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SOLUTION MANUAL CHAPTER 2 PROBLEM 2.1 Two forces are applied at point B of beam AB. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: R = 3.30 kN, α = 66.6° R = 3.30 kN 66.6° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 3 PROBLEM 2.2 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule. SOLUTION We measure: (a) Parallelogram law: (b) Triangle rule: We measure: α = 51.3° β = 59.0° R = 139.1 lb, γ = 67.0° R = 139.1 lb 67.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 4 PROBLEM 2.3 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: α = 21.2° R = 20.1 kN, R = 20.1 kN 21.2° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 5 PROBLEM 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: R = 8.03 kips, α = 3.8° R = 8.03 kips 3.8° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 6 PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant. SOLUTION Using the triangle rule and the law of sines: (a) (b) 120 N P = sin 30° sin 25° P = 101.4 N 30° + β + 25° = 180° β = 180° − 25° − 30° = 125° 120 N R = sin 30° sin125° R = 196.6 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 7 PROBLEM 2.6 A trolley that moves along a horizontal beam is acted upon by two forces as shown. (a) Knowing that α = 25°, determine by trigonometry the magnitude of the force P so that the resultant force exerted on the trolley is vertical. (b) What is the corresponding magnitude of the resultant? SOLUTION Using the triangle rule and the law of sines: (a) (b) 1600 N P = sin 25° sin 75° P = 3660 N 25° + β + 75° = 180° β = 180° − 25° − 75° = 80° 1600 N R = sin 25° sin 80° R = 3730 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 8 PROBLEM 2.7 A trolley that moves along a horizontal beam is acted upon by two forces as shown. Determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 2500 N. 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° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 2.8 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the left-hand portion of the cable is T1 = 800 lb, determine by trigonometry (a) the required tension T2 in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) 75° + 40° + α = 180° α = 180° − 75° − 40° = 65° (b) T2 800 lb = sin 65° sin 75° T2 = 853 lb 800 lb R = sin 65° sin 40° R = 567 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10 PROBLEM 2.9 A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and the law of sines: (a) 75° + 40° + β = 180° β = 180° − 75° − 40° = 65° (b) T1 1000 lb = sin 75° sin 65° T1 = 938 lb 1000 lb R = sin 75° sin 40° R = 665 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 11 PROBLEM 2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R. SOLUTION Using the triangle rule and law of sines: (a) sin α sin 25° = 50 N 35 N sin α = 0.60374 α = 37.138° (b) α = 37.1° α + β + 25° = 180° β = 180° − 25° − 37.138° = 117.862° R 35 N = sin117.862° sin 25° R = 73.2 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 12 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° (b) 425 lb P = sin 70° sin 60° P = 392 lb 425 lb R = sin 70° sin 50° R = 346 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 13 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 = 47.402° (b) R 500 lb = sin (42.598° + 30°) sin 60° α = 42.6° R = 551 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 14 PROBLEM 2.13 A steel tank is to be positioned in an excavation. Determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R. SOLUTION The smallest force P will be perpendicular to R. (a) P = (425 lb) cos 30° (b) R = (425 lb)sin 30° P = 368 lb R = 213 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 15 PROBLEM 2.14 For the hook support of Prob. 2.10, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R. SOLUTION The smallest force P will be perpendicular to R. (a) P = (50 N)sin 25° (b) R = (50 N) cos 25° P = 21.1 N R = 45.3 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 16 PROBLEM 2.15 Solve Problem 2.2 by trigonometry. PROBLEM 2.2 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule. SOLUTION 8 10 α = 38.66° tan α = 6 10 β = 30.96° tan β = Using the triangle rule: Using the law of cosines: α + β + ψ = 180° 38.66° + 30.96° + ψ = 180° ψ = 110.38° R 2 = (120 lb)2 + (40 lb) 2 − 2(120 lb)(40 lb) cos110.38° R = 139.08 lb Using the law of sines: sin γ sin110.38° = 40 lb 139.08 lb γ = 15.64° φ = (90° − α ) + γ φ = (90° − 38.66°) + 15.64° φ = 66.98° R = 139.1 lb 67.