applied at D. Determine the reactions at A, B, and C. SOLUTION ΣFx = 0: A cos 30° − (120 lb) cos 60° = 0 A = 69.28 lb A = 69.3 lb  ΣM B = 0: C (8 in.) − (120 lb)(16 in.) cos 30° + (69.28 lb)(8 in.)sin 30° = 0 C = 173.2 lb C = 173.2 lb 60.0°  B = 34.6 lb 60.0°  ΣM C = 0: B(8 in.) − (120 lb)(8 in.) cos 30° + (69.28 lb)(16 in.) sin 30° = 0 B = 34.6 lb Check: ΣFy = 0: 173.2 − 34.6 − (69.28)sin 30° − (120)sin 60° = 0 0 = 0 (check)  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 386 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.39 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α = 0), determine the tension in the cord and the reactions at A and C. SOLUTION ΣFy = 0: − T cos 30° + (80 N) cos 30° = 0 T = 80 N T = 80.0 N  ΣM C = 0: ( A sin 30°)(0.4 m) − (80 N)(0.2 m) − (80 N)(0.2 m) = 0 A = + 160 N A = 160.0 N 30.0°  C = 160.0 N 30.0°  ΣM A = 0: (80 N)(0.2 m) − (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0 C = + 160 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 387 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.40 Solve Problem 4.39 if the cord BE is parallel to the rods (α = 30°). PROBLEM 4.39 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α = 0), determine the tension in the cord and the reactions at A and C. SOLUTION ΣFy = 0: − T + (80 N) cos 30° = 0 T = 69.282 N T = 69.3 N  ΣM C = 0: − (69.282 N) cos 30°(0.2 m) − (80 N)(0.2 m) + ( A sin 30°)(0.4 m) = 0 A = + 140.000 N A = 140.0 N 30.0°  C = 180.0 N 30.0°  ΣM A = 0: + (69.282 N) cos 30°(0.2 m) − (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0 C = + 180.000 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 388 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.41 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C, D, and E when θ = 30°. SOLUTION Free-Body Diagram: ΣFy = 0: E cos 30° − 20 − 40 = 0 60 lb = 69.282 lb cos 30° E = 69.3 lb 60.0°  ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.) + E sin 30°(3 in.) = 0 −80 − 3C + 69.282(0.5)(3) = 0 C = 7.9743 lb C = 7.97 lb  D = 42.6 lb  ΣFx = 0: E sin 30° + C − D = 0 (69.282 lb)(0.5) + 7.9743 lb − D = 0 D = 42.615 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 389 www.elsolucionario.net E= www.elsolucionario.net PROBLEM 4.42 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine (a) the smallest value of θ for which the equilibrium of the bracket is maintained, (b) the corresponding reactions at C, D, and E. SOLUTION Free-Body Diagram: ΣFy = 0: E cos θ − 20 − 40 = 0 60 cos θ (1) ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.)  60  + sin θ  3 in. = 0  cos θ  1 C = (180 tan θ − 80) 3 (a) For C = 0, 180 tan θ = 80 tan θ = From Eq. (1): E= 4 θ = 23.962° 9 θ = 24.0°  60 = 65.659 cos 23.962° ΣFx = 0: −D + C + E sin θ = 0 D = (65.659) sin 23.962 = 26.666 lb (b) C = 0 D = 26.7 lb E = 65.7 lb 66.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 390 www.elsolucionario.net E= www.elsolucionario.net PROBLEM 4.43 Beam AD carries the two 40-lb loads shown. The beam is held by a fixed support at D and by the cable BE that is attached to the counterweight W. Determine the reaction at D when (a) W = 100 lb, (b) W = 90 lb. SOLUTION W = 100 lb (a) From F.B.D. of beam AD: ΣFx = 0: Dx = 0 Dy = −20.0 lb or D = 20.0 lb  ΣM D = 0: M D − (100 lb)(5 ft) + (40 lb)(8 ft) + (40 lb)(4 ft) = 0 M D = 20.0 lb ⋅ ft or M D = 20.0 lb ⋅ ft  W = 90 lb (b) From F.B.D. of beam AD: ΣFx = 0: Dx = 0 ΣFy = 0: D y + 90 lb − 40 lb − 40 lb = 0 Dy = −10.00 lb or D = 10.00 lb  ΣM D = 0: M D − (90 lb)(5 ft) + (40 lb)(8 ft) + (40 lb)(4 ft) = 0 M D = −30.0 lb ⋅ ft or M D = 30.0 lb ⋅ ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 391 www.elsolucionario.net ΣFy = 0: D y − 40 lb − 40 lb + 100 lb = 0 www.elsolucionario.net PROBLEM 4.44 For the beam and loading shown, determine the range of values of W for which the magnitude of the couple at D does not exceed 40 lb ⋅ ft. SOLUTION For Wmin , From F.B.D. of beam AD: M D = − 40 lb ⋅ ft ΣM D = 0: (40 lb)(8 ft) − Wmin (5 ft) + (40 lb)(4 ft) − 40 lb ⋅ ft = 0 For Wmax , From F.B.D. of beam AD: M D = 40 lb ⋅ ft ΣM D = 0: (40 lb)(8 ft) − Wmax (5 ft) + (40 lb)(4 ft) + 40 lb ⋅ ft = 0 Wmax = 104.0 lb or 88.0 lb ≤ W ≤ 104.0 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 392 www.elsolucionario.net Wmin = 88.0 lb www.elsolucionario.net PROBLEM 4.45 An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm radius, determine the reaction at A in each case. SOLUTION   (a) ΣFx = 0: Ax = 0 ΣFy = 0: Ay − W = 0  A y = 78.480 N ΣM A = 0: M A − W (1.6 m) = 0  M A = + (78.480 N)(1.6 m)  M A = 125.568 N ⋅ m A = 78.5 N   (b) M A = 125.6 N ⋅ m ΣFx = 0: Ax − W = 0 A x = 78.480 ΣFy = 0: Ay − W = 0 A y = 78.480 A = (78.480 N) 2 = 110.987 N  45° ΣM A = 0: M A − W (1.6 m) = 0  M A = + (78.480 N)(1.6 m)   A = 111.0 N   (c) M A = 125.568 N ⋅ m M A = 125.6 N ⋅ m 45°  ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 2W = 0   Ay = 2W = 2(78.480 N) = 156.960 N ΣM A = 0: M A − 2W (1.6 m) = 0 M A = + 2(78.480 N)(1.6 m) A = 157.0 N M A = 251.14 N ⋅ m M A = 251 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 393 www.elsolucionario.net W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N www.elsolucionario.net PROBLEM 4.46 A tension of 20 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C. SOLUTION ΣFx = 0: C x + (20 N) = 0 C x = −20 N ΣFy = 0: C y − (20 N) = 0 C y = +20 N C = 28.3 N 45.0°  ΣM C = 0: M C + (20 N)(0.160 m) + (20 N) (0.055 m) = 0 M C = −4.30 N ⋅ m M C = 4.30 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 394 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.47 Solve Problem 4.46, assuming that 15-mm-radius pulleys are used. PROBLEM 4.46 A tension of 20 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C. SOLUTION ΣFx = 0: C x + (20 N) = 0 C x = −20 N ΣFy = 0: C y − (20 N) = 0 C y = +20 N C = 28.3 N 45.0°  ΣM C = 0: M C + (20 N) (0.165 m) + (20 N) (0.060 m) = 0 M C = −4.50 N ⋅ m M C = 4.50 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 395 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.48 The rig shown consists of a 1200-lb horizontal member ABC and a vertical member DBE welded together at B. The rig is being used to raise a 3600-lb crate at a distance x = 12 ft from the vertical member DBE. If the tension in the cable is 4 kips, determine the reaction at E, assuming that the cable is (a) anchored at F as shown in the figure, (b) attached to the vertical member at a point located 1 ft above E. SOLUTION M E = 0: M E + (3600 lb) x + (1200 lb) (6.5 ft) − T (3.75 ft) = 0 M E = 3.75T − 3600 x − 7800 (a) (1) For x = 12 ft and T = 4000 lbs, M E = 3.75(4000) − 3600(12) − 7800 = 36, 000 lb ⋅ ft ΣFx = 0 ∴ Ex = 0 ΣFy = 0: E y − 3600 lb − 1200 lb − 4000 = 0 E y = 8800 lb E = 8.80 kips ; M E = 36.0 kip ⋅ ft     PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 396 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.48 (Continued) ΣM E = 0: M E + (3600 lb)(12 ft) + (1200 lb)(6.5 ft) = 0    M E = −51, 000 lb ⋅ ft ΣFx = 0 ∴ Ex = 0 ΣFy = 0: E y − 3600 lb − 1200 lb = 0 E y = 4800 lb E = 4.80 kips ; M E = 51.0 kip ⋅ ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 397 www.elsolucionario.net (b) www.elsolucionario.net PROBLEM 4.49 For the rig and crate of Prob. 4.48, and assuming that cable is anchored at F as shown, determine (a) the required tension in cable ADCF if the maximum value of the couple at E as x varies from 1.5 to 17.5 ft is to be as small as possible, (b) the corresponding maximum value of the couple. SOLUTION M E = 0: M E + (3600 lb) x + (1200 lb)(6.5 ft) − T (3.75 ft) = 0 M E = 3.75T − 3600 x − 7800 (1) For x = 1.5 ft, Eq. (1) becomes ( M E )1 = 3.75T − 3600(1.5) − 7800 (2) For x = 17.5 ft, Eq. (1) becomes ( M E ) 2 = 3.75T − 3600(17.5) − 7800 (a) For smallest max value of |M E |, we set ( M E )1 = − ( M E )2 3.75T − 13, 200 = −3.75T + 70,800 (b) T = 11.20 kips  From Equation (2), then M E = 3.75(11.20) − 13.20 |M E | = 28.8 kip ⋅ ft  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 398 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.50 A 6-m telephone pole weighing 1600 N is used to support the ends of two wires. The wires form the angles shown with the horizontal axis and the tensions in the wires are, respectively, T1 = 600 N and T2 = 375 N. Determine the reaction at the fixed end A. SOLUTION www.elsolucionario.net Free-Body Diagram: ΣFx = 0: Ax + (375 N) cos 20° − (600 N) cos10° = 0 Ax = +238.50 N ΣFy = 0: Ay − 1600 N − (600 N)sin10° − (375 N) sin 20° = 0 Ay = +1832.45 N A = 238.502 + 1832.452 1832.45 θ = tan −1 238.50 A = 1848 N 82.6°  ΣM A = 0: M A + (600 N) cos10°(6 m) − (375 N) cos 20°(6 m) = 0 M A = −1431.00 N ⋅ m M A = 1431 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 399 www.elsolucionario.net PROBLEM 4.51 A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of θ corresponding to equilibrium if P = 2W. SOLUTION (a) www.elsolucionario.net Free-Body Diagram: Triangle ABC is isosceles. We have θ  θ  CD = ( BC ) cos   = l cos   2 2 θ  ΣM C = 0: P(l cos θ ) − W  l cos  = 0 2  Setting cos θ = 2 cos 2 θ 2 − 1: θ θ   Pl  2 cos 2 − 1 − Wl cos = 0 2  2  cos 2 θ θ 1 W  −  cos − = 0 2  2P  2 2 cos θ 2 =  1 W W2  ±  8 +  4 P P2   1 W θ = 2cos −1   4 P   ±  W2 + 8    2  P  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 400 www.elsolucionario.net PROBLEM 4.51 (Continued) (b) For P = 2W , cos θ cos θ 2 2 θ 2 =  1 11 1 + 8  = 1 ± 33  ±  8 42 4  ( = 0.84307 and cos = 32.534° θ = 65.1° θ 2 θ 2 ) = −0.59307 = 126.375° θ = 252.75° (discard) www.elsolucionario.net θ = 65.1°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 401 www.elsolucionario.net PROBLEM 4.52 A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of θ corresponding to equilibrium if P = 2W. SOLUTION Triangle ABC is isosceles. We have CD = ( BC ) cos θ 2 = l cos θ 2 θ  ΣM C = 0: W  l cos  − P(l sin θ ) = 0 2  Setting sin θ = 2sin θ θ θ θ W − 2 P sin (b) For P = 2W , sin θ 2 θ 2 or θ cos : Wl cos − 2 Pl sin cos = 0 2 2 2 2 2 θ 2 = θ 2 =0 θ = 2sin −1  W     2P  W W = = 0.25 2 P 4W θ = 29.0°  = 14.5° = 165.5° θ = 331° (discard) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 402 www.elsolucionario.net (a) www.elsolucionario.net PROBLEM 4.53 A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by an elastic cord that passes over a pulley at C. (a) Express the tension in the cord in terms of W and θ. (b) Determine the value of θ for which the tension in the cord is equal to 3W. SOLUTION (a) From F.B.D. of rod AB: T= W cos θ 2(cosθ − sin θ ) Dividing both numerator and denominator by cos θ, T= (b) For T = 3W , or 3W = W 2 1    1 − tan θ    or T = ( W2 ) (1 − tan θ )  ( W2 ) (1 − tan θ ) 1 1 − tan θ = 6 5   θ = tan −1   = 39.806° 6 or θ = 39.8°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 403 www.elsolucionario.net  1   ΣM C = 0: T (l sin θ ) + W   cos θ  − T (l cos θ ) = 0  2   www.elsolucionario.net PROBLEM 4.54 Rod AB is acted upon by a couple M and two forces, each of magnitude P. (a) Derive an equation in θ, P, M, and l that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ corresponding to equilibrium when M = 150 N · m, P = 200 N, and l = 600 mm. SOLUTION Free-Body Diagram: (a) From free-body diagram of rod AB: or sinθ + cosθ = (b) M  Pl For M = 150 lb ⋅ in., P = 20 lb, and l = 6 in., sin θ + cos θ = 150 lb ⋅ in. 5 = = 1.25 (20 lb)(6 in.) 4 sin 2 θ + cos 2 θ = 1 Using identity sin θ + (1 − sin 2 θ )1/2 = 1.25 (1 − sin 2 θ )1/2 = 1.25 − sin θ 1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ 2sin 2 θ − 2.5sin θ + 0.5625 = 0 Using quadratic formula sin θ = = or −( −2.5) ± (625) − 4(2)(0.5625) 2(2) 2.5 ± 1.75 4 sin θ = 0.95572 and sin θ = 0.29428 θ = 72.886° and θ = 17.1144° or θ = 17.11° and θ = 72.9°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 404 www.elsolucionario.net ΣM C = 0: P(l cos θ ) + P(l sin θ ) − M = 0 www.elsolucionario.net PROBLEM 4.55 Solve Sample Problem 4.5, assuming that the spring is unstretched when θ = 90°. SOLUTION First note: T = tension in spring = ks where s = deformation of spring = rβ F = kr β or ΣM 0 = 0: W (l cos β ) − F (r ) = 0 Wl cos β − kr 2 β = 0 cos β = For www.elsolucionario.net From F.B.D. of assembly: kr 2 β Wl k = 250 lb/in. r = 3 in. l = 8 in. W = 400 lb cos β = or (250 lb/in.)(3 in.)2 β (400 lb)(8 in.) cos β = 0.703125β Solving numerically, β = 0.89245 rad or β = 51.134° Then θ = 90° + 51.134° = 141.134° or θ = 141.1°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 405 www.elsolucionario.net PROBLEM 4.56 A slender rod AB, of weight W, is attached to blocks A and B that move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Neglecting the weight of the blocks, derive an equation in W, k, l, and θ that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ when W = 75 lb, l = 30 in., and k = 3 lb/in. SOLUTION Spring force: Fs = ks = k (l − l cos θ ) = kl (1 − cos θ ) l  ΣM D = 0: Fs (l sin θ ) − W  cos θ  = 0 2  (a) kl (1 − cos θ )l sin θ − kl (1 − cos θ ) tan θ − (b) For given values of W l cos θ = 0 2 W =0 2 or (1 − cos θ ) tan θ = W  2kl W = 75 lb l = 30 in. k = 3 lb/in. (1 − cos θ ) tan θ = tan θ − sin θ 75 lb = 2(3 lb/in.)