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# Solucionario de Mecanica vectorial para ingenieros Beer Johnston Cap 2-5-estatica

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• 1. PROBLEM 2.1 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 75 N and Q = 125 N, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: R = 179 N, α = 75.1° R = 179 N 75.1° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 2. PROBLEM 2.2 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 60 lb and Q = 25 lb, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: R = 77.1 lb, α = 85.4° R = 77.1 lb 85.4° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 3. PROBLEM 2.3 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. SOLUTION We measure: (a) Parallelogram law: (b) α = 51.3° β = 59.0° Triangle rule: We measure: γ = 67.0° R = 139.1 lb, R = 139.1 lb 67.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 4. PROBLEM 2.4 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. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: R = 3.30 kN, α = 66.6° R = 3.30 kN 66.6° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 5. PROBLEM 2.5 The 300-lb force is to be resolved into components along lines a-a′ and b-b′. (a) Determine the angle α by trigonometry knowing that the component along line a-a′ is to be 240 lb. (b) What is the corresponding value of the component along b-b′? SOLUTION (a) Using the triangle rule and law of sines: sin β sin 60° = 240 lb 300 lb sin β = 0.69282 β = 43.854° α + β + 60° = 180° α = 180° − 60° − 43.854° = 76.146° (b) Law of sines: Fbb′ 300 lb = sin 76.146° sin 60° α = 76.1° W Fbb′ = 336 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 6. PROBLEM 2.6 The 300-lb force is to be resolved into components along lines a-a′ and b-b′. (a) Determine the angle α by trigonometry knowing that the component along line b-b′ is to be 120 lb. (b) What is the corresponding value of the component along a-a′? SOLUTION Using the triangle rule and law of sines: (a) sin α sin 60° = 120 lb 300 lb sin α = 0.34641 α = 20.268° (b) α = 20.3° W α + β + 60° = 180° β = 180° − 60° − 20.268° = 99.732° Faa′ 300 lb = sin 99.732° sin 60° Faa′ = 341 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 7. PROBLEM 2.7 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: (a) sin α sin 25° = 50 N 35 N sin α = 0.60374 α = 37.138° (b) α = 37.1° W α + β + 25° = 180° β = 180° − 25° − 37.138° = 117.86° R 35 N = sin117.86 sin 25° R = 73.2 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 8. PROBLEM 2.8 For the hook support of Problem 2.1, knowing that the magnitude of P is 75 N, determine by trigonometry (a) the required magnitude of the force Q if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R. PROBLEM 2.1 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 75 N and Q = 125 N, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the triangle rule and law of sines: (a) (b) Q 75 N = sin 20° sin 35° Q = 44.7 N W α + 20° + 35° = 180° α = 180° − 20° − 35° = 125° R 75 N = sin125° sin 35° R = 107.1 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 9. PROBLEM 2.9 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? SOLUTION Using the triangle rule and the law of sines: (a) (b) 1600 N P = sin 25° sin 75° P = 3660 N W 25° + β + 75° = 180° β = 180° − 25° − 75° = 80° 1600 N R = sin 25° sin 80° R = 3730 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 10. PROBLEM 2.10 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. SOLUTION 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 11. 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. SOLUTION Using the triangle rule and the law of sines: (a) β + 50° + 60° = 180° β = 180° − 50° − 60° = 70° 425 lb P = sin 70° sin 60° (b) P = 392 lb W 425 lb R = sin 70° sin 50° R = 346 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 12. 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. SOLUTION 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 = 47.40° (b) α = 42.6° W R 500 lb = sin (42.6° + 30°) sin 60° R = 551 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 13. PROBLEM 2.13 For the hook support of Problem 2.7, 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 The smallest force P will be perpendicular to R. (a) P = (50 N)sin 25° (b) P = 21.1 N R = (50 N) cos 25° W R = 45.3 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 14. PROBLEM 2.14 For the steel tank of Problem 2.11, 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. 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. SOLUTION The smallest force P will be perpendicular to R. (a) P = (425 lb) cos 30° (b) P = 368 lb R = (425 lb)sin 30° W R = 213 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 15. PROBLEM 2.15 Solve Problem 2.2 by trigonometry. PROBLEM 2.2 Two forces P and Q are applied as shown at Point A of a hook support. Knowing that P = 60 lb and Q = 25 lb, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the triangle rule and the law of cosines: 20° + 35° + α = 180° α = 125° R 2 = P 2 + Q 2 − 2 PQ cos α R 2 = (60 lb)2 + (25 lb) 2 − 2(60 lb)(25 lb) cos125° R 2 = 3600 + 625 + 3000(0.5736) R = 77.108 lb Using the law of sines: sin β sin125° = 25 lb 77.108 lb β = 15.402° 70° + β = 85.402° R = 77.1 lb 85.4° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 16. PROBLEM 2.16 Solve Problem 2.3 by trigonometry. PROBLEM 2.3 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. SOLUTION 8 10 α = 38.66° 6 tan β = 10 β = 30.96° tan α = Using the triangle rule: Using the law of cosines: Using the law of sines: 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 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 17. PROBLEM 2.17 Solve Problem 2.4 by trigonometry. PROBLEM 2.4 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. SOLUTION Using the law of cosines: R 2 = (2 kN)2 + (3 kN)2 − 2(2 kN)(3 kN) cos80° R = 3.304 kN Using the law of sines: sin γ sin 80° = 2 kN 3.304 kN γ = 36.59° β + γ + 80° = 180° γ = 180° − 80° − 36.59° γ = 63.41° φ = 180° − β + 50° φ = 66.59° R = 3.30 kN 66.6° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 18. PROBLEM 2.18 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 − 2(15 kN)(10 kN) cos120° = 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 ¹ = 0.39737 α = 23.414 φ = α + 50° = 73.414 R = 21.8 kN 73.4° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 19. PROBLEM 2.19 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 10 kN in member A and 15 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 = (10 kN) 2 + (15 kN)2 − 2(10 kN)(15 kN) cos120° = 475 kN 2 R = 21.794 kN and Hence: 15 kN 21.794 kN = sin α sin120° § 15 kN · sin α = ¨ ¸ sin120° © 21.794 kN ¹ = 0.59605 α = 36.588° φ = α + 50° = 86.588° R = 21.8 kN 86.6° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 20. PROBLEM 2.20 For the hook support of Problem 2.7, 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 2.7 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°) = 105° Then R 2 = (75 N) 2 + (50 N) 2 − 2(75 N)(50 N) cos 105° 2 R = 10066.1 N 2 R = 100.330 N 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 21. PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: OA = (600) 2 + (800) 2 = 1000 mm OB = (560)2 + (900) 2 = 1060 mm OC = (480) 2 + (900)2 = 1020 mm Fx = +640 N W 600 1000 Fy = +480 N W Fx = −(424 N) 560 1060 Fx = −224 N W 900 1060 Fy = −360 N W Fx = + (408 N) 480 1020 Fx = +192.0 N W Fy = −(408 N) 408-N Force: 800 1000 Fy = −(424 N) 424-N Force: Fx = + (800 N) Fy = +(800 N) 800-N Force: 900 1020 Fy = −360 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 22. PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: OA = (84) 2 + (80) 2 = 116 in. OB = (28)2 + (96)2 = 100 in. OC = (48)2 + (90)2 = 102 in. Fx = +21.0 lb W 80 116 Fy = +20.0 lb W Fx = −(50 lb) 28 100 Fx = −14.00 lb W 96 100 Fy = + 48.0 lb W Fx = + (51 lb) 48 102 Fx = +24.0 lb W Fy = −(51 lb) 51-lb Force: 84 116 Fy = +(50 lb) 50-lb Force: Fx = + (29 lb) Fy = +(29 lb) 29-lb Force: 90 102 Fy = −45.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 23. PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION Fy = −34.6 lb W Fx = −(50 lb)sin 50° Fx = −38.3 lb W Fy = −32.1 lb W Fx = + (60 lb) cos 25° Fx = 54.4 lb W Fy = +(60 lb)sin 25° 60-lb Force: Fx = 20.0 lb W Fy = −(50 lb) cos 50° 50-lb Force: Fx = + (40 lb) cos 60° Fy = −(40 lb)sin 60° 40-lb Force: Fy = 25.4 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 24. PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION Fy = 51.4 N W Fx = + (120 N) cos 70° Fx = 41.0 N W Fy = 112.8 N W Fx = −(150 N) cos 35° Fx = −122. 9 N W Fy = +(150 N) sin 35° 150-N Force: Fx = 61.3 N W Fy = +(120 N) sin 70° 120-N Force: Fx = + (80 N) cos 40° Fy = + (80 N) sin 40° 80-N Force: Fy = 86.0 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 25. PROBLEM 2.25 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. SOLUTION P sin 35° = 300 lb (a) P= (b) Vertical component 300 lb sin 35° P = 523 lb W Pv = P cos 35° = (523 lb) cos 35° Pv = 428 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 26. PROBLEM 2.26 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. SOLUTION (a) 750 N = P sin 20° P = 2193 N (b) P = 2190 N W PABC = P cos 20° = (2193 N) cos 20° PABC = 2060 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 27. PROBLEM 2.27 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 120-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC. SOLUTION (a) Px sin 38° 120 N = sin 38° P= = 194.91 N (b) or P = 194.9 N W or Py = 153.6 N W Px tan 38° 120 N = tan 38° Py = = 153.59 N PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 28. PROBLEM 2.28 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P has a 180-N component along line AC, determine (a) the magnitude of the force P, (b) its component in a direction perpendicular to AC. SOLUTION (a) P= Py cos 38° 180 N = cos 38° = 228.4 N (b) P = 228 N W Px = Py tan 38° = (180 N) tan 38° = 140.63 N Px = 140.6 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 29. PROBLEM 2.29 Member CB of the vise shown exerts on block B a force P directed along line CB. Knowing that P must have a 1200-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION We note: CB exerts force P on B along CB, and the horizontal component of P is Px = 1200 N: Then (a) Px = P sin 55° Px sin 55° 1200 N = sin 55° = 1464.9 N P= (b) P = 1465 N W Px = Py tan 55° Px tan 55° 1200 N = tan 55° = 840.2 N Py = Py = 840 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 30. PROBLEM 2.30 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. SOLUTION (a) P= Py cos 55° 350 lb cos 55° = 610.2 lb = (b) P = 610 lb W Px = P sin 55° = (610.2 lb) sin 55° = 499.8 lb Px = 500 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 31. PROBLEM 2.31 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 Components of the forces were determined in Problem 2.22: Force x Comp. (lb) y Comp. (lb) 29 lb +21.0 +20.0 50 lb –14.00 +48.0 51 lb +24.0 –45.0 Rx = +31.0 Ry = + 23.0 R = Rx i + R y j = (31.0 lb) i + (23.0 lb) j Ry tan α = Rx 23.0 31.0 α = 36.573° 23.0 lb R= sin (36.573°) = = 38.601 lb R = 38.6 lb 36.6° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 32. PROBLEM 2.32 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 Components of the forces were determined in Problem 2.24: Force 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 R = Rx i + Ry j = ( −20.6 N)i + (250.2 N) j Ry tan α = Rx 250.2 N 20.6 N tan α = 12.1456 α = 85.293° tan α = R= 250.2 N sin 85.293° R = 251 N 85.3° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 33. PROBLEM 2.33 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. (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 Ry tan α = Rx 41.42 lb 36.08 lb tan α = 1.14800 α = 48.942° tan α = R= 41.42 lb sin 48.942° R = 54.9 lb 48.9° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 34. PROBLEM 2.34 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: Force x Comp. (N) 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 tan α = Ry Rx 240 608 α = 21.541° = 240 N sin(21.541°) = 653.65 N R= R = 654 N 21.5° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 35. 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: 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 36. PROBLEM 2.36 Knowing that the tension in cable BC is 725 N, determine the resultant of the three forces exerted at Point B of beam AB. SOLUTION Cable BC Force: 500-N Force: 780-N Force: and 840 = −525 N 1160 840 Fy = (725 N) = 500 N 1160 Fx = −(725 N) 3 Fx = −(500 N) = −300 N 5 4 Fy = −(500 N) = −400 N 5 12 = 720 N 13 5 Fy = −(780 N) = −300 N 13 Fx = (780 N) Rx = ΣFx = −105 N R y = ΣFy = −200 N R = (−105 N)2 + (−200 N) 2 = 225.89 N Further: tan α = 200 105 α = tan −1 200 105 = 62.3° R = 226 N Thus: 62.3° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 37. PROBLEM 2.37 Knowing that α = 40°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: Fx = (60 lb) cos 20° = 56.38 lb Fy = (60 lb)sin 20° = 20.52 lb 80-lb Force: Fx = (80 lb) cos 60° = 40.00 lb Fy = (80 lb)sin 60° = 69.28 lb 120-lb Force: Fx = (120 lb) cos 30° = 103.92 lb Fy = −(120 lb)sin 30° = −60.00 lb and Rx = ΣFx = 200.30 lb R y = ΣFy = 29.80 lb R = (200.30 lb) 2 + (29.80 lb) 2 = 202.50 lb Further: tan α = 29.80 200.30 α = tan −1 29.80 200.30 R = 203 lb = 8.46° 8.46° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 38. PROBLEM 2.38 Knowing that α = 75°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: Fx = (60 lb) cos 20° = 56.38 lb Fy = (60 lb) sin 20° = 20.52 lb 80-lb Force: Fx = (80 lb) cos 95 ° = −6.97 lb Fy = (80 lb)sin 95° = 79.70 lb 120-lb Force: Fx = (120 lb) cos 5 ° = 119.54 lb Fy = (120 lb) sin 5° = 10.46 lb Then Rx = ΣFx = 168.95 lb R y = ΣFy = 110.68 lb and R = (168.95 lb) 2 + (110.68 lb) 2 = 201.98 lb 110.68 168.95 tan α = 0.655 α = 33.23° tan α = R = 202 lb 33.2° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 39. 