1. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 125 N,Q = 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: 179 N, 75.1R = = 179 N=R 75.1 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. 4 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 25 lb,Q = 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: 77.1lb, 85.4R = = 77.1lb=R 85.4 W 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. 5 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: 51.3 59.0 = = (a) Parallelogram law: (b) Triangle rule: We measure: 139.1 lb,R = 67.0 = 139.1lbR = 67.0 W 4. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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: 3.30 kN, 66.6R = = 3.30 kN=R 66.6 W 5. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.5 The 300-lb force is to be resolved into components along lines a-a and b-b. (a) Determine the angle 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 sin 60 240 lb 300 lb sin 0.69282 43.854 60 180 180 60 43.854 76.146 = = = + + = = = 76.1 = W (b) Law of sines: 300 lb sin76.146 sin 60 bbF = 336 lbbbF = W 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. 8 PROBLEM 2.6 The 300-lb force is to be resolved into components along lines a-a and b-b. (a) Determine the angle 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 sin 60 120 lb 300 lb = sin 0.34641 20.268 = = 20.3 = W (b) 60 180 180 60 20.268 99.732 + + = = = 300 lb sin99.732 sin 60 aaF = 341 lbaaF = W 7. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 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 sin 25 50 N 35 N sin 0.60374 = = 37.138 = 37.1 = W (b) 25 180 180 25 37.138 117.86 + + = = = 35 N sin117.86 sin 25 R = 73.2 NR = W 8. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 75 NP = and 125 N,Q = 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) 75 N sin 20 sin35 Q = 44.7 NQ = W (b) 20 35 180 180 20 35 125 + + = = = 75 N sin125 sin35 R = 107.1 NR = W 9. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.9 A trolley that moves along a horizontal beam is acted upon by two forces as shown. (a) Knowing that 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) 1600 N sin 25 sin 75 P = 3660 NP = W (b) 25 75 180 180 25 75 80 + + = = = 1600 N sin 25 sin80 R = 3730 NR = W 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. 12 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: 2 2 2 (1600 N) (2500 N) 2(1600 N)(2500 N)cos 75 2596 N P P = + = Using the law of sines: sin sin75 1600 N 2596 N 36.5 = = P is directed 90 36.5 or 53.5 below the horizontal. 2600 N=P 53.5 W 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. 13 PROBLEM 2.11 A steel tank is to be positioned in an excavation. Knowing that = 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) 50 60 180 180 50 60 70 + + = = = 425 lb sin 70 sin 60 P = 392 lbP = W (b) 425 lb sin 70 sin50 R = 346 lbR = W 12. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 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) ( 30 ) 60 180 180 ( 30 ) 60 90 sin (90 ) sin60 425 lb 500 lb + + + = = + = = 90 47.40 = 42.6 = W (b) 500 lb sin (42.6 30 ) sin 60 R = + 551 lbR = W 13. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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) (50 N)sin 25P = 21.1 N=P W (b) (50 N)cos 25R = 45.3 NR = W 14. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 = 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) (425 lb)cos30P = 368 lb=P W (b) (425 lb)sin30R = 213 lbR = W 15. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 60 lbP = and 25 lb,Q = 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: 2 2 2 2 2 2 2 20 35 180 125 2 cos (60 lb) (25 lb) 2(60 lb)(25 lb)cos125 3600 625 3000(0.5736) 77.108 lb R P Q PQ R R R + + = = = + = + = + + = Using the law of sines: sin sin125 25 lb 77.108 lb 15.402 = = 70 85.402 + = 77.1lb=R 85.4 W 16. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 tan 10 38.66 6 tan 10 30.96 = = = = Using the triangle rule: 180 38.66 30.96 180 110.38 + + = + + = = Using the law of cosines: 2 22 (120 lb) (40 lb) 2(120 lb)(40 lb)cos110.38 139.08 lb R R = + = Using the law of sines: sin sin110.38 40 lb 139.08 lb = 15.64 (90 ) (90 38.66 ) 15.64 66.98 = = + = + = 139.1 lb=R 67.0 W 17. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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: 2 2 2 (2 kN) (3 kN) 2(2 kN)(3 kN)cos80 3.304 kN R R = + = Using the law of sines: sin sin80 2 kN 3.304 kN = 36.59 80 180 180 80 36.59 63.41 180 50 66.59 = + + = = = = + = 3.30 kN=R 66.6 W 18. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 180 (40 20 ) 120 = + = Then 2 2 2 2 (15 kN) (10 kN) 2(15 kN)(10 kN)cos120 475 kN 21.794 kN R R = + = = and 10 kN 21.794 kN sin sin120 10 kN sin sin120 21.794 kN 0.39737 23.414 = = = = Hence: 50 73.414 = + = 21.8 kN=R 73.4 W 19. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 180 (40 20 ) 120 = + = Then 2 2 2 2 (10 kN) (15 kN) 2(10 kN)(15 kN)cos120 475 kN 21.794 kN R R = + = = and 15 kN 21.794 kN sin sin120 15 kN sin sin120 21.794 kN 0.59605 36.588 = = = = Hence: 50 86.588 = + = 21.8 kN=R 86.6 W 20. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.20 For the hook support of Problem 2.7, knowing that 75P = N and 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 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 180 (50 25 ) 105 = + = Then 2 2 2 2 2 (75 N) (50 N) 2(75 N)(50 N)cos105 10066.1 N 100.330 N R R R = + = = and sin sin105 75 N 100.330 N sin 0.72206 46.225 = = = Hence: 25 46.225 25 21.225 = = 100.3 N=R 21.2 W 21. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: 2 2 2 2 2 2 (600) (800) 1000 mm (560) (900) 1060 mm (480) (900) 1020 mm OA OB OC = + = = + = = + = 800-N Force: 800 (800 N) 1000 xF = + 640 NxF = + W 600 (800 N) 1000 yF = + 480 NyF = + W 424-N Force: 560 (424 N) 1060 xF = 224 NxF = W 900 (424 N) 1060 yF = 360 NyF = W 408-N Force: 480 (408 N) 1020 xF = + 192.0 NxF = + W 900 (408 N) 1020 yF = 360 NyF = W 22. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: 2 2 2 2 2 2 (84) (80) 116 in. (28) (96) 100 in. (48) (90) 102 in. OA OB OC = + = = + = = + = 29-lb Force: 84 (29 lb) 116 xF = + 21.0 lbxF = + W 80 (29 lb) 116 yF = + 20.0 lbyF = + W 50-lb Force: 28 (50 lb) 100 xF = 14.00 lbxF = W 96 (50 lb) 100 yF = + 48.0 lbyF = + W 51-lb Force: 48 (51 lb) 102 xF = + 24.0 lbxF = + W 90 (51 lb) 102 yF = 45.0 lbyF = W 23. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION 40-lb Force: (40 lb)cos60xF = + 20.0 lbxF = W (40 lb)sin60yF = 34.6 lbyF = W 50-lb Force: (50 lb)sin50xF = 38.3 lbxF = W (50 lb)cos50yF = 32.1 lbyF = W 60-lb Force: (60 lb)cos25xF = + 54.4 lbxF = W (60 lb)sin 25yF = + 25.4 lbyF = W 24. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION 80-N Force: (80 N)cos40xF = + 61.3 NxF = W (80 N)sin 40yF = + 51.4 NyF = W 120-N Force: (120 N)cos70xF = + 41.0 NxF = W (120 N)sin 70yF = + 112.8 NyF = W 150-N Force: (150 N)cos35xF = 122. 9 NxF = W (150 N)sin35yF = + 86.0 NyF = W 25. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 (a) sin35 300 lbP = 300 lb sin35 P = 523 lbP = W (b) Vertical component cos35vP P= (523 lb)cos35= 428 lbvP = W 26. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 sin 20P= 2193 NP = 2190 NP = W (b) cos20ABCP P= (2193 N)cos20= 2060 NABCP = W 27. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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) sin38 120 N sin38 xP P = = 194.91 N= or 194.9 NP = W (b) tan38 120 N tan38 x y P P = = 153.59 N= or 153.6 NyP = W 28. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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) cos38 180 N cos 38 yP P = = 228.4 N= 228 NP = W (b) tan38 (180 N)tan38 x yP P= = 140.63 N= 140.6 NxP = W 29. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 1200 N:xP = Then (a) sin55xP P= sin55 1200 N sin55 1464.9 N xP P = = = 1465 NP = W (b) tan55x yP P= tan55 1200 N tan55 840.2 N x y P P = = = 840 Ny =P W 30. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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) cos 55 yP P = 350 lb cos 55 610.2 lb = = 610 lbP = W (b) sin 55xP P= (610.2 lb)sin 55 499.8 lb = = 500 lbxP = W 31. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 31.0xR = + 23.0yR = + (31.0 lb) (23.0 lb) tan 23.0 31.0 36.573 23.0 lb sin (36.573 ) x y y x R R R R R = + = + = = = = i j i j R 38.601 lb= 38.6 lb=R 36.6 W 32. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 20.6xR = 250.2yR = + ( 20.6 N) (250.2 N) tan 250.2 N tan 20.6 N tan 12.1456 85.293 x y y x R R R R = + = + = = = = R i j i j 250.2 N sin85.293 R = 251 N=R 85.3 W 33. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 36.08xR = + 41.42yR = ( 36.08 lb) ( 41.42 lb) tan 41.42 lb tan 36.08 lb tan 1.14800 48.942 x y y x R R R R = + = + + = = = = R i j i j 41.42 lb sin 48.942 R = 54.9 lb=R 48.9 W 34. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 608xR = + 240yR = (608 lb) ( 240 lb) tan 240 608 21.541 x y y x R R R R = + = + = = = R i j i j 240 N sin(21.541) 653.65 N R = = 654 N=R 21.5 W 35. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.35 Knowing that = 35, determine the resultant of the three forces shown. SOLUTION 100-N Force: (100 N)cos35 81.915 N (100 N)sin35 57.358 N x y F F = + = + = = 150-N Force: (150 N)cos65 63.393 N (150 N)sin 65 135.946 N x y F F = + = + = = 200-N Force: (200 N)cos35 163.830 N (200 N)sin35 114.715 N x y F F = = = = 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 18.522xR = 308.02yR = ( 18.522 N) ( 308.02 N) tan 308.02 18.522 86.559 x y y x R R R R = + = + = = = R i j i j 308.02 N sin86.559 R = 309 N=R 86.6 W 36. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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: 840 (725 N) 525 N 1160 840 (725 N) 500 N 1160 x y F F = = = = 500-N Force: 3 (500 N) 300 N 5 4 (500 N) 400 N 5 x y F F = = = = 780-N Force: 12 (780 N) 720 N 13 5 (780 N) 300 N 13 x y F F = = = = and 2 2 105 N 200 N ( 105 N) ( 200 N) 225.89 N x x y y R F R F R = = = = = + = Further: 1 200 tan 105 200 tan 105 62.3 = = = Thus: 226 N=R 62.3 W 37. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.37 Knowing that = 40, determine the resultant of the three forces shown. SOLUTION 60-lb Force: (60 lb)cos20 56.38 lb (60 lb)sin 20 20.52 lb x y F F = = = = 80-lb Force: (80 lb)cos60 40.00 lb (80 lb)sin 60 69.28 lb x y F F = = = = 120-lb Force: (120 lb)cos30 103.92 lb (120 lb)sin30 60.00 lb x y F F = = = = and 2 2 200.30 lb 29.80 lb (200.30 lb) (29.80 lb) 202.50 lb x x y y R F R F R = = = = = + = Further: 29.80 tan 200.30 = 1 29.80 tan 200.30 8.46 = = 203 lb=R 8.46 W 38. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.38 Knowing that = 75, determine the resultant of the three forces shown. SOLUTION 60-lb Force: (60 lb)cos 20 56.38 lb (60 lb)sin 20 20.52 lb x y F F = = = = 80-lb Force: (80 lb)cos 95 6.97 lb (80 lb)sin 95 79.70 lb x y F F = = = = 120-lb Force: (120 lb)cos 5 119.54 lb (120 lb)sin 5 10.46 lb x y F F = = = = Then 168.95 lb 110.68 lb x x y y R F R F = = = = and 2 2 (168.95 lb) (110.68 lb) 201.98 lb R = + = 110.68 tan 168.95 tan 0.655 33.23 = = = 202 lb=R 33.2 W 39. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.39 For the collar of Problem 2.35, determine (a) the required value of if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION (100 N)cos (150 N)cos( 30 ) (200 N)cos (100 N)cos (150 N)cos( 30 ) x x x R F R = = + + = + + (1) (100 N)sin (150 N)sin( 30 ) (200 N)sin (300 N)sin (150 N)sin( 30 ) y y y R F R = = + = + (2) (a) For R to be vertical, we must have 0.xR = We make 0xR = in Eq. (1): 100cos 150cos( 30 ) 0 100cos 150(cos cos 30 sin sin 30 ) 0 29.904cos 75sin + + = + = = 29.904 tan 75 0.3988 21.74 = = = 21.7 = W (b) Substituting for in Eq. (2): 300sin 21.74 150sin51.74 228.9 N yR = = | | 228.9 NyR R= = 229 NR = W 40. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 840 12 3 (780 N) (500 N) 1160 13 5 21 420 N 29 x x BC x BC R F T R T = = + = + (1) 800 5 4 (780 N) (500 N) 1160 13 5 20 700 N 29 y y BC y BC R F T R T = = = (2) (a) For R to be vertical, we must have 0xR = Set 0xR = in Eq. (1) 21 420 N 0 29 BCT + = 580 NBCT = W (b) Substituting for BCT in Eq. (2): 20 (580 N) 700 N 29 300 N y y R R = = | | 300 NyR R= = 300 NR = W 41. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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: sin10 (50 lb)cos35 (75 lb)cos60 sin10 78.46 lb x x AC AC R F T T = = + + = + (1) (50 lb)sin35 (75 lb)sin 60 cos10 93.63 lb cos10 y y AC y AC R F T R T = = + = (2) (a) Set 0yR = in Eq. (2): 93.63 lb cos10 0 95.07 lb AC AC T T = = 95.1 lbACT = W (b) Substituting for ACT in Eq. (1): (95.07 lb)sin10 78.46 lb 94.97 lb x x R R R = + = = 95.0 lbR = W 42. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.42 For the block of Problems 2.37 and 2.38, determine (a) the required value of 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 (60 lb) (80 lb)cos (120 lb)sinx xR F = = + + (1) and (80 lb)sin (120 lb)cosy yR F = = (2) (a) Set 0yR = in Eq. (2). (80 lb)sin (120 lb)cos 0 = Dividing each term by cos gives: (80 lb)tan 120 lb 120 lb tan 80 lb 56.310 = = = 56.3 = W (b) Substituting for in Eq. (1) gives: 60 lb (80 lb)cos56.31 (120lb)sin56.31 204.22 lbxR = + + = 204 lbxR = W 43. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Knowing that = 20, determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: 1962 N sin70 sin50 sin 60 AC BCT T = = (a) 1962 N sin 70 2128.9 N sin60 ACT = = 2.13 kNACT = W (b) 1962 N sin50 1735.49 N sin60 BCT = = 1.735 kNBCT = W 44. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 Force Triangle Law of sines: 500 N sin60 sin 40 sin80 AC BCT T = = (a) 500 N sin 60 439.69 N sin80 ACT = = 440 NACT = W (b) 500 N sin 40 326.35 N sin80 BCT = = 326 NBCT = W 45. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.45 Two cables are tied together at C and are loaded as shown. Knowing that P = 500 N and = 60, determine the tension in (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: 500 N sin35 sin 75 sin70 AC BCT T = = (a) 500 N sin35 sin 70 ACT = 305 NACT = W (b) 500 N sin 75 sin 70 BCT = 514 NBCT = W 46. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 2 mg (200 kg)(9.81m/s ) 1962 N W = = = Law of sines: 1962 N sin 15 sin 105 sin 60 AC BCT T = = (a) (1962 N) sin 15 sin 60 ACT = 586 NACT = W (b) (1962 N)sin 105 sin 60 BCT = 2190 NBCT = W 47. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.47 Knowing that 20 , = determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Free-Body Diagram Force Triangle Law of sines: 1200 lb sin 110 sin 5 sin 65 AC BCT T = = (a) 1200 lb sin 110 sin 65 ACT = 1244 lbACT = W (b) 1200 lb sin 5 sin 65 BCT = 115.4 lbBCT = W 48. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.48 Knowing that 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 Force Triangle Law of sines: 300 lb sin 35 sin 50 sin 95 AC BCF T = = (a) 300 lb sin 35 sin 95 ACF = 172.7 lbACF = W (b) 300 lb sin 50 sin 95 BCT = 231 lbBCT = W 49. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 500P = lb and 650Q = 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: 0A B= + + + =R P Q F F Substituting components: (500 lb) [(650 lb)cos50 ] [(650 lb)sin50 ] ( cos50 ) ( sin50 ) 0B A AF F F = + + + = R j i j i i j In the y-direction (one unknown force) 500 lb (650 lb)sin50 sin50 0AF + = Thus, 500 lb (650 lb)sin50 sin50 AF + = 1302.70 lb= 1303 lbAF = W In the x-direction: (650 lb)cos50 cos50 0B AF F + = Thus, cos50 (650 lb)cos50 (1302.70 lb)cos50 (650 lb)cos50 B AF F= = 419.55 lb= 420 lbBF = W 50. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 750AF = lb and 400BF = lb, determine the magnitudes of P and Q. SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: 0A B= + + + =R P Q F F Substituting components: cos 50 sin 50 [(750 lb)cos 50 ] [(750 lb)sin 50 ] (400 lb) P Q Q= + + + R j i j i j i In the x-direction (one unknown force) cos 50 [(750 lb)cos 50 ] 400 lb 0Q + = (750 lb)cos 50 400 lb cos 50 127.710 lb Q = = In the y-direction: sin 50 (750 lb)sin 50 0P Q + = sin 50 (750 lb)sin 50 (127.710 lb)sin 50 (750 lb)sin 50 476.70 lb P Q= + = + = 477 lb; 127.7 lbP Q= = W 51. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.51 A welded connection is in equilibrium under the action of the four forces shown. Knowing that 8AF = kN and 16BF = kN, determine the magnitudes of the other two forces. SOLUTION Free-Body Diagram of Connection 3 3 0: 0 5 5 x B C AF F F F = = With 8 kN 16 kN A B F F = = 4 4 (16 kN) (8 kN) 5 5 CF = 6.40 kNCF = W 3 3 0: 0 5 5 y D B AF F F F = + = With FA and FB as above: 3 3 (16 kN) (8 kN) 5 5 DF = 4.80 kNDF = W 52. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.52 A welded connection is in equilibrium under the action of the four forces shown. Knowing that 5AF = kN and 6DF = kN, determine the magnitudes of the other two forces. SOLUTION Free-Body Diagram of Connection 3 3 0: 0 5 5 y D A BF F F F = + = or 3 5 B D AF F F= + With 5 kN, 8 kNA DF F= = 5 3 6 kN (5 kN) 3 5 BF = + 15.00 kNBF = W 4 4 0: 0 5 5 x C B AF F F F = + = 4 ( ) 5 4 (15 kN 5 kN) 5 C B AF F F= = 8.00 kNCF = W 53. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 0: cos30 0y CAF T Q = = With 60 lbQ = (a) (60 lb)(0.866)CAT = 52.0 lbCAT = W (b) 0: sin30 0x CBF P T Q = = With 75 lbP = 75 lb (60 lb)(0.50)CBT = or 45.0 lbCBT = W 54. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 0: cos60 75 lb 0x BCF T Q = + = 75 lb cos60BCT Q= (1) 0: sin 60 0y ACF T Q = = sin 60ACT Q= (2) Requirement 60 lb:ACT From Eq. (2): sin60 60 lbQ 69.3 lbQ Requirement 60 lb:BCT From Eq. (1): 75 lb sin 60 60 lbQ 30.0 lbQ 30.0 lb 69.3 lbQ W 55. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.55 A sailor is being rescued using a boatswains 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 30 = and 10 = and that the combined weight of the boatswains 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 0: cos 10 cos 30 cos 30 0x ACB ACB CDF T T T = = 0.137158CD ACBT T= (1) 0: sin 10 sin 30 sin 30 900 0y ACB ACB CDF T T T = + + = 0.67365 0.5 900ACB CDT T+ = (2) (a) Substitute (1) into (2): 0.67365 0.5(0.137158 ) 900ACB ACBT T+ = 1212.56 NACBT = 1213 NACBT = W (b) From (1): 0.137158(1212.56 N)CDT = 166.3 NCDT = W 56. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.56 A sailor is being rescued using a boatswains 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 25 = and 15 = and that the tension in cable CD is 80 N, determine (a) the combined weight of the boatswains chair and the sailor, (b) in tension in the support cable ACB. SOLUTION Free-Body Diagram 0: cos 15 cos 25 (80 N)cos 25 0x ACB ACBF T T = = 1216.15 NACBT = 0: (1216.15 N)sin 15 (1216.15 N)sin 25yF = + (80 N)sin 25 0 862.54 N W W + = = (a) 863 NW = W (b) 1216 NACBT = W 57. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 . SOLUTION Free-Body Diagram Force Triangle (a) Law of cosines 2 2 2 (600) (750) 2(600)(750)cos(25 45 )P = + + 784.02 NP = 784 NP = W (b) Law of sines sin sin (25 45 ) 600 N 784.02 N + = 46.0 = 46.0 25 71.0 = + = W 58. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.58 For the situation described in Figure P2.47, determine (a) the value of for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension. PROBLEM 2.47 Knowing that 20 , = determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Free-Body Diagram Force Triangle To be smallest, BCT must be perpendicular to the direction of .ACT (a) Thus, 5 = 5.00 = W (b) (1200 lb)sin 5BCT = 104.6 lbBCT = W 59. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.59 For the structure and loading of Problem 2.48, determine (a) the value of for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension. SOLUTION BCT must be perpendicular to ACF to be as small as possible. Free-Body Diagram: C Force Triangle is a right triangle To be a minimum, BCT must be perpendicular to .ACF (a) We observe: 90 30 = 60.0 = W (b) (300 lb)sin 50BCT = or 229.81lbBCT = 230 lbBCT = W 60. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 725 N)T = 0: 2 1200 N 0y yF T = = 2 2 2 2 2 2 600 N (600 N) (870 N) 630 N y x y x x T T T T T T = + = + = = By similar triangles: 2.1 m 870 N 630 N 2.90 m BC BC = = 2( ) 5.80 m L BC= = 5.80 mL = W 61. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 . SOLUTION Free-Body Diagram: C Force Triangle Force triangle is isosceles with 2 180 85 47.5 = = (a) 2(800 N)cos 47.5 1081 NP = = Since 0,P the solution is correct. 1081 NP = W (b) 180 50 47.5 82.5 = = 82.5 = W 62. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 . SOLUTION Free-Body Diagram Force Triangle (a) Law of cosines: 2 2 2 (1200 N) (600 N) 2(1200 N)(600 N)cos 85 1294 N P P = + = Since 1200 N,P the solution is correct. 1294 NP = W (b) Law of sines: sin sin 85 1200 N 1294 N 67.5 180 50 67.5 = = = 62.5 = W 63. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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) 4.5 in.,x = (b) 15 in.x = SOLUTION (a) Free Body: Collar A Force Triangle 50 lb 4.5 20.5 P = 10.98 lbP = W (b) Free Body: Collar A Force Triangle 50 lb 15 25 P = 30.0 lbP = W 64. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 2 2 2 (50) (48) 196 14.00 lb N N = = = Similar Triangles 48 lb 20 in. 14 lb x = 68.6 in.x = W 65. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.65 A 160-kg load is supported by the rope-and-pulley arrangement shown. Knowing that 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 0: 2 sin 20 cos 0xF P P = = and cos 0.8452 or 46.840 46.840 = = = + For 0: 2 cos20 sin 46.840 1569.60 N 0yF P P = + = or 602 N=P 46.8 W For 46.840 = 0: 2 cos20 sin( 46.840 ) 1569.60 N 0yF P P = + = or 1365 N=P 46.8 W 66. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.66 A 160-kg load is supported by the rope-and-pulley arrangement shown. Knowing that 40 , = determine (a) the angle , (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) 0: 2 sin sin cos 40 0xF P P = = 1 sin cos 40 2 22.52 = = 22.5 = W (b) 0: sin 40 2 cos 22.52 1569.60 N 0yF P P = + = 630 NP = W 67. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 (a) 0: 2 (600 lb) 0 1 (600 lb) 2 yF T T = = = 300 lbT = W (b) 0: 2 (600 lb) 0 1 (600 lb) 2 yF T T = = = 300 lbT = W (c) 0: 3 (600 lb) 0 1 (600 lb) 3 yF T T = = = 200 lbT = W (d) 0: 3 (600 lb) 0 1 (600 lb) 3 yF T T = = = 200 lbT = W (e) 0: 4 (600 lb) 0 1 (600 lb) 4 yF T T = = = 150.0 lbT = W 68. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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) 0: 3 (600 lb) 0 1 (600 lb) 3 yF T T = = = 200 lbT = W (d) 0: 4 (600 lb) 0 1 (600 lb) 4 yF T T = = = 150.0 lbT = W 69. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 750 N,P = determine (a) the tension in cable ACB, (b) the magnitude of load Q. SOLUTION Free-Body Diagram: Pulley C (a) 0: (cos25 cos55 ) (750 N)cos55 0x ACBF T = = Hence: 1292.88 NACBT = 1293 NACBT = W (b) 0: (sin 25 sin55 ) (750 N)sin55 0 (1292.88 N)(sin 25 sin55 ) (750 N)sin55 0 y ACBF T Q Q = + + = + + = or 2219.8 NQ = 2220 NQ = W 70. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 0: (cos25 cos55 ) cos55 0x ACBF T P = = or 0.58010 ACBP T= (1) 0: (sin 25 sin55 ) sin55 1800 N 0y ACBF T P = + + = or 1.24177 0.81915 1800 NACBT P+ = (2) (a) Substitute Equation (1) into Equation (2): 1.24177 0.81915(0.58010 ) 1800 NACB ACBT T+ = Hence: 1048.37 NACBT = 1048 NACBT = W (b) Using (1), 0.58010(1048.37 N) 608.16 NP = = 608 NP = W 71. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.71 Determine (a) the x, y, and z components of the 750-N force, (b) the angles x, y, and z that the force forms with the coordinate axes. SOLUTION sin 35 (750 N)sin 35 430.2 N h h F F F = = = (a) cos 25x hF F= cos35yF F= sin 25z hF F= (430.2 N)cos25= (750 N)cos 35= (430.2 N)sin 25= 390 N,xF = + 614 N,yF = + 181.8 NzF = + W (b) 390 N cos 750 N x x F F + = = 58.7x = W 614 N cos 750 N y y F F + = = 35.0y = W 181.8 N cos 750 N z z F F + = = 76.0z = W 72. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.72 Determine (a) the x, y, and z components of the 900-N force, (b) the angles x, y, and z that the force forms with the coordinate axes. SOLUTION cos 65 (900 N)cos 65 380.4 N h h F F F = = = (a) sin 20x hF F= sin 65yF F= cos 20z hF F= (380.4 N)sin 20= (900 N)sin 65= (380.4 N)cos20= 130.1 N,xF = 816 N,yF = + 357 NzF = + W (b) 130.1 N cos 900 N x x F F = = 98.3x = W 816 N cos 900 N y y F F + = = 25.0y = W 357 N cos 900 N z z F F + = = 66.6z = W 73. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 x, y, and z that the force exerted at A forms with the coordinate axes. SOLUTION (a) sin30 sin50 110.3 N (Given)xF F= = 110.3 N 287.97 N sin 30 sin 50 F = = 288 NF = W (b) 110.3 N cos 0.38303 287.97 N x x F F = = = 67.5x = W cos30 249.39 249.39 N cos 0.86603 287.97 N y y y F F F F = = = = = 30.0y = W sin30 cos50 (287.97 N)sin 30cos 50 92.552 N zF F= = = 92.552 N cos 0.32139 287.97 N z z F F = = = 108.7z = W 74. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 x, y, and z that the force exerted at B forms with the coordinate axes. SOLUTION (a) sin30 sin 40 32.14 N (Given)zF F= = 32.14 100.0 N sin30 sin 40 F = = 100.0 NF = W (b) sin30 cos40 (100.0 N)sin 30cos 40 38.302 N xF F= = = 38.302 N cos 0.38302 100.0 N x x F F = = = 112.5x = W cos30 86.603 NyF F= = 86.603 N cos 0.86603 100 N y y F F = = = 30.0y = W 32.14 NzF = 32.14 N cos 0.32140 100 N z z F F = = = 108.7z = W 75. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 , ,x y and z that the force forms with the coordinate axes. SOLUTION (a) (60 lb)sin30cos 60 15 lbxF = = 15.00 lbxF = W (60 lb)cos30 51.96 lbyF = = 52.0 lbyF = + W (60 lb)sin30 sin 60 25.98 lbzF = = 26.0 lbzF = + W (b) 15.0 lb cos 0.25 60 lb x x F F = = = 104.5x = W 51.96 lb cos 0.866 60 lb y y F F = = = 30.0y = W 25.98 lb cos 0.433 60 lb z z F F = = = 64.3z = W 76. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 x, y, and z that the force exerted at C forms with the coordinate axes. SOLUTION (a) sin30 cos60 20 lb (Given)xF F= = 20 lb 80 lb sin 30cos60 F = = 80.0 lbF = W (b) 20 lb cos 0.25 80 lb x x F F = = = 104.5x = W (80 lb)cos30 69.282 lb 69.282 lb cos 0.86615 80 lb y y y F F F = = = = = 30.0y = W (80 lb)sin30 sin 60 34.641lb 34.641 cos 0.43301 80 z z z F F F = = = = = 64.3z = W 77. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 x, y, and z that the force forms with the coordinate axes. SOLUTION (a) (120 lb)cos 60 cos 20xF = 56.382 lbxF = 56.4 lbxF = + W (120 lb)sin 60 103.923 lb y y F F = = 103.9 lbyF = W (120 lb)cos 60 sin 20 20.521 lb z z F F = = 20.5 lbzF = W (b) 56.382 lb cos 120 lb x x F F = = 62.0x = W 103.923 lb cos 120 lb y y F F = = 150.0y = W 20.52 lb cos 120 lb z z F F = = 99.8z = W 78. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 x, y, and z that the force forms with the coordinate axes. SOLUTION (a) (85 lb)sin 36 sin 48xF = 37.129 lb= 37.1 lbxF = W (85 lb)cos 36 68.766 lb yF = = 68.8 lbyF = W (85 lb)sin 36 cos 48 33.431 lb zF = = 33.4 lbzF = W (b) 37.129 lb cos 85 lb x x F F = = 64.1x = W 68.766 lb cos 85 lb y y F F = = 144.0y = W 33.431 lb cos 85 lb z z F F = = 66.8z = W 79. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.79 Determine the magnitude and direction of the force F = (320 N)i + (400 N)j (250 N)k. SOLUTION 2 2 2 2 2 2 (320 N) (400 N) ( 250 N) x y zF F F F F = + + = + + 570 NF = W 320 N cos 570 N x x F F = = 55.8x = W 400 N cos 570 N y y F F = = 45.4y = W 250 N cos 570 N z y F F = = 116.0z = W 80. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.80 Determine the magnitude and direction of the force F = (240 N)i (270 N)j + (680 N)k. SOLUTION 2 2 2 2 2 (240 N) ( 270 N) (680 N) x y zF F F F F = + + = + + 770 NF = W 240 N cos 770 N x x F F = = 71.8x = W 270 N cos 770 N y y F F = = 110.5y = W 680 N cos 770 N z z F F = = 28.0z = W 81. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.81 A force acts at the origin of a coordinate system in a direction defined by the angles x = 70.9 and y = 144.9. Knowing that the z component of the force is 52.0 lb, determine (a) the angle z, (b) the other components and the magnitude of the force. SOLUTION (a) We have 2 2 2 2 2 2 (cos ) (cos ) (cos ) 1 (cos ) 1 (cos ) (cos )x y z y y z + + = = Since 0zF we must have cos 0z Thus, taking the negative square root, from above, we have: 2 2 cos 1 (cos70.9 ) (cos144.9 ) 0.47282z = = 118.2z = W (b) Then: 52.0 lb 109.978 lb cos 0.47282 z z F F = = = and cos (109.978 lb)cos70.9x xF F = = 36.0 lbxF = W cos (109.978 lb)cos144.9y yF F = = 90.0 lbyF = W 110.0 lbF = W 82. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.82 A force acts at the origin of a coordinate system in a direction defined by the angles y = 55 and z = 45. Knowing that the x component of the force is 500 lb, determine (a) the angle x, (b) the other components and the magnitude of the force. SOLUTION (a) We have 2 2 2 2 2 2 (cos ) (cos ) (cos ) 1 (cos ) 1 (cos ) (cos )x y z y y z + + = = Since 0xF we must have cos 0x Thus, taking the negative square root, from above, we have: 2 2 cos 1 (cos55) (cos45) 0.41353x = = 114.4x = W (b) Then: 500 lb 1209.10 lb cos 0.41353 x x F F = = = 1209 lbF = W and cos (1209.10 lb)cos55y yF F = = 694 lbyF = W cos (1209.10 lb)cos45z zF F = = 855 lbzF = W 83. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.83 A force F of magnitude 210 N acts at the origin of a coordinate system. Knowing that Fx = 80 N, z = 151.2, and Fy < 0, determine (a) the components Fy and Fz, (b) the angles x and y. SOLUTION (a) cos (210 N)cos151.2z zF F = = 184.024 N= 184.0 NzF = W Then: 2 2 2 2 x y zF F F F= + + So: 2 2 2 2 (210 N) (80 N) ( ) (184.024 N)yF= + + Hence: 2 2 2 (210 N) (80 N) (184.024 N)yF = 61.929 N= 62.0 lbyF = W (b) 80 N cos 0.38095 210 N x x F F = = = 67.6x = W 61.929 N cos 0.29490 210 N y y F F = = = 107.2y = W 84. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 2.84 A force F of magnitude 230 N acts at the origin of a coordinate system. Knowing that x = 32.5, Fy = 60 N, and Fz > 0, determine (a) the components Fx and Fz, (b) the angles y and z. SOLUTION (a) We have cos (230 N)cos32.5x xF F = = 194.0 NxF = W Then: 193.980 NxF = 2 2 2 2 x y zF F F F= + + So: 2 2 2 2 (230 N) (193.980 N) ( 60 N) zF= + + Hence: 2 2 2 (230 N) (193.980 N) ( 60 N)zF = + 108.0 NzF = W (b) 108.036 NzF = 60 N cos 0.26087 230 N y y F F = = = 105.1y = W 108.036 N cos 0.46972 230 N z z F F = = = 62.0z = W 85. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 2 2 2 (20 ft) (100 ft) (25 ft) (20 ft) (100 ft) ( 25 ft) 105 ft 525 lb [(20 ft) (100 ft) (25 ft) ] 105 ft (100.0 lb) (500 lb) (125.0 lb) BA BA BA F BA F BA = + = + + = = = = + = + i j k F i j k F i j k JJJG JJJG 100.0 lb, 500 lb, 125.0 lbx y zF F F= + = + = W 86. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 2 2 2 (20 ft) (100 ft) (70 ft) (20 ft) (100 ft) ( 70 ft) 126 ft 315 lb [(20 ft) (100 ft) (74 ft) ] 126 ft (50 lb) (250 lb) (185 lb) DA DA DA F DA F DA = + + = + + + = = = = + + = + + i j k F i j k F i j k JJJG JJJG 50 lb, 250 lb, 185.0 lbx y zF F F= + = + = + W 87. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 2 2 2 (480 mm) (510 mm) (320 mm) (480 mm) (510 mm ) (320 mm) 770 mm 385 N [(480 mm) (510 mm) (320 mm) ] 770 mm (240 N) (255 N) (160 N) DB DB DB F DB F DB = + = + + = = = = + = + i j k F i j k i j k JJJG JJJG 240 N, 255 N, 160.0 Nx y zF F F= + = = + W 88. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 2 2 2 (270 mm) (400 mm) (600 mm) (270 mm) (400 mm) (600 mm) 770 mm 385 N [(270 mm) (400 mm) (600 mm) ] 770 mm (135 N) (200 N) (300 N) EB EB EB F EB F EB = + = + + = = = = + = + i j k F i j k F i j k JJJG JJJG 135.0 N, 200 N, 300 Nx y zF F F= + = = + W 89. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 2 2 2 (900 mm) (600 mm) (360 mm) (900 mm) (600 mm) (360 mm) 1140 mm 1425 N [ (900 mm) (600 mm) (360 mm) ] 1140 mm (1125 N) (750 N) (450 N) BA BA BA BA BA BA BA T BA T BA = + + = + + = = = = + + = + + i j k T T i j k i j k JJJG JJJG ( ) 1125 N, ( ) 750 N, ( ) 450 NBA x BA y BA zT T T= = = W 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. 92 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 2 2 2 (900 mm) (600 mm) (920 mm) (900 mm) (600 mm) (920 mm) 1420 mm 2130 N [ (900 mm) (600 mm) (920 mm) ] 1420 mm (1350 N) (900 N) (1380 N) CA CA CA CA CA CA CA T CA T CA = + = + + = = = = + = + i j k T T i j k i j k JJJG JJJG ( ) 1350 N, ( ) 900 N, ( ) 1380 NCA x CA y CA zT T T= = = W 91. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 (300 N)[ cos30 sin15 sin30 cos30 cos15 ] (67.243 N) (150 N) (250.95 N) (400 N)[cos50 cos20 sin50 cos50 sin 20 ] (400 N)[0.60402 0.76604 0.21985] (241.61 N) (306.42 N) (87.939 N) (174. = + + = + + = + = + = + = + = P i j k i j k Q i j k i j i j k R P Q 2 2 2 367 N) (456.42 N) (163.011 N) (174.367 N) (456.42 N) (163.011 N) 515.07 N R + + = + + = i j k 515 NR = W 174.367 N cos 0.33853 515.07 N x x R R = = = 70.2x = W 456.42 N cos 0.88613 515.07 N y y R R = = = 27.6y = W 163.011 N cos 0.31648 515.07 N z z R R = = = 71.5z = W 92. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 (400 N)[ cos30 sin15 sin30 cos30 cos15 ] (89.678 N) (200 N) (334.61 N) (300 N)[cos50 cos20 sin50 cos50 sin 20 ] (181.21 N) (229.81 N) (65.954 N) (91.532 N) (429.81 N) (268.66 N) (91.5R = + + = + + = + = + = + = + + = P i j k i j k Q i j k i j k R P Q i j k 2 2 2 32 N) (429.81 N) (268.66 N) 515.07 N + + = 515 NR = W 91.532 N cos 0.177708 515.07 N x x R R = = = 79.8x = W 429.81 N cos 0.83447 515.07 N y y R R = = = 33.4y = W 268.66 N cos 0.52160 515.07 N z z R R = = = 58.6z = W 93. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 2 2 2 2 2 2 (40 in.) (45 in.) (60 in.) (40 in.) (45 in.) (60 in.) 85 in. (100 in.) (45 in.) (60 in.) (100 in.) (45 in.) (60 in.) 125 in. (40 in.) (45 in.) (60 in.) (425 lb) 85 in. AB AB AB AB AB AB AC AC AB T T AB = + = + + = = + = + + = + = = = i j k i j k i j k T JJJG JJJG JJJG (200 lb) (225 lb) (300 lb) (100 in.) (45 in.) (60 in.) (510 lb) 125 in. (408 lb) (183.6 lb) (244.8 lb) (608) (408.6 lb) (544.8 lb) AB AC AC AC AC AC AB AC AC T T AC = + + = = = = + = + = + T i j k i j k T T i j k R T T i j k JJJG Then: 912.92 lbR = 913 lbR = W and 608 lb cos 0.66599 912.92 lb x = = 48.2x = W 408.6 lb cos 0.44757 912.92 lb y = = 116.6y = W 544.8 lb cos 0.59677 912.92 lb z = = 53.4z = W 94. PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 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 2 2 2 2 2 2 (40 in.) (45 in.) (60 in.) (40 in.) (45 in.) (60 in.) 85 in. (100 in.) (45 in.) (60 in.) (100 in.) (45 in.) (60 in.) 125 in. (40 in.) (45 in.) (60 in.) (510 lb) 85 in. AB AB AB AB AB AB AC AC AB T T AB = + = + + = = + = + + = + = = = i j k i j k i j k T JJJG JJJG JJJG (240 lb) (270 lb) (360 lb) (100 in.) (45 in.) (60 in.) (425 lb) 125 in. (340 lb) (153 lb) (204 lb) (580 lb) (423 lb) (564 lb) AB AC AC AC AC AC AB AC AC T T AC