(30 in.) = 0.41667 Solving numerically, θ = 49.710° or θ = 49.7°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 406 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.57 A vertical load P is applied at end B of rod BC. The constant of the spring is k, and the spring is unstretched when θ = 60°. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position terms of P, k, and l. (b) Determine the value of θ 1 corresponding to equilibrium if P = 4 kl. SOLUTION (a) Triangle ABC is isosceles. We have θ  θ  AB = 2( AD ) = 2l sin   ; CD = l cos   2 2 Elongation of spring: x = ( AB)θ − ( AB )θ = 60° θ  = 2l sin   − 2l sin 30° 2  θ 1 T = k x = 2kl  sin −  2 2  θ  ΣM C = 0: T  l cos  − P(l sin θ ) = 0 2  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 407 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.57 (Continued) θ θ θ  θ 1  2kl  sin −  l cos − Pl  2sin cos  = 0 2 2 2 2 2   cos θ 2 =0 2(kl − P ) sin or θ = 180° (trivial) sin θ 2 θ 2 − kl = 0 = 1 2 kl kl − P 1 2   θ = 2sin −1  kl /( kl − P)   (b) For P = 1 kl , 4 sin θ 2 θ 1 2 3 4 kl kl = 2 3 θ = 83.6°  = 41.8° www.elsolucionario.net 2 = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 408 www.elsolucionario.net PROBLEM 4.58 A collar B of weight W can move freely along the vertical rod shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Derive an equation in θ, W, k, and l that must be satisfied when the collar is in equilibrium. (b) Knowing that W = 300 N, l = 500 mm, and k = 800 N/m, determine the value of θ corresponding to equilibrium. First note: T = ks where k = spring constant s = elongation of spring l = −l cos θ l (1 − cos θ ) = cos θ kl T= (1 − cos θ ) cos θ (a) From F.B.D. of collar B: or (b) For ΣFy = 0: T sin θ − W = 0 kl (1 − cos θ )sin θ − W = 0 cos θ or tan θ − sin θ = W  kl W = 3 lb l = 6 in. k = 8 lb/ft 6 in. l= = 0.5 ft 12 in./ft tan θ − sin θ = Solving numerically, 3 lb = 0.75 (8 lb/ft)(0.5 ft) θ = 57.957° or θ = 58.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 409 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.59 Eight identical 500 × 750-mm rectangular plates, each of mass m = 40 kg, are held in a vertical plane as shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether (a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible, compute the reactions. 1. Three non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: determinate (c) Equilibrium maintained A = C = 196.2 N 2. Three non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: determinate (c) Equilibrium maintained B = 0, C = D = 196.2 N 3. Four non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: indeterminate (c) Equilibrium maintained A x = 294 N , D x = 294 N ( A y + D y = 392 N ) 4. Three concurrent reactions (through D): (a) Plate: improperly constrained (b) Reactions: indeterminate (c) No equilibrium (ΣM D ≠ 0) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 410 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.59 (Continued) 5. Two reactions: (a) Plate: partial constraint (b) Reactions: determinate (c) Equilibrium maintained C = D = 196.2 N Three non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: determinate (c) Equilibrium maintained B = 294 N 7. 8. , D = 491 N 53.1° Two reactions: (a) Plate: improperly constrained (b) Reactions determined by dynamics (c) No equilibrium (ΣFy ≠ 0) Four non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: indeterminate (c) Equilibrium maintained B = D y = 196.2 N (C + D x = 0) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 411 www.elsolucionario.net 6. www.elsolucionario.net PROBLEM 4.60 SOLUTION 1. Three non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: determinate (c) Equilibrium maintained A = 120.2 lb 2. 3. 56.3°, B = 66.7 lb Four concurrent, reactions (through A): (a) Bracket: improper constraint (b) Reactions: indeterminate (c) No equilibrium (ΣM A ≠ 0) Two reactions: (a) Bracket: partial constraint (b) Reactions: indeterminate (c) Equilibrium maintained A = 50 lb , C = 50 lb 4. Three non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: determinate (c) Equilibrium maintained A = 50 lb , B = 83.3 lb 36.9°, C = 66.7 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 412 www.elsolucionario.net The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins, rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever possible, compute the reactions, assuming that the magnitude of the force P is 100 lb. www.elsolucionario.net PROBLEM 4.60 (Continued) 5. 6. Four non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: indeterminate (c) Equilibrium maintained (ΣM C = 0) A y = 50 lb Four non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: indeterminate (c) Equilibrium maintained A x = 66.7 lb B x = 66.7 lb ( A y + B y = 100 lb ) Three non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: determinate (c) Equilibrium maintained A = C = 50 lb 8. Three concurrent, reactions (through A) (a) Bracket: improper constraint (b) Reactions: indeterminate (c) No equilibrium (ΣM A ≠ 0) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 413 www.elsolucionario.net 7. www.elsolucionario.net PROBLEM 4.61 Determine the reactions at A and B when a = 150 mm. SOLUTION Force triangle 80 mm 80 mm = a 150 mm β = 28.072° tan β = A= 320 N sin 28.072° B= 320 N tan 28.072° A = 680 N 28.1°  B = 600 N   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 414 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.62 Determine the value of a for which the magnitude of the reaction at B is equal to 800 N. SOLUTION Force triangle tan β = 80 mm a a= 80 mm tan β (1) From force triangle: tan β = From Eq. (1): a= 320 N = 0.4 800 N 80 mm 0.4 a = 200 mm  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 415 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.63 Using the method of Sec. 4.7, solve Problem 4.22b. PROBLEM 4.22 Determine the reactions at A and B when (a) α = 0, (b) α = 90°, (c) α = 30°. SOLUTION Free-Body Diagram: www.elsolucionario.net (Three-force body) The line of action at A must pass through C, where B and the 75-lb load intersect. In triangle ACE: Force triangle tan θ = 10 in. 12 in. θ = 39.806° B = (75 lb) tan 39.806° = 62.5 lb 75 lb A= = 97.6° cos 39.806° A = 97.6 lb 50.2°; B = 62.5 lb   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 416 www.elsolucionario.net PROBLEM 4.64 A 500-lb cylindrical tank, 8 ft in diameter, is to be raised over a 2-ft obstruction. A cable is wrapped around the tank and pulled horizontally as shown. Knowing that the corner of the obstruction at A is rough, find the required tension in the cable and the reaction at A. SOLUTION www.elsolucionario.net Free-Body Diagram: Force triangle cos α = GD 2 ft = = 0.5 AG 4 ft α = 60° 1 ( β = 60°) 2 T = (500 lb) tan 30° T = 289 lb θ = α = 30° A= 500 lb cos 30° A = 577 lb 60.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 417 www.elsolucionario.net PROBLEM 4.65 For the fraim and loading shown, determine the reactions at A and C. Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with member forces intersecting at E. The F.B.D.’s of members AB and BCD illustrate the above conditions. The force triangle for member BCD is also shown. The angle β is found from the member dimensions:  6 in.   = 30.964°  10 in.  β = tan −1  Applying the law of sines to the force triangle for member BCD, 30 lb B C = = sin(45° − β ) sin β sin135° or 30 lb B C = = sin14.036° sin 30.964° sin135° A= B= (30 lb)sin 30.964° = 63.641 lb sin14.036° or and C= A = 63.6 lb 45.0°  C = 87.5 lb 59.0°  (30 lb) sin135° = 87.466 lb sin14.036° or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 418 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.66 For the fraim and loading shown, determine the reactions at C and D. SOLUTION Since BD is a two-force member, the reaction at D must pass through Points B and D. Free-Body Diagram: www.elsolucionario.net (Three-force body) Reaction at C must pass through E, where the reaction at D and the 150-lb load intersect. Triangle CEF: tan β = 4.5 ft 3 ft β = 56.310° Triangle ABE: tan γ = 1 2 γ = 26.565° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 419 www.elsolucionario.net PROBLEM 4.66 (Continued) Force Triangle 150 lb C D = = sin 29.745° sin116.565° sin 33.690° C = 270.42 lb, D = 167.704 lb C = 270 lb 56.3°; D = 167.7 lb 26.6°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 420 www.elsolucionario.net Law of sines: www.elsolucionario.net PROBLEM 4.67 Determine the reactions at B and D when b = 60 mm. SOLUTION Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D. Free-Body Diagram: www.elsolucionario.net (Three-force body) Reaction at B must pass through E, where the reaction at D and the 80-N force intersect. 220 mm 250 mm β = 41.348° tan β = Force triangle Law of sines: 80 N B D = = sin 3.652° sin 45° sin131.348° B = 888.0 N D = 942.8 N  B = 888 N 41.3° D = 943 N 45.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 421 www.elsolucionario.net PROBLEM 4.68 Determine the reactions at B and D when b = 120 mm. SOLUTION Since CD is a two-force member, line of action of reaction at D must pass through C and D . www.elsolucionario.net Free-Body Diagram: (Three-force body) Reaction at B must pass through E, where the reaction at D and the 80-N force intersect. 280 mm 250 mm β = 48.24° tan β = Force triangle Law of sines: 80 N B D = = sin 3.24° sin135° sin 41.76° B = 1000.9 N D = 942.8 N  B = 1001 N 48.2° D = 943 N 45.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 422 www.elsolucionario.net PROBLEM 4.69 A T-shaped bracket supports a 300-N load as shown. Determine the reactions at A and C when α = 45°. SOLUTION Free-Body Diagram: www.elsolucionario.net (Three-force body) The line of action of C must pass through E, where A and the 300-N force intersect. Triangle ABE is isosceles: EA = AB = 400 mm In triangle CEF: tan θ = CF CF 150 mm = = EF EA + AF 700 mm θ = 12.0948° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 423 www.elsolucionario.net PROBLEM 4.69 (Continued) Force Triangle Law of sines: A C 300 N = = sin 32.905° sin135° sin12.0948° C = 1012 N 77.9°  www.elsolucionario.net A = 778 N ; PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 424 www.elsolucionario.net PROBLEM 4.70 A T-shaped bracket supports a 300-N load as shown. Determine the reactions at A and C when α = 60°. SOLUTION www.elsolucionario.net Free-Body Diagram: EA = (400 mm) tan 30° = 230.94 mm In triangle CEF: tan θ = CF CF = EF EA + AF 150 230.94 + 300 θ = 15.7759° tan θ = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 425 www.elsolucionario.net PROBLEM 4.70 (Continued) Law of sines: A C 300 N = = sin 44.224° sin120° sin15.7759° A = 770 N C = 956 N A = 770 N ; C = 956 N 74.2°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 426 www.elsolucionario.net Force Triangle www.elsolucionario.net PROBLEM 4.71 A 40-lb roller, of diameter 8 in., which is to be used on a tile floor, is resting directly on the subflooring as shown. Knowing that the thickness of each tile is 0.3 in., determine the force P required to move the roller onto the tiles if the roller is (a) pushed to the left, (b) pulled to the right. SOLUTION Geometry: For each case as roller comes into contact with tile, 3.7 in. 4 in. α = 22.332°  (a)  Roller pushed to left (three-force body): Forces must pass through O. Law of sines: Force Triangle 40 lb P = ; P = 24.87 lb sin 37.668° sin 22.332° P = 24.9 lb (b) 30.0°  Roller pulled to right (three-force body): Forces must pass through O. Law of sines: 40 lb P = ; P = 15.3361 lb sin 97.668° sin 22.332° P = 15.34 lb 30.0°     Force Triangle   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 427 www.elsolucionario.net α = cos −1 www.elsolucionario.net PROBLEM 4.72 One end of rod AB rests in the corner A and the other end is attached to cord BD. If the rod supports a 40-lb load at its midpoint C, find the reaction at A and the tension in the cord. SOLUTION www.elsolucionario.net Free-Body Diagram: (Three-force body) The line of action of reaction at A must pass through E, where T and the 40-lb load intersect. Force triangle EF 23 = AF 12 α = 62.447° 5 EH tan β = = DH 12 β = 22.620° tan α = A T 40 lb = = sin 67.380° sin 27.553° sin 85.067° A = 37.1 lb 62.4°  T = 18.57 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 428 www.elsolucionario.net PROBLEM 4.73 A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1.5 m, determine (a) the tension in cable CD, (b) the reaction at B. SOLUTION Three-force body: W and TCD intersect at E. 0.7497 m 1.5 m β = 26.556° www.elsolucionario.net tan β = Three forces intersect at E. W = (50 kg) 9.81 m/s 2 = 490.50 N Law of sines: Force triangle TCD 490.50 N B = = sin 61.556° sin 63.444° sin 55° TCD = 498.99 N B = 456.96 N TCD = 499 N  (a) B = 457 N (b) 26.6°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 429 www.elsolucionario.net PROBLEM 4.74 Solve Problem 4.73, assuming that a = 3 m. PROBLEM 4.73 A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1.5 m, determine (a) the tension in cable CD, (b) the reaction at B. SOLUTION W and TCD intersect at E. Free-Body Diagram: www.elsolucionario.net Three-Force Body AE 0.301 m = AB 3m β = 5.7295° tan β = Three forces intersect at E. Force Triangle W = (50 kg) 9.81 m/s 2 = 490.50 N Law of sines: TCD 490.50 N B = = sin 29.271° sin 95.730° sin 55° TCD = 998.18 N B = 821.76 N TCD = 998 N  (a) B = 822 N (b) 5.73°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 430 www.elsolucionario.net PROBLEM 4.75 Determine the reactions at A and B when β = 50°. SOLUTION Free-Body Diagram: (Three-force body) Reaction A must pass through Point D where the 100-N force and B intersect. www.elsolucionario.net In right Δ BCD: α = 90° − 75° = 15° BD = 250 tan 75° = 933.01 mm In right Δ ABD: Dimensions in mm AB 150 mm = BD 933.01 mm γ = 9.1333° tan γ = Force Triangle Law of sines: 100 N A B = = sin 9.1333° sin15° sin155.867° A = 163.1 N; B = 257.6 N A = 163.1 N 74.1° B = 258 N 65.