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 = (100 N) cos α + (150 N) cos (α + 30°) − (200 N) cos α Rx = −(100 N) cos α + (150 N) cos (α + 30°) (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 = 0.3988 α = 21.74° tan α = (b) α = 21.7° W Substituting for α in Eq. (2): Ry = −300sin 21.74° − 150sin 51.74° = −228.9 N R = | Ry | = 228.9 N R = 229 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 40. PROBLEM 2.40 For the beam of Problem 2.36, determine (a) the required tension in cable BC if the resultant of the three forces exerted at Point B is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION Rx = ΣFx = − Rx = − 21 TBC + 420 N 29 Ry = ΣFy = Ry = (a) (1) 800 5 4 TBC − (780 N) − (500 N) 1160 13 5 20 TBC − 700 N 29 (2) For R to be vertical, we must have Rx = 0 Set Rx = 0 in Eq. (1) (b) 840 12 3 TBC + (780 N) − (500 N) 1160 13 5 − 21 TBC + 420 N = 0 29 TBC = 580 N W Substituting for TBC in Eq. (2): 20 (580 N) − 700 N 29 Ry = −300 N Ry = R = | Ry | = 300 N R = 300 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 41. PROBLEM 2.41 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. SOLUTION Using the x and y axes shown: Rx = ΣFx = TAC sin10° + (50 lb) cos 35° + (75 lb) cos 60° = TAC sin10° + 78.46 lb (1) Ry = ΣFy = (50 lb)sin 35° + (75 lb)sin 60° − TAC cos10° Ry = 93.63 lb − TAC cos10° (a) (2) Set Ry = 0 in Eq. (2): 93.63 lb − TAC cos10° = 0 TAC = 95.07 lb (b) TAC = 95.1 lb W Substituting for TAC in Eq. (1): Rx = (95.07 lb) sin10° + 78.46 lb = 94.97 lb R = Rx R = 95.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 42. PROBLEM 2.42 For the block of Problems 2.37 and 2.38, determine (a) the required value of α if the resultant of the three forces shown is to be parallel to the incline, (b) the corresponding magnitude of the resultant. SOLUTION Select the x axis to be along a a′. Then Rx = ΣFx = (60 lb) + (80 lb) cos α + (120 lb)sin α (1) Ry = ΣFy = (80 lb)sin α − (120 lb) cos α (2) and (a) Set Ry = 0 in Eq. (2). (80 lb) sin α − (120 lb) cos α = 0 Dividing each term by cos α gives: (80 lb) tan α = 120 lb 120 lb 80 lb α = 56.310° tanα = (b) α = 56.3° W Substituting for α in Eq. (1) gives: Rx = 60 lb + (80 lb) cos 56.31° + (120 lb)sin 56.31° = 204.22 lb Rx = 204 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 43. PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Knowing that α = 20°, determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Law of sines: Force Triangle TAC T 1962 N = BC = sin 70° sin 50° sin 60° (a) TAC = 1962 N sin 70° = 2128.9 N sin 60° TAC = 2.13 kN W (b) TBC = 1962 N sin 50° = 1735.49 N sin 60° TBC = 1.735 kN W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 44. 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. SOLUTION Free-Body Diagram Law of sines: Force Triangle TAC T 500 N = BC = sin 60° sin 40° sin 80° (a) TAC = 500 N sin 60° = 439.69 N sin 80° TAC = 440 N W (b) TBC = 500 N sin 40° = 326.35 N sin 80° TBC = 326 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 45. PROBLEM 2.45 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 Free-Body Diagram 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 W (b) TBC = 500 N sin 75° sin 70° TBC = 514 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 46. PROBLEM 2.46 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram 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 W (b) TBC = (1962 N) sin 105° sin 60° TBC = 2190 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 47. PROBLEM 2.47 Knowing that α = 20°, determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Free-Body Diagram 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 W (b) TBC = 1200 lb sin 5° sin 65° TBC = 115.4 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 48. PROBLEM 2.48 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 Free-Body Diagram Law of sines: Force Triangle FAC T 300 lb = BC = sin 35° sin 50° sin 95° (a) FAC = 300 lb sin 35° sin 95° FAC = 172.7 lb W (b) TBC = 300 lb sin 50° sin 95° TBC = 231 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 49. PROBLEM 2.49 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 Free-Body Diagram 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 W (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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 50. PROBLEM 2.50 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 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 Q= (750 lb) cos 50° − 400 lb cos 50° = 127.710 lb 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 51. PROBLEM 2.51 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 Free-Body Diagram of Connection ΣFx = 0: With 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 W 3 3 FB − FA = 0 5 5 3 3 FD = (16 kN) − (8 kN) 5 5 FD = 4.80 kN W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 52. PROBLEM 2.52 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 5 kN and FD = 6 kN, determine the magnitudes of the other two forces. SOLUTION Free-Body Diagram of Connection 3 3 ΣFy = 0: − FD − FA + FB = 0 5 5 3 FA 5 or FB = FD + With FA = 5 kN, FD = 8 kN FB = 5ª 3 º 6 kN + (5 kN) » 3« 5 ¬ ¼ ΣFx = 0: − FC + FB = 15.00 kN W 4 4 FB − FA = 0 5 5 4 ( FB − FA ) 5 4 = (15 kN − 5 kN) 5 FC = FC = 8.00 kN W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 53. PROBLEM 2.53 Two cables tied together at C are loaded as shown. Knowing that Q = 60 lb, determine the tension (a) in cable AC, (b) in cable BC. SOLUTION ΣFy = 0: TCA − Q cos 30° = 0 With Q = 60 lb (a) TCA = (60 lb)(0.866) (b) ΣFx = 0: P − TCB − Q sin 30° = 0 With TCA = 52.0 lb W P = 75 lb TCB = 75 lb − (60 lb)(0.50) or TCB = 45.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 54. PROBLEM 2.54 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. SOLUTION Free-Body Diagram ΣFx = 0: −TBC − Q cos 60° + 75 lb = 0 TBC = 75 lb − Q cos 60° (1) ΣFy = 0: TAC − Q sin 60° = 0 TAC = Q sin 60° (2) TAC Յ 60 lb: Requirement Q sin 60° Յ 60 lb From Eq. (2): Q Յ 69.3 lb TBC Յ 60 lb: Requirement From Eq. (1): 75 lb − Q sin 60° Յ 60 lb Q Ն 30.0 lb 30.0 lb Յ Q Յ 69.3 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 55. PROBLEM 2.55 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. SOLUTION 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) TCD = 0.137158(1212.56 N) TACB = 1213 N W TCD = 166.3 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 56. PROBLEM 2.56 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 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 W (b) TACB = 1216 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 57. PROBLEM 2.57 For the cables of Problem 2.45, 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 Free-Body Diagram (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 W sin β sin (25° + 45°) = 600 N 784.02 N β = 46.0° α = 46.0° + 25° = 71.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 58. PROBLEM 2.58 For the situation described in Figure P2.47, 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 2.47 Knowing that α = 20°, determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Free-Body Diagram Force Triangle To be smallest, TBC must be perpendicular to the direction of TAC . (a) (b) Thus, α = 5° Į = 5.00° TBC = (1200 lb) sin 5° W TBC = 104.6 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 59. PROBLEM 2.59 For the structure and loading of Problem 2.48, 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. SOLUTION 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° W TBC = (300 lb)sin 50° (b) or TBC = 229.81 lb TBC = 230 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 60. PROBLEM 2.60 Knowing that portions AC and BC of cable ACB must be equal, determine the shortest length of cable that can be used to support the load shown if the tension in the cable is not to exceed 870 N. SOLUTION Free-Body Diagram: C (For T = 725 N) ΣFy = 0: 2Ty − 1200 N = 0 Ty = 600 N Tx2 + Ty2 = T 2 Tx2 + (600 N)2 = (870 N) 2 Tx = 630 N By similar triangles: BC 2.1 m = 870 N 630 N BC = 2.90 m L = 2( BC ) = 5.80 m L = 5.80 m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 61. PROBLEM 2.61 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 Free-Body Diagram: C 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 W Since P Ͼ 0, the solution is correct. (b) α = 180° − 50° − 47.5° = 82.5° α = 82.5° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 62. PROBLEM 2.62 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 Free-Body Diagram (a) Law of cosines: Force Triangle P 2 = (1200 N) 2 + (600 N) 2 − 2(1200 N)(600 N) cos 85° P = 1294 N Since P Ͼ 1200 N, the solution is correct. P = 1294 N W (b) Law of sines: sin β sin 85° = 1200 N 1294 N β = 67.5° α = 180° − 50° − 67.5° α = 62.5° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 63. 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 (a) Free Body: Collar A Force Triangle P 50 lb = 4.5 20.5 (b) Free Body: Collar A P = 10.98 lb W Force Triangle P 50 lb = 15 25 P = 30.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 64. 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 Free Body: Collar A 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. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 65. PROBLEM 2.65 A 160-kg load is supported by the rope-and-pulley arrangement shown. Knowing that β = 20°, 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 Chapter 4.) SOLUTION Free-Body Diagram: Pulley A ΣFx = 0: 2P sin 20° − P cos α = 0 and cos α = 0.8452 or α = ± 46.840° α = + 46.840 For ΣFy = 0: 2P cos 20° + P sin 46.840° − 1569.60 N = 0 or P = 602 N 46.8° W α = −46.840 For ΣFy = 0: 2P cos 20° + P sin( −46.840°) − 1569.60 N = 0 or P = 1365 N 46.8° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 66. PROBLEM 2.66 A 160-kg load is supported by the rope-and-pulley arrangement shown. Knowing that α = 40°, determine (a) the angle β, (b) the magnitude of the force P that must be exerted on the free end of the rope to maintain equilibrium. (See the hint for Problem 2.65.) SOLUTION Free-Body Diagram: Pulley A (a) ΣFx = 0: 2 P sin sin β − P cos 40° = 0 1 cos 40° 2 β = 22.52° sin β = β = 22.5° W (b) ΣFy = 0: P sin 40° + 2 P cos 22.52° − 1569.60 N = 0 P = 630 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 67. 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 2.65.) SOLUTION Free-Body Diagram of Pulley (a) ΣFy = 0: 2T − (600 lb) = 0 T= 1 (600 lb) 2 T = 300 lb W (b) ΣFy = 0: 2T − (600 lb) = 0 T= 1 (600 lb) 2 T = 300 lb W (c) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb W (d) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb W (e) ΣFy = 0: 4T − (600 lb) = 0 T= 1 (600 lb) 4 T = 150.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 68. 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 2.65.) SOLUTION Free-Body Diagram of Pulley and Crate (b) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb W (d) ΣFy = 0: 4T − (600 lb) = 0 T= 1 (600 lb) 4 T = 150.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 69. 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: TACB = 1293 N W 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 70. 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) ΣFy = 0: TACB (sin 25° + sin 55°) + P sin 55° − 1800 N = 0 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 W (b) P = 0.58010(1048.37 N) = 608.16 N Using (1), P = 608 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 71. PROBLEM 2.71 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° = (750 N)sin 35° Fh = 430.2 N (a) Fx = Fh cos 25° = (430.2 N) cos 25° Fx = +390 N, (b) Fy = F cos 35° = (750 N) cos 35° Fy = +614 N, cos θ x = cos θ y = cos θ z = Fx +390 N = 750 N F Fy F = +614 N 750 N Fz +181.8 N = 750 N F Fz = Fh sin 25° = (430.2 N) sin 25° Fz = +181.8 N W θ x = 58.7° W θ y = 35.0° W θ z = 76.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 72. PROBLEM 2.72 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° = (900 N) cos 65° Fh = 380.4 N (a) Fx = Fh sin 20° = (380.4 N)sin 20° Fx = −130.1 N, (b) Fy = F sin 65° = (900 N)sin 65° Fy = +816 N, cos θ x = cos θ y = cos θ z = Fx −130.1 N = 900 N F Fy Fz = Fh cos 20° = (380.4 N) cos 20° Fz = +357 N W θ x = 98.3° W +816 N 900 N θ y = 25.0° W Fz +357 N = 900 N F θ z = 66.6° W F = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 73. PROBLEM 2.73 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. SOLUTION (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 W Fx 110.3 N = = 0.38303 F 287.97 N θ x = 67.5° W Fy = F cos 30° = 249.39 cos θ y = Fy F = 249.39 N = 0.86603 287.97 N θ y = 30.0° W 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 74. PROBLEM 2.74 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 z component of the force exerted by wire BD on the plate is –32.14 N, determine (a) the tension in wire BD, (b) the angles θx, θy, and θz that the force exerted at B forms with the coordinate axes. SOLUTION (a) Fz = − F sin 30° sin 40° = 32.14 N (Given) F= (b) 32.14 = 100.0 N sin 30° sin 40° F = 100.0 N W Fx = − F sin 30° cos 40° = −(100.0 N)sin 30°cos 40° = −38.302 N cos θ x = Fx 38.302 N = = −0.38302 100.0 N F θ x = 112.5° W Fy = F cos 30° = 86.603 N cos θ y = Fy F = 86.603 N = 0.86603 100 N θ y = 30.0° W Fz = −32.14 N cos θ z = Fz −32.14 N = = −0.32140 F 100 N θ z = 108.7° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 75. PROBLEM 2.75 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 tension in wire CD is 60 lb, determine (a) the components of the force exerted by this wire on the plate, (b) the angles θ x , θ y , and θ z that the force forms with the coordinate axes. SOLUTION (a) Fx = −(60 lb) sin 30°cos 60° = −15 lb Fx = −15.00 lb W Fy = (60 lb) cos 30° = 51.96 lb Fz = (60 lb) sin 30° sin 60° = 25.98 lb (b) Fy = +52.0 lb W Fz = +26.0 lb W cos θ x = cos θ y = cos θ z = Fx −15.0 lb = = −0.25 60 lb F Fy θ x = 104.5° W 51.96 lb = 0.866 60 lb θ y = 30.0° W Fz 25.98 lb = = 0.433 F 60 lb θ z = 64.3° W F = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 76. PROBLEM 2.76 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 CD on the plate is –20.0 lb, determine (a) the tension in wire CD, (b) the angles θx, θy, and θz that the force exerted at C forms with the coordinate axes. SOLUTION (a) Fx = − F sin 30° cos 60° = −20 lb (Given) F= (b) cos θ x = 20 lb = 80 lb sin 30°cos 60° F = 80.0 lb W Fx −20 lb = = − 0.25 80 lb F θ x = 104.5° W Fy = (80 lb) cos 30° = 69.282 lb cos θ y = Fy F = 69.282 lb = 0.86615 80 lb θ y = 30.0° W Fz = (80 lb)sin 30° sin 60° = 34.641 lb cos θ z = Fz 34.641 = = 0.43301 80 F θ z = 64.3° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 77. 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. SOLUTION (a) Fx = (120 lb) cos 60° cos 20° Fx = 56.382 lb Fx = +56.4 lb W Fy = −(120 lb)sin 60° Fy = −103.923 lb Fy = −103.9 lb W Fz = −(120 lb) cos 60° sin 20° Fz = −20.521 lb (b) cos θ x = cos θ y = cos θ z = Fz = −20.5 lb W Fx 56.382 lb = F 120 lb Fy F = −103.923 lb 120 lb Fz −20.52 lb = F 120 lb θ x = 62.0° W θ y = 150.0° W θ z = 99.8° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 78. 