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 17 PROBLEM 2.16 Solve Problem 2.4 by trigonometry. PROBLEM 2.4 Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule. SOLUTION Using the force triangle and the laws of cosines and sines: We have: γ = 180° − (50° + 25°) = 105° Then R 2 = (4 kips) 2 + (6 kips)2 − 2(4 kips)(6 kips) cos105° = 64.423 kips 2 R = 8.0264 kips And 4 kips 8.0264 kips = sin(25° + α ) sin105° sin(25° + α ) = 0.48137 25° + α = 28.775° α = 3.775° R = 8.03 kips 3.8° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 18 PROBLEM 2.17 For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N. PROBLEM 2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant. SOLUTION Using the laws of cosines and sines: P 2 = (120 N) 2 + (160 N)2 − 2(120 N)(160 N) cos 25° P = 72.096 N And sin α sin 25° = 120 N 72.096 N sin α = 0.70343 α = 44.703° P = 72.1 N 44.7° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 19 PROBLEM 2.18 For the hook support of Prob. 2.10, knowing that P = 75 N and α = 50°, determine by trigonometry the magnitude and direction of the resultant of the two forces applied to the support. PROBLEM 2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R. SOLUTION Using the force triangle and the laws of cosines and sines: We have β = 180° − (50° + 25°) = 105° Then R 2 = (75 N) 2 + (50 N) 2 − 2(75 N)(50 N) cos 105° R = 10, 066.1 N 2 R = 100.330 N 2 and Hence: sin γ sin105° = 75 N 100.330 N sin γ = 0.72206 γ = 46.225° γ − 25° = 46.225° − 25° = 21.225° R = 100.3 N 21.2° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 20 PROBLEM 2.19 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces. SOLUTION Using the force triangle and the laws of cosines and sines: We have γ = 180° − (20° + 10°) = 150° Then R 2 = (48 N)2 + (60 N)2 − 2(48 N)(60 N) cos150° R = 104.366 N and Hence: 48 N 104.366 N = sin α sin150° sin α = 0.22996 α = 13.2947° φ = 180° − α − 80° = 180° − 13.2947° − 80° = 86.705° R = 104.4 N 86.7° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 21 PROBLEM 2.20 Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 60 N and Q = 48 N, determine by trigonometry the magnitude and direction of the resultant of the two forces. SOLUTION Using the force triangle and the laws of cosines and sines: We have γ = 180° − (20° + 10°) = 150° Then R 2 = (60 N)2 + (48 N) 2 − 2(60 N)(48 N) cos 150° R = 104.366 N and Hence: 60 N 104.366 N = sin α sin150° sin α = 0.28745 α = 16.7054° φ = 180° − α − 180° = 180° − 16.7054° − 80° = 83.295° R = 104.4 N 83.3° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 22 PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION 80-N Force: 120-N Force: 150-N Force: Fx = + (80 N) cos 40° Fx = 61.3 N Fy = + (80 N) sin 40° Fy = 51.4 N Fx = + (120 N) cos 70° Fx = 41.0 N Fy = +(120 N) sin 70° Fy = 112.8 N Fx = −(150 N) cos 35° Fx = −122. 9 N Fy = +(150 N) sin 35° Fy = 86.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 23 PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION 40-lb Force: 50-lb Force: 60-lb Force: Fx = + (40 lb) cos 60° Fx = 20.0 lb Fy = −(40 lb)sin 60° Fy = −34.6 lb Fx = −(50 lb)sin 50° Fx = −38.3 lb Fy = −(50 lb) cos 50° Fy = −32.1 lb Fx = + (60 lb) cos 25° Fx = 54.4 lb Fy = +(60 lb)sin 25° Fy = 25.4 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 24 PROBLEM 2.23 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 800-N Force: 424-N Force: 408-N Force: Fx = + (800 N) 800 1000 Fx = +640 N Fy = +(800 N) 600 1000 Fy = +480 N Fx = −(424 N) 560 1060 Fx = −224 N Fy = −(424 N) 900 1060 Fy = −360 N Fx = + (408 N) 480 1020 Fx = +192.0 N Fy = −(408 N) 900 1020 Fy = −360 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 25 PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION Compute the following distances: OA = (24 in.)2 + (45 in.)2 = 51.0 in. OB = (28 in.) 2 + (45 in.) 2 = 53.0 in. OC = (40 in.) 2 + (30 in.) 2 = 50.0 in. 102-lb Force: 106-lb Force: 200-lb Force: Fx = −102 lb 24 in. 51.0 in. Fx = −48.0 lb Fy = +102 lb 45 in. 51.0 in. Fy = +90.0 lb Fx = +106 lb 28 in. 53.0 in. Fx = +56.0 lb Fy = +106 lb 45 in. 53.0 in. Fy = +90.0 lb Fx = −200 lb 40 in. 50.0 in. Fx = −160.0 lb Fy = −200 lb 30 in. 50.0 in. Fy = −120.