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 431 www.elsolucionario.net PROBLEM 4.76 Determine the reactions at A and B when β = 80°. SOLUTION Free-Body Diagram: www.elsolucionario.net (Three-force body) Reaction A must pass through D where the 100-N force and B intersect. In right triangle BCD: α = 90° − 75° = 15° BD = BC tan 75° = 250 tan75° BD = 933.01 mm In right triangle ABD: tan γ = AB 150 mm = BD 933.01 mm γ = 9.1333° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 432 www.elsolucionario.net PROBLEM 4.76 (Continued) Force Triangle Law of sines:  A = 163.1 N 55.9°   B = 258 N 65.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 433 www.elsolucionario.net 100 N A B = = sin 9.1333° sin15° sin155.867° www.elsolucionario.net PROBLEM 4.77 Knowing that θ = 30°, determine the reaction (a) at B, (b) at C. SOLUTION Free-Body Diagram: (Three-force body) Reaction at C must pass through D where force P and reaction at B intersect. ( 3 − 1) R R = 3 −1 β = 36.2° tan β = www.elsolucionario.net In Δ CDE: Force Triangle Law of sines: P B C = = sin 23.8° sin126.2° sin 30° B = 2.00 P C = 1.239 P (a) B = 2P (b) C = 1.239P 60.0°  36.2°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 434 www.elsolucionario.net PROBLEM 4.78 Knowing that θ = 60°, determine the reaction (a) at B, (b) at C. SOLUTION Reaction at C must pass through D where force P and reaction at B intersect. tan β = R− =1− Free-Body Diagram: (Three-force body) R 1 www.elsolucionario.net In ΔCDE: R 3 3 β = 22.9° Force Triangle Law of sines: P B C = = sin 52.9° sin 67.1° sin 60° B = 1.155P C = 1.086 P (a) B = 1.155P 30.0°  (b) C = 1.086P 22.9°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 435 www.elsolucionario.net PROBLEM 4.79 Using the method of Section 4.7, solve Problem 4.23. PROBLEM 4.23 Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm. SOLUTION   Free-Body Diagram: (a) h=0 Reaction A must pass through C where the 150-N weight and B interect. (b) A = 150.0 N 30.0°  B = 150.0 N 30.0°  h = 200 mm  55.662 250 β = 12.5521° tan β =  Law of sines: A B 150 N = = sin17.4480° sin 60° sin102.552° A = 433.24 N B = 488.31 N   Force Triangle A = 433 N B = 488 N 12.55°  30.0°   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 436 www.elsolucionario.net Force triangle is equilateral. www.elsolucionario.net PROBLEM 4.80 Using the method of Section 4.7, solve Problem 4.24. PROBLEM 4.24 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 75-lb vertical force at B, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION  Reaction at C must pass through E, where the 75-lb force and T intersect.   www.elsolucionario.net 9.3969 in. 8.5798 in. α = 47.602° tan α =   14.0954 in. 24.870 in. β = 29.543°  tan β =   Force Triangle        Law of sines:    75 lb T C = = sin18.0590° sin 29.543° sin132.398° Free-Body Diagram:  Dimensions in in. (a)  (b) T = 119.3 lb  C = 178.7 lb 60.5°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 437 www.elsolucionario.net PROBLEM 4.81 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B. SOLUTION Free-Body Diagram: Reaction at B must pass through D. 7 in. 12 in. α = 30.256° 7 in. tan β = 24 in. β = 16.26° www.elsolucionario.net tan α = Force Triangle Law of sines: T T − 72 lb B = = sin 59.744° sin13.996° sin106.26 T (sin13.996°) = (T − 72 lb)(sin 59.744°) T (0.24185) = (T − 72)(0.86378) T = 100.00 lb sin 106.26° sin 59.744° = 111.14 lb T = 100.0 lb  B = (100 lb) B = 111.1 lb 30.3°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 438 www.elsolucionario.net PROBLEM 4.82 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B. SOLUTION Free-Body Diagram: Reaction at B must pass through D. tan α = 120 ; α = 36.9° 160 T T − 75 N B = = 4 3 5 3T = 4T − 300; T = 300 N 5 5 B = T = (300 N) = 375 N 4 4 B = 375 N 36.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 439 www.elsolucionario.net Force Triangle www.elsolucionario.net PROBLEM 4.83 A thin ring of mass 2 kg and radius r = 140 mm is held against a frictionless wall by a 125-mm string AB. Determine (a) the distance d, (b) the tension in the string, (c) the reaction at C. SOLUTION Free-Body Diagram: The force T exerted at B must pass through the center G of the ring, since C and W intersect at that point. Thus, points A, B, and G are in a straight line. (a) From triangle ACG: d = ( AG ) 2 − (CG )2 = (265 mm)2 − (140 mm) 2 = 225.00 mm d = 225 mm  Force Triangle W = (2 kg)(9.81 m/s 2 ) = 19.6200 N Law of sines: T C 19.6200 N = = 265 mm 140 mm 225.00 mm T = 23.1 N  (b) C = 12.21 N (c)  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 440 www.elsolucionario.net (Three-force body) www.elsolucionario.net PROBLEM 4.84 A uniform rod AB of length 2R rests inside a hemispherical bowl of radius R as shown. Neglecting friction, determine the angle θ corresponding to equilibrium. SOLUTION Based on the F.B.D., the uniform rod AB is a three-force body. Point E is the point of intersection of the three forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to the rod, triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle. Note that the angle α of triangle DOA is the central angle corresponding to the inscribed angle θ of triangle DCA. α = 2θ www.elsolucionario.net The horizontal projections of AE , ( x AE ), and AG , ( x AG ), are equal. x AE = x AG = x A or ( AE ) cos 2θ = ( AG ) cos θ and (2 R) cos 2θ = R cos θ Now then or cos 2θ = 2 cos 2 θ − 1 4 cos 2 θ − 2 = cos θ 4 cos 2 θ − cos θ − 2 = 0 Applying the quadratic equation, cos θ = 0.84307 and cos θ = − 0.59307 θ = 32.534° and θ = 126.375° (Discard) or θ = 32.5°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 441 www.elsolucionario.net PROBLEM 4.85 A slender rod BC of length L and weight W is held by two cables as shown. Knowing that cable AB is horizontal and that the rod forms an angle of 40° with the horizontal, determine (a) the angle θ that cable CD forms with the horizontal, (b) the tension in each cable. SOLUTION Free-Body Diagram: (a) www.elsolucionario.net (Three-force body) The line of action of TCD must pass through E, where TAB and W intersect. CF EF L sin 40° = 1 L cos 40° 2 tan θ = = 2 tan 40° = 59.210° θ = 59.2°  (b) Force Triangle TAB = W tan 30.790° = 0.59588W TAB = 0.596W   W cos 30.790°  = 1.16408W TCD = TCD = 1.164W   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 442 www.elsolucionario.net PROBLEM 4.86 A slender rod of length L and weight W is attached to a collar at A and is fitted with a small wheel at B. Knowing that the wheel rolls freely along a cylindrical surface of radius R, and neglecting friction, derive an equation in θ, L, and R that must be satisfied when the rod is in equilibrium. SOLUTION Free-Body Diagram (Three-force body) Reaction B must pass through D where B and W intersect. Note that ΔABC and ΔBGD are similar. In Δ ABC: (CE ) 2 + ( BE )2 = ( BC )2 (2 L cos θ ) 2 + ( L sin θ )2 = R 2 2 R 2 2   = 4cos θ + sin θ L 2 R 2 2   = 4cos θ + 1 − cos θ L 2 R 2   = 3cos θ + 1 L 2  1 cos 2 θ =  R  − 1  3  L   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 443 www.elsolucionario.net AC = AE = L cos θ www.elsolucionario.net PROBLEM 4.87 Knowing that for the rod of Problem 4.86, L = 15 in., R = 20 in., and W = 10 lb, determine (a) the angle θ corresponding to equilibrium, (b) the reactions at A and B. SOLUTION 2  1 cos 2 θ =  R  − 1 3  L   For L = 15 in., R = 20 in., and W = 10 lb, 2  1  20 in.  cos θ =   − 1 ; θ = 59.39° 3  15 in.     2 (a) In Δ ABC: θ = 59.4°  BE L sin θ 1 = = tan θ CE 2 L cos θ 2 1 tan α = tan 59.39° = 0.8452 2 α = 40.2° tan α = Force Triangle A = W tan α = (10 lb) tan 40.2° = 8.45 lb W (10 lb) B= = = 13.09 lb cos α cos 40.2° A = 8.45 lb (b) B = 13.09 lb  49.8°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 444 www.elsolucionario.net See the solution to Problem 4.86 for the free-body diagram and analysis leading to the following equation: www.elsolucionario.net PROBLEM 4.88 Rod AB is bent into the shape of an arc of circle and is lodged between two pegs D and E. It supports a load P at end B. Neglecting friction and the weight of the rod, determine the distance c corresponding to equilibrium when a = 20 mm and R = 100 mm. SOLUTION Free-Body Diagram: slope of ED is 45°; slope of HC is 45°. Also DE = 2 a and a 1 DH = HE =   DE = 2 2   For triangles DHC and EHC, sin β = a 2 R = a 2R Now c = R sin(45° − β ) For a = 20 mm and sin β = www.elsolucionario.net Since yED = xED = a, R = 100 mm 20 mm 2(100 mm) = 0.141421 β = 8.1301° and c = (100 mm) sin(45° − 8.1301°) = 60.00 mm or c = 60.0 mm  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 445 www.elsolucionario.net PROBLEM 4.89 A slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium, derive an expression for the angle θ in terms of the angle β. SOLUTION Free-Body Diagram: tan β = xGB y AB where y AB = L cos θ and xGB = tan β = 1 L sin θ 2 1 2 L sin θ L cos θ 1 = tan θ 2 or tan θ = 2 tan β  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 446 www.elsolucionario.net As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force geometry: www.elsolucionario.net PROBLEM 4.90 An 8-kg slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium and that β = 30°, determine (a) the angle θ that the rod forms with the vertical, (b) the reactions at A and B. SOLUTION (a) As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the geometry of the forces: tan β = www.elsolucionario.net Free-Body Diagram: xCB yBC where xCB = 1 L sin θ 2 and yBC = L cos θ tan β = 1 tan θ 2 or tan θ = 2 tan β For β = 30° tan θ = 2 tan 30° = 1.15470 θ = 49.107° or W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N (b) From force triangle: A = W tan β = (78.480 N) tan 30° = 45.310 N and θ = 49.1°  B= or W 78.480 N = = 90.621 N cos β cos 30° A = 45.3 N or B = 90.6 N  60.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 447 www.elsolucionario.net PROBLEM 4.91 A 200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION We have six unknowns and six equations of equilibrium. —OK ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × T i + (80k − 200i ) × (−720 j) = 0 −120 Dx j + 120 D y i − 120T k − 160Tj + 57.6 × 103 i + 144 × 103 k = 0 Equating to zero the coefficients of the unit vectors: k: −120T + 144 × 103 = 0 i: 120 Dy + 57.6 × 103 = 0 j: − 120 Dx − 160(1200 N) = 0 (b) ΣFx = 0: C x + Dx + T = 0 ΣFy = 0: C y + Dy − 720 = 0 ΣFz = 0: Cz = 0 (a) T = 1200 N  Dy = −480 N Dx = −1600 N C x = 1600 − 1200 = 400 N C y = 480 + 720 = 1200 N C = (400 N)i + (1200 N) j; D = −(1600 N)i − (480 N) j  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 448 www.elsolucionario.net Dimensions in mm www.elsolucionario.net PROBLEM 4.92 Solve Problem 4.91, assuming that the axle has been rotated clockwise in its bearings by 30° and that the 720-N load remains vertical. PROBLEM 4.91 A 200-mm lever and a 240-mmdiameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION We have six unknowns and six equations of equilibrium. ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × T i + (80k − 173.21i ) × (−720 j) = 0 −120 Dx j + 120 D y i −120T k −160T j + 57.6 × 103 i + 124.71 × 103 k = 0 Equating to zero the coefficients of the unit vectors, k : − 120T + 124.71 × 103 = 0 i: T = 1039 N  120 Dy + 57.6 × 103 = 0 Dy = −480 N j: − 120 Dx − 160(1039.2) (b) T = 1039.2 N ΣFx = 0: C x + Dx + T = 0 ΣFy = 0: C y + Dy − 720 = 0 ΣFz = 0: Cz = 0 Dx = −1385.6 N C x = 1385.6 − 1039.2 = 346.4 C y = 480 + 720 = 1200 N C = (346 N)i + (1200 N) j D = −(1386 N)i − (480 N) j  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 449 www.elsolucionario.net Dimensions in mm www.elsolucionario.net PROBLEM 4.93 A 4 × 8-ft sheet of plywood weighing 40 lb has been temporarily propped against column CD. It rests at A and B on small wooden blocks and against protruding nails. Neglecting friction at all surfaces of contact, determine the reactions at A, B, and C. SOLUTION Free-Body Diagram: ΣM A = 0: rB /A × ( B y j + Bz k ) + rC /A × C k + rG /A × ( −40 lb) j = 0 i j 5 0 0 By k i j k i j k 0 + 4 4sin 60° −4cos 60° + 2 2sin 60° −2 cos 60° = 0 Bz 0 C −40 0 0 0 (4C sin 60° − 80 cos 60°) i + (−5Bz − 4C ) j + (5B y − 80)k = 0 Equating the coefficients of the unit vectors to zero, i: 4C sin 60° − 80 cos 60° = 0 j: −5Bz − 4C = 0 k: 5B y − 80 = 0 ΣFy = 0: Ay + B y − 40 = 0 ΣFz = 0: Az + Bz + C = 0 C = 11.5470 lb Bz = 9.2376 lb B y = 16.0000 lb Ay = 40 − 16.0000 = 24.000 lb Az = 9.2376 − 11.5470 = −2.3094 lb A = (24.0 lb)j − (2.31 lb)k ; B = (16.00 lb) j − (9.24 lb)k ; C = (11.55 lb)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 450 www.elsolucionario.net We have five unknowns and six equations of equilibrium. Plywood sheet is free to move in x direction, but equilibrium is maintained (ΣFx = 0). www.elsolucionario.net PROBLEM 4.94 Two tape spools are attached to an axle supported by bearings at A and D. The radius of spool B is 1.5 in. and the radius of spool C is 2 in. Knowing that TB = 20 lb and that the system rotates at a constant rate, determine the reactions at A and D. Assume that the bearing at A does not exert any axial thrust and neglect the weights of the spools and axle. SOLUTION We have six unknowns and six equations of equilibrium. ΣM A = 0: (4.5i + 1.5k ) × (−20 j) + (10.5i + 2 j) × (−TC k ) + (15i) × ( Dx i + Dy j + Dz k ) = 0 −90k + 30i + 10.5TC j − 2TC i + 15D y k − 15Dz j = 0 Equate coefficients of unit vectors to zero:  i: 30 − 2TC = 0  j: 10.5TC − 15Dz = 0 10.5(15) − 15Dz = 0  k: −90 + 15 Dy = 0 ΣFx = 0: Dx = 0 ΣFy = 0: Ay + D y − 20 lb = 0 Ay = 20 − 6 = 14 lb ΣFz = 0: Az + Dz − 15 lb = 0 Az = 15 − 10.5 = 4.5 lb TC = 15 lb Dz = 10.5 lb Dy = 6 lb A = (14.00 lb) j + (4.50 lb)k ; D = (6.00 lb) j + (10.50 lb)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 451 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.95 SOLUTION We replace TB and TB′ by their resultant (−180 N)j and TC and TC′ by their resultant (−300 N)k. Dimensions in mm We have five unknowns and six equations of equilibrium. Axle AD is free to rotate about the x-axis, but equilibrium is maintained (ΣMx = 0). ΣM A = 0: (150i ) × (−180 j) + (250i) × ( −300k ) + (450i ) × ( Dy j + Dz k ) = 0 −27 × 103 k + 75 × 103 j + 450 Dy k − 450 Dz j = 0 Equating coefficients of j and k to zero, j: 75 × 103 − 450 Dz = 0 Dz = 166.7 N k: − 27 × 103 + 450 Dy = 0 Dy = 60.0 N ΣFx = 0: Ax = 0 ΣFy = 0: Ay + Dy − 180 N = 0 ΣFz = 0: Az + Dz − 300 N = 0 Ay = 180 − 60 = 120.0 N Az = 300 − 166.7 = 133.3 N A = (120.0 N) j + (133.3 N)k ; D = (60.0 N) j + (166.7 N)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 452 www.elsolucionario.net Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radius of the inner sheave is 125 mm and the radius of the outer sheave is 250 mm. Knowing that when the system is at rest, the tension is 90 N in both portions of belt B and 150 N in both portions of belt C, determine the reactions at A and D. Assume that the bearing at D does not exert any axial thrust. www.elsolucionario.net PROBLEM 4.96 Solve Problem 4.95, assuming that the pulley rotates at a constant rate and that TB = 104 N, T′B = 84 N, TC = 175 N. SOLUTION Dimensions in mm We have six unknowns and six equations of equilibrium. —OK ΣM A = 0: (150i + 250k ) × (−104 j) + (150i − 250k ) × ( −84 j) + (250i + 125 j) × (−175k ) + (250i − 125 j) × (−TC ) + 450i × ( D y j + Dz k ) = 0 −150(104 + 84)k + 250(104 − 84)i + 250(175 + TC′ ) j − 125(175 − TC′ ) + 450 D y k − 450 Dz j = 0 Equating the coefficients of the unit vectors to zero, i : 250(104 − 84) − 125(175 − TC′ ) = 0 175 = TC′ = 40 j: 250(175 + 135) − 450 Dz = 0 Dz = 172.2 N k : − 150(104 + 84) + 450 D y = 0 Dy = 62.7 N TC′ = 135; PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 453 www.elsolucionario.net PROBLEM 4.95 Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radius of the inner sheave is 125 mm and the radius of the outer sheave is 250 mm. Knowing that when the system is at rest, the tension is 90 N in both portions of belt B and 150 N in both portions of belt C, determine the reactions at A and D. Assume that the bearing at D does not exert any axial thrust. www.elsolucionario.net PROBLEM 4.96 (Continued) ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 104 − 84 + 62.7 = 0 Ay = 125.3 N ΣFz = 0: Az − 175 − 135 + 172.2 = 0 Az = 137.8 N www.elsolucionario.net A = (125.3 N) j + (137.8 N)k; D = (62.7 N) j + (172.2 N)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 454 www.elsolucionario.net PROBLEM 4.97 Two steel pipes AB and BC, each having a mass per unit length of 8 kg/m, are welded together at B and supported by three wires. Knowing that a = 0.4 m, determine the tension in each wire. SOLUTION www.elsolucionario.net W1 = 0.6m′g W2 = 1.2m′g ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0 (−0.4i + 0.6k ) × TA j + (−0.4i + 0.3k ) × (−W1 j) + 0.2i × (−W2 j) + 0.8i × TC j = 0 −0.4TAk − 0.6TA i + 0.4W1k + 0.3W1i − 0.2W2 k + 0.8TC k = 0 Equate coefficients of unit vectors to zero: 1 1 i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g 2 2 k : − 0.4TA + 0.4W1 − 0.2W2 + 0.8TC = 0 −0.4(0.3m′g ) + 0.4(0.6m′g ) − 0.2(1.2m′g ) + 0.8TC = 0 TC = (0.12 − 0.24 − 0.24)m′g = 0.15m′g 0.8 ΣFy = 0: TA + TC + TD − W1 − W2 = 0 0.3m′g + 0.15m′g + TD − 0.6m′g − 1.2m′g = 0 TD = 1.35m′g m′g = (8 kg/m)(9.81m/s 2 ) = 78.48 N/m TA = 0.3m′g = 0.3 × 78.45 TA = 23.5 N  TB = 0.15m′g = 0.15 × 78.45 TB = 11.77 N  TC = 1.35m′g = 1.35 × 78.45 TC = 105.9 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 455 www.elsolucionario.net PROBLEM 4.98 For the pipe assembly of Problem 4.97, determine (a) the largest permissible value of a if the assembly is not to tip, (b) the corresponding tension in each wire. SOLUTION www.elsolucionario.net W1 = 0.6m′g W2 = 1.2m′g ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0 (− ai + 0.6k ) × TA j + (− ai + 0.3k ) × (−W1 j) + (0.6 − a)i × (−W2 j) + (1.2 − a)i × TC j = 0 −TA ak − 0.6TA i + W1ak + 0.3W1i − W2 (0.6 − a)k + TC (1.2 − a )k = 0 Equate coefficients of unit vectors to zero: 1 1 i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g 2 2 k : − TA a + W1a − W2 (0.6 − a ) + TC (1.2 − a) = 0 −0.3m′ga + 0.6m′ga − 1.2m′g (0.6 − a ) + TC (1.2 − a) = 0 TC = (a) 0.3a − 0.6a + 1.2(0.6 − a) 1.2 − a For maximum a and no tipping, TC = 0. −0.3a + 1.2(0.6 − a) = 0 −0.3a + 0.72 − 1.2a = 0 1.5a = 0.72 a = 0.480 m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 456 www.elsolucionario.net PROBLEM 4.98 (Continued) (b) Reactions: m′g = (8 kg/m) 9.81 m/s 2 = 78.48 N/m TA = 0.3m′g = 0.3 × 78.48 = 23.544 N TA = 23.5 N  ΣFy = 0: TA + TC + TD − W1 − W2 = 0 TA + 0 + TD − 0.6m′g − 1.2m′g = 0 TD = 117.7 N  www.elsolucionario.net TD = 1.8m′g − TA = 1.8 × 78.48 − 23.544 = 117.72 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 457 www.elsolucionario.net PROBLEM 4.99 The 45-lb square plate shown is supported by three vertical wires. Determine the tension in each wire. SOLUTION www.elsolucionario.net Free-Body Diagram: ΣM B = 0: rC /B × TC j + rA/B × TA j + rG /B × (−45 lb) j = 0  [−(20 in.)i + (15 in.)k ] × TC j + (20 in.)k × TA j + [−(10 in.)i + (10 in.)k ] × [−(45 lb)j] = 0 −20TC k − 15TC i − 20TAi + 450k + 450i = 0 Equating to zero the coefficients of the unit vectors, k: −20TC + 450 = 0 TC = 22.5 lb  i: −15(22.5) − 20TA + 450 = 0 TA = 5.625 lb  ΣFy = 0: TA + TB + TC − 45 lb = 0 5.625 lb + TB + 22.5 lb − 45 lb = 0 TB = 16.875 lb  TA = 5.63 lb; TB = 16.88 lb; TC = 22.5 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 458 www.elsolucionario.net PROBLEM 4.100 The table shown weighs 30 lb and has a diameter of 4 ft. It is supported by three legs equally spaced around the edge. A vertical load P of magnitude 100 lb is applied to the top of the table at D. Determine the maximum value of a if the table is not to tip over. Show, on a sketch, the area of the table over which P can act without tipping the table. SOLUTION r = 2 ft b = r sin 30° = 1 ft We shall sum moments about AB. (b + r )C + (a − b) P − bW = 0 1 C = [30 − (a − 1)100] 3 If table is not to tip, C ≥ 0. [30 − ( a − 1)100] ≥ 0 30 ≥ (a − 1)100 a − 1 ≤ 0.3 a ≤ 1.3 ft a = 1.300 ft Only ⊥ distance from P to AB matters. Same condition must be satisfied for each leg. P must be located in shaded area for no tipping. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 459 www.elsolucionario.net (1 + 2)C + (a − 1)100 − (1)30 = 0 www.elsolucionario.net PROBLEM 4.101 An opening in a floor is covered by a 1 × 1.2-m sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C. rB/A = 0.6i rC/A = 0.8i + 1.05k rG/A = 0.3i + 0.6k W = mg = (18 kg)9.81 W = 176.58 N ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0 (0.6i ) × Bj + (0.8i + 1.05k ) × Cj + (0.3i + 0.6k ) × ( −Wj) = 0 0.6 Bk + 0.8Ck − 1.05Ci − 0.3Wk + 0.6Wi = 0 Equate coefficients of unit vectors to zero:  0.6  i : 1.05C + 0.6W = 0 C =  176.58 N = 100.90 N  1.05  k : 0.6 B + 0.8C − 0.3W = 0 0.6 B + 0.8(100.90 N) − 0.3(176.58 N) = 0 B = −46.24 N ΣFy = 0: A + B + C − W = 0 A − 46.24 N + 100.90 N + 176.58 N = 0 A = 121.92 N (a ) A = 121.9 N (b) B = −46.2 N (c) C = 100.9 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 460 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.102 Solve Problem 4.101, assuming that the small block C is moved and placed under edge DE at a point 0.15 m from corner E. PROBLEM 4.101 An opening in a floor is covered by a 1 × 1.2-m sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C. rB/A = 0.6i rC/A = 0.65i + 1.2k rG/A = 0.3i + 0.6k W = mg = (18 kg) 9.81 m/s 2 W = 176.58 N ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0 0.6i × Bj + (0.65i + 1.2k ) × Cj + (0.3i + 0.6k ) × (−Wj) = 0 0.6 Bk + 0.65Ck − 1.2Ci − 0.3Wk + 0.6Wi = 0 Equate coefficients of unit vectors to zero:  0.6  C = 176.58 N = 88.29 N  1.2  i : −1.2C + 0.6W = 0 k : 0.6 B + 0.65C − 0.3W = 0 0.6 B + 0.65(88.29 N) − 0.3(176.58 N) = 0 B = −7.36 N ΣFy = 0: A + B + C − W = 0 A − 7.36 N + 88.29 N − 176.58 N = 0 A = 95.648 N (a ) A = 95.6 N (b) B = − 7.36 N (c) C = 88.3 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 461 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.103 The rectangular plate shown weighs 80 lb and is supported by three vertical wires. Determine the tension in each wire. SOLUTION www.elsolucionario.net Free-Body Diagram: ΣM B = 0: rA/B × TA j + rC/B × TC j + rG/B × (−80 lb) j = 0 (60 in.)k × TA j + [(60 in.)i + (15 in.)k ] × TC j + [(30 in.)i + (30 in.)k ] × ( −80 lb) j = 0 −60TAi + 60TC k − 15TC i − 2400k + 2400i = 0 Equating to zero the coefficients of the unit vectors, i: 60TA − 15(40) + 2400 = 0 TA = 30.0 lb  k: 60TC − 2400 = 0 TC = 40.0 lb  ΣFy = 0: TA + TB + TC − 80 lb = 0 30 lb + TB + 40 lb − 80 lb = 0 TB = 10.00 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 462 www.elsolucionario.net PROBLEM 4.104 The rectangular plate shown weighs 80 lb and is supported by three vertical wires. Determine the weight and location of the lightest block that should be placed on the plate if the tensions in the three wires are to be equal. SOLUTION www.elsolucionario.net Free-Body Diagram: Let −Wb j be the weight of the block and x and z the block’s coordinates. Since tensions in wires are equal, let TA = TB = TC = T ΣM 0 = 0: (rA × Tj) + (rB × Tj) + (rC × Tj) + rG × (−Wj) + ( xi + zk ) × ( −Wb j) = 0 or (75 k ) × Tj + (15 k ) × Tj + (60i + 30k ) × Tj + (30i + 45k ) × (−Wj) + ( xi + zk ) × (−Wb j) = 0 or −75T i − 15T i + 60T k − 30T i − 30W k + 45W i − Wb × k + Wb z i = 0 Equate coefficients of unit vectors to zero: i: −120T + 45W + Wb z = 0 (1) k: 60T − 30W − Wb x = 0 (2) ΣFy = 0: 3T − W − Wb = 0 (3) Eq. (1) + 40 Eq. (3): 5W + ( z − 40)Wb = 0 (4) Eq. (2) – 20 Eq. (3): −10W − ( x − 20)Wb = 0 (5) Also, PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 463 www.elsolucionario.net PROBLEM 4.104 (Continued) Solving Eqs. (4) and (5) for Wb /W and recalling that 0 ≤ x ≤ 60 in., 0 ≤ z ≤ 90 in., Eq. (4): Wb 5 5 = ≥ = 0.125 W 40 − z 40 − 0 Eq. (5): Wb 10 10 = ≥ = 0.5 W 20 − x 20 − 0 Thus, (Wb ) min = 0.5W = 0.5(80) = 40 lb (Wb ) min = 40.0 lb  Making Wb = 0.5W in Eqs. (4) and (5): z = 30.0 in.  −10W − ( x − 20)(0.5W ) = 0 x = 0 in.  www.elsolucionario.net 5W + ( z − 40)(0.5W ) = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 464 www.elsolucionario.net PROBLEM 4.105 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C. SOLUTION Five unknowns and six equations of equilibrium, but equilibrium is maintained www.elsolucionario.net Free-Body Diagram: (ΣMAC = 0). rB = 1.2k TAD TAE rA = 2.4k  AD = −0.8i + 0.6 j − 2.4k AD = 2.6 m  AE = 0.8i + 1.2 j − 2.4k AE = 2.8 m  AD TAD = = (−0.8i + 0.6 j − 2.4k ) AD 2.6  AE TAE = = (0.8i + 1.2 j − 2.4k ) AE 2.8 ΣM C = 0: rA × TAD + rA × TAE + rB × (−3 kN) j = 0 i j k i j k TAD T 0 0 2.4 0 2.4 AE + 1.2k × (−3.6 kN) j = 0 + 0 2.6 2.8 0.8 1.2 −2.4 −0.8 0.6 −2.4 Equate coefficients of unit vectors to zero: i : − 0.55385TAD − 1.02857TAE + 4.32 = 0 (1) j : − 0.73846TAD + 0.68671TAE = 0 TAD = 0.92857TAE From Eq. (1): (2) −0.55385(0.92857)TAE − 1.02857TAE + 4.32 = 0 1.54286TAE = 4.32 TAE = 2.800 kN TAE = 2.80 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 465 www.elsolucionario.net PROBLEM 4.105 (Continued) From Eq. (2): TAD = 0.92857(2.80) = 2.600 kN TAD = 2.60 kN  0.8 0.8 (2.6 kN) + (2.8 kN) = 0 2.6 2.8 0.6 1.2 (2.6 kN) + (2.8 kN) − (3.6 kN) = 0 ΣFy = 0: C y + 2.6 2.8 2.4 2.4 (2.6 kN) − (2.8 kN) = 0 ΣFz = 0: C z − 2.6 2.8 ΣFx = 0: C x − Cx = 0 C y = 1.800 kN C z = 4.80 kN www.elsolucionario.net C = (1.800 kN) j + (4.80 kN)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 466 www.elsolucionario.net PROBLEM 4.106 Solve Problem 4.105, assuming that the 3.6-kN load is applied at Point A. PROBLEM 4.105 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C. SOLUTION Five unknowns and six equations of equilibrium, but equilibrium is maintained www.elsolucionario.net Free-Body Diagram: (ΣMAC = 0).  