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. SOLUTION (a) Fx = (85 lb)sin 36° sin 48° = 37.129 lb Fx = 37.1 lb W Fy = −(85 lb) cos 36° = −68.766 lb Fy = −68.8 lb W Fz = (85 lb)sin 36° cos 48° Fz = 33.4 lb W = 33.431 lb (b) cos θ x = cos θ y = cos θ z = Fx 37.129 lb = F 85 lb Fy F = −68.766 lb 85 lb Fz 33.431 lb = F 85 lb θ x = 64.1° W θ y = 144.0° W θ z = 66.8° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 79. PROBLEM 2.79 Determine the magnitude and direction of the force F = (320 N)i + (400 N)j − (250 N)k. SOLUTION F = Fx2 + Fy2 + Fz2 F = (320 N)2 + (400 N) 2 + (−250 N) 2 cos θ x = cos θ y = cos θ y = Fx 320 N = F 570 N Fy F = F = 570 N W θ x = 55.8° W 400 N 570 N θ y = 45.4° W Fz −250 N = F 570 N θ z = 116.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 80. PROBLEM 2.80 Determine the magnitude and direction of the force F = (240 N)i − (270 N)j + (680 N)k. SOLUTION F = Fx2 + Fy2 + Fz2 F = (240 N) 2 + (−270 N) 2 + (680 N) cos θ x = cos θ y = cos θ z = Fx 240 N = F 770 N Fy F = F = 770 N W θ x = 71.8° W −270 N 770 N Fz 680 N = F 770 N θ y = 110.5° W θ z = 28.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 81. PROBLEM 2.81 A force acts at the origin of a coordinate system in a direction defined by the angles θx = 70.9° and θy = 144.9°. Knowing that the z component of the force is –52.0 lb, determine (a) the angle θz, (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 Fz Ͻ 0 we must have cos θ z Ͻ 0 Thus, taking the negative square root, from above, we have: cos θ z = − 1 − (cos 70.9°) 2 − (cos144.9°) 2 = 0.47282 (b) θ z = 118.2° W Then: F= Fz 52.0 lb = = 109.978 lb cos θ z 0.47282 Fx = F cos θ x = (109.978 lb) cos 70.9° Fx = 36.0 lb W Fy = F cos θ y = (109.978 lb) cos144.9° and Fy = −90.0 lb W F = 110.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 82. PROBLEM 2.82 A force acts at the origin 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 Ͻ 0 we must have cos θ x Ͻ 0 Thus, taking the negative square root, from above, we have: cos θ x = − 1 − (cos 55) 2 − (cos 45) 2 = 0.41353 (b) θ x = 114.4° W Then: Fx 500 lb = = 1209.10 lb cos θ x 0.41353 F = 1209 lb W Fy = F cos θ y = (1209.10 lb) cos 55° Fy = 694 lb W Fz = F cos θ z = (1209.10 lb) cos 45° Fz = 855 lb W F= and PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 83. PROBLEM 2.83 A force F of magnitude 210 N acts at the origin 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° (a) = −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 W cos θ x = cos θ y = Fy = −62.0 lb W Fx 80 N = = 0.38095 F 210 N Fy F = 61.929 N = −0.29490 210 N θ x = 67.6° W θ y = 107.2 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 84. PROBLEM 2.84 A force F of magnitude 230 N acts at the origin 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 W Fx = 193.980 N F 2 = Fx2 + Fy2 + Fz2 So: 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 W Fz = 108.036 N Fy −60 N = − 0.26087 F 230 N F 108.036 N cos θ z = z = = 0.46972 F 230 N cos θ y = = θ y = 105.1° W θ z = 62.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 85. PROBLEM 2.85 A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AB is 525 lb, determine the components of the force exerted by the wire on the bolt at B. SOLUTION JJJ G BA = (20 ft)i + (100 ft) j − (25 ft)k BA = (20 ft) 2 + (100 ft) 2 + (−25 ft)2 = 105 ft F = F Ȝ BA JJJ G BA =F BA 525 lb [(20 ft)i + (100 ft) j − (25 ft)k ] = 105 ft F = (100.0 lb)i + (500 lb) j − (125.0 lb)k Fx = +100.0 lb, Fy = +500 lb, Fz = −125.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 86. PROBLEM 2.86 A transmission tower is held by three guy wires anchored by bolts at B, C, and D. If the tension in wire AD is 315 lb, determine the components of the force exerted by the wire on the bolt at D. SOLUTION JJJ G DA = (20 ft)i + (100 ft) j + (70 ft)k DA = (20 ft) 2 + (100 ft) 2 + ( +70 ft) 2 = 126 ft F = F Ȝ DA JJJ G DA =F DA 315 lb [(20 ft)i + (100 ft) j + (74 ft)k ] = 126 ft F = (50 lb)i + (250 lb) j + (185 lb)k Fx = +50 lb, Fy = +250 lb, Fz = +185.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 87. PROBLEM 2.87 A frame 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 JJJ G DB = (480 mm)i − (510 mm) j + (320 mm)k DB = (480 mm) 2 + (510 mm 2 ) + (320 mm) 2 = 770 mm F = F Ȝ DB JJJ G 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 88. PROBLEM 2.88 For the frame and cable of Problem 2.87, determine the components of the force exerted by the cable on the support at E. PROBLEM 2.87 A frame 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 JJJ G EB = (270 mm)i − (400 mm) j + (600 mm)k EB = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm F = F Ȝ EB JJJ G 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 89. PROBLEM 2.89 Knowing that the tension in cable AB is 1425 N, determine the components of the force exerted on the plate at B. SOLUTION JJJ G BA = −(900 mm)i + (600 mm) j + (360 mm)k BA = (900 mm) 2 + (600 mm) 2 + (360 mm) 2 = 1140 mm TBA = TBA Ȝ BA JJJ G BA = TBA BA 1425 N TBA = [ −(900 mm)i + (600 mm) j + (360 mm)k ] 1140 mm = −(1125 N)i + (750 N) j + (450 N)k (TBA ) x = −1125 N, (TBA ) y = 750 N, (TBA ) z = 450 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 90. PROBLEM 2.90 Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C. SOLUTION JJJ G CA = −(900 mm)i + (600 mm) j − (920 mm)k CA = (900 mm)2 + (600 mm)2 + (920 mm) 2 = 1420 mm TCA = TCA λCA JJJ G 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 91. PROBLEM 2.91 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 300 N and Q = 400 N. SOLUTION 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 ] = (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 cos θ x = cos θ y = cos θ z = R = 515 N W Rx 174.367 N = = 0.33853 515.07 N R Ry θ x = 70.2° W 456.42 N = 0.88613 515.07 N θ y = 27.6° W Rz 163.011 N = = 0.31648 R 515.07 N θ z = 71.5° W R = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 92. PROBLEM 2.92 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 400 N and Q = 300 N. SOLUTION P = (400 N)[ − cos 30° sin15°i + sin 30° j + cos 30° cos15°k ] = − (89.678 N)i + (200 N) j + (334.61 N)k Q = (300 N)[cos 50° cos 20°i + sin 50° j − cos 50° sin 20°k ] = (181.21 N)i + (229.81 N)j − (65.954 N)k R = P+Q = (91.532 N)i + (429.81 N) j + (268.66 N)k R = (91.532 N)2 + (429.81 N) 2 + (268.66 N) 2 = 515.07 N cos θ x = cos θ y = cos θ z = R = 515 N W Rx 91.532 N = = 0.177708 R 515.07 N Ry θ x = 79.8° W 429.81 N = 0.83447 515.07 N θ y = 33.4° W Rz 268.66 N = = 0.52160 R 515.07 N θ z = 58.6° W R = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 93. 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. SOLUTION JJJ G AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in. JJJG AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.) 2 + (45 in.) 2 + (60 in.) 2 = 125 in. JJJ G ª (40 in.)i − (45 in.) j + (60 in.)k º AB = (425 lb) « TAB = TAB ȜAB = TAB » 85 in. AB ¬ ¼ TAB = (200 lb)i − (225 lb) j + (300 lb)k JJJG ª (100 in.)i − (45 in.) j + (60 in.)k º AC = (510 lb) « TAC = TAC ȜAC = TAC » AC 125 in. ¬ ¼ 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: R = 912.92 lb R = 913 lb W cos θ x = 608 lb = 0.66599 912.92 lb cos θ y = 408.6 lb = −0.44757 912.92 lb θ y = 116.6° W cos θ z = and 544.8 lb = 0.59677 912.92 lb θ z = 53.4° W θ x = 48.2° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 94. 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. SOLUTION JJJ G AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.) 2 + (45 in.) 2 + (60 in.) 2 = 85 in. JJJG AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.) 2 + (45 in.) 2 + (60 in.) 2 = 125 in. JJJ G ª (40 in.)i − (45 in.) j + (60 in.)k º AB = (510 lb) « TAB = TAB ȜAB = TAB » 85 in. AB ¬ ¼ TAB = (240 lb)i − (270 lb) j + (360 lb)k JJJG ª (100 in.)i − (45 in.) j + (60 in.)k º AC = (425 lb) « TAC = TAC ȜAC = TAC » AC 125 in. ¬ ¼ TAC = (340 lb)i − (153 lb) j + (204 lb)k R = TAB + TAC = (580 lb)i − (423 lb) j + (564 lb)k Then: R = 912.92 lb R = 913 lb W cos θ x = 580 lb = 0.63532 912.92 lb cos θ y = −423 lb = −0.46335 912.92 lb cos θ z = and 564 lb = 0.61780 912.92 lb θ x = 50.6° W θ y = 117.6° W θ z = 51.8° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 95. PROBLEM 2.95 For the frame of Problem 2.87, 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 2.87 A frame 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 JJJ G BD = −(480 mm)i + (510 mm) j − (320 mm)k BD = (480 mm) 2 + (510 mm) 2 + (320 mm) 2 = 770 mm JJJ G BD FBD = TBD Ȝ 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 JJJ G BE = −(270 mm)i + (400 mm) j − (600 mm)k BE = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm JJJ G BE FBE = TBE Ȝ 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 W cos θ x = −375 N 747.83 N θ x = 120.1° W cos θ y = 455 N 747.83 N θ y = 52.5° W cos θ z = −460 N 747.83 N θ z = 128.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 96. PROBLEM 2.96 For the cables of Problem 2.89, 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 TAB = −TBA (use results of Problem 2.89) (TAB ) x = +1125 N (TAB ) y = −750 N (TAB ) z = − 450 N TAC = −TCA (use results of Problem 2.90) (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 2 2 2 R = Rx + Ry + Rz = (+2475) 2 + (−1650) 2 + ( +930)2 = 3116.6 N cos θ x = cos θ y = cos θ z = R = 3120 N W Rx +2475 = R 3116.6 Ry θ x = 37.4° W −1650 3116.6 θ y = 122.0° W Rz + 930 = R 3116.6 θ z = 72.6° W R = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 98. PROBLEM 2.98 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 AD is 125 lb and that the resultant of the forces exerted at A by wires AC and AD must be contained in the xy plane, determine (a) the tension in AC, (b) the magnitude and direction of the resultant of the two forces. SOLUTION R = TAC + TAD = TAC (cos 60° cos 20°i − sin 60° j − cos 60° sin 20°k ) + (125 lb)(sin 36° sin 48°i − cos 36° j + sin 36° cos 48°k ) (a) (1) Since Rz = 0, The coefficient of k must be zero. TAC (− cos 60° sin 20°) + (125 lb)(sin 36° cos 48°) = 0 TAC = 287.49 lb (b) TAC = 287 lb W Substituting for TAC into Eq. (1) gives: R = [(287.49 lb) cos 60° cos 20° + (125 lb) sin 36° sin 48°]i − [(287.49 lb) sin 60° + (125 lb) cos 36°]j + 0 R = (189.677 lb)i − (350.10 lb) j R = (189.677 lb)2 + (350.10 lb) 2 R = 398 lb W = 398.18 lb cos θ x = 189.677 lb 398.18 lb θ x = 61.6° W cos θ y = 350.10 lb 398.18 lb θ y = 151.6° W cos θ z = 0 θ z = 90.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 101. PROBLEM 2.100 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. SOLUTION See Problem 2.99 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.6TAB + 0.32432TAC = 0 − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 0.56757TAC − 0.50769TAD = 0 (1) (2) (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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 102. PROBLEM 2.101 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. SOLUTION See Problem 2.99 for the figure and the 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) Substituting TAD = 481 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives TAC = 430.26 N TAB = 232.57 N P = 926 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 106. 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. SOLUTION 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 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 −0.36TAB + 0.47059TAC − 0.35065TAD = 0 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 107. 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. SOLUTION 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 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 − 0.36TAB + 0.47059TAC − 0.35065TAD = 0 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 108. PROBLEM 2.106 A 1600-lb crate is supported by three cables as shown. Determine the tension in each cable. SOLUTION 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 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 −0.36TAB + 0.47059TAC − 0.35065TAD = 0 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 W TAC = 830 lb W TAD = 528 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 110. PROBLEM 2.108 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. SOLUTION We assume that TAD = 0 and write ΣFA = 0: TAB + TAC + Qj + (1200 N)i = 0 JJJ G AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm JJJG AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm JJJ G AB § 48 12 19 · TAB = TAB ȜAB = TAB = ¨ − i − j + k ¸ TAB AB © 53 53 53 ¹ JJJG AC § 12 3 4 · TAC = TAC ȜAC = TAC = ¨ − i − j − k ¸ TAC AC © 13 13 13 ¹ 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 Յ Q Ͻ 300 N W Note: This solution assumes that Q is directed upward as shown (Q Ն 0), if negative values of Q are considered, cable AD remains taut, but AC becomes slack for Q = −460 N. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 119. PROBLEM 2.115 For the rectangular plate of Problems 2.111 and 2.112, determine the tension in each of the three cables knowing that the weight of the plate is 792 N. SOLUTION See Problem 2.111 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 W TAC = 56.250 N TAC = 56.2 N W TAD = 536.25 N TAD = 536 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 121. 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 W TAC = 1025 N W TAD = 915 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 124. PROBLEM 2.119 Using two ropes and a roller chute, two workers are unloading a 200-lb cast-iron counterweight from a truck. Knowing that at the instant shown the counterweight is kept from moving and that the positions of Points A, B, and C are, respectively, A(0, –20 in., 40 in.), B(–40 in., 50 in., 0), and C(45 in., 40 in., 0), and assuming that no friction exists between the counterweight and the chute, determine the tension in each rope. (Hint: Since there is no friction, the force exerted by the chute on the counterweight must be perpendicular to the chute.) SOLUTION From the geometry of the chute: N= N (2 j + k ) 5 = N (0.8944 j + 0.4472k ) The force in each rope can be written as the product of the magnitude of the force and the unit vector along the cable. Thus, with JJJ G AB = (40 in.)i + (70 in.)j − (40 in.)k and AB = (40 in.) 2 + (70 in.) 2 + (40 in.)2 = 90 in. JJJ G AB TAB = T Ȝ AB = TAB AB TAB [(−40 in.)i + (70 in.)j − (40 in.)k ] = 90 in. 7 4 · § 4 TAB = TAB ¨ − i + j − k ¸ 9 9 ¹ © 9 JJJG AC = (45 in.)i + (60 in.)j − (40 in.)k AC = (45 in.) 2 + (60 in.)2 + (40 in.)2 = 85 in. JJJG AC TAC = T ȜAC = TAC AC TAC [(45 in.)i + (60 in.)j − (40 in.)k ] = 85 in. 12 8 · § 9 TAC = TAC ¨ i + j − k ¸ © 17 17 17 ¹ Then: ΣF = 0: N + TAB + TAC + W = 0 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 125. PROBLEM 2.119 (Continued) With W = 200 lb, and equating the factors of i, j, and k to zero, we obtain the linear algebraic equations: i: j: k: 4 9 − TAB + TAC = 0 9 17 7 12 2 TAB + TAC + − 200 lb = 0 9 17 5 4 8 1 − TAB − TAC + N =0 9 17 5 (1) (2) (3) Using conventional methods for solving linear algebraic equations we obtain: TAB = 65.6 lb W TAC = 55.1 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 126. PROBLEM 2.120 Solve Problem 2.119 assuming that a third worker is exerting a force P = −(40 lb)i on the counterweight. PROBLEM 2.119 Using two ropes and a roller chute, two workers are unloading a 200-lb cast-iron counterweight from a truck. Knowing that at the instant shown the counterweight is kept from moving and that the positions of Points A, B, and C are, respectively, A(0, –20 in., 40 in.), B(–40 in., 50 in., 0), and C(45 in., 40 in., 0), and assuming that no friction exists between the counterweight and the chute, determine the tension in each rope. (Hint: Since there is no friction, the force exerted by the chute on the counterweight must be perpendicular to the chute.) SOLUTION See Problem 2.119 for the analysis leading to the vectors describing the tension in each rope. 7 4 · § 4 TAB = TAB ¨ − i + j − k ¸ 9 9 ¹ © 9 12 8 · § 9 TAC = TAC ¨ i + j − k ¸ 17 17 17 ¹ © Then: Where and ΣFA = 0: N + TAB + TAC + P + W = 0 P = −(40 lb)i W = (200 lb)j Equating the factors of i, j, and k to zero, we obtain the linear equations: 4 9 i : − TAB + TAC − 40 lb = 0 9 17 j: k: 2 7 12 N + TAB + TAC − 200 lb = 0 9 17 5 1 5 4 8 N − TAB − TAC = 0 9 17 Using conventional methods for solving linear algebraic equations we obtain TAB = 24.8 lb W TAC = 96.4 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 127. PROBLEM 2.121 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.) SOLUTION Free Body A: TAB = T Ȝ AB JJJ K AB =T AB (−130 mm)i + (400 mm) j + (160 mm)k =T 450 mm 40 16 · § 13 =T ¨− i + j+ k¸ 45 45 ¹ © 45 TAC = T Ȝ AC JJJK 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 ¹ © 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. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 128. PROBLEM 2.121 (Continued) Data: W = 376 N 1.70521T = 376 N T = 220.50 N 0.59501(220.50 N) = P P = 131.2 N W 0.134240(220.50 N) = Q Q = 29.6 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 129. PROBLEM 2.122 For the system of Problem 2.121, determine W and Q knowing that P = 164 N. PROBLEM 2.121 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.) SOLUTION Refer to Problem 2.121 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 T = 275.63 N Eq. (2): 1.70521(275.63 N) = W W = 470 N W Eq. (3): 0.134240(275.63 N) = Q Q = 37.0 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 131. PROBLEM 2.123 (Continued) JJJ G AE = −(0.4 m)i + (1.6 m) j − (0.86 m)k Finally, AE = (−0.4 m) 2 + (1.6 m) 2 + (−0.86 m) 2 = 1.86 m JJJ G AE TAE = T ȜAE = TAE AE T = AE [−(0.4 m)i + (1.6 m) j − (0.86 m)k ] 1.86 m TAE = TAE (−0.2151i + 0.8602 j − 0.4624k ) With the weight of the container W = −W j, at A we have: ΣF = 0: TAB + TAC + TAD − Wj = 0 Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations: −0.4382TAB + 0.6190TAD − 0.2151TAE = 0 (1) 0.8989TAB + 0.8TAC + 0.7619TAD + 0.8602TAE − W = 0 (2) 0.6TAC + 0.1905TAD − 0.4624TAE = 0 (3) Knowing that W = 1000 N and that because of the pulley system at B TAB = TAD = P, where P is the externally applied (unknown) force, we can solve the system of linear Equations (1), (2) and (3) uniquely for P. P = 378 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 132. PROBLEM 2.124 Knowing that the tension in cable AC of the system described in Problem 2.123 is 150 N, determine (a) the magnitude of the force P, (b) the weight W of the container. PROBLEM 2.123 A container of weight W is suspended from ring A, to which cables AC and AE are attached. A force P is applied to the end F of a third cable that passes over a pulley at B and through ring A and that is attached to a support at D. Knowing that W = 1000 N, determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.) SOLUTION Here, as in Problem 2.123, the support of the container consists of the four cables AE, AC, AD, and AB, with the condition that the force in cables AB and AD is equal to the externally applied force P. Thus, with the condition TAB = TAD = P and using the linear algebraic equations of Problem 2.131 with TAC = 150 N, we obtain (a) P = 454 N W (b) W = 1202 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 133. PROBLEM 2.125 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 Free Body Diagrams of Collars: A: B: ȜAB JJJ G 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 ȜAB 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 ȜAB 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 W P = 45.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 134. PROBLEM 2.126 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 See Problem 2.125 for the diagrams and analysis leading to Equations (1) and (2) below: 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 = 6.708 in. From Eq. (3) x = 2 z = 2(6.708 in.) = 13.416 in. x = 13.42 in., z = 6.71 in. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 135. PROBLEM 2.127 The direction of the 75-lb forces may vary, but the angle between the forces is always 50°. Determine the value of Į for which the resultant of the forces acting at A is directed horizontally to the left. SOLUTION We must first replace the two 75-lb forces by their resultant R1 using the triangle rule. R1 = 2(75 lb) cos 25° = 135.946 lb R1 = 135.946 lb α + 25° Next we consider the resultant R 2 of R1 and the 240-lb force where R 2 must be horizontal and directed to the left. Using the triangle rule and law of sines, sin (α + 25°) sin (30°) = 240 lb 135.946 sin (α + 25°) = 0.88270 α + 25° = 61.970° α = 36.970° α = 37.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 136. PROBLEM 2.128 A stake is being pulled out of the ground by means of two ropes as shown. Knowing the magnitude and direction of the force exerted on one rope, determine the magnitude and direction of the force P that should be exerted on the other rope if the resultant of these two forces is to be a 40-lb vertical force. SOLUTION Triangle rule: Law of cosines: Law of sines: P 2 = (30) 2 + (40) 2 − 2(30)(40) cos 25° P = 18.0239 lb sin α sin 25° = 30 lb 18.0239 lb α = 44.703° 90° − α = 45.297° P = 18.02 lb 45.3° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 137. PROBLEM 2.129 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. SOLUTION (a) P= (b) Px = Py sin 35° Py tan 40° = 240 lb sin 40° or P = 373 lb W = 240 lb tan 40° or Px = 286 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 138. PROBLEM 2.130 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 at C: ΣFx = 0: − 12 ft 7.5 ft TAC + TBC = 0 12.5 ft 8.5 ft TBC = 1.08800TAC ΣFy = 0: (a) 3.5 ft 4 ft TAC + TBC − 396 lb = 0 12 ft 8.5 ft 3.5 ft 4 ft (1.08800TAC ) − 396 lb = 0 TAC + 12.5 ft 8.5 ft (0.28000 + 0.51200)TAC = 396 lb TAC = 500.0 lb (b) TBC = (1.08800)(500.0 lb) TAC = 500 lb W TBC = 544 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 139. PROBLEM 2.131 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 Free Body: C (a) ΣFx = 0: − (b) ΣFy = 0: 12 4 TAC + (360 N) = 0 13 5 TAC = 312 N W 5 3 (312 N) + TBC + (360 N) − 480 N = 0 13 5 TBC = 480 N − 120 N − 216 N TBC = 144 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 140. PROBLEM 2.132 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 ΣFx = 0: − 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 © ¹© ¹ TBC = 480 N − 14 P 15 (2) From (1), TAC Ͼ 0 requires P Ͼ 0. From (2), TBC Ͼ 0 requires 14 P Ͻ 480 N, P Ͻ 514.29 N 15 0 Ͻ P Ͻ 514 N W Allowable range: PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 141. PROBLEM 2.133 A force acts at the origin of a coordinate system in a direction defined by the angles θ x = 69.3° and θ z = 57.9°. Knowing that the y component of the force is –174.0 lb, determine (a) the angle θ y , (b) the other components and the magnitude of the force. SOLUTION (a) To determine θ y , use the relation cos 2 θ y + cos 2 θ y + cos 2 θ z = 1 or cos 2 θ y = 1 − cos 2 θ x − cos 2 θ y Since Fy Ͻ 0, we must have cos θ y Ͻ 0 cos θ y = − 1 − cos 2 69.3° − cos 2 57.9° θ y = 140.3° W = − 0.76985 Fy −174.0 lb = 226.02 lb −0.76985 F = 226 lb W Fx = F cos θ x = (226.02 lb) cos 69.3° Fx = 79.9 lb W Fz = F cos θ z = (226.02 lb) cos 57.9° (b) Fz = 120.1 lb W F= cos θ y = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 142. PROBLEM 2.134 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. 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 W 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 W Fz = +677 lb W θ x = 118.5° W θ y = 30.5° W θ z = 80.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 143. PROBLEM 2.135 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 is 10 kN in cable AB and 7.5 kN in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION JJJ G AB = −15.588i + 15 j + 12k AB = 24.739 m JJJG AC = −15.588i + 18.60 j − 15k AC = 28.530 m JJJ G AB TAB = TAB ȜAB = TAB AB −15.588i + 15 j + 12k TAB = (10 kN) 24.739 TAB = (6.301 kN)i + (6.063 kN) j + (4.851 kN)k JJJG −15.588i + 18.60 j − 15k AC TAC = TAC ȜAC = TAC (7.5 kN) 28.530 AC TAC = −(4.098 kN)i + (4.890 kN) j − (3.943 kN)k R = TAB + TAC = −(10.399 kN)i + (10.953 kN) j + (0.908 kN)k R = (10.399)2 + (10.953)2 + (0.908) 2 R = 15.13 kN W = 15.130 kN cos θ x = cos θ y = cos θ z = Rx −10.399 kN = = − 0.6873 15.130 kN R Ry θ z = 133.4° W 10.953 kN = 0.7239 15.130 kN θ y = 43.6° W Rz 0.908 kN = = 0.0600 R 15.130 kN θ z = 86.6° W R = PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 145. PROBLEM 2.136 (Continued) Substitute for TAB and TAC from (1) and (3) into (2): 15 30 · § ¨ 0.8 × 0.96899 + 17 × 0.88953 + 53 ¸ TAD − 1165 N = 0 © ¹ 2.2578TAD − 1165 N = 0 TAD = 516 N W From (1): TAB = 0.96899(516 N) TAB = 500 N W From (3): TAC = 0.88953(516 N) TAC = 459 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 146. PROBLEM 2.137 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 For both Problems 2.137 and 2.138: Free Body Diagrams of Collars: ( AB) 2 = x 2 + y 2 + z 2 Here (0.525 m) 2 = (.20 m) 2 + y 2 + z 2 y 2 + z 2 = 0.23563 m 2 or Thus, why y given, z is determined, Now ȜAB JJJ G 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 ȜAB = 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 ) = y (1.90476) y PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 147. PROBLEM 2.137 (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 = 1.155 kN W or and 341 N(0.46 m)(0.866) (0.155 m) = (1012.00 N) Q= (b) Q = 1.012 kN W or PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 148. PROBLEM 2.138 Solve Problem 2.137 assuming that y = 275 mm. PROBLEM 2.137 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 From the analysis of Problem 2.137, 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 W or and Q= (b) 341 N(0.40 m) (0.275 m) Q = 496 N W or PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 149. CHAPTER 3
• 150. PROBLEM 3.1 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28°, determine the moment of the 16-N force about Point B by resolving the force into horizontal and vertical components. SOLUTION Note that and θ = α − 20° = 28° − 20° = 8° Fx = (16 N) cos8° = 15.8443 N Fy = (16 N) sin 8° = 2.2268 N Also x = (0.17 m) cos 20° = 0.159748 m y = (0.17 m) sin 20° = 0.058143 m Noting that the direction of the moment of each force component about B is counterclockwise, MB = xFy + yFx = (0.159748 m)(2.2268 N) + (0.058143 m)(15.8443 N) = 1.277 N ⋅ m or M B = 1.277 N ⋅ m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 151. PROBLEM 3.2 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28°, determine the moment of the 16-N force about Point B by resolving the force into components along ABC and in a direction perpendicular to ABC. SOLUTION First resolve the 4-lb force into components P and Q, where Q = (16 N) sin 28° = 7.5115 N Then M B = rA/B Q = (0.17 m)(7.5115 N) or M B = 1.277 N ⋅ m = 1.277 N ⋅ m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 152. PROBLEM 3.3 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° = 271.89 N Fy = (300 N) sin 25° = 126.785 N F = (271.89 N)i + (126.785 N) j JJJ G 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) W The smallest force Q at B must be perpendicular to JJJ G DB at 45° JJJ G M D = Q ( DB ) 41.700 N ⋅ m = Q (0.28284 m) Q = 147.4 N 45° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 153. PROBLEM 3.4 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the magnitude and sense of the horizontal force applied at C that creates the same moment about D, (c) the smallest force applied at C that creates the same moment about D. SOLUTION (a) See Problem 3.3 for the figure and analysis leading to the determination of MD M D = 41.7 N ⋅ m (b) W Since C is horizontal C = C i JJJG r = DC = (0.2 m)i − (0.125 m) j M D = r × C i = C (0.125 m)k 41.7 N ⋅ m = (0.125 m)(C ) C = 333.60 N (c) C = 334 N W The smallest force C must be perpendicular to DC; thus, it forms α with the vertical 0.125 m 0.2 m α = 32.0° tan α = M D = C ( DC ); DC = (0.2 m) 2 + (0.125 m) 2 = 0.23585 m 41.70 N ⋅ m = C (0.23585 m) C = 176.8 N 58.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 154. PROBLEM 3.5 An 8-lb force P is applied to a shift lever. Determine the moment of P about B when a is equal to 25°. SOLUTION First note Px = (8 lb) cos 25° = 7.2505 lb Py = (8 lb)sin 25° = 3.3809 lb Noting that the direction of the moment of each force component about B is clockwise, have M B = − xPy − yPx = −(8 in.)(3.3809 lb) − (22 in.)(7.2505 lb) = −186.6 lb ⋅ in. or M B = 186.6 lb ⋅ in. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 155. PROBLEM 3.6 For the shift lever shown, determine the magnitude and the direction of the smallest force P that has a 210-lb ⋅ in. clockwise moment about B. SOLUTION For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, α =θ 8 in. 22 in. = 19.98° = tan −1 and Where M B = dPmin d = rA/B = (8 in.) 2 + (22 in.)2 = 23.409 in. Then 210 lb ⋅ in. 23.409 in. = 8.97 lb Pmin = Pmin = 8.97 lb 19.98° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 156. PROBLEM 3.7 An 11-lb force P is applied to a shift lever. The moment of P about B is clockwise and has a magnitude of 250 lb ⋅ in. Determine the value of a. SOLUTION By definition M B = rA/B P sin θ where θ = α + (90° − φ ) and φ = tan −1 also 8 in. = 19.9831° 22 in. rA/B = (8 in.) 2 + (22 in.) 2 = 23.409 in. Then or 250 lb ⋅ in = (23.409 in.)(11 lb) x sin(α + 90° − 19.9831°) sin (α + 70.0169°) = 0.97088 or α + 70.0169° = 76.1391° and α + 70.0169° = 103.861° α = 6.12° 33.8° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 159
• 157. 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 a = 10°, (c) the smallest force P that creates the same moment about B. SOLUTION (a) M B = rC/B FN We have = (4 in.)(200 lb) = 800 lb ⋅ in. or MB = 800 lb ⋅ in. W (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 W (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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 160
• 158. PROBLEM 3.9 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 EC = 0.875 m 5 = 1.90 m + 0.2 m 12 12 (TAB ) 13 12 = (1040 N) 13 = 960 N Then TABx = and TABy = Then 5 (1040 N) 13 = 400 N M D = TABx (0.875 m) − TABy (0.2 m) = (960 N)(0.875 m) − (400 N)(0.2 m) = 760 N ⋅ m We have W or M D = 760 N ⋅ m (b) or M D = 760 N ⋅ m W M D = TABx ( y ) + TABx ( x) = (960 N)(0) + (400 N)(1.90 m) = 760 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 159. PROBLEM 3.10 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 160. PROBLEM 3.11 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. SOLUTION 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 = 0.875m (d + 0.20) 2 + (0.875) 2 m ª 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 º » (d ) » ¼ (d + 0.20) 2 + (0.875) 2 = 4.7852d 2 or 3.7852d 2 − 0.40d − .8056 = 0 Using the quadratic equation, the minimum values of d are 0.51719 m and − .41151 m. Since only the positive value applies here, d = 0.51719 m or d = 517 mm W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 161. PROBLEM 3.12 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 = 12.2241 in. Then and 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 162. PROBLEM 3.13 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 = (17.2 in.) 2 + (7.62 in.) 2 = 18.8123 in. Then and 17.2 in. 18.8123 in. 7.62 in. sin θ = 18.8123 in. cos θ = FCB = ( FCB cos θ )i − ( FCB sin θ ) j = 125 lb [(17.2 in.)i + (7.62 in.) j] 18.8123 in. Now M A = rB/A × FCB where rB/A = (20.5 in.)i − (4.38 in.) j Then MA = [(20.5 in.)i − (4.38 in.) j] × = (1538.53 lb ⋅ in.)k = (128.2 lb ⋅ ft)k 125 lb (17.2i − 7.62 j) 18.8123 in. or M A = 128.2 lb ⋅ ft W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 163. 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 We have 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 2 (65) + (72) 2 72 2 (65) + (72) 2 (485 N) = 325 N (485 N) = 360 N M C = (55 mm)(360 N) + (162)(325 N) = 72450 N ⋅ mm = 72.450 N ⋅ m or M C = 72.5 N ⋅ m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 164. 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 a cos β = 1 1 sin ( a + β ) + sin ( a − β ). 2 2 SOLUTION Note: B = B(cos β i + sin β j) B′ = B(cos β i − sin β j) C = C (cos α i + sin α j) Now | B × C | = BC sin (α − β ) (1) | B′ × C | = BC sin (α + β ) By definition: (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 ) W 2 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 165. PROBLEM 3.16 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 origin 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. Then, Where F = F λ AB λ AB = = By definition Where Then rB − rA d AB 1 (21i + 20 j) 29 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 ¹ = Finally § 64 · ¨ 29 F ¸ = d ( F ) © ¹ 64 d= m 29 d = 2.21 m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 166. PROBLEM 3.17 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 We have A = |P × Q| where (a) P = −7i + 3j − 3k Q = 2i + 2 j + 5k Then i j k P × Q = −7 3 −3 2 2 5 = [(15 + 6)i + ( −6 + 35) j + ( −14 − 6)k ] = (21)i + (29) j(−20)k A = (20) 2 + (29) 2 + (−20) 2 We have A = |P × Q| where (b) or A = 41.0 W P = 6i − 5 j − 2k Q = −2i + 5 in. 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 167. PROBLEM 3.18 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×B |A × B| We have Ȝ= where (a) A = 1i + 2 j − 5k B = 4i − 7 j − 5k Then i j k A × B = 1 +2 −5 4 −7 −5 = (−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 Ȝ= 15(−3i − 1j − 1k ) 15 11 1 11 (−3i − j − k ) W A×B |A × B| We have Ȝ= where (b) or Ȝ = 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 Ȝ= 4(2 j + 3k ) 4 13 or Ȝ = 1 13 (2 j + 3k ) W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 168. PROBLEM 3.19 Determine the moment about the origin O of the force F = 4i + 5j − 3k that acts at a Point A. Assume that the position vector of A is (a) r = 2i − 3j + 4k, (b) r = 2i + 2.5j − 1.5k, (c) r = 2i + 5j + 6k. SOLUTION (a) i j k M O = 2 −3 4 4 5 −3 = (9 − 20)i + (16 + 6) j + (10 + 12)k (b) i j k M O = 2 2.5 −1.5 4 5 −3 = (−7.5 + 7.5)i + (−6 + 6) j + (10 − 10)k (c) M O = −11i + 22 j + 22k W MO = 0 W i j k MO = 2 5 6 4 5 −3 = (−15 − 30)i + (24 + 6) j + (10 − 20)k M O = −45i + 30 j − 10k W 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. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 169. PROBLEM 3.20 Determine the moment about the origin O of the force F = −2i + 3j + 5k that acts at a Point A. Assume that the position vector of A is (a) r = i + j + k, (b) r = 2i + 3j − 5k, (c) r = −4i + 6j + 10k. SOLUTION (a) i j k MO = 1 1 1 −2 3 5 = (5 − 3)i + (−2 − 5) j + (3 + 2)k (b) i j k MO = 2 3 − 5 −2 3 5 = (15 + 15)i + (10 − 10) j + (6 + 6)k (c) M O = 2i − 7 j + 5k W M O = 30i + 12k W i j k M O = −4 6 10 −2 3 5 = (30 − 30)i + ( −20 + 20) j + (−12 + 12)k MO = 0 W 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. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 170. PROBLEM 3.21 A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A. SOLUTION 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 − cos 30° sin 30° 0 = 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 171. PROBLEM 3.22 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B. SOLUTION We have M O = rB/O × FB where rB/O = (7 m) j FB = TAB + TBC TAB = Ȝ BATAB = −(0.75 m)i − (7 m) j + (6 m)k (.75) 2 + (7) 2 + (6) 2 m (555 N) TBC = Ȝ BC TBC = (4.25 m)i − (7 m) j + (1 m)k (4.25) 2 + (7) 2 + (1) 2 m (660 N) FB = [−(45.00 N)i − (420.0 N) j + (360.0 N)k ] + [(340.0 N)i − (560.0 N) j + (80.00 N)k ] = (295.0 N)i − (980.0 N) j + (440.0 N)k and i j k MO = 0 7 0 N⋅m 295 980 440 = (3080 N ⋅ m)i − (2070 N ⋅ m)k or M O = (3080 N ⋅ m)i − (2070 N ⋅ m)k W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 172. PROBLEM 3.23 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 = 7.6 m 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 173. PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 57-lb force directed along BA. Determine the moment about C of that force. SOLUTION We have M C = rA/C × FBA where rA/C = (48 in.)i − (6 in.) j + (36 in.)k and FBA = Ȝ BA FBA ª −(5 in.)i + (90 in.) j − (30 in.)k º » (57 lb) =« « (5)2 + (90) 2 + (30) 2 in. » ¬ ¼ = −(3 lb)i + (54 lb) j − (18 lb)k i j k M C = 48 6 36 lb ⋅ in. 3 54 18 = −(1836 lb ⋅ in.)i + (756 lb ⋅ in.) j + (2574 lb ⋅ in.) or M C = −(153.0 lb ⋅ ft)i + (63.0 lb ⋅ ft) j + (215 lb ⋅ ft)k W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 174. PROBLEM 3.25 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. SOLUTION We have M A = rE/A × TDE where (a) rE/A = (2.3 m) j TDE = Ȝ 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 W We have M A = rG/A × TCG where (b) rG/A = (2.7 m)i + (2.3 m) j TCG = Ȝ 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 176. PROBLEM 3.27 In Problem 3.22, determine the perpendicular distance from Point O to cable AB. PROBLEM 3.22 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B. SOLUTION We have where | M O | = TBA d d = perpendicular distance from O to line AB. Now M O = rB/O × TBA and rB/O = (7 m) j TBA = Ȝ BATAB = −(0.75 m)i − (7 m) j + (6 m)k (0.75) 2 + (7) 2 + (6) 2 m (555 N) = −(45.0 N)i − (420 N) j + (360 N)k i j k MO = 0 7 0 N⋅m −45 −420 360 = (2520.0 N ⋅ m)i + (315.00 N ⋅ m)k and | M O | = (2520.0) 2 + (315.00) 2 = 2539.6 N ⋅ m 2539.6 N ⋅ m = (555 N)d or d = 4.5759 m or d = 4.58 m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 177. PROBLEM 3.28 In Problem 3.22, determine the perpendicular distance from Point O to cable BC. PROBLEM 3.22 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B. SOLUTION We have where | M O | = TBC d d = perpendicular distance from O to line BC. M O = rB/O × TBC rB/O = 7 mj TBC = Ȝ BC TBC = (4.25 m)i − (7 m) j + (1 m)k (4.25)2 + (7) 2 + (1) 2 m (660 N) = (340 N)i − (560 N) j + (80 N)k i j k MO = 0 7 0 340 −560 80 = (560 N ⋅ m)i − (2380 N ⋅ m)k and | M O | = (560)2 + (2380) 2 = 2445.0 N ⋅ m 2445.0 N ⋅ m = (660 N)d d = 3.7045 m or d = 3.70 m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 178. PROBLEM 3.29 In Problem 3.24, determine the perpendicular distance from Point D to a line drawn through Points A and B. PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 57-lb force directed along BA. Determine the moment about C of that force. SOLUTION We have where | M D | = FBA d d = perpendicular distance from D to line AB. M D = rA/D × FBA rA/D = −(6 in.) j + (36 in.)k FBA = Ȝ BA FBA = (−(5 in.)i + (90 in.) j − (30 in.)k ) (5)2 + (90) 2 + (30) 2 in. (57 lb) = −(3 lb)i + (54 lb) j − (18 lb)k i j k M D = 0 −6 36 lb ⋅ in. −3 54 −18 = −(1836.00 lb ⋅ in.)i − (108.000 lb ⋅ in.) j − (18.0000 lb ⋅ in.)k and | M D | = (1836.00) 2 + (108.000)2 + (18.0000)2 = 1839.26 lb ⋅ in. 1839.26 lb ⋅ in = (57 lb)d d = 32.268 in. or d = 32.3 in. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 179. PROBLEM 3.30 In Problem 3.24, determine the perpendicular distance from Point C to a line drawn through Points A and B. PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 57-lb force directed along BA. Determine the moment about C of that force. SOLUTION We have where | M C | = FBA d d = perpendicular distance from C to line AB. M C = rA/C × FBA rA/C = (48 in.)i − (6 in.) j + (36 in.)k FBA = Ȝ BA FBA = ( −(5 in.)i + (90 in.) j − (30 in.)k ) (5) 2 + (90) 2 + (30) 2 in. (57 lb) = −(3 lb)i + (54 lb) j − (18 lb)k i j k M C = 48 −6 36 lb ⋅ in. −3 54 −18 = −(1836.00lb ⋅ in.)i − (756.00 lb ⋅ in.) j + (2574.0 lb ⋅ in.)k and | M C | = (1836.00) 2 + (756.00) 2 + (2574.0)2 = 3250.8 lb ⋅ in. 3250.8 lb ⋅ in. = 57 lb d = 57.032 in. or d = 57.0 in. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 180. PROBLEM 3.31 In Problem 3.25, determine the perpendicular distance from Point A to portion DE of cable DEF. PROBLEM 3.25 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. SOLUTION We have where | M A | = TDE d d = perpendicular distance from A to line DE. M A = rE/A × TDE rE/A = (2.3 m) j TDE = Ȝ 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 = − (1242.00 N ⋅ m)i − (248.00 N ⋅ m)k and |M A | = (1242.00)2 + (248.00) 2 = 1266.52 N ⋅ m 1266.52 N ⋅ m = (810 N)d d = 1.56360 m or d = 1.564 m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 181. PROBLEM 3.32 In Problem 3.25, determine the perpendicular distance from Point A to a line drawn through Points C and G. PROBLEM 3.25 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. SOLUTION We have where |M A | = TCG d d = perpendicular distance from A to line CG. M A = rG/A × TCG rG/A = (2.7 m)i + (2.3 m) j TCG = ȜCGTCG = −(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 0 N⋅m M A = 2.7 2.3 −108 594 −540 = −(1242.00 N ⋅ m)i + (1458.00 N ⋅ m) j + (1852.00 N ⋅ m)k and |M A | = (1242.00) 2 + (1458.00) 2 + (1852.00)2 = 2664.3 N ⋅ m 2664.3 N ⋅ m = (810 N)d d = 3.2893 m or d = 3.29 m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 182. PROBLEM 3.33 In Problem 3.26, determine the perpendicular distance from Point C to portion AD of the line ABAD. PROBLEM 3.26 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 First compute the moment about C of the force FDA exerted by the line on D: From Problem 3.26: 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 183. 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 ) d = perpendicular distance from C to line AB Where F = Ȝ AB F = (24 ft) i + (24 ft) j − (28) k (24) 2 + (24)2 + (18) 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 184. PROBLEM 3.35 Given the vectors P = 3i − j + 2k , Q = 4i + 5 j − 3k , and S = −2i + 3j − k , compute the scalar products P ⋅ Q, P ⋅ S, and Q ⋅ S. SOLUTION P ⋅ Q = (3i − 1j + 2k ) ⋅ (4i − 5 j − 3k ) = (3)(4) + (−1)(−5) + (2)(−3) =1 or P ⋅Q =1 W or P ⋅ S = −11 W or Q ⋅ S = 10 W P ⋅ S = (3i − 1j + 2k ) ⋅ (−2i + 3j − 1k ) = (3)(−2) + (−1)(3) + (2)( −1) = −11 Q ⋅ S = (4i − 5 j − 3k ) ⋅ ( −2i + 3j − 1k ) = (4)(−2) + (5)(3) + ( −3)(−1) = 10 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 185. PROBLEM 3.36 Form the scalar products B ⋅ C and B′ ⋅ C, where B = B ′, and use the results obtained to prove the identity cos a cos β = 1 1 cos (a + β ) + cos (a − β ). 2 2 SOLUTION By definition B ⋅ C = BC cos(α − β ) B = B [(cos β )i + (sin β ) j] C = C [(cos α )i + (sin α ) j] where (B cos β )( C cos α ) + ( B sin β )(C sin α ) = BC cos (α − β ) cos β cos α + sin β sin α = cos(α − β ) or (1) B′ ⋅ C = BC cos (α + β ) By definition B′ = [(cos β )i − (sin β ) j] where (B cos β ) (C cos α ) + (− B sin β )(C sin α ) = BC cos (α + β ) or cos β cos α − sin β sin α = cos (α + β ) (2) Adding Equations (1) and (2), 2 cos β cos α = cos (α − β ) + cos (α + β ) or cos α cos β = 1 1 cos (α + β ) + cos (α − β ) W 2 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 186. PROBLEM 3.37 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. SOLUTION First note JJJ G AB = AB (sin 37° j − cos 37°k ) CD = CD(− cos 40° cos 55° j + sin 40° j − cos 40° sin 55°k ) Now JJJ JJJ G G 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 187. PROBLEM 3.38 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 EF. SOLUTION First note Now or JJJ G AB = AB (sin 37° j − cos 37°k ) JJJ G EF = EF (cos 32° cos 45°i + sin 32° j − cos 32° sin 45°k ) JJJ JJJ G G AB ⋅ EF = ( AB )( EF ) cos θ AB (sin 37° j − cos 37°k ) ⋅ EF (cos 32° cos 45° j + sin 32° j − cos 32° sin 45°k ) = ( AB )( EF ) cos θ or cos θ = (sin 37°)(sin 32°) + ( − cos 37°)( − cos 32° sin 45°) = 0.79782 or θ = 37.1° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 188. PROBLEM 3.39 Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC. SOLUTION 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 JJJ G AB = −(6.5 ft)i − (8 ft) j + (2 ft)k JJJG AC = −(8 ft) j + (6 ft)k JJJ JJJG G 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 189. PROBLEM 3.40 Consider the volleyball net shown. Determine the angle formed by guy wires AC and AD. SOLUTION First note AC = (0)2 + (−8) 2 + (6) 2 = 10 ft AD = (4) 2 + ( −8) 2 + (1) 2 and By definition or = 9 ft JJJG AC = −(8 ft)j + (6 ft)k JJJG AD = (4 ft) j − (8 ft) j + (1 ft)k JJJG JJJG 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 = 0.777 78 or θ = 38.9° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 190. PROBLEM 3.41 Knowing that the tension in cable AC is 1260 N, determine (a) the angle between cable AC and the boom AB, (b) the projection on AB of the force exerted by cable AC at Point A. SOLUTION (a) First note AC = (−2.4) 2 + (0.8)2 + (1.2) 2 = 2.8 m AB = (−2.4) 2 + (−1.8)2 + (0) 2 and By definition or = 3.0 m JJJG AC = −(2.4 m)i + (0.8 m) j + (1.2 m)k JJJ G AB = −(2.4 m)i − (1.8 m) j JJJG JJJ G AC ⋅ AB = ( AC )( AB) cos θ (−2.4i + 0.8 j + 1.2k ) ⋅ (−2.4i − 1.8 j) = (2.8)(30) × cos θ or or (b) (−2.4)(−2.4) + (0.8)(−1.8) + (1.2)(0) = 8.4cos θ cos θ = 0.514 29 We have or θ = 59.0° W (TAC ) AB = TAC ⋅ λ AB = TAC cos θ = (1260 N)(0.51429) or (TAC ) AB = 648 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 192. PROBLEM 3.43 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Knowing that the distance from O to P is 6 in. and that the tension in the cord is 3 lb, determine (a) the angle between the elastic cord and the rod OA, (b) the projection on OA of the force exerted by cord PC at Point P. SOLUTION First note OA = (12)2 + (12)2 + (−6) 2 = 18 in. OA 1 = (12i + 12 j − 6k ) OA 18 1 = (2i + 2 j − k ) 3 Then λ OA = Now 1 OP = 6 in.  OP = (OA) 3 The coordinates of Point P are (4 in., 4 in., −2 in.) JJJ G PC = (5 in.)i + (11 in.) j + (14 in.)k so that PC = (5) 2 + (11) 2 + (14)2 = 342 in. and (a) We have or or JJJ G PC ⋅ λ OA = ( PC ) cos θ 1 (5i + 11j + 14k ) ⋅ (2i + 2 j − k ) = 342 cos θ 3 1 [(5)(2) + (11)(2) + (14)( −1)] 3 342 = 0.324 44 cos θ = or θ = 71.1° W (b) We have (TPC ) OA = TPC ⋅ λ OA = (TPC λ PC ) ⋅ λ OA PC ⋅ λ OA PC = TPC cos θ = TPC = (3 lb)(0.324 44) or (TPC )OA = 0.973 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 193. PROBLEM 3.44 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Determine the distance from O to P for which cord PC and rod OA are perpendicular. SOLUTION First note OA = (12)2 + (12)2 + (−6)2 = 18 in. Then λ OA = OA 1 = (12i + 12 j − 6k ) OA 18 1 = (2i + 2 j − k ) 3 Let the coordinates of Point P be (x in., y in., z in.). Then JJJ G PC = [(9 − x)in.]i + (15 − y )in.] j + [(12 − z )in.]k Also, and JJJ G d OP = dOP λ OA = OP (2i + 2 j − k ) 3 JJJ G OP = ( x in.)i + ( y in.) j + ( z in.)k 2 2 1 x = dOP y = dOP z = dOP 3 3 3 The requirement that OA and PC be perpendicular implies that JJJ G λ OA ⋅ PC = 0 or 1 (2 j + 2 j − k ) ⋅ [(9 − x)i + (15 − y ) j + (12 − z )k ] = 0 3 or ª 2 2 § · § · § 1 ·º (2) ¨ 9 − dOP ¸ + (2) ¨15 − dOP ¸ + (−1) «12 − ¨ − dOP ¸ » = 0 3 3 © ¹ © ¹ © 3 ¹¼ ¬ or dOP = 12.00 in. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 194. PROBLEM 3.45 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. (a) 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 W (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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 195. PROBLEM 3.46 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). −2 3 4 −5 = 0 −1 2 Then 4 2 Sx or 32 + 10S x − 6 − 20 + 8 − 12S x = 0 Sx = 7 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 196. PROBLEM 3.47 The 0.61 × 1.00-m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 66 N, determine the moment about each of the coordinate axes of the force exerted by the cord at D. SOLUTION First note z = (0.61) 2 − (0.11)2 = 0.60 m Then d DE = (0.3) 2 + (0.6) 2 + (−0.6) 2 = 0.9 m 66 N (0.3i + 0.6 j − 0.6k ) 0.9 = 22[(1 N)i + (2 N) j − (2 N)k ] and TDE = Now M A = rD/A × TDE where rD/A = (0.11 m) j + (0.60 m)k Then i j k M A = 22 0 0.11 0.60 −2 1 2 = 22[(−0.22 − 1.20)i + 0.60 j − 0.11k ] = − (31.24 N ⋅ m)i + (13.20 N ⋅ m) j − (2.42 N ⋅ m)k M x = −31.2 N ⋅ m, M y = 13.20 N ⋅ m, M z = −2.42 N ⋅ m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 197. PROBLEM 3.48 The 0.61 × 1.00-m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 66 N, determine the moment about each of the coordinate axes of the force exerted by the cord at C. SOLUTION First note z = (0.61) 2 − (0.11)2 = 0.60 m Then dCE = (−0.7) 2 + (0.6) 2 + (−0.6) 2 = 1.1 m 66 N (−0.7i + 0.6 j − 0.6k ) 1.1 = 6[−(7 N)i + (6 N) j − (6 N)k ] and TCE = Now M A = rE/A × TCE where rE/A = (0.3 m)i + (0.71 m) j Then i j k M A = 6 0.3 0.71 0 −7 −6 6 = 6[ −4.26i + 1.8 j + (1.8 + 4.97)k ] = − (25.56 N ⋅ m)i + (10.80 N ⋅ m) j + (40.62 N ⋅ m)k M x = −25.6 N ⋅ m, M y = 10.80 N ⋅ m, M z = 40.6 N ⋅ m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 198. PROBLEM 3.49 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 JJJ G 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 TBA (2.2i − 3.2 j − ak ) (N) d BA MD = i j k TBA 2.2 1.6 0 d BA 2.2 −3.2 − a = Thus 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 TBA (N ⋅ m) d or a = 1.252 m W BA PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 199. PROBLEM 3.50 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 man applies a 195-N force to end A of the rope and that the moment of that force about the y axis is 132 N ⋅ m, determine the distance a. SOLUTION d BA = (2.2) 2 + (−3.2) 2 + (−a ) 2 First note = 15.08 + a 2 m 195 N (2.2i − 3.2 j − a k ) d BA and TBA = Now M y = j ⋅ (rA/D × TBA ) where rA/0 = (2.2 m)i + (1.6 m) j Then 0 1 0 195 My = 2.2 1.6 0 d BA 2.2 −3.2 − a = 195 (2.2a ) (N ⋅ m) d BA Substituting for My and dBA 132 N ⋅ m = or 195 15.08 + a 2 (2.2a ) 0.30769 15.08 + a 2 = a Squaring both sides of the equation 0.094675(15.08 + a 2 ) = a 2 or a = 1.256 m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 201. PROBLEM 3.52 For the davit of Problem 3.51, determine the largest allowable distance x when the tension in line ABAD is 60 lb. SOLUTION From the solution of Problem 3.51, TAD is now TAD = T = AD AD 60 lb x 2 + ( −7.75) 2 + (−3)2 ( xi − 7.75 j − 3k ) Then M z = k ⋅ (rA / C × TAD ) becomes 279 = 279 = 60 x 2 + (−7.75) 2 + ( −3) 2 60 0 0 1 0 7.75 3 x −7.75 −3 | − (1)(7.75)( x) | 2 x + 69.0625 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 202. PROBLEM 3.53 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. SOLUTION 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. 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 − 9.9694cos φ ) in. = −43 lb ⋅ ft(12 in./ft) and (3) From Equation (3) § 634.33 · φ = cos −1 ¨ ¸ = 24.636° © 697.86 ¹ or From Equation (1) § 1022.90 · d =¨ ¸ = 34.577 in. © 29.583 ¹ or d = 34.6 in. W φ = 24.6° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 203. PROBLEM 3.54 When a force F is applied to the handle of the valve shown, its moments about the x and z axes are, respectively, M x = −77 lb ⋅ ft and M z = −81 lb ⋅ ft. For d = 27 in., determine the moment My of F about the y axis. SOLUTION We have ΣM O : rA/O × F = M O Where rA/O = −(4 in.)i + (11 in.) j − (27 in.)k F = F (cos θ cos φ i − sin θ j + cos θ sin φ k ) MO = F i −4 cos θ cos φ j 11 − sin θ k −27 lb ⋅ in. cos θ sin φ = F [(11cos θ sin φ − 27 sin θ )i + (−27 cos θ cos φ + 4 cos θ sin φ ) j + (4sin θ − 11cos θ cos φ )k ](lb ⋅ in.) M x = F (11cos θ sin φ − 27sin θ )(lb ⋅ in.) (1) M y = F (−27 cos θ cos φ + 4cos θ sin φ ) (lb ⋅ in.) (2) M z = F (4sin θ − 11cos θ cos φ ) (lb ⋅ in.) and (3) Now, Equation (1) cos θ sin φ = 1 § Mx · + 27sin θ ¸ 11 ¨ F © ¹ (4) and Equation (3) cos θ cos φ = Mz · 1§ ¨ 4sin θ − F ¸ 11 © ¹ (5) Substituting Equations (4) and (5) into Equation (2), ­ ª1 §M ª1§ M ·º ·º ½ ° ° M y = F ®−27 « ¨ 4sin θ − z ¸ » + 4 « ¨ x + 27 sin θ ¸ » ¾ 11 © 11 © F F ¹¼ ¹¼ ° ° ¬ ¬ ¯ ¿ or My = 1 (27 M z + 4 M x ) 11 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 204. PROBLEM 3.54 (Continued) 4 Noting that the ratios 27 and 11 are the ratios of lengths, have 11 27 4 (−81 lb ⋅ ft) + (−77 lb ⋅ ft) 11 11 = 226.82 lb ⋅ ft My = or M y = −227 lb ⋅ ft W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 207. PROBLEM 3.57 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 First note d AE = (0.9) 2 + (−0.6) 2 + (0.2) 2 = 1.1 m Then TAE = 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 ] DB = (1.2) 2 + ( −0.35) 2 + (0)2 λ DB = 1.25 m JJJ G DB = DB 1 (1.2i − 0.35 j) = 1.25 1 = (24i − 7 j) 25 M DB = λ DB ⋅ (rA/D × TAE ) TDA = −(0.1 m) j + (0.2 m)k M DB 24 −7 0 1 = (5) 0 −0.1 0.2 25 −6 9 2 1 = (−4.8 − 12.6 + 28.8) 5 or M DB = 2.28 N ⋅ m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 208. PROBLEM 3.58 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 First note dCF = (0.6) 2 + (−0.9) 2 + (−0.2) 2 = 1.1 m Then TCF = Also DB = (1.2) 2 + ( −0.35) 2 + (0)2 33 N (0.6i − 0.9 j + 0.2k ) 1.1 = 3[(6 N)i − (9 N) j − (2 N)k ] = 1.25 m JJJ G 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 209. PROBLEM 3.59 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 We have M OA = λOA ⋅ (rC/O × P) where From triangle OBC (OA) x = a 2 (OA) z = (OA) x tan 30° = Since a§ 1 · a ¨ ¸= 2© 3 ¹ 2 3 (OA) 2 = (OA) 2 + (OA) 2 + (OAz )2 x y 2 or § a · §a· a 2 = ¨ ¸ + (OA) 2 + ¨ ¸ y ©2¹ ©2 3¹ (OA) y = a 2 − 2 a2 a2 2 − =a 4 12 3 Then rA/O = 2 a a i+a j+ k 2 3 2 3 and Ȝ OA = 1 2 1 i+ j+ k 2 3 2 3 P = λBC P = (a sin 30°)i − (a cos 30°)k P ( P) = (i − 3k ) 2 a rC/O = ai 2 3 0 2 3 §P· (a) ¨ ¸ 0 ©2¹ 1 M OA 1 2 = 1 0 − 3 = 1 aP § 2 · aP (1)(− 3) = ¨− ¨ 3¸ ¸ 2 © 2 ¹ M OA = aP W 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 210. PROBLEM 3.60 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 3.59 to determine the perpendicular distance between edges OA and BC. SOLUTION (a) For edge OA to be perpendicular to edge BC, JJJ JJJ G G OA ⋅ BC = 0 where From triangle OBC (OA) x = a 2 a§ 1 · a ¨ ¸= 2© 3 ¹ 2 3 JJJ § a · G § a · OA = ¨ ¸ i + (OA) y j + ¨ ¸k ©2¹ ©2 3¹ JJJ G BC = ( a sin 30°) i − (a cos 30°) k (OA) z = (OA) x tan 30° = and = Then or so that (b) Have M OA a a 3 a i− k = (i − 3 k ) 2 2 2 ªa § a · º a « i + (OA) y j + ¨ ¸ k » ⋅ (i − 3k ) = 0 2 ©2 3¹ ¼ ¬2 a2 a2 + (OA) y (0) − =0 4 4 JJJ JJJ G G OA ⋅ BC = 0 JJJ G JJJ G OA is perpendicular to BC . W JJJ G JJJ G = 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 or d = = Pd a W 2 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 212. PROBLEM 3.61 (Continued) where Then rE/A = (36 in.)i + (96 in.) j + (27 in.)k M AD = = −1 3 4 (2) 36 96 27 26 −3 −22 6 1 2 26 (2304 + 81 − 2376 + 864 + 216 + 2376) or M AD = 1359 lb ⋅ in. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 214. PROBLEM 3.62 (Continued) where Then rE/A = (36 in.)i + (96 in.) j + (27 in.)k M AD = = 4 −1 3 (6) 36 96 27 26 1 − 8 −4 1 6 26 (−1536 − 27 − 864 − 288 − 144 + 864) or M AD = −2350 lb ⋅ in. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 215. 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 Let M1 = moment of F2 about the line of action of M1 and M 2 = moment of F1 about the line of action of M 2 Now, by definition M1 = λ1 ⋅ (rB/A × F2 ) = λ1 ⋅ (rB/A × λ 2 ) F2 M 2 = λ2 ⋅ (rA/B × F1 ) = λ2 ⋅ (rA/B × λ1 ) F1 Since F1 = F2 = F and rA/B = −rB/A M1 = λ1 ⋅ (rB/A × λ 2 ) F M 2 = λ 2 ⋅ (−rB/A × λ1 ) F Using Equation (3.39) so that λ1 ⋅ (rB/A × λ 2 ) = λ 2 ⋅ ( −rB/A × λ1 ) M 2 = λ 1⋅ (rB/A × λ 2 ) F M12 = M 21 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 218. PROBLEM 3.66 In Problem 3.57, determine the perpendicular distance between cable AE and the line joining Points D and B. PROBLEM 3.57 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 From the solution to Problem 3.57 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,JJJ follows that only the perpendicular component of TAE will itG 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 219. PROBLEM 3.67 In Problem 3.58, determine the perpendicular distance between cable CF and the line joining Points D and B. PROBLEM 3.58 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 From the solution to Problem 3.58 ΤCF = 33 N 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,JJJ follows that only the perpendicular component of TCF will itG 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 222. PROBLEM 3.70 Two parallel 60-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 (a) We have where ΣM B : − d1C x + d 2 C y = M d1 = (0.360 m) sin 55° = 0.29489 m d 2 = (0.360 m) sin 55° = 0.20649 m C x = (60 N) cos 20° = 56.382 N C y = (60 N)sin 20° = 20.521 N M = −(0.29489 m)(56.382 N)k + (0.20649 m)(20.521 N)k = −(12.3893 N ⋅ m)k (b) We have We have W or M = 12.39 N ⋅ m W or M = 12.39 N ⋅ m W M = Fd (−k ) = 60 N[(0.360 m) sin(55° − 20°)]( −k ) = −(12.3893 N ⋅ m)k (c) or M = 12.39 N ⋅ m ΣM A : Σ(rA × F) = rB/A × FB + rC/A × FC = M i j k sin 55° 0 M = (0.520 m)(60 N) cos 55° − cos 20° − sin 20° 0 i j k + (0.800 m)(60 N) cos 55° sin 55° 0 cos 20° sin 20° 0 = (17.8956 N ⋅ m − 30.285 N ⋅ m)k = −(12.3892 N ⋅ m)k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 223. PROBLEM 3.71 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) = 336 lb ⋅ in. (b) (c) 336 lb ⋅ in. − d 2 (12 lb) = 0 d 2 = 28.0 in. W M total = M1 + M 2 We have or W 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 224. PROBLEM 3.72 A couple M of magnitude 18 N ⋅ m is applied to the handle of a screwdriver to tighten a screw into a block of wood. Determine the magnitudes of the two smallest horizontal forces that are equivalent to M if they are applied (a) at corners A and D, (b) at corners B and C, (c) anywhere on the block. SOLUTION (a) M = Pd We have or 18 N ⋅ m = P(.24 m) P = 75.0 N or Pmin = 75.0 N W d BC = ( BE ) 2 + ( EC ) 2 (b) = (.24 m) 2 + (.08 m) 2 = 0.25298 m M = Pd We have 18 N ⋅ m = P(0.25298 m) P = 71.152 N or P = 71.2 N W d AC = ( AD) 2 + ( DC ) 2 (c) = (0.24 m) 2 + (0.32 m) 2 = 0.4 m M = Pd AC We have 18 N ⋅ m = P(0.4 m) P = 45.0 N or P = 45.0 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 225. PROBLEM 3.73 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 = (35 lb)(7 in.) + (25 lb)(9 in.) = 245 lb ⋅ in. + 225 lb ⋅ in. (a) M = 470 lb ⋅ in. (b) W 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: With pegs A and D: With pegs B and C: (c) θ = 53.1° W θ = 53.1° W 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 ¹ = 11 in. M = Fd We must have 470 lb ⋅ in. = F (11 in.) F = 42.7 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 226. PROBLEM 3.74 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 485 lb ⋅ in. = [(6 + d )in.](35 lb) + [(8 + d )in.](25 lb) d = 1.250 in. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 227. PROBLEM 3.75 The shafts of an angle drive are acted upon by the two couples shown. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION Based on where M = M1 + M 2 M1 = −(8 lb ⋅ ft)j M 2 = −(6 lb ⋅ ft)k M = −(8 lb ⋅ ft)j − (6 lb ⋅ ft)k and |M| = (8) 2 + (6) 2 = 10 lb ⋅ ft or M = 10.00 lb ⋅ ft W M |M| −(8 lb ⋅ ft)j − (6 lb ⋅ ft)k = 10 lb ⋅ ft = −0.8 j − 0.6k Ȝ= or M = |M| Ȝ = (10 lb ⋅ ft)(−0.8 j − 0.6k ) cos θ x = 0 cos θ y = −0.8 θ x = 90° θ y = 143.130° cos θ z = −0.6 θ z = 126.870° or θ x = 90.0° θ y = 143.1° θ z = 126.9° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 228. PROBLEM 3.76 If P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION We have where M = M1 + M 2 M1 = rG/C × F1 rG/C = −(0.3 m)i F1 = (18 N)k M1 = −(0.3 m)i × (18 N)k = (5.4 N ⋅ m) j Also, M 2 = rD/F × F2 rD/F = −(.15 m)i + (.08 m) j F2 = λED F2 = (.15 m)i + (.08 m) j + (.17 m)k (.15)2 + (.08) 2 + (.17) 2 m (34 N) = 141.421 N ⋅ m(.15i + .08j + .17k ) i j k M 2 = 141.421 N ⋅ m −.15 .08 0 −.15 .08 .17 = 141.421(.0136i + 0.0255 j)N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 229. PROBLEM 3.76 (Continued) and M = [(5.4 N ⋅ m)j] + [141.421(.0136i + .0255 j) N ⋅ m] = (1.92333 N ⋅ m)i + (9.0062 N ⋅ m)j | M | = (M x )2 + (M y )2 = (1.92333)2 + (9.0062) 2 = 9.2093 N ⋅ m λ= or M = 9.21 N ⋅ m W M (1.92333 N ⋅ m)i + (9.0062 N ⋅ m) j = |M| 9.2093 N ⋅ m = 0.20885 + 0.97795 cos θ x = 0.20885 θ x = 77.945° or θ x = 77.9° W or θ y = 12.05° W or θ z = 90.0° W cos θ y = 0.97795 θ y = 12.054° cos θ z = 0.0 θ z = 90° PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 230. PROBLEM 3.77 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 M1 = rC × F1 = (30 in.)i × [−(16 lb) j] = −(480 lb ⋅ in.)k 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. W 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° W cos θ z = −0.35045 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 231. PROBLEM 3.78 If P = 20 lb, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION From the solution to Problem. 3.77 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 = 1169.96 lb ⋅ in M = 1170 lb ⋅ in. W 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° W cos θ z = −0.180921 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 232. PROBLEM 3.79 If P = 20 N, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION We have M = M1 + M 2 + M 3 where i j k M1 = rG/C × F1 = −0.3 0 0 N ⋅ m = (5.4 N ⋅ m)j 0 0 18 M 2 = rD/F i j k × F2 = −.15 .08 0 141.421 N ⋅ m −.15 .08 .17 = 141.421(.0136i + .0255 j)N ⋅ m (See Solution to Problem 3.76.) i j k M 3 = rC/A × F3 = 0.3 0 0.17 N ⋅ m 0 20 0 = −(3.4 N ⋅ m)i + (6 N ⋅ m)k M = [(1.92333 − 3.4)i + (5.4 + 3.6062) j + (6)k ] N ⋅ m = −(1.47667 N ⋅ m)i + (9.0062 N ⋅ m) j + (6 N ⋅ m) 2 2 2 |M| = M x + M y + M z = (1.47667) + (9.0062) + (6)2 or M = 10.92 N ⋅ m W = 10.9221 N ⋅ m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 233. PROBLEM 3.79 (Continued) M −1.47667 + 9.0062 + 6 = |M| 10.9221 = −0.135200i + 0.82459 j + 0.54934k Ȝ= cos θ x = −0.135200 θ x = 97.770 or θ x = 97.8° W cos θ y = 0.82459 θ y = 34.453 or θ y = 34.5° W cos θ z = 0.54934 θ z = 56.678 or θ z = 56.7° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 234. PROBLEM 3.80 Shafts A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and B lie in the vertical yz plane, while shaft C is directed along the x axis. Replace the couples applied to the shafts with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION Represent the given couples by the following couple vectors: M A = −1600sin 20° j + 1600cos 20°k = −(547.232 N ⋅ m) j + (1503.51 N ⋅ m)k M B = 1200sin 20° j + 1200 cos 20°k = (410.424 N ⋅ m) j + (1127.63 N ⋅ m)k M C = −(1120 N ⋅ m)i The single equivalent couple is M = M A + M B + MC = −(1120 N ⋅ m)i − (136.808 N ⋅ m) j + (2631.1 N ⋅ m)k M = (1120) 2 + (136.808)2 + (2631.1)2 = 2862.8 N ⋅ m −1120 cos θ x = 2862.8 −136.808 cos θ y = 2862.8 2631.1 cos θ z = 2862.8 M = 2860 N ⋅ m θ x = 113.0° θ y = 92.7° θ z = 23.2° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 235. PROBLEM 3.81 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 (a) Based on ΣF : FA = T = 560 lb FA = 560 lb or 20° W 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 W 20° W 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 236. PROBLEM 3.82 A 160-lb force P is applied at Point A of a structural member. Replace P with (a) an equivalent force-couple system at C, (b) an equivalent system consisting of a vertical force at B and a second force at D. SOLUTION (a) Based on ΣF : PC = P = 160 lb or PC = 160 lb 60° W ΣM C : M C = − Px d cy + Py dCx where Px = (160 lb) cos 60° = 80 lb Py = (160 lb)sin 60° dCx = 138.564 lb = 4 ft dCy = 2.75 ft M C = (80 lb)(2.75 ft) + (138.564 lb)(4 ft) = 220 lb ⋅ ft + 554.26 lb ⋅ ft = 334.26 lb ⋅ ft (b) Based on or M C = 334 lb ⋅ ft W ΣFx : PDx = P cos 60° = (160 lb) cos 60° = 80 lb ΣM D : ( P cos 60°)( d DA ) = PB (d DB ) [(160 lb) cos 60°](1.5 ft) = PB (6 ft) PB = 20.0 lb or PB = 20.0 lb W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 237. PROBLEM 3.82 (Continued) ΣFy : P sin 60° = PB + PDy (160 lb)sin 60° = 20.0 lb + PDy PDy = 118.564 lb PD = ( PDx ) 2 + ( PDy ) 2 = (80) 2 + (118.564) 2 = 143.029 lb § PDy · ¸ © PDx ¹ § 118.564 · = tan −1 ¨ ¸ © 80 ¹ = 55.991° θ = tan −1 ¨ or PD = 143.0 lb 56.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 238. PROBLEM 3.83 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 W ΣM : M B = Fd B = 80 N (.05 m) = 4.0000 N ⋅ m M B = 4.00 N ⋅ m or (b) W 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 (.04 m) FC = 100.000 N or FC = 100.0 N W ΣFy : 0 = FD − FC FD = 100.000 N or FD = 100.0 N W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 239. PROBLEM 3.84 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. SOLUTION 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° W 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 240. PROBLEM 3.85 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 F = 250 N (b) M = 75.0 N ⋅ m 60° W Require 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 241. PROBLEM 3.86 Solve Problem 3.85, assuming α = β = 25°. SOLUTION (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 W 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° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 242. PROBLEM 3.87 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. SOLUTION (a) We have ΣF : F = (360 N) j − (360 N) j − (600 N)k or F = −(600 N)k W ΣM D : (360 N)(0.15 m) = (600 N)(d ) and d = 0.09 m or d = 90.0 mm below ED W (b) We have from Part a F = −(600 N)k W ΣM D : −(360 N)(0.15 m) = −(600 N)( d ) and d = 0.09 m or d = 90.0 mm above ED W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 243. PROBLEM 3.88 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 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 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 244. PROBLEM 3.89 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. SOLUTION 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. Have FB = 2.9 lb − 2.65 lb = 0.25 lb, where the 2.65 lb force be 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. = [(.25 lb) cos 25°](a ) a = 63.600 in. F′ = (.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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 245. PROBLEM 3.90 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 (a) First note that the two 20-lb forces form A couple. Then F = 48 lb θ where θ = 180° − (60° + 55°) = 65° and M = ΣM B = (30 in.)(48 lb) cos55° − (70 in.)(20 lb) cos 20° = −489.62 lb ⋅ in The equivalent force-couple system at B is F = 48.0 lb (b) 65° M = 490 lb ⋅ in. W 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 −489.62 lb ⋅ in. = −a (48.0 lb) cos 55° or a = 17.78 in. F ′ = 48.0 lb 65.0° W and is applied to the lever 17.78 in. W To the left of pin B PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 246. PROBLEM 3.91 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, must require that P be perpendicular to rB/E . Then ΣM E : (0.2 sin 30° + 0.2)m × 300 N + (0.2 m)sin 30° × 300 N − (0.4 m) Pmin = 0 or Pmin = 300 N Pmin = 300 N 30.0° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 247. PROBLEM 3.92 A rectangular plate is acted upon by the force and couple shown. This system is to be replaced with a single equivalent force. (a) For α = 40°, specify the magnitude and the line of action of the equivalent force. (b) Specify the value of α if the line of action of the equivalent force is to intersect line CD 300 mm to the right of D. SOLUTION (a) The given force-couple system (F, M) at B is F = 48 N and M = ΣM B = (0.4 m)(15 N) cos 40° + (0.24 m)(15 N)sin 40° or M = 6.9103 N ⋅ m The single equivalent force F′ is equal to F. Further for equivalence ΣM B : M = dF ′ 6.9103 N ⋅ m = d × 48 N or d = 0.14396 m or and the line of action of F′ intersects line AB 144 mm to the right of A. (b) F ′ = 48 N W W Following the solution to Part a but with d = 0.1 m and α unknown, have ΣM B : (0.4 m)(15 N) cos α + (0.24 m)(15 N) sin α = (0.1 m)(48 N) or 5cos α + 3sin α = 4 Rearranging and squaring 25 cos 2 α = (4 − 3 sin α )2 Using cos 2 α = 1 − sin 2 α and expanding 25(1 − sin 2 α ) = 16 − 24 sin α + 9 sin 2 α or Then 34 sin 2 α − 24 sin α − 9 = 0 24 ± (−24) 2 − 4(34)(−9) 2(34) sin α = 0.97686 or sin α = −0.27098 sin α = α = 77.7° or α = −15.72° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 248. PROBLEM 3.93 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 F = − (1220 N)i W Also, we have ΣM G : rA/G × P = M i j k 1220 0 −.1 −.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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 249. PROBLEM 3.94 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. SOLUTION We have ΣF : PAB = FC where PAB = ȜAB PAB = (33 mm)i + (990 mm) j − (594 mm)k (175 N) 1155.00 mm or FC = (5.00 N)i + (150 N) j − (90.0 N)k W 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 250. PROBLEM 3.95 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. 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 W M = (4.10 kip ⋅ ft)i + (16.38 kip ⋅ ft)k W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 251. PROBLEM 3.96 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. 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 W M = −(23.0 kip ⋅ ft)i + (11.52 kip ⋅ ft)k W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 252. PROBLEM 3.97 Replace the 150-N force with an equivalent force-couple system at A. SOLUTION Equivalence requires 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 0.1 N⋅m −0.12 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 W M = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 253. PROBLEM 3.98 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. 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 = Ȝ 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 = ȜEJ 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) 0.31 −0.0233 0 + =0 d 2 + 0.0058 42 42 −49 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 254. PROBLEM 3.98 (Continued) (b) F2 = F1 = (42i + 42 j − 49k )N or F2 = (42 N)i + (42 N) j − (49 N)k W 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 −49 42 42 = (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 or M 2 = − (25.9 N ⋅ m)i + (21.2 N ⋅ m) j W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 255. PROBLEM 3.99 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. SOLUTION 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 0 M ′ = (−1800i − 1120k ) + 45 14 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 W The equivalent force-couple system at H is M ′ = −(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 256. PROBLEM 3.100 The handpiece for a miniature industrial grinder weighs 0.6 lb, and its center of gravity is located on the y axis. The head of the handpiece is offset in the xz plane in such a way that line BC forms an angle of 25° with the x direction. Show that the weight of the handpiece and the two couples M1 and M2 can be replaced with a single equivalent force. Further, assuming that M1 = 0.68 lb ⋅ in. and M2 = 0.65 lb ⋅ in., determine (a) the magnitude and the direction of the equivalent force, (b) the point where its line of action intersects the xz plane. SOLUTION First assume that the given force W and couples M1 and M2 act at the origin. Now W = − Wj and M = M1 + M 2 = − ( M 2 cos 25°)i + ( M 1 − M 2 sin 25°)k Note that since W and M are perpendicular, it follows that they can be replaced with a single equivalent force. F = W or F = − Wj = − (0.6 lb) j or F = − (0.600 lb)j W (a) We have (b) Assume that the line of action of F passes through Point P(x, 0, z). Then for equivalence M = rP/0 × F where rP/0 = xi + zk − ( M 2 cos 25°)i + ( M1 − M 2 sin 25°)k i j = x 0 0 −W Equating the i and k coefficients, (b) For z= k z = (Wz )i − (Wx)k 0 − M z cos 25° W and § M − M 2 sin 25° · x = −¨ 1 ¸ W © ¹ W = 0.6 lb M1 = 0.68 lb ⋅ in. M 2 = 0.65 lb ⋅ in. x= 0.68 − 0.65sin 25° = 0.67550 in. − 0.6 or z= − 0.65cos 25° = − 0.98183 in. 0.6 or z = − 0.982 in. W x = 0.675 in. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 257. PROBLEM 3.101 A 4-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? SOLUTION (a) (a) We have ΣFy : −400 N − 200 N = Ra or and R a = 600 N W ΣM A : 1800 N ⋅ m − (200 N)(4 m) = M a or M a = 1000 N ⋅ m (b) We have ΣFy : − 600 N = Rb or and We have M b = 900 N ⋅ m W ΣFy : 300 N − 900 N = Rc or and R b = 600 N W ΣM A : − 900 N ⋅ m = M b or (c) W R c = 600 N W ΣM A : 4500 N ⋅ m − (900 N)(4 m) = M c or M c = 900 N ⋅ m W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 258. PROBLEM 3.101 (Continued) (d) We have ΣFy : − 400 N + 800 N = Rd or and ΣM A : (800 N)(4 m) − 2300 N ⋅ m = M d or (e) We have R e = 600 N W M e = 200 N ⋅ m R f = 600 N W ΣM A : − 300 N ⋅ m + 300 N ⋅ m + (200 N)(4 m) = M f or M f = 800 N ⋅ m (g) We have R g = 1000 N W ΣM A : 200 N ⋅ m + 4000 N ⋅ m − (800 N)(4 m) = M g or M g = 1000 N ⋅ m (h) We have R h = 600 N W ΣM A : 2400 N ⋅ m − 300 N ⋅ m − (300 N)(4 m) = M h or (b) W ΣFy : − 300 N − 300 N = Rh or and W ΣFy : − 200 N − 800 N = Rg or and W ΣFy : − 800 N + 200 N = R f or and W ΣM A : 200 N ⋅ m + 400 N ⋅ m − (200 N)(4 m) = M e or ( f ) We have M d = 900 N ⋅ m ΣFy : − 400 N − 200 N = Re or and R d = 400 N W M h = 900 N ⋅ m W W Therefore, loadings (c) and (h) are equivalent. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 259. PROBLEM 3.102 A 4-m-long beam is loaded as shown. Determine the loading of Problem 3.101 which is equivalent to this loading. SOLUTION We have and ΣFy : − 200 N − 400 N = R or R = 600 N W ΣM A : −400 N ⋅ m + 2800 N ⋅ m − (400 N)(4 m) = M M = 800 N ⋅ m or Equivalent to case ( f ) of Problem 3.101 W Problem 3.101 Equivalent force-couples at A Case R (a) 600 N 1000 N ⋅ m (b) 600 N 900 N ⋅ m (c) 600 N 900 N ⋅ m (d) 400 N 900 N ⋅ m (e) 600 N 200 N ⋅ m (f ) 600 N 800 N ⋅ m (g) 1000 N 1000 N ⋅ m (h) 600 N 900 N ⋅ m M W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 261. PROBLEM 3.104 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 origin. 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 origin 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) j + (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. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 262. PROBLEM 3.105 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. SOLUTION (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 W (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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 263. 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. 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. W L = 42 in. (b) We have 375.4 + 3.5d = 11.7(42) or d = 33.1 in. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 265. PROBLEM 3.107 (Continued) 184 80 64 2 80 32 a− a+ a + a + 24 − a 2 = 0 3 3 9 3 9 or 230 − or 16a 2 − 276a + 1143 = 0 276 ± (−276) 2 − 4(16)(1143) 2(16) Then a= or a = 10.3435 m and a = 6.9065 m Since AB = 9 m, a must be less than 9 m a = 6.91 m W 6.9065 1.5 Using Eq. (1) R = 2300 − 400 and using Eq. (2) (b) or R = 458 N W 4 10(6.9065) + 9 − (6.9065)2 3 = 3.16 m L= 8 23 − (6.9065) 3 R is applied 3.16 m to the right of A. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 266. PROBLEM 3.108 Gear C is rigidly attached to arm AB. If the forces and couple shown can be reduced to a single equivalent force at A, determine the equivalent force and the magnitude of the couple M. SOLUTION We have For equivalence ΣFx : −18sin 30° + 25cos 40° = Rx or Rx = 10.1511 lb ΣFy : −18cos 30° − 40 − 25sin 40° = Ry or Then Ry = −71.658 lb R = (10.1511) 2 + (71.658) 2 = 72.416 tanθ = or Also 71.658 10.1511 θ = 81.9° R = 72.4 lb 81.9° W ΣM A : M − (22 in.)(18 lb)sin 35° − (32 in.)(40 lb) cos 25° − (48 in.)(25 lb) sin 65° = 0 M = 2474.8 lb ⋅ in. or M = 206 lb ⋅ ft W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 267. PROBLEM 3.109 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 (b) 28.0° W 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. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 268. PROBLEM 3.110 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, must have M iR = 0 (a) B M A = ΣM A = M + (12 in.)[(30 lb) sin 60°] − (8 in.)(45 lb) = 0 M = +48.231 lb ⋅ in. (b) M = 48.2 lb ⋅ in. M = 240 lb ⋅ in. W R M B = ΣM B = M + (12 in.)(10 lb) − (8 in.)(45 lb) = 0 M = +240 lb ⋅ in. (c) W R M C = ΣM C = M + (12 in.)(10 lb) − (8 in.)[(30 lb) cos 60°] = 0 M =0 M=0 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 269. PROBLEM 3.111 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 (a) 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 = −1.20000 θ = −50.194° (b) R = 1562 N 50.2° W R M C = 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 270. PROBLEM 3.112 Solve Problem 3.111, assuming that the 760-N force is directed to the right. PROBLEM 3.111 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 (a) = ( −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 ¹ θ = 66.5714° R = 1308 N 66.6° W R M C = 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 271. 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 + (300 lb)(− cos 40°i − sin 40° j) − (180 lb) j R = −(147.728 lb)i − (758.36 lb) j 2 2 R = Rx + Ry = (147.728) 2 + (758.36) 2 = 772.62 lb § Ry · ¸ © Rx ¹ § −758.36 · = tan −1 ¨ ¸ © −147.728 ¹ = 78.977° θ = tan −1 ¨ or R = 773 lb We have ΣM A = dRy where 79.0° W ΣM A = −[240 lb cos 70°](6 ft) − [240 lbsin 70°](4 ft) − (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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 272. PROBLEM 3.114 Pulleys A and B are mounted on bracket CDEF. The tension on each side of the two belts is as shown. Replace the four forces with a single equivalent force, and determine where its line of action intersects the bottom edge of the bracket. SOLUTION Equivalent force-couple at A due to belts on pulley A We have ΣF : − 120 lb − 160 lb = RA R A = 280 lb We have ΣM A : −40 lb(2 in.) = M A M A = 80 lb ⋅ in. Equivalent force-couple at B due to belts on pulley B We have ΣF: (210 lb + 150 lb) 25° = R B R B = 360 lb We have 25° ΣM B : − 60 lb(1.5 in.) = M B M B = 90 lb ⋅ in. Equivalent force-couple at F We have ΣF: R F = ( − 280 lb) j + (360 lb)(cos 25°i + sin 25° j) = (326.27 lb)i − (127.857 lb) j R = RF 2 2 = RFx + RFy = (326.27)2 + (127.857)2 = 350.43 lb § RFy · ¸ © RFx ¹ § −127.857 · = tan −1 ¨ ¸ © 326.27 ¹ = −21.399° θ = tan −1 ¨ or R F = R = 350 lb 21.4° W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 273. PROBLEM 3.114 (Continued) We have ΣM F : M F = − (280 lb)(6 in.) − 80 lb ⋅ in. − [(360 lb) cos 25°](1.0 in.) + [(360 lb) sin 25°](12 in.) − 90 lb ⋅ in. M F = − (350.56 lb ⋅ in.)k To determine where a single resultant force will intersect line FE, M F = dRy d= MF Ry −350.56 lb ⋅ in. −127 ⋅ 857 lb = 2.7418 in. = or d = 2.74 in. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 274. PROBLEM 3.115 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 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. W (b) The rivet hole is 0.227 m to the right of G. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 275. PROBLEM 3.116 Solve Problem 3.115, assuming that P = 60 N. PROBLEM 3.115 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 3.115 leading to the development of Equation (1) 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. W (b) The rivet hole is 0.259 m to the right of G. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 276. PROBLEM 3.117 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 ΣF : (60 lb)i − (32 lb) j + (140 lb)(cos 30°i + sin 30° j) = R R = (181.244 lb)i + (38.0 lb) j or R = 185.2 lb We have 11.84° W 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. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 278. 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 ª1 º 1 + 16 x 2 (1 − 3x 2 ) − ( x − x3 ) « (32 x)(1 + 16 x 2 ) −1/ 2 » dM ¬2 ¼ =0 = 8F 2 dx (1 + 16 x ) (1 + 16 x 2 )(1 − 3x 2 ) − 16 x( x − x3 ) = 0 32 x 4 + 3x 2 − 1 = 0 or 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 279. PROBLEM 3.119 Four forces are applied to the machine component ABDE as shown. Replace these forces by an equivalent force-couple system at A. SOLUTION 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 R = rB × [−(300 N)i − (50 N) j] A + 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 R = (30.8 N ⋅ m) j − (220 N ⋅ m)k W A PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 280. PROBLEM 3.120 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. 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 W 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 281. PROBLEM 3.121 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 force-couple system at A consisting of the force R = (2.6 lb)i + Ry j − (0.7 lb)k and the couple M R = M x i + (1.0 lb · ft)j − (0.72 lb · ft)k. (b) Find A the corresponding values of Ry and M x . SOLUTION (a) From the statement of the problem, equivalence requires ΣF : B + C = R ΣFx : Bx + C x = 2.6 lb (1) ΣFy : − C y = R y or (2) ΣFz : − C z = −0.7 lb or C z = 0.7 lb and R ΣM A : (rB/A × B + M B ) + rC/A × C = M A or § 1.75 · ΣM x : (1 lb ⋅ ft) + ¨ ft ¸ (C y ) = M x © 12 ¹ (3) § 3.75 · § 1.75 · § 3.5 · ΣM y : ¨ ft ¸ ( Bx ) + ¨ ft ¸ (C x ) + ¨ ft ¸ (0.7 lb) = 1 lb ⋅ ft © 12 ¹ © 12 ¹ © 12 ¹ 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 ¹ C y = 2.4686 lb or B = (2.5 lb)i C = (0.1000 lb)i − (2.47 lb) j − (0.700 lb)k W (b) Eq. (2)  Using Eq. (3) Ry = −2.47 lb W § 1.75 · 1+ ¨ ¸ (2.4686) = M x © 12 ¹ or M x = 1.360 lb ⋅ ft W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 282. PROBLEM 3.122 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. SOLUTION We have ΣF : or Fx : Ax + Bx = 8 lb A+B=C 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 j 8 Bx 0 By k i j 2 + 8 2 Ax 0 Ay k 8 lb ⋅ in. = (360 lb ⋅ in.)i 2 (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. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 283. PROBLEM 3.122 (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 W B = −(9.60 lb)i + (36.0 lb) j + (2.00 lb)k W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 284. PROBLEM 3.123 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. SOLUTION We have ΣF : R = R A = RλBC or We have where ȜBC = (42 in.)i − (96 in.) j − (16 in.)k 106 in. RA = where 21.2 lb (42i − 96 j − 16k ) 106 R A = (8.40 lb)i − (19.20 lb) j − (3.20 lb)k W 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 + (2lb ⋅ 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 285. PROBLEM 3.124 A mechanic replaces a car’s exhaust system by firmly clamping the catalytic converter FG to its mounting brackets H and I and then loosely assembling the mufflers and the exhaust pipes. To position the tailpipe AB, he pushes in and up at A while pulling down at B. (a) Replace the given force system with an equivalent force-couple system at D. (b) Determine whether pipe CD tends to rotate clockwise or counterclockwise relative to muffler DE, as viewed by the mechanic. SOLUTION (a) Equivalence requires ΣF : R = A + B = (100 N)(cos 30° j − sin 30° k ) − (115 N) j = −(28.4 N) j − (50 N)k and where ΣM D : M D = rA/D × FA + rB/D × FB rA/D = −(0.48 m)i − (0.225 m) j + (1.12 m)k rB/D = −(0.38 m)i + (0.82 m)k Then i j k i j k M D = 100 −0.48 −0.225 1.12 + 115 −0.38 0 0.82 0 cos 30° − sin 30° 0 −1 0 = 100[(0.225sin 30° − 1.12cos 30°)i + (−0.48sin 30°) j + (−0.48cos 30°)k ] + 115[(0.82)i + (0.38)k ] = 8.56i − 24.0 j + 2.13k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 286. PROBLEM 3.124 (Continued) The equivalent force-couple system at D is R = −(28.4 N) j − (50.0 N)k M D = (8.56 N ⋅ m)i − (24.0 N ⋅ m) j + (2.13 N ⋅ m)k (b) Since ( M D ) z is positive, pipe CD will tend to rotate counterclockwise relative to muffler DE. W W W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 287. PROBLEM 3.125 For the exhaust system of Problem 3.124, (a) replace the given force system with an equivalent force-couple system at F, where the exhaust pipe is connected to the catalytic converter, (b) determine whether pipe EF tends to rotate clockwise or counterclockwise, as viewed by the mechanic. SOLUTION (a) Equivalence requires ΣF : R = A + B = (100 N)(cos 30° j − sin 30° k ) − (115 N) j = −(28.4 N) j − (50 N)k and where M F : M F = rA/F × A + rB/F × B rA/F = −(0.48 m)i − (0.345 m) j + (2.10 m)k rB/F = −(0.38 m)i − (0.12 m) j + (1.80 m)k Then i j k i j k M F = 100 −0.48 −0.345 2.10 + 115 −0.38 0.12 1.80 0 cos 30° − sin 30° 0 −1 0 M F = 100[(0.345 sin 30° − 2.10 cos 30°)i + (−0.48 sin 30°) j + (−0.48 cos 30°)k ] + 115[(1.80)i + (0.38)k ] = 42.4i − 24.0 j + 2.13k PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 288. PROBLEM 3.125 (Continued) The equivalent force-couple system at F is R = −(28.4 N) j − (50 N)k M F = (42.4 N ⋅ m)i − (24.0 N ⋅ m) j + (2.13 N ⋅ m)k (b) Since ( M F ) z is positive, pipe EF will tend to rotate counterclockwise relative to the mechanic. W W W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 289. PROBLEM 3.126 The head-and-motor assembly of a radial drill press was originally positioned with arm AB parallel to the z axis and the axis of the chuck and bit parallel to the y axis. The assembly was then rotated 25° about the y axis and 20° about the centerline of the horizontal arm AB, bringing it into the position shown. The drilling process was started by switching on the motor and rotating the handle to bring the bit into contact with the workpiece. Replace the force and couple exerted by the drill press with an equivalent forcecouple system at the center O of the base of the vertical column. SOLUTION We have R = F = (11 lb)[( sin 20° cos 25°)]i − (cos 20°) j − (sin 20° sin 25°)k ] = (3.4097 lb)i − (10.3366 lb)j − (1.58998 lb)k or R = (3.41 lb)i − (10.34 lb) j − (1.590 lb)k We have W M O = rB/O × F × M C where rB/O = [(14 in.) sin 25°]i + (15 in.) j + [(14 in.) cos 25°]k = (5.9167 in.)i + (15 in.) j + (12.6883 in.)k M C = (90 lb ⋅ in.)[(sin 20° cos 25°)i − (cos 20°) j − (sin 20° sin 25°)k ] = (27.898 lb ⋅ in.)i − (84.572 lb ⋅ in.) j − (13.0090 lb ⋅ in.)k i j k 15 12.6883 lb ⋅ in. M O = 5.9167 3.4097 −10.3366 1.58998 + (27.898 − 84.572 − 13.0090) lb ⋅ in. = (135.202 lb ⋅ in.)i − (31.901 lb ⋅ in.) j − (125.313 lb ⋅ in.)k or M O = (135.2 lb ⋅ in.)i − (31.9 lb ⋅ in.) j − (125.3 lb ⋅ in.)k W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 290. 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. SOLUTION 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 W 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 W 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 291. 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. SOLUTION We have 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 W 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 W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 292. PROBLEM 3.129 Four signs are mounted on a frame 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 W 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. W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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
• 293. PROBLEM 3.130 Four signs are mounted on a frame 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. 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 W ΣM y : − (9 ft)(105 ft) − (14.5 − b) ft × (90 lb) + (8 ft)(50 lb) = 0 or b = 20.6 ft W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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