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 26 PROBLEM 2.25 The hydraulic cylinder BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 750-N component perpendicular to member ABC, determine (a) the magnitude of the force P, (b) its component parallel to ABC. SOLUTION (a) 750 N = P sin 20° P = 2192.9 N (b) P = 2190 N PABC = P cos 20° = (2192.9 N) cos 20° PABC = 2060 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 27 PROBLEM 2.26 Cable AC exerts on beam AB a force P directed along line AC. Knowing that P must have a 350-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component. SOLUTION (a) P= Py cos 55° 350 lb cos 55° = 610.21 lb = (b) P = 610 lb Px = P sin 55° = (610.21 lb)sin 55° = 499.85 lb Px = 500 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 28 PROBLEM 2.27 Member BC exerts on member AC a force P directed along line BC. Knowing that P must have a 325-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION BC = (650 mm)2 + (720 mm) 2 = 970 mm 650 Px = P 970 (a) or 970 P = Px 650 970 = 325 N 650 = 485 N P = 485 N (b) 720 Py = P 970 720 = 485 N 970 = 360 N Py = 970 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 29 PROBLEM 2.28 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 240-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component. SOLUTION (a) P= (b) Px = Py sin 40° = Py tan 40° 240 lb sin 40° = 240 lb tan 40° or P = 373 lb or Px = 286 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 30 PROBLEM 2.29 The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 720-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC. SOLUTION (a) 37 Px 12 37 = (720 N) 12 = 2220 N P= P = 2.22 kN (b) 35 Px 12 35 = (720 N) 12 = 2100 N Py = Py = 2.10 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 31 PROBLEM 2.30 The hydraulic cylinder BC exerts on member AB a force P directed along line BC. Knowing that P must have a 600-N component perpendicular to member AB, determine (a) the magnitude of the force P, (b) its component along line AB. SOLUTION 180° = 45° + α + 90° + 30° α = 180° − 45° − 90° − 30° = 15° (a) Px P P P= x cos α 600 N = cos15° = 621.17 N cos α = P = 621 N (b) tan α = Py Px Py = Px tan α = (600 N) tan15° = 160.770 N Py = 160.8 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 32 PROBLEM 2.31 Determine the resultant of the three forces of Problem 2.23. PROBLEM 2.23 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.23: 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 Ry tan α = Rx 240 608 α = 21.541° 240 N R= sin(21.541°) = 653.65 N = R = 654 N 21.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 33 PROBLEM 2.32 Determine the resultant of the three forces of Problem 2.21. PROBLEM 2.21 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.21: 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 tan α = Ry Rx 250.2 N 20.6 N tan α = 12.1456 tan α = α = 85.293° R= 250.2 N sin 85.293° R = 251 N 85.3° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 34 PROBLEM 2.33 Determine the resultant of the three forces of Problem 2.22. PROBLEM 2.22 Determine the x and y components of each of the forces shown. SOLUTION 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 tan α = Ry Rx 41.42 lb 36.08 lb tan α = 1.14800 tan α = α = 48.942° R= 41.42 lb sin 48.942° R = 54.9 lb 48.