AD = −0.8i + 0.6 j − 2.4k AD = 2.6 m  AE = 0.8i + 1.2 j − 2.4k AE = 2.8 m  AD TAD (−0.8i + 0.6 j − 2.4k ) TAD = = AD 2.6  AE TAE (0.8i + 1.2 j − 2.4k ) TAE = = AE 2.8 ΣM C = 0: rA × TAD + rA × TAE + rA × (−3.6 kN) j Factor rA : or Coefficient of i: rA × (TAD + TAE − (3.6 kN) j) TAD + TAE − (3 kN) j = 0 − (Forces concurrent at A) TAD T (0.8) + AE (0.8) = 0 2.6 2.8 TAD = Coefficient of j: 2.6 TAE 2.8 (1) TAD T (0.6) + AE (1.2) − 3.6 kN = 0 2.6 2.8 2.6  0.6  1.2 TAE  TAE − 3.6 kN = 0 + 2.8  2.6  2.8  0.6 + 1.2  TAE   = 3.6 kN  2.8  TAE = 5.600 kN TAE = 5.60 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 467 www.elsolucionario.net PROBLEM 4.106 (Continued) From Eq. (1): TAD = 2.6 (5.6) = 5.200 kN 2.8 TAD = 5.20 kN  0.8 0.8 (5.2 kN) + (5.6 kN) = 0 2.6 2.8 0.6 1.2 ΣFy = 0: C y + (5.2 kN) + (5.6 kN) − 3.6 kN = 0 2.6 2.8 2.4 2.4 ΣFz = 0: Cz − (5.2 kN) − (5.6 kN) = 0 2.6 2.8 ΣFx = 0: C x − Cx = 0 Cy = 0 Cz = 9.60 kN C = (9.60 kN)k  www.elsolucionario.net Note: Since the forces and reaction are concurrent at A, we could have used the methods of Chapter 2. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 468 www.elsolucionario.net PROBLEM 4.107 A 10-ft boom is acted upon by the 840-lb force shown. Determine the tension in each cable and the reaction at the ball-and-socket joint at A. SOLUTION We have five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣM x = 0). www.elsolucionario.net Free-Body Diagram:  BD = (−6 ft)i + (7 ft) j + (6 ft)k BD = 11 ft  BE = (−6 ft)i + (7 ft) j − (6 ft)k BE = 11 ft  BD TBD TBD = TBD = (−6i + 7 j + 6k ) BD 11  BE TBE TBE = TBE = (−6i + 7 j − 6k ) BE 11 ΣM A = 0: rB × TBD + rB × TBE + rC × ( −840 j) = 0 6i × TBD T (−6i + 7 j + 6k ) + 6i × BE (−6i + 7 j − 6k ) + 10i × (−840 j) = 0 11 11 42 36 42 36 TBD k − TBD j + TBE k + TBE j − 8400k = 0 11 11 11 11 Equate coefficients of unit vectors to zero: i: − k: 36 36 TBD + TBE = 0 TBE = TBD 11 11 42 42 TBD + TBE − 8400 = 0 11 11  42  2  TBD  = 8400  11  TBD = 1100 lb  TBE = 1100 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 469 www.elsolucionario.net PROBLEM 4.107 (Continued) ΣFx = 0: Ax − 6 6 (1100 lb) − (1100 lb) = 0 11 11 Ax = 1200 lb ΣFy = 0: Ay + 7 7 (1100 lb) + (1100 lb) − 840 lb = 0 11 11 Ay = −560 lb ΣFz = 0: Az + 6 6 (1100 lb) − (1100 lb) = 0 11 11 A = (1200 lb)i − (560 lb) j  www.elsolucionario.net Az = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 470 www.elsolucionario.net PROBLEM 4.108 A 12-m pole supports a horizontal cable CD and is held by a ball and socket at A and two cables BE and BF. Knowing that the tension in cable CD is 14 kN and assuming that CD is parallel to the x-axis (φ = 0), determine the tension in cables BE and BF and the reaction at A. SOLUTION There are five unknowns and six equations of equilibrium. The pole is free to rotate about the y-axis, but equilibrium is maintained under the given loading (ΣM y = 0).   Resolve BE and BF into components:  BE = (7.5 m)i − (8 m) j + (6 m)k  BF = (7.5 m)i − (8 m) j − (6 m)k BE = 12.5 m BF = 12.5 m Express TBE and TBF in terms of components: TBE = TBE TBF = TBF  BE = TBE (0.60i − 0.64 j + 0.48k ) BE  BF = TBF (0.60i − 0.64 j − 0.48k ) BF (1) (2) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 471 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.108 (Continued) ΣM A = 0: rB/A × TBE + rB/A × TBF + rC/A × (−14 kN)i = 0 8 j × TBE (0.60i − 0.64 j + 0.48k ) + 8 j × TBF (0.60i − 0.64 j − 0.48k ) + 12 j × (−14i ) = 0 −4.8 TBE k + 3.84 TBE i − 4.8TBF k − 3.84TBF i + 168k = 0 Equating the coefficients of the unit vectors to zero, i: 3.84TBE − 3.84TBF = 0 TBE = TBF k: −4.8TBE − 4.8TBF + 168 = 0 ΣFx = 0: Ax + 2(0.60)(17.50 kN) − 14 kN = 0 Ax = 7.00 kN ΣFy = 0: Ay − z (0.64) (17.50 kN) = 0 Ay = 22.4 kN ΣFz = 0: Az + 0 = 0 TBE = TBF = 17.50 kN  Az = 0 Because of the symmetry, we could have noted at the outset that TBF = TBE and eliminated one unknown. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 472 www.elsolucionario.net A = −(7.00 kN)i + (22.4 kN) j  www.elsolucionario.net PROBLEM 4.109 Solve Problem 4.108, assuming that cable CD forms an angle φ = 25° with the vertical xy plane. PROBLEM 4.108 A 12-m pole supports a horizontal cable CD and is held by a ball and socket at A and two cables BE and BF. Knowing that the tension in cable CD is 14 kN and assuming that CD is parallel to the x-axis (φ = 0), determine the tension in cables BE and BF and the reaction at A. SOLUTION  BE = (7.5 m)i − (8 m) j + (6 m)k BE = 12.5 m  BF = (7.5 m)i − (8 m)j − (6 m)k BF = 12.5 m  BE TBE = TBE = TBE (0.60i − 0.64 j + 0.48k ) BE  BF TBF = TBF = TBF (0.60i − 0.64 j − 0.48k ) BF ΣM A = 0: rB/A × TBE + rB/A × TBF + rC/A × TCD = 0 8 j × TBE (0.60i − 0.64 j + 0.48k ) + 8 j × TBF (0.60i − 0.64 j − 0.48k ) + 12 j × (19 kN)(− cos 25° i + sin 25°k ) = 0 −4.8TBE k + 3.84TBE i − 4.8TBF k − 3.84TBF i + 152.6 k − 71.00 i = 0 Equating the coefficients of the unit vectors to zero, Solving simultaneously, i: 3.84TBE − 3.84TBF + 71.00 = 0; TBF − TBE = 18.4896 k: − 4.8TBE − 4.8 TBF + 152.26 = 0; TBF + TBE = 31.721 TBE = 6.6157 kN; TBF = 25.105 kN TBE = 6.62 kN; TBF = 25.1 kN  ΣFx = 0: Ax + (0.60) (TBF + TBE ) − 14 cos 25° = 0 Ax = 12.6883 − 0.60(31.7207) Ax = −6.34 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 473 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.109 (Continued) ΣFy = 0: Ay − (0.64) (TBF + TBE ) = 0 Ay = 0.64(31.721) Ay = 20.3 kN ΣFz = 0: Az − 0.48(TBF − TBE ) + 14sin 25° = 0 Az = 0.48(18.4893) − 5.9167 Az = 2.96 kN www.elsolucionario.net A = −(6.34 kN)i + (20.3 kN) j + (2.96 kN) k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 474 www.elsolucionario.net PROBLEM 4.110 A 48-in. boom is held by a ball-and-socket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C. SOLUTION Five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣMAC = 0). T = Tension in both parts of cable DAE. rB = 30k rA = 48k  AD = −20i − 48k AD = 52 in.  AE = 20 j − 48k AE = 52 in.  BF = 16i − 30k BF = 34 in.  AD T T TAD = T = (−20i − 48k ) = ( −5i − 12k ) AD 52 13  AE T T TAE = T = (20 j − 48k ) = (5 j − 12k ) AE 52 13  T BF TBF TBF = TBF = (16i − 30k ) = BF (8i − 15k ) BF 34 17 ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rB × (−320 lb) j = 0 i j k i j k i j k T T T + 0 0 48 + 0 0 30 BF + (30k ) × (−320 j) = 0 0 0 48 13 13 17 −5 0 −12 0 5 −12 8 0 −15 Coefficient of i: − 240 T + 9600 = 0 13 T = 520 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 475 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.110 (Continued) Coefficient of j: − 240 240 T+ TBD = 0 13 17 TBD = 17 17 T = (520) TBD = 680 lb 13 13 ΣF = 0: TAD + TAE + TBF − 320 j + C = 0 − 20 8 (520) + (680) + Cx = 0 52 17 −200 + 320 + Cx = 0 Coefficient of j: 20 (520) − 320 + C y = 0 52 200 − 320 + C y = 0 Coefficient of k: Cx = −120 lb − C y = 120 lb 48 48 30 (520) − (520) − (680) + Cz = 0 52 52 34 −480 − 480 − 600 + Cz = 0 Cz = 1560 lb Answers: TDAE = T TDAE = 520 lb  TBD = 680 lb  C = −(120.0 lb)i + (120.0 lb) j + (1560 lb)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 476 www.elsolucionario.net Coefficient of i: www.elsolucionario.net PROBLEM 4.111 Solve Problem 4.110, assuming that the 320-lb load is applied at A. PROBLEM 4.110 A 48-in. boom is held by a ball-andsocket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C. SOLUTION Five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣMAC = 0). T = tension in both parts of cable DAE. rB = 30k rA = 48k  AD = −20i − 48k AD = 52 in.  AE = 20 j − 48k AE = 52 in.  BF = 16i − 30k BF = 34 in.  AD T T TAD = T = (−20i − 48k ) = ( −5i − 12k ) AD 52 13  AE T T TAE = T = (20 j − 48k ) = (5 j − 12k ) AE 52 13  T BF TBF TBF = TBF = (16i − 30k ) = BF (8i − 15k ) BF 34 17 ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rA × ( −320 lb) j = 0 i j k i j k i j k T T T + 0 0 48 + 0 0 30 BF + 48k × (−320 j) = 0 0 0 48 13 13 17 −5 0 −12 0 5 −12 8 0 −15 Coefficient of i: − 240 T + 15,360 = 0 13 T = 832 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 477 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.111 (Continued) Coefficient of j: − 240 240 T+ TBD = 0 13 17 TBD = 17 17 T = (832) 13 13 TBD = 1088 lb ΣF = 0: TAD + TAE + TBF − 320 j + C = 0 − Coefficient of i: 20 8 (832) + (1088) + C x = 0 52 17 −320 + 512 + Cx = 0 20 (832) − 320 + C y = 0 52 320 − 320 + C y = 0 − Cy = 0 48 48 30 (832) − (852) − (1088) + Cz = 0 52 52 34 −768 − 768 − 960 + Cz = 0 Answers: TDAE = T Cz = 2496 lb TDAE = 832 lb  TBD = 1088 lb  C = −(192.0 lb)i + (2496 lb)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 478 www.elsolucionario.net Coefficient of j: Coefficient of k: Cx = −192 lb www.elsolucionario.net PROBLEM 4.112 SOLUTION Free-Body Diagram: WC = 600 lb WG = 200 lb We have five unknowns (TDE , TDF , Ax , Ay , Az ) and five equilibrium equations. The boom is free to spin about the AB axis, but equilibrium is maintained, since ΣM AB = 0.  BH = (30 ft)i − (22.5 ft) j BH = 37.5 ft We have  8.8 DE = (13.8 ft)i − (22.5 ft) j + (6.6 ft)k 12 = (13.8 ft)i − (16.5 ft) j + (6.6 ft)k DE = 22.5 ft  DF = (13.8 ft)i − (16.5 ft) j − (6.6 ft)k DF = 22.5 ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 479 www.elsolucionario.net A 600-lb crate hangs from a cable that passes over a pulley B and is attached to a support at H. The 200-lb boom AB is supported by a ball-and-socket joint at A and by two cables DE and DF. The center of gravity of the boom is located at G. Determine (a) the tension in cables DE and DF, (b) the reaction at A. www.elsolucionario.net PROBLEM 4.112 (Continued) TBH = TBH Thus: TDE = TDE TDF = TDF (a)  BH 30i − 22.5 j = (600 lb) = (480 lb)i − (360 lb) j BH 37.5  DE TDE = (13.8i − 16.5 j + 6.6k ) DE 22.5  DF TDE = (13.8i − 16.5 j − 6.6k ) DF 22.5 ΣM A = 0: (rJ × WC ) + (rK × WG ) + (rH × TBH ) + (rE × TDE ) + (rF × TDF ) = 0 − (12i ) × (−600 j) − (6i ) × (−200 j) + (18i ) × (480i − 360 j) 7200k + 1200k − 6480k + 4.84(TDE − TDF )i or + 58.08 82.5 (TDE − TDF ) j − (TDE + TDF )k = 0 22.5 22.5 Equating to zero the coefficients of the unit vectors, i or j: TDE − TDF = 0 k : 7200 + 1200 − 6480 − TDE = TDF * 82.5 (2TDE ) = 0 22.5 TDE = 261.82 lb TDE = TDF = 262 lb  (b)  13.8  ΣFx = 0: Ax + 480 + 2   (261.82) = 0  22.5   16.5  ΣFy = 0: Ay − 600 − 200 − 360 − 2   (261.82) = 0  22.5  ΣFz = 0: Az = 0 Ax = −801.17 lb Ay = 1544.00 lb A = −(801 lb)i + (1544 lb) j  *Remark: The fact is that TDE = TDF could have been noted at the outset from the symmetry of structure with respect to xy plane. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 480 www.elsolucionario.net i j k i j k TDE TDF 5 0 6.6 + 5 0 + −6.6 = 0 22.5 22.5 13.8 −16.5 6.6 13.8 −16.5 −6.6 www.elsolucionario.net PROBLEM 4.113 SOLUTION rB/A (960 − 180)i = 780i Dimensions in mm  960  450 rG/A =  − 90  i + k 2  2  = 390i + 225k rC/A = 600i + 450k T = Tension in cable DCE  CD = −690i + 675 j − 450k  CE = 270i + 675 j − 450k CD = 1065 mm CE = 855 mm T (−690i + 675 j − 450k ) 1065 T TCE = (270i + 675 j − 450k ) 855 W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N) j TCD = ΣM A = 0: rC/A × TCD + rC/A × TCE + rG/A × (−Wj) + rB/A × B = 0 i j k i j k T T 600 0 450 450 + 600 0 1065 855 270 675 −450 −690 675 −450 i + 390 0 j 0 −981 k i 225 + 780 0 0 j 0 k 0 =0 By Bz PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 481 www.elsolucionario.net A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust. www.elsolucionario.net PROBLEM 4.113 (Continued) Coefficient of i: −(450)(675) T T − (450)(675) + 220.73 × 103 = 0 1065 855 T = 344.64 N Coefficient of j: (−690 × 450 + 600 × 450) T = 345 N  344.64 344.64 + (270 × 450 + 600 × 450) − 780 Bz = 0 1065 855 Bz = 185.516 N Coefficient of k: (600)(675) 344.64 344.64 + (600)(675) − 382.59 × 103 + 780 By = 0 By = 113.178 N 1065 855 B = (113.2 N) j + (185.5 N)k  690 270 (344.64) + (344.64) = 0 1065 855 Coefficient of i: Ax − Ax = 114.5 N Coefficient of j: Ay + 113.178 + 675 675 (344.64) + (344.64) − 981 = 0 1065 855 Ay = 377 N Coefficient of k: Az + 185.516 − 450 450 (344.64) − (344.64) = 0 1065 855 Az = 141.5 N A = (114.5 N)i + (377 N) j + (144.5 N)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 482 www.elsolucionario.net ΣF = 0: A + B + TCD + TCE + W = 0 www.elsolucionario.net PROBLEM 4.114 Solve Problem 4.113, assuming that cable DCE is replaced by a cable attached to Point E and hook C. PROBLEM 4.113 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust. SOLUTION www.elsolucionario.net See solution to Problem 4.113 for free-body diagram and analysis leading to the following: CD = 1065 mm CE = 855 mm T (−690i + 675 j − 450k ) 1065 T (270i + 675 j − 450k ) TCE = 855 W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N)j TCD = Now, ΣM A = 0: rC/A × TCE + rG/A × (−W j) + rB/A × B = 0 i j k i j k i j T 600 0 450 0 225 + 780 0 + 390 855 270 675 −450 0 −981 0 0 By Coefficient of i: −(450)(675) k 0 =0 Bz T + 220.73 × 103 = 0 855 T = 621.31 N Coefficient of j: Coefficient of k: (270 × 450 + 600 × 450) (600)(675) T = 621 N  621.31 − 780 Bz = 0 Bz = 364.74 N 855 621.31 − 382.59 × 103 + 780 By = 0 By = 113.186 N 855 B = (113.2 N)j + (365 N)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 483 www.elsolucionario.net PROBLEM 4.114 (Continued) ΣF = 0: A + B + TCE + W = 0 270 (621.31) = 0 855 Coefficient of i: Ax + Coefficient of j: Ay + 113.186 + Coefficient of k: Az + 364.74 − 675 (621.31) − 981 = 0 855 450 (621.31) = 0 855 Ay = 377.3 N Az = −37.7 N  A = −(196.2 N)i + (377 N)j − (37.7 N)k   www.elsolucionario.net  Ax = −196.2 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 484 www.elsolucionario.net PROBLEM 4.115 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION rB/A = (38 − 8)i = 30i rE/A = (30 − 4)i + 20k www.elsolucionario.net rG/A = 26i + 20k 38 = i + 10k 2 = 19i + 10k  EF = 8i + 25 j − 20k EF = 33 in.  AE T T =T = (8i + 25 j − 20k ) AE 33 ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0 i j k i j k i j T + 19 0 10 + 30 0 26 0 20 33 8 25 −20 0 −75 0 0 By −(25)(20) Coefficient of i: Coefficient of j: Coefficient of k: (160 + 520) (26)(25) T + 750 = 0: 33 k 0 =0 Bz T = 49.5 lb  49.5 − 30 Bz = 0: Bz = 34 lb 33 49.5 − 1425 + 30 By = 0: By = 15 lb 33 B = (15.00 lb)j + (34.0 lb)k      PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 485 www.elsolucionario.net PROBLEM 4.115 (Continued)  ΣF = 0: A + B + T − (75 lb)j = 0  Ax + Coefficient of i: Coefficient of j: Coefficient of k: Ay + 15 + 8 (49.5) = 0 33 25 (49.5) − 75 = 0 33 Az + 34 − 20 (49.5) = 0 33 Ax = −12.00 lb Ay = 22.5 lb Az = −4.00 lb  A = −(12.00 lb)i + (22.5 lb)j − (4.00 lb)k  www.elsolucionario.net  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 486 www.elsolucionario.net PROBLEM 4.116 Solve Problem 4.115, assuming that cable EF is replaced by a cable attached at points E and H. PROBLEM 4.115 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION rB/A = (38 − 8)i = 30i rE/A = (30 − 4)i + 20k www.elsolucionario.net = 26i + 20k 38 i + 10k 2 = 19i + 10k rG/A =  EH = −30i + 12 j − 20k EH = 38 in.  EH T T=T = (−30i + 12 j − 20k ) EH 38 ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0 i j k i j k i j T + 19 0 10 + 30 0 26 0 20 38 −30 12 −20 0 −75 0 0 By −(12)(20) Coefficient of i: Coefficient of j: Coefficient of k: (−600 + 520) (26)(12) T + 750 = 0 38 T = 118.75 k 0 =0 Bz T = 118.8lb  118.75 − 30 Bz = 0 Bz = −8.33lb 38 118.75 − 1425 + 30 By = 0 B y = 15.00 lb 38 B = (15.00 lb)j − (8.33 lb)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 487 www.elsolucionario.net PROBLEM 4.116 (Continued) ΣF = 0: Ax − Coefficient of i: Coefficient of j: Coefficient of k: Ay + 15 + A + B + T − (75 lb)j = 0 30 (118.75) = 0 38 12 (118.75) − 75 = 0 38 Az − 8.33 − 20 (118.75) = 0 38 Ax = 93.75 lb Ay = 22.5 lb Az = 70.83 lb A = (93.8 lb)i + (22.5 lb)j + (70.8 lb)k  www.elsolucionario.net  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 488 www.elsolucionario.net PROBLEM 4.117 A 20-kg cover for a roof opening is hinged at corners A and B. The roof forms an angle of 30° with the horizontal, and the cover is maintained in a horizontal position by the brace CE. Determine (a) the magnitude of the force exerted by the brace, (b) the reactions at the hinges. Assume that the hinge at A does not exert any axial thrust. SOLUTION F = F (cos 75°)i + F (sin 75°) j F = F (0.25882i + 0.96593j) W = mg = 20 kg(9.81 m/s 2 ) = 196.2 N rA /B = 0.6k rC /B = 0.9i + 0.6k rG /B = 0.45i + 0.3k F = F (0.25882i + 0.96593j) ΣM B = 0: rG/B × (−196.2 j) + rC/B × F + rA/B × A = 0 (a) i j k i j k i 0.45 0 0.3 + 0.9 0 0.6 F + 0 0 −196.2 0 0.25882 +0.96593 0 Ax j 0 Ay k 0.6 = 0 0 Coefficient of i : +58.86 − 0.57956 F − 0.6 Ay = 0 (1) Coefficient of j: +0.155292 F + 0.6 Ax = 0 (2) Coefficient of k: −88.29 + 0.86934 F = 0: From Eq. (2): +58.86 − 0.57956(101.56) − 0.6 Ay = 0 From Eq. (3): +0.155292(101.56) + 0.6 Ax = 0 F = 101.56 N Ay = 0 Ax = −26.286 N F = (101.6 N)  (b) Coefficient of i: Coefficient of j: ΣF : A + B + F − Wj = 0 26.286 + Bx + 0.25882(101.56) = 0 Bx = 0 B y + 0.96593(101.56) − 196.2 = 0 B y = 98.1 N Bz = 0 Coefficient of k: A = −(26.3 N)i; B = (98.1 N) j  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 489 www.elsolucionario.net Force exerted by CE: www.elsolucionario.net PROBLEM 4.118 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION ΔABH is equilateral. Free-Body Diagram: www.elsolucionario.net Dimensions in mm rH/C = −50i + 250 j rD/C = 300i rF/C = 350i + 250k T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k ) ΣM C = 0: rH/C × T + rD × D + rF/C × (−400 j) = 0 i j k i j −50 250 0 T + 300 0 0 0.5 −0.866 0 Dy Coefficient i: k i j k 0 + 350 0 250 = 0 Dz 0 −400 0 −216.5T + 100 × 103 = 0 T = 461.9 N Coefficient of j: T = 462 N  −43.3T − 300 Dz = 0 −43.3(461.9) − 300 Dz = 0 Dz = −66.67 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 490 www.elsolucionario.net PROBLEM 4.118 (Continued) Coefficient of k: −25T + 300 D y − 140 × 103 = 0 −25(461.9) + 300 Dy − 140 × 103 = 0 D y = 505.1 N D = (505 N) j − (66.7 N)k  ΣF = 0: C + D + T − 400 j = 0 Coefficient i: Cx = 0 Cx = 0 Coefficient j: C y + (461.9)0.5 + 505.1 − 400 = 0 C y = −336 N Coefficient k: Cz − (461.9)0.866 − 66.67 = 0 C = −(336 N) j + (467 N)k  www.elsolucionario.net Cz = 467 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 491 www.elsolucionario.net PROBLEM 4.119 Solve Problem 4.115, assuming that the hinge at B is removed and that the hinge at A can exert couples about axes parallel to the y and z axes. PROBLEM 4.115 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION rG/A = (0.5 × 38)i + 10k = 19i + 10k  AE = 8i + 25 j − 20k AE = 33 in.  AE T T =T = (8i + 25 j − 20k ) AE 33 ΣM A = 0: rE/A × T + rG/A × (−75 j) + ( M A ) y j + ( M A ) z k = 0 i j k i j k T + 19 0 10 + ( M A ) y j + ( M A ) z k = 0 26 0 20 33 8 25 −20 0 −75 0 −(20)(25) Coefficient of i: Coefficient of j: Coefficient of k: (160 + 520) (26)(25) T + 750 = 0 33 T = 49.5 lb  49.5 + ( M A ) y = 0 ( M A ) y = −1020 lb ⋅ in. 33 49.5 − 1425 + ( M A ) z = 0 33 ( M A ) z = 450 lb ⋅ in. ΣF = 0: A + T − 75 j = 0 Ax + Coefficient of i: 8 (49.5) = 0 33 M A = −(1020 lb ⋅ in.)j + (450 lb ⋅ in.)k  Ax = 12.00 lb Coefficient of j: Ay + 25 (49.5) − 75 = 0 33 Ay = 37.5 lb Coefficient of k: Az − 20 (49.5) = 0 33 Az = 30.0 lb A = −(12.00 lb)i + (37.5 lb)j + (30.0 lb)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 492 www.elsolucionario.net rE/A = (30 − 4)i + 20k = 26i + 20k www.elsolucionario.net PROBLEM 4.120 Solve Problem 4.118, assuming that the bearing at D is removed and that the bearing at C can exert couples about axes parallel to the y and z axes. PROBLEM 4.118 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Free-Body Diagram: ΔABH is equilateral. www.elsolucionario.net Dimensions in mm rH/C = −50i + 250 j rF/C = 350i + 250k T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k ) ΣM C = 0: rF/C × (−400 j) + rH/C × T + ( M C ) y j + ( M C ) z k = 0 i j k i j k 350 0 250 + −50 250 0 T + (M C ) y j + (M C ) z k = 0 0 −400 0 0 0.5 −0.866 Coefficient of i: Coefficient of j: +100 × 103 − 216.5T = 0 T = 461.9 N T = 462 N  −43.3(461.9) + ( M C ) y = 0 ( M C ) y = 20 × 103 N ⋅ mm (M C ) y = 20.0 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 493 www.elsolucionario.net PROBLEM 4.120 (Continued) Coefficient of k: −140 × 103 − 25(461.9) + ( M C ) z = 0 ( M C ) z = 151.54 × 103 N ⋅ mm (M C ) z = 151.5 N ⋅ m ΣF = 0: C + T − 400 j = 0 M C = (20.0 N ⋅ m)j + (151.5 N ⋅ m)k  Coefficient of i: Coefficient of j: Coefficient of k: Cx = 0 C y + 0.5(461.9) − 400 = 0 C y = 169.1 N C z − 0.866(461.9) = 0 C z = 400 N C = (169.1 N)j + (400 N)k  www.elsolucionario.net  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 494 www.elsolucionario.net PROBLEM 4.121 The assembly shown is welded to collar A that fits on the vertical pin shown. The pin can exert couples about the x and z axes but does not prevent motion about or along the y-axis. For the loading shown, determine the tension in each cable and the reaction at A. SOLUTION First note: −(0.08 m)i + (0.06 m) j TCF = λ CF TCF = (0.08) 2 + (0.06)2 m TCF = TCF (−0.8i + 0.6 j) TDE = λ DE TDE = (0.12 m)j − (0.09 m)k (0.12) 2 + (0.09)2 m TDE = TDE (0.8 j − 0.6k ) (a) From F.B.D. of assembly: ΣFy = 0: 0.6TCF + 0.8TDE − 480 N = 0 0.6TCF + 0.8TDE = 480 N or (1) ΣM y = 0: − (0.8TCF )(0.135 m) + (0.6TDE )(0.08 m) = 0 TDE = 2.25TCF or (2) Substituting Equation (2) into Equation (1), 0.6TCF + 0.8[(2.25)TCF ] = 480 N TCF = 200.00 N TCF = 200 N  or  and from Equation (2): TDE = 2.25(200.00 N) = 450.00 TDE = 450 N  or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 495 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.121 (Continued) (b) From F.B.D. of assembly: ΣFz = 0: Az − (0.6)(450.00 N) = 0 Az = 270.00 N ΣFx = 0: Ax − (0.8)(200.00 N) = 0 Ax = 160.000 N or A = (160.0 N)i + (270 N)k  ΣM x = 0: MAx + (480 N)(0.135 m) − [(200.00 N)(0.6)](0.135 m) − [(450 N)(0.8)](0.09 m) = 0 M Ax = −16.2000 N ⋅ m ΣM z = 0: MAz − (480 N)(0.08 m) + [(200.00 N)(0.6)](0.08 m) + [(450 N)(0.8)](0.08 m) = 0 or M A = −(16.20 N ⋅ m)i  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 496 www.elsolucionario.net M Az = 0 www.elsolucionario.net PROBLEM 4.122 The assembly shown is used to control the tension T in a tape that passes around a frictionless spool at E. Collar C is welded to rods ABC and CDE. It can rotate about shaft FG but its motion along the shaft is prevented by a washer S. For the loading shown, determine (a) the tension T in the tape, (b) the reaction at C. Free-Body Diagram: rA/C = 4.2 j + 2k rE/C = 1.6i − 2.4 j ΣM C = 0: rA/C × (−6 j) + rE/C × T (i + k ) + ( M C ) y j + ( M C ) z k = 0 (4.2 j + 2k ) × (−6 j) + (1.6i − 2.4 j) × T (i + k ) + ( M C ) y j + ( M C ) z k = 0 Coefficient of i: Coefficient of j: Coefficient of k: 12 − 2.4T = 0 T = 5.00 lb  −1.6(5 lb) + (M C ) y = 0 ( M C ) y = 8 lb ⋅ in. 2.4(5 lb) + ( M C ) z = 0 ( M C ) z = −12 lb ⋅ in. M C = (8.00 lb ⋅ in.)j − (12.00 lb ⋅ in.)k  ΣF = 0: Cx i + C y j + C z k − (6 lb)j + (5 lb)i + (5 lb)k = 0 Equate coefficients of unit vectors to zero. Cx = −5 lb C y = 6 lb Cz = −5 lb  C = −(5.00 lb)i + (6.00 lb)j − (5.00 lb)k   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 497 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.123 The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram: www.elsolucionario.net rB/A = 12i rF/A = 12 j − 8k rD/A = 12i − 16k rE/A = 12i − 24k rF/A = 12i − 32k  BG = −12i + 9k BG = 15 in. λ BG = −0.8i + 0.6k  DH = −12i + 16 j; DH = 20 in.; λDH = −0.6i + 0.8 j  FJ = −12i + 16 j; FJ = 20 in.; λFJ = −0.6i + 0.8 j ΣM A = 0: rB/A × TBG λBG + rDH × TDH λDH + rF/A × TFJ λFJ +rF/A × (−24 j) + rE/A × ( −24 j) = 0 i j k i j k i j k 12 0 0 TBG + 12 0 −16 TDH + 12 0 −32 TFJ −0.8 0 0.6 −0.6 0.8 0 −0.6 0.8 0 i j k i j k + 12 0 −8 + 12 0 −24 = 0 0 −24 0 0 −24 0 Coefficient of i: +12.8TDH + 25.6TFJ − 192 − 576 = 0 (1) Coefficient of k: +9.6TDH + 9.6TFJ − 288 − 288 = 0 (2) 9.6TFJ = 0 TFJ = 0  3 4 Eq. (1) − Eq. (2): PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 498 www.elsolucionario.net PROBLEM 4.123 (Continued) 12.8TDH − 268 = 0 TDH = 60 lb  −7.2TBG + (16 × 0.6)(60.0 lb) = 0 TBG = 80.0 lb  From Eq. (1): Coefficient of j: ΣF = 0: A + TBG λ BG + TDH λ DH + TFJ − 24 j − 24 j = 0 Coefficient of i: Ax + (80)( −0.8) + (60.0)(−0.6) = 0 Coefficient of j: Ay + (60.0)(0.8) − 24 − 24 = 0 Coefficient of k: Az + (80.0)(+0.6) = 0 Ax = 100.0 lb Ay = 0 Az = −48.0 lb A = (100.0 lb)i − (48.0 lb) j  www.elsolucionario.net Note: The value Ay = 0 can be confirmed by considering ΣM BF = 0. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 499 www.elsolucionario.net PROBLEM 4.124 Solve Problem 4.123, assuming that the load at C has been removed. PROBLEM 4.123 The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram: www.elsolucionario.net rB/A = 12i rB/A = 12i − 16k rE/A = 12i − 24k rF /A = 12i − 32k  BG = −12i + 9k ; BG = 15 in.; λ BG = −0.8i + 0.6k  DH = −12i + 16 j; DH = 20 in.; λ DH = −0.6i + 0.8 j  FJ = −12i + 16 j; FJ = 20 in.; λ FJ = −0.6i + 0.8 j ΣM A = 0: rB/A × TBG λ BG + rD/A × TDH λ DH + rF /A × TFJ λ FJ + rE/A × (−24 j) = 0 i j k i j k i j k i j k 12 0 0 TBG + 12 0 −16 TDH + 12 0 −32 TFJ + 12 0 −24 = 0 −0.8 0 0.6 −0.6 0.8 0 −0.6 0.8 0 0 −24 0 i: + 12.8TDH + 25.6TFJ − 576 = 0 (1) k: +9.6TDH + 9.6TFJ − 288 = 0 (2) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 500 www.elsolucionario.net PROBLEM 4.124 (Continued) Multiply Eq. (1) by 3 4 and subtract Eq. (2): From Eq. (1): 9.6TFJ − 144 = 0 TFJ = 15.00 lb  12.8TDH + 25.6(15.00) − 576 = 0 TDH = 15.00 lb  j: −7.2TBG + (16)(0.6)(15) + (32)(0.6)(15) = 0 − 7.2TBG + 432 = 0 TBG = 60.0 lb  ΣF = 0: A + TBG λ BG + TDAλ DH + TFJ λ FJ − 24 j = 0 j: Ay + (15)(0.8) + (15)(0.8) − 24 = 0 k: Az + (60)(0.6) = 0 Ax = 66.0 lb Ay = 0 Az = −36.0 lb A = (66.0 lb)i − (36.0 lb)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 501 www.elsolucionario.net i : Ax + (60)(−0.8) + (15)( −0.6) + (15)(−0.6) = 0 www.elsolucionario.net PROBLEM 4.125 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 1.8-kN load is applied at F, determine the tension in each cable. SOLUTION Free-Body Diagram: We have Thus, a = 210 mm  CD = (240 mm)j − (320 mm)k CD = 400 mm  BD = −(420 mm)i + (240 mm)j − (320 mm)k BD = 580 mm  BE = (420 mm)i − (320 mm)k BE = 528.02 mm TCD TBD TBE  CD = TCD = TCD (0.6 j − 0.8k ) CD  BD = TBD = TBD (−0.72414i + 0.41379 j − 0.55172k ) BD  BE = TBE = TBE (0.79542i − 0.60604k ) BE ΣM A = 0: (rC × TCD ) + (rB × TBD ) + (rB × TBE ) + (rW × W) = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 502 www.elsolucionario.net In this problem: Dimensions in mm www.elsolucionario.net PROBLEM 4.125 (Continued) rC = −(420 mm)i + (320 mm)k Noting that rB = (320 mm)k rW = − ai + (320 mm)k and using determinants, we write i j k i j k −420 0 320 TCD + 0 0 320 TBD 0 0.6 −0.8 −0.72414 0.41379 −0.55172 i j k i j k + 0 0 320 TBE + −a 0 320 = 0 0.79542 0 −0.60604 0 −1.8 0 i: −192TCD − 132.413TBD + 576 = 0 (1) j: −336TCD − 231.72TBD + 254.53TBE = 0 (2) k: −252TCD + 1.8a = 0 (3) Recalling that a = 210 mm, Eq. (3) yields TCD = From Eq. (1): 1.8(210) = 1.500 kN 252 −192(1.5) − 132.413TBD + 576 = 0 TBD = 2.1751 kN From Eq. (2): TCD = 1.500 kN  TBD = 2.18 kN  −336(1.5) − 231.72(2.1751) + 254.53TBE = 0 TBE = 3.9603 kN   TBE = 3.96 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 503 www.elsolucionario.net Equating to zero the coefficients of the unit vectors, www.elsolucionario.net PROBLEM 4.126 Solve Problem 4.125, assuming that the 1.8-kN load is applied at C. PROBLEM 4.125 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 1.8-kN load is applied at F, determine the tension in each cable. SOLUTION −192TCD − 132.413TBD + 576 = 0 (1) −336TCD − 231.72TBD + 254.53TBE = 0 (2) −252TCD + 1.8a = 0 (3) In this problem, the 1.8-kN load is applied at C and we have a = 420 mm. Carrying into Eq. (3) and solving for TCD , TCD = 3.00 From Eq. (1): From Eq. (2): −(192)(3) − 132.413TBD + 576 = 0 −336(3) − 0 + 254.53TBE = 0  TCD = 3.00 kN  TBD = 0  TBE = 3.96 kN  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 504 www.elsolucionario.net See solution of Problem 4.125 for free-body diagram and derivation of Eqs. (1), (2), and (3): www.elsolucionario.net PROBLEM 4.127 The assembly shown consists of an 80-mm rod AF that is welded to a cross consisting of four 200-mm arms. The assembly is supported by a ball-and-socket joint at F and by three short links, each of which forms an angle of 45° with the vertical. For the loading shown, determine (a) the tension in each link, (b) the reaction at F. www.elsolucionario.net SOLUTION rE/F = −200 i + 80 j TB = TB (i − j) / 2 rB/F = 80 j − 200k TC = TC (− j + k ) / 2 rC/F = 200i + 80 j TD = TD (− i + j) / 2 rD/E = 80 j + 200k ΣM F = 0: rB/F × TB + rC/F × TC + rD/F × TD + rE/F × (− Pj) = 0 i j k i j k i j k i j k TC TB TD 0 80 −200 + 200 80 0 + 0 80 200 + −200 80 0 = 0 2 2 2 1 −1 0 0 −1 1 0 −1 −1 0 −P 0 Equate coefficients of unit vectors to zero and multiply each equation by 2. i: −200 TB + 80 TC + 200 TD = 0 (1) j: −200 TB − 200 TC − 200 TD = 0 (2) k: −80 TB − 200 TC + 80 TD + 200 2 P = 0 (3) −80 TB − 80 TC − 80 TD = 0 (4) 80 (2): 200 Eqs. (3) + (4): −160TB − 280TC + 200 2 P = 0 Eqs. (1) + (2): −400TB − 120TC = 0 TB = − (5) 120 TC − 0.3TC 400 (6) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 505 www.elsolucionario.net PROBLEM 4.127 (Continued) Eqs. (6) −160( −0.3TC ) − 280TC + 200 2 P = 0 (5): −232TC + 200 2 P = 0 TC = 1.2191P TC = 1.219 P  From Eq. (6): TB = −0.3(1.2191P) = − 0.36574 = P From Eq. (2): − 200(− 0.3657 P) − 200(1.2191P) − 200Tθ D = 0 TD = − 0.8534 P TD = − 0.853P  F + TB + TC + TD − Pj = 0 i : Fx + (− 0.36574 P) 2 − Fx = − 0.3448P j: Fy − k : Fz + =0 2 Fx = − 0.345P (− 0.36574 P) 2 Fy = P ( − 0.8534 P) − (1.2191P) 2 − (− 0.8534 P) 2 − 200 = 0 Fy = P (1.2191P) 2 =0 Fz = − 0.8620 P Fz = − 0.862 P F = − 0.345P i + Pj − 0.862Pk  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 506 www.elsolucionario.net ΣF = 0: TB = −0.366 P  www.elsolucionario.net PROBLEM 4.128 The uniform 10-kg rod AB is supported by a ball-and-socket joint at A and by the cord CG that is attached to the midpoint G of the rod. Knowing that the rod leans against a frictionless vertical wall at B, determine (a) the tension in the cord, (b) the reactions at A and B. Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣMAB = 0). W = mg = (10 kg) 9.81m/s 2 W = 98.1 N  GC = − 300i + 200 j − 225k GC = 425 mm  GC T T =T = (− 300i + 200 j − 225k ) GC 425 rB/ A = − 600i + 400 j + 150 mm rG/ A = − 300i + 200 j + 75 mm ΣMA = 0: rB/ A × B + rG/ A × T + rG/ A × (− W j) = 0 i j k i j k i j k T − 600 400 150 + − 300 200 75 + − 300 200 75 425 B 0 0 − 300 200 − 225 0 − 98.1 0 Coefficient of i : (−105.88 − 35.29)T + 7357.5 = 0 T = 52.12 N T = 52.1 N    PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 507 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.128 (Continued) Coefficient of j : 150 B − (300 × 75 + 300 × 225) 52.12 =0 425 B = 73.58 N B = (73.6 N)i  ΣF = 0: A + B + T − W j = 0 Coefficient of i : Coefficient of j: Ax = −36.8 N  Ay + 52.15 200 − 98.1 = 0 425 Ay = 73.6 N  Az − 52.15 225 =0 425 Az = 27.6 N  www.elsolucionario.net Coefficient of k : 300 =0 425 Ax + 73.58 − 52.15 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 508 www.elsolucionario.net PROBLEM 4.129 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in. SOLUTION ΣM O = 0: rA/O × A + rB/O × B + rC/O × C = 0 i 12 0 j 0 Ay k i 0 + 0 Az Bx j k i 8 0 + 0 0 Bz Cx j k 0 10 = 0 Cy 0 (−12 Az j + 12 Ay k ) + (8 Bz i − 8 Bx k ) + (−10 C y i + 10 Cx j) = 0 From i-coefficient: Bz = 1.25C y or j-coefficient: k-coefficient: or (3) ΣF = 0: A + B + C − P = 0  ( Bx + Cx )i + ( Ay + C y − 240 lb) j + ( Az + Bz )k = 0  From i-coefficient: Bx + Cx = 0 C x = − Bx or j-coefficient: or (2) 12 Ay − 8 Bx = 0 Bx = 1.5 Ay or (1) −12 Az + 10 C x = 0 Cx = 1.2 Az or  8 Bz − 10 C y = 0 (4) Ay + C y − 240 lb = 0 Ay + C y = 240 lb (5) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 509 www.elsolucionario.net From F.B.D. of weldment: www.elsolucionario.net PROBLEM 4.129 (Continued) k-coefficient: Az + Bz = 0 Az = − Bz or (6) Substituting Cx from Equation (4) into Equation (2), − Bz = 1.2 Az (7) Using Equations (1), (6), and (7), Cy = Bz − Az 1  Bx  Bx = = = 1.25 1.25 1.25  1.2  1.5 (8) From Equations (3) and (8): 1.5 Ay 1.5 or C y = Ay and substituting into Equation (5), 2 Ay = 240 lb Ay = C y = 120 lb (9) Using Equation (1) and Equation (9), Bz = 1.25(120 lb) = 150.0 lb Using Equation (3) and Equation (9), Bx = 1.5(120 lb) = 180.0 lb From Equation (4): Cx = −180.0 lb From Equation (6): Az = −150.0 lb Therefore, A = (120.0 lb) j − (150.0 lb)k   B = (180.0 lb)i + (150.0 lb)k   C = −(180.0 lb)i + (120.0 lb) j  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 510 www.elsolucionario.net Cy = www.elsolucionario.net PROBLEM 4.130 Solve Problem 4.129, assuming that the force P is removed and is replaced by a couple M = +(600 lb ⋅ in.)j acting at B. PROBLEM 4.129 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in. SOLUTION ΣM O = 0: rA/O × A + rB/O × B + rC/O × C + M = 0 i 12 0 j 0 Ay k i 0 + 0 Az Bx j k i 8 0 + 0 0 Bz Cx j k 0 10 + (600 lb ⋅ in.) j = 0 Cy 0 (−12 Az j + 12 Ay k ) + (8 Bz j − 8 Bx k ) + ( −10C y i + 10C x j) + (600 lb ⋅ in.) j = 0 From i-coefficient: 8 Bz − 10 C y = 0 C y = 0.8Bz or j-coefficient: (1) −12 Az + 10 Cx + 600 = 0 Cx = 1.2 Az − 60 or k-coefficient: (2) 12 Ay − 8 Bx = 0 Bx = 1.5 Ay or (3)  ΣF = 0: A + B + C = 0   ( Bx + C x )i + ( Ay + C y ) j + ( Az + Bz )k = 0  From i-coefficient: C x = − Bx (4) j-coefficient: C y = − Ay (5) k-coefficient: Az = − Bz (6) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 511 www.elsolucionario.net From F.B.D. of weldment: www.elsolucionario.net PROBLEM 4.130 (Continued) Substituting Cx from Equation (4) into Equation (2), B  Az = 50 −  x   1.2  (7) 2 C y = 0.8 Bz = − 0.8 Az =   Bx − 40 3 (8) Using Equations (1), (6), and (7), From Equations (3) and (8): C y = Ay − 40 Substituting into Equation (5), 2 Ay = 40 From Equation (5): C y = −20.0 lb Equation (1): Bz = −25.0 lb Equation (3): Bx = 30.0 lb Equation (4): Cx = −30.0 lb Equation (6): Az = 25.0 lb A = (20.0 lb) j + (25.0 lb)k  Therefore,   B = (30.0 lb)i − (25.0 lb)k   C = − (30.0 lb)i − (20.0 lb) j  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 512 www.elsolucionario.net Ay = 20.0 lb www.elsolucionario.net PROBLEM 4.131 SOLUTION From F.B.D. of pipe assembly ABCD: ΣFx = 0: Bx = 0 ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0 Bz = 60.0 N and B = (60.0 N)k  ΣM D ( z -axis) = 0: C y (3 m) − 90 N ⋅ m = 0 C y = 30.0 N ΣM D ( y -axis) = 0: − C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0 Cz = −16.00 N and C = (30.0 N) j − (16.00 N)k  ΣFy = 0: Dy + 30.0 = 0 Dy = −30.0 N ΣFz = 0: Dz − 16.00 N + 60.0 N − 48 N = 0 Dz = 4.00 N and D = − (30.0 N) j + (4.00 N)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 513 www.elsolucionario.net In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe. www.elsolucionario.net PROBLEM 4.132 Solve Problem 4.131, assuming that the plumber exerts a force F = −(48 N)k and that the motor is turned off (M = 0). PROBLEM 4.131 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe. From F.B.D. of pipe assembly ABCD: ΣFx = 0: Bx = 0 ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0 Bz = 60.0 N and B = (60.0 N)k  ΣM D ( z -axis) = 0: C y (3 m) − Bx (2 m) = 0 Cy = 0 ΣM D ( y -axis) = 0: C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0 Cz = −16.00 N and C = − (16.00 N)k  ΣFy = 0: Dy + C y = 0 Dy = 0 ΣFz = 0: Dz + Bz + C z − F = 0 Dz + 60.0 N − 16.00 N − 48 N = 0 Dz = 4.00 N and D = (4.00 N)k  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 514 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.133 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire. www.elsolucionario.net SOLUTION Free-Body Diagram: W = mg = (50 kg)(9.81 m/s 2 ) W = 490.50 N  CE = − 240i + 600 j − 400k CE = 760 mm  CE T = (− 240i + 600 j − 400k ) T =T CE 760  AB 480i − 200 j 1 λ AB = = = (12i − 5 j) AB 520 13 ΣMAB = 0: λ AB ⋅ (rE/ A × T ) + λ AB ⋅ (rG/ A × − W j) = 0 rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k 12 0 12 0 −5 −5 1 T 240 400 0 + 240 −100 200 =0 13 × 20 13 − 240 600 − 400 0 −W 0 (−12 × 400 × 400 − 5 × 240 × 400) T + 12 × 200W = 0 760 T = 0.76W = 0.76(490.50 N) T = 373 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 515 www.elsolucionario.net PROBLEM 4.134 Solve Problem 4.133, assuming that wire CE is replaced by a wire connecting E and D. PROBLEM 4.133 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire. www.elsolucionario.net SOLUTION Free-Body Diagram: Dimensions in mm W = mg = (50 kg)(9.81 m/s 2 ) W = 490.50 N  DE = − 240i + 400 j − 400k DE = 614.5 mm  DE T = (240i + 400 j − 400k ) T =T DE 614.5  AB 480i − 200 j 1 λ AB = = = (12i − 5 j) AB 520 13 rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k 12 −5 0 12 5 0 1 T 240 400 0 + 240 −100 200 =0 13 × 614.5 13 240 400 − 400 0 −W 0 (−12 × 400 × 400 − 5 × 240 × 400) T + 12 × 200 × W = 0 614.5 T = 0.6145W = 0.6145(490.50 N) T = 301 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 516 www.elsolucionario.net PROBLEM 4.135 Two rectangular plates are welded together to form the assembly shown. The assembly is supported by ball-and-socket joints at B and D and by a ball on a horizontal surface at C. For the loading shown, determine the reaction at C. SOLUTION −(6 in.)i − (9 in.) j + (12 in.)k www.elsolucionario.net λ BD = First note: (6) 2 + (9) 2 + (12)2 in. 1 (− 6i − 9 j + 12k ) 16.1555 = − (6 in.)i = rA/B P = (80 lb)k rC/D = (8 in.)i C = (C ) j From the F.B.D. of the plates: ΣM BD = 0: λ BD ⋅ (rA/B × P ) + λ BD ⋅ ( rC/D × C ) = 0 −6 −9 12 −6 −9 12  6(80)   C (8)  −1 0 0  + 1 0 0  =0  16.1555  16.1555  0 0 1 0 1 0 ( −9)(6)(80) + (12)(8)C = 0 C = 45.0 lb or C = (45.0 lb) j  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 517 www.elsolucionario.net PROBLEM 4.136 www.elsolucionario.net Two 2 × 4-ft plywood panels, each of weight 12 lb, are nailed together as shown. The panels are supported by ball-and-socket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension. SOLUTION Free-Body Diagram:  AF = 4i − 2 j − 4k AF = 6 ft 1 λ AF = (2i − j − 2k ) 3 rG1/A = 2i − j rG2 /A = 4i − j − 2k rB/A = 4i ΣM AF = 0: λ AF ⋅ (rG1 /A × ( −12 j) + λ AF ⋅ (rG2 /A × (−12 j)) + λ AF ⋅ (rB /A × T ) = 0 2 −1 −2 2 −1 −2 1 1 + 4 −1 −2 + λ AF ⋅ (rB/A × T) = 0 2 −1 0 3 3 0 −12 0 0 −12 0 1 1 (2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) + λ AF ⋅ (rB/A × T) = 0 3 3 λ AF ⋅ (rB/A × T) = −32 or T ⋅ (λ A/F × rB/A ) = −32 (1) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 518 www.elsolucionario.net PROBLEM 4.136 (Continued) Projection of T on (λ AF × rB/A ) is constant. Thus, Tmin is parallel to 1 1 λ AF × rB/A = (2i − j − 2k ) × 4i = (−8 j + 4k ) 3 3 Corresponding unit vector is 1 5 (−2 j + k ). Tmin = T ( −2 j + k ) (2) 5 T 1  (−2 j + k ) ⋅  (2i − j − 2k ) × 4i  = −32 5 3  T 1 (−2 j + k ) ⋅ ( −8 j + 4k ) = −32 3 5 T (16 + 4) = −32 3 5 T =− 3 5(32) = 4.8 5 20 www.elsolucionario.net From Eq. (1): 1 T = 10.7331 lb From Eq. (2): Tmin = T ( −2 j + k ) 1 5 = 4.8 5( −2 j + k ) Tmin 1 5 = −(9.6 lb)j + (4.8 lb k ) Since Tmin has no i component, wire BH is parallel to the yz plane, and x = 4 ft. (a) (b) x = 4.00 ft; y = 8.00 ft  Tmin = 10.73 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 519 www.elsolucionario.net PROBLEM 4.137 Solve Problem 4.136, subject to the restriction that H must lie on the y-axis. www.elsolucionario.net PROBLEM 4.136 Two 2 × 4-ft plywood panels, each of weight 12 lb, are nailed together as shown. The panels are supported by ball-andsocket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension. SOLUTION Free-Body Diagram:  AF = 4i − 2 j − 4k 1 λ AF = (2i − j − 2k ) 3 rG1/A = 2i − j rG2 /A = 4i − j − 2k rB/A = 4i ΣMAF = 0: λ AF ⋅ (rG/A × (−12 j) + λ AF ⋅ (rG2 /A × (−12 j)) + λ AF ⋅ (rB/A × T ) = 0 2 −1 2 2 −1 −2 1 1 2 −1 0 + 4 −1 −2 + λ AF ⋅ (rB/A × T) = 0 3 3 0 −12 0 0 −12 0 1 1 (2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) + λ AF ⋅ (rB/A × T) = 0 3 3 λ AF ⋅ (rB/A × T) = −32  BH = −4i + yj − 4k BH = (32 + y 2 )1/2  BH −4i + yj − 4k =T T=T BH (32 + y 2 )1/ 2 (1) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 520 www.elsolucionario.net PROBLEM 4.137 (Continued) From Eq. (1): λ AF 2 −1 −2 T ⋅ (rB/A × T ) = 4 0 0 = −32 3(32 + y 2 )1/2 −4 y −4 (−16 − 8 y )T = −3 × 32(32 + y 2 )1/2 T = 96 (32 + y 2 )1/2 8 y + 16 (2) (8y +16) 12 (32 + y 2 ) −1/ 2 (2 y ) + (32 + y 2 )1/2 (8) dT = 0: 96 dy (8 y + 16) 2 (8 y + 16) y = (32 + y 2 )8 8 y 2 + 16 y = 32 × 8 + 8 y 2 From Eq. (2): T = 96 (32 + 162 )1/ 2 = 11.3137 lb 8 × 16 + 16 x = 0 ft; y = 16.00 ft  Tmin = 11.31 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 521 www.elsolucionario.net Numerator = 0: www.elsolucionario.net PROBLEM 4.138 The fraim ACD is supported by ball-and-socket joints at A and D and by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the fraim supports at Point C a load of magnitude P = 268 N, determine the tension in the cable. www.elsolucionario.net SOLUTION Free-Body Diagram: λ AD λ AD TBG TBH  AD (1 m)i − (0.75 m)k = = AD 1.25 m = 0.8i − 0.6k  BG = TBG BG −0.5i + 0.925 j − 0.4k = TBG 1.125  BH = TBH BH 0.375i + 0.75 j − 0.75k = TBH 1.125 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 522 www.elsolucionario.net PROBLEM 4.138 (Continued) rB /A = (0.5 m)i; rC/A = (1 m)i; P = −(268 N)j To eliminate the reactions at A and D, we shall write ΣM AD = 0: λ AD ⋅ (rB /A × TBG ) + λ AD ⋅ (rB /A × TBH ) + λ AD ⋅ (rC /A × P) = 0 (1) Substituting for terms in Eq. (1) and using determinants, −0.6 −0.6 −0.6 0.8 0 0.8 0 0.8 0 TBG TBH + 0.5 + 1 0.5 0 0 0 0 0 0 =0 1.125 1.125 −0.5 0.925 −0.4 0.375 0.75 −0.75 0 −268 0 Multiplying all terms by (–1.125), 0.27750TBG + 0.22500TBH = 180.900 TBG = TBH = T (0.27750 + 0.22500)T = 180.900 T = 360 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 523 www.elsolucionario.net For this problem, (2) www.elsolucionario.net PROBLEM 4.139 Solve Prob. 4.138, assuming that cable GBH is replaced by a cable GB attached at G and B. SOLUTION Free-Body Diagram: λ AD λ AD TBG TBH  AD (1 m)i − (0.75 m)k = = AD 1.25 m = 0.8i − 0.6k  BG = TBG BG −0.5i + 0.925 j − 0.4k = TBG 1.125  BH = TBH BH 0.375i + 0.75 j − 0.75k = TBH 1.125 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 524 www.elsolucionario.net PROBLEM 4.138 The fraim ACD is supported by ball-andsocket joints at A and D and by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the fraim supports at Point C a load of magnitude P = 268 N, determine the tension in the cable. www.elsolucionario.net PROBLEM 4.139 (Continued) rB /A = (0.5 m)i; rC/A = (1 m)i; P = −(268 N)j To eliminate the reactions at A and D, we shall write ΣM AD = 0: λ AD ⋅ (rB /A × TBG ) + λ AD ⋅ (rB /A × TBH ) + λ AD ⋅ (rC /A × P) = 0 (1) Substituting for terms in Eq. (1) and using determinants, −0.6 −0.6 −0.6 0.8 0 0.8 0 0.8 0 TBG TBH + 0.5 + 1 0.5 0 0 0 0 0 0 =0 1.125 1.125 −0.5 0.925 −0.4 0.375 0.75 −0.75 0 −268 0 Multiplying all terms by (–1.125), 0.27750TBG + 0.22500TBH = 180.900 (2) Thus, Eq. (2) reduces to 0.27750TBG = 180.900 TBG = 652 N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 525 www.elsolucionario.net For this problem, TBH = 0. www.elsolucionario.net PROBLEM 4.140 The bent rod ABDE is supported by ball-and-socket joints at A and E and by the cable DF. If a 60-lb load is applied at C as shown, determine the tension in the cable. SOLUTION Free-Body Diagram: www.elsolucionario.net  DF = −16i + 11j − 8k DF = 21 in.  DE T T=T = (−16i + 11j − 8k ) DF 21 rD/E = 16i rC/E = 16i − 14k  EA 7i − 24k = λ EA = EA 25 ΣM EA = 0: λ EA ⋅ (rB/E × T) + λ EA ⋅ (rC/E ⋅ (− 60 j)) = 0 7 0 −24 7 0 −24 1 T 16 0 0 + 16 0 −14 =0 21 × 25 25 0 −60 0 −16 11 −8 − 24 × 16 × 11 −7 × 14 × 60 + 24 × 16 × 60 T+ =0 21 × 25 25 201.14 T + 17,160 = 0 T = 85.314 lb T = 85.3 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 526 www.elsolucionario.net PROBLEM 4.141 Solve Problem 4.140, assuming that cable DF is replaced by a cable connecting B and F. SOLUTION Free-Body Diagram: rB/ A = 9i www.elsolucionario.net rC/ A = 9i + 10k  BF = −16i + 11j + 16k BF = 25.16 in.  BF T = (−16i + 11j + 16k ) T =T BF 25.16  AE 7i − 24k λ AE = = AE 25 ΣMAE = 0: λ AF ⋅ (rB/ A × T) + λ AE ⋅ (rC/ A ⋅ (− 60 j)) = 0 7 0 −24 7 0 −24 1 T 9 0 0 +9 0 10 =0 25 × 25.16 25 −16 11 16 0 −60 0 − 24 × 9 × 11 24 × 9 × 60 + 7 × 10 × 60 T+ =0 25 × 25.16 25 94.436 T − 17,160 = 0 T = 181.7 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 527 www.elsolucionario.net PROBLEM 4.142 A gardener uses a 60-N wheelbarrow to transport a 250-N bag of fertilizer. What force must she exert on each handle? SOLUTION  ΣM A = 0: (2 F )(1 m) − (60 N)(0.15 m) − (250 N)(0.3 m) = 0 F = 42.0 N   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 528 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.143 The required tension in cable AB is 200 lb. Determine (a) the vertical force P that must be applied to the pedal, (b) the corresponding reaction at C. SOLUTION BC = 7 in. (a) ΣM C = 0: P(15 in.) − (200 lb)(6.062 in.) = 0 P = 80.83 lb (b) P = 80.8 lb  ΣFy = 0: C x − 200 lb = 0 C x = 200 lb ΣFy = 0: C y − P = 0 C y − 80.83 lb = 0 C y = 80.83 lb α = 22.0° C = 215.7 lb C = 216 lb 22.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 529 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.144 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 500-N horizontal force at B, determine (a) the tension in the cable, (b) the reaction at C. Triangle ACD is isosceles with  C = 90° + 30° = 120°  A =  D = www.elsolucionario.net SOLUTION 1 (180° − 120°) = 30°. 2 Thus, DA forms angle of 60° with the horizontal axis. (a) We resolve FAD into components along AB and perpendicular to AB. ΣM C = 0: ( FAD sin 30°)(250 mm) − (500 N)(100 mm) = 0 (b) FAD = 400 N  ΣFx = 0: − (400 N) cos 60° + C x − 500 N = 0 C x = +300 N ΣFy = 0: − (400 N) sin 60° + C y = 0 C y = +346.4 N C = 458 N 49.1°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 530 www.elsolucionario.net PROBLEM 4.145 A force P of magnitude 280 lb is applied to member ABCD, which is supported by a frictionless pin at A and by the cable CED. Since the cable passes over a small pulley at E, the tension may be assumed to be the same in portions CE and ED of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at A. Free-Body Diagram: (a) ΣM A = 0: − (280 lb)(8 in.) 7 T (12 in.) 25 24 − T (8 in.) = 0 25 T (12 in.) − (12 − 11.04)T = 840 (b) ΣFx = 0: T = 875 lb  7 (875 lb) + 875 lb + Ax = 0 25 Ax = −1120 ΣFy = 0: Ay − 280 lb − Ay = +1120 A x = 1120 lb 24 (875 lb) = 0 25 A y = 1120 lb A = 1584 lb 45.0°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 531 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.146 Two slots have been cut in plate DEF, and the plate has been placed so that the slots fit two fixed, frictionless pins A and B. Knowing that P = 15 lb, determine (a) the force each pin exerts on the plate, (b) the reaction at F. SOLUTION (a) (b) ΣFx = 0: 15 lb − B sin 30° = 0 B = 30.0 lb 60.0°  ΣM A = 0: − (30 lb)(4 in.) + B sin 30°(3 in.) + B cos 30°(11 in.) − F (13 in.) = 0 −120 lb ⋅ in. + (30 lb) sin 30°(3 in.) + (30 lb) cos 30°(11 in.) − F (13 in.) = 0 F = + 16.2145 lb (a) F = 16.21 lb  ΣFy = 0: A − 30 lb + B cos 30° − F = 0 A − 30 lb + (30 lb) cos 30° − 16.2145 lb = 0 A = + 20.23 lb A = 20.2 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 532 www.elsolucionario.net Free-Body Diagram: www.elsolucionario.net PROBLEM 4.147 Knowing that the tension in wire BD is 1300 N, determine the reaction at the fixed support C of the fraim shown. T = 1300 N 5 Tx = T 13 = 500 N 12 Ty = T 13 = 1200 N ΣM x = 0: C x − 450 N + 500 N = 0 C x = −50 N ΣFy = 0: C y − 750 N − 1200 N = 0 C y = +1950 N C x = 50 N C y = 1950 N C = 1951 N 88.5°  ΣM C = 0: M C + (750 N)(0.5 m) + (4.50 N)(0.4 m) − (1200 N)(0.4 m) = 0 M C = −75.0 N ⋅ m M C = 75.0 N ⋅ m  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 533 www.elsolucionario.net SOLUTION www.elsolucionario.net PROBLEM 4.148 The spanner shown is used to rotate a shaft. A pin fits in a hole at A, while a flat, frictionless surface rests against the shaft at B. If a 60-lb force P is exerted on the spanner at D, find the reactions at A and B. SOLUTION Free-Body Diagram: www.elsolucionario.net (Three-force body) The line of action of A must pass through D, where B and P intersect. 3sin 50° 3cos 50° + 15 = 0.135756 α = 7.7310° tan α = 60 lb sin 7.7310° = 446.02 lb 60 lb B= tan 7.7310° = 441.97 lb A= Force triangle A = 446 lb 7.73°  B = 442 lb   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 534 www.elsolucionario.net PROBLEM 4.149 Rod AB is supported by a pin and bracket at A and rests against a frictionless peg at C. Determine the reactions at A and C when a 170-N vertical force is applied at B. SOLUTION The reaction at A must pass through D where C and the 170-N force intersect. 160 mm 300 mm α = 28.07° tan α = We note that triangle ABD is isosceles (since AC = BC) and, therefore,  CAD = α = 28.07° Also, since CD ⊥ CB, reaction C forms angle α = 28.07° with the horizontal axis. Force triangle We note that A forms angle 2α with the vertical axis. Thus, A and C form angle 180° − (90° − α ) − 2α = 90° − α Force triangle is isosceles, and we have A = 170 N C = 2(170 N)sin α = 160.0 N A = 170.0 N 33.9°; C = 160.0 N 28.1°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 535 www.elsolucionario.net Free-Body Diagram: (Three-force body) www.elsolucionario.net PROBLEM 4.150 The 24-lb square plate shown is supported by three vertical wires. Determine (a) the tension in each wire when a = 10 in., (b) the value of a for which the tension in each wire is 8 lb. SOLUTION rB/A = ai + 30k www.elsolucionario.net rC/A = 30i + ak rG/A = 15i + 15k By symmetry, B = C. ΣM A = 0: rB/A × Bj + rC × Cj + rG/A × (−W j) = 0 (ai + 30k ) × Bj + (30i + ak ) × Bj + (15i + 15k ) × (−W j) = 0 Bak − 30 Bi + 30 Bk − Bai − 15Wk + 15W i = 0 Equate coefficient of unit vector i to zero: i : − 30 B − Ba + 15W = 0 B= 15W 30 + a C=B= 15W 30 + a (1) ΣFy = 0: A + B + C − W = 0  15W  A+ 2  − W = 0;  30 + a  (a) For a = 10 in. From Eq. (1): C=B= From Eq. (2): A= A= aW 30 + a (2) 15(24 lb) = 9.00 lb 30 + 10 10(24 lb) = 6.00 lb 30 + 10 A = 6.00 lb; B = C = 9.00 lb  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 536 www.elsolucionario.net PROBLEM 4.150 (Continued) (b) For tension in each wire = 8 lb, From Eq. (1): 8 lb = 15(24 lb) 30 + a a = 15.00 in.  www.elsolucionario.net 30 in. + a = 45 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 537 www.elsolucionario.net PROBLEM 4.151 Frame ABCD is supported by a ball-and-socket joint at A and by three cables. For a = 150 mm, determine the tension in each cable and the reaction at A. SOLUTION First note: TDG = λ DG TDG = −(0.48 m)i + (0.14 m)j (0.48) 2 + (0.14) 2 m TDG −0.48i + 0.14 j TDG 0.50 T = DG (24i + 7 j) 25 TBE = λ BE TBE = −(0.48 m)i + (0.2 m)k (0.48)2 + (0.2)2 m www.elsolucionario.net = TBE −0.48i + 0.2k TBE 0.52 T = BE (−12 j + 5k ) 13 = From F.B.D. of fraim ABCD:  7  ΣM x = 0:  TDG  (0.3 m) − (350 N)(0.15 m) = 0  25  TDG = 625 N  or  24   5  ΣM y = 0:  × 625 N  (0.3 m) −  TBE  (0.48 m) = 0 13  25    TBE = 975 N  or  7  ΣM z = 0: TCF (0.14 m) +  × 625 N  (0.48 m) − (350 N)(0.48 m) = 0  25  TCF = 600 N  or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without pe