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 35 PROBLEM 2.34 Determine the resultant of the three forces of Problem 2.24. PROBLEM 2.24 Determine the x and y components of each of the forces shown. SOLUTION Components of the forces were determined in Problem 2.24: Force x Comp. (lb) y Comp. (lb) 102 lb −48.0 +90.0 106 lb +56.0 +90.0 200 lb −160.0 −120.0 Rx = −152.0 Ry = 60.0 R = Rx i + Ry j = ( −152 lb)i + (60.0 lb) j tan α = Ry Rx 60.0 lb 152.0 lb tan α = 0.39474 tan α = α = 21.541° R= 60.0 lb sin 21.541° R = 163.4 lb 21.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 36 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° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 2.36 Knowing that the tension in rope AC is 365 N, determine the resultant of the three forces exerted at point C of post BC. SOLUTION Determine force components: Cable force AC: 500-N Force: 200-N Force: and 960 = −240 N 1460 1100 = −275 N Fy = −(365 N) 1460 Fx = −(365 N) 24 = 480 N 25 7 Fy = (500 N) = 140 N 25 Fx = (500 N) 4 = 160 N 5 3 Fy = −(200 N) = −120 N 5 Fx = (200 N) Rx = ΣFx = −240 N + 480 N + 160 N = 400 N Ry = ΣFy = −275 N + 140 N − 120 N = −255 N R = Rx2 + Ry2 = (400 N) 2 + (−255 N) 2 = 474.37 N Further: 255 400 α = 32.5° tan α = R = 474 N 32.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 38 PROBLEM 2.37 Knowing that α = 40°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: Fx = (60 lb) cos 20° = 56.382 lb Fy = (60 lb)sin 20° = 20.521 lb 80-lb Force: Fx = (80 lb) cos 60° = 40.000 lb Fy = (80 lb)sin 60° = 69.282 lb 120-lb Force: Fx = (120 lb) cos 30° = 103.923 lb Fy = −(120 lb)sin 30° = −60.000 lb and Rx = ΣFx = 200.305 lb Ry = ΣFy = 29.803 lb R = (200.305 lb)2 + (29.803 lb) 2 = 202.510 lb Further: tan α = 29.803 200.305 α = tan −1 29.803 200.305 R = 203 lb = 8.46° 8.46° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 39 PROBLEM 2.38 Knowing that α = 75°, determine the resultant of the three forces shown. SOLUTION 60-lb Force: Fx = (60 lb) cos 20° = 56.382 lb Fy = (60 lb) sin 20° = 20.521 lb 80-lb Force: Fx = (80 lb) cos 95° = −6.9725 lb Fy = (80 lb)sin 95° = 79.696 lb 120-lb Force: Fx = (120 lb) cos 5 ° = 119.543 lb Fy = (120 lb)sin 5° = 10.459 lb Then Rx = ΣFx = 168.953 lb Ry = ΣFy = 110.676 lb and R = (168.953 lb) 2 + (110.676 lb)2 = 201.976 lb 110.676 168.953 tan α = 0.65507 α = 33.228° tan α = R = 202 lb 33.2° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 40 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 = 0.39872 tan α = α = 21.738° (b) α = 21.7° Substituting for α in Eq. (2): Ry = −300sin 21.738° − 150sin 51.738° = −228.89 N R = | Ry | = 228.89 N R = 229 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 41 PROBLEM 2.40 For the post of Prob. 2.36, determine (a) the required tension in rope AC if the resultant of the three forces exerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant. SOLUTION Rx = ΣFx = − Rx = − 48 TAC + 640 N 73 R y = ΣFy = − Ry = − (a) 960 24 4 TAC + (500 N) + (200 N) 1460 25 5 1100 7 3 TAC + (500 N) − (200 N) 1460 25 5 55 TAC + 20 N 73 (2) For R to be horizontal, we must have R y = 0. Set R y = 0 in Eq. (2): − 55 TAC + 20 N = 0 73 TAC = 26.545 N (b) (1) TAC = 26.5 N Substituting for TAC into Eq. (1) gives 48 (26.545 N) + 640 N 73 Rx = 622.55 N Rx = − R = Rx = 623 N R = 623 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 42 PROBLEM 2.41 A hoist trolley is subjected to the three forces shown. Knowing that α = 40°, determine (a) the required magnitude of the force P if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION Rx = ΣFx = P + (200 lb)sin 40° − (400 lb) cos 40° Rx = P − 177.860 lb Ry = ΣFy = (200 lb) cos 40° + (400 lb) sin 40° Ry = 410.32 lb (a) (2) For R to be vertical, we must have Rx = 0. Set Rx = 0 in Eq. (1) 0 = P − 177.860 lb P = 177.860 lb (b) (1) P = 177.9 lb Since R is to be vertical: R = Ry = 410 lb R = 410 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 43 PROBLEM 2.42 A hoist trolley is subjected to the three forces shown. Knowing that P = 250 lb, determine (a) the required value of α if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant. SOLUTION Rx = ΣFx = 250 lb + (200 lb)sin α − (400 lb) cos α Rx = 250 lb + (200 lb)sin α − (400 lb) cos α Ry = (a) (1) ΣFy = (200 lb) cos α + (400 lb)sin α For R to be vertical, we must have Rx = 0. Rx = 0 in Eq. (1) Set 0 = 250 lb + (200 lb)sin α − (400 lb) cos α (400 lb) cos α = (200 lb) sin α + 250 lb 2 cos α = sin α + 1.25 4cos 2 α = sin 2 α + 2.5sin α + 1.5625 4(1 − sin 2 α ) = sin 2 α + 2.5sin α + 1.5625 0 = 5sin 2 α + 2.5sin α − 2.4375 Using the quadratic formula to solve for the roots gives sin α = 0.49162 α = 29.447° or (b) α = 29.4° Since R is to be vertical: R = Ry = (200 lb) cos 29.447° + (400 lb) sin 29.447° R = 371 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 44 PROBLEM 2.43 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram 1100 960 α = 48.888° 400 tan β = 960 β = 22.620° tan α = Force Triangle Law of sines: TAC TBC 15.696 kN = = sin 22.620° sin 48.888° sin108.492° (a) TAC = 15.696 kN (sin 22.620°) sin108.492° TAC = 6.37 kN (b) TBC = 15.696 kN (sin 48.888°) sin108.492° TBC = 12.47 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 45 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 3 2.25 α = 53.130° 1.4 tan β = 2.25 β = 31.891° tan α = Free-Body Diagram Law of sines: Force-Triangle TAC TBC 660 N = = sin 31.891° sin 53.130° sin 94.979° (a) TAC = 660 N (sin 31.891°) sin 94.979° TAC = 350 N (b) TBC = 660 N (sin 53.130°) sin 94.979° TBC = 530 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 46 PROBLEM 2.45 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 (b) TBC = 1200 lb sin 5° sin 65° TBC = 115.4 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 47 PROBLEM 2.46 Knowing that α = 55° and that boom AC exerts on pin C a force directed along line AC, determine (a) the magnitude of that force, (b) the tension in cable BC. SOLUTION 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° (b) TBC = 300 lb sin 50° sin 95° FAC = 172.7 lb TBC = 231 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 48 PROBLEM 2.47 Two cables are tied together at C and loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free-Body Diagram 1.4 4.8 α = 16.2602° tan α = 1.6 3 β = 28.073° tan β = Force Triangle Law of sines: TAC TBC 1.98 kN = = sin 61.927° sin 73.740° sin 44.333° (a) TAC = 1.98 kN sin 61.927° sin 44.333° TAC = 2.50 kN (b) TBC = 1.98 kN sin 73.740° sin 44.333° TBC = 2.72 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 49 PROBLEM 2.48 Two cables are tied together at C and are loaded as shown. Knowing that P = 500 N and α = 60°, determine the tension in (a) in cable AC, (b) in cable BC. SOLUTION 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 (b) TBC = 500 N sin 75° sin 70° TBC = 514 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 50 PROBLEM 2.49 Two forces of magnitude TA = 8 kips and TB = 15 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, determine the magnitudes of the forces TC and TD. SOLUTION Free-Body Diagram ΣFx = 0 15 kips − 8 kips − TD cos 40° = 0 TD = 9.1379 kips ΣFy = 0 TD sin 40° − TC = 0 (9.1379 kips) sin 40° − TC = 0 TC = 5.8737 kips TC = 5.87 kips TD = 9.14 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 51 PROBLEM 2.50 Two forces of magnitude TA = 6 kips and TC = 9 kips are applied as shown to a welded connection. Knowing that the connection is in equilibrium, determine the magnitudes of the forces TB and TD. SOLUTION Free-Body Diagram Σ Fx = 0 TB − 6 kips − TD cos 40° = 0 Σ Fy = 0 TD sin 40° − 9 kips = 0 (1) 9 kips sin 40° TD = 14.0015 kips TD = Substituting for TD into Eq. (1) gives: TB − 6 kips − (14.0015 kips) cos 40° = 0 TB = 16.7258 kips TB = 16.73 kips TD = 14.00 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 52 PROBLEM 2.51 Two cables are tied together at C and loaded as shown. Knowing that P = 360 N, determine the tension (a) in cable AC, (b) in cable BC. SOLUTION Free Body: C (a) ΣFx = 0: − (b) ΣFy = 0: 12 4 TAC + (360 N) = 0 13 5 TAC = 312 N 5 3 (312 N) + TBC + (360 N) − 480 N = 0 13 5 TBC = 480 N − 120 N − 216 N TBC = 144 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 53 PROBLEM 2.52 Two cables are tied together at C and loaded as shown. Determine the range of values of P for which both cables remain taut. SOLUTION Free Body: C 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 TBC = 480 N − 14 P 15 (2) From (1), TAC ⬎ 0 requires P ⬎ 0. From (2), TBC ⬎ 0 requires 14 P ⬍ 480 N, P ⬍ 514.29 N 15 0 ⬍ P ⬍ 514 N Allowable range: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 54 PROBLEM 2.53 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 30° and β = 10° and that the combined weight of the boatswain’s chair and the sailor is 900 N, determine the tension (a) in the support cable ACB, (b) in the traction cable CD. 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) TACB = 1213 N TCD = 0.137158(1212.56 N) TCD = 166.3 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 55 PROBLEM 2.54 A sailor is being rescued using a boatswain’s chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that α = 25° and β = 15° and that the tension in cable CD is 80 N, determine (a) the combined weight of the boatswain’s chair and the sailor, (b) in tension in the support cable ACB. SOLUTION Free-Body Diagram ΣFx = 0: TACB cos 15° − TACB cos 25° − (80 N) cos 25° = 0 TACB = 1216.15 N ΣFy = 0: (1216.15 N) sin 15° + (1216.15 N) sin 25° + (80 N) sin 25° − W = 0 W = 862.54 N (a) W = 863 N (b) TACB = 1216 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 56 PROBLEM 2.55 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that P = 500 lb and Q = 650 lb, determine the magnitudes of the forces exerted on the rods A and B. SOLUTION 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 (650 lb) cos 50° + FB − FA cos 50° = 0 FB = FA cos 50° − (650 lb) cos50° = (1302.70 lb) cos 50° − (650 lb) cos 50° = 419.55 lb FB = 420 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 57 PROBLEM 2.56 Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magnitudes of the forces exerted on rods A and B are FA = 750 lb and FB = 400 lb, determine the magnitudes of P and Q. SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: R = P + Q + FA + FB = 0 Substituting components: R = − Pj + Q cos 50°i − Q sin 50° j − [(750 lb) cos 50°]i + [(750 lb)sin 50°] j + (400 lb)i In the x-direction (one unknown force): Q cos 50° − [(750 lb) cos 50°] + 400 lb = 0 (750 lb) cos 50° − 400 lb cos 50° = 127.710 lb Q= In the y-direction: − P − Q sin 50° + (750 lb) sin 50° = 0 P = −Q sin 50° + (750 lb) sin 50° = −(127.710 lb)sin 50° + (750 lb) sin 50° = 476.70 lb P = 477 lb; Q = 127.7 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 58 PROBLEM 2.57 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension in each cable is 800 N, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α. SOLUTION 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 Since P ⬎ 0, the solution is correct. (b) α = 180° − 50° − 47.5° = 82.5° α = 82.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 59 PROBLEM 2.58 Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension is 1200 N in cable AC and 600 N in cable BC, determine (a) the magnitude of the largest force P that can be applied at C, (b) the corresponding value of α. SOLUTION 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 = 1294.02 N Since P ⬎ 1200 N, the solution is correct. P = 1294 N (b) Law of sines: sin β sin 85° = 1200 N 1294.02 N β = 67.5° α = 180° − 50° − 67.5° α = 62.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 60 PROBLEM 2.59 For the situation described in Figure P2.45, determine (a) the value of α for which the tension in rope BC is as small as possible, (b) the corresponding value of the tension. PROBLEM 2.45 Knowing that α = 20°, determine the tension (a) in cable AC, (b) in rope BC. SOLUTION Free-Body Diagram Force Triangle To be smallest, TBC must be perpendicular to the direction of TAC . (a) (b) Thus, α = 5° α = 5.00° TBC = (1200 lb) sin 5° TBC = 104.6 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 61 PROBLEM 2.60 For the structure and loading of Problem 2.46, determine (a) the value of α for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension. 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° TBC = (300 lb)sin 50° (b) or TBC = 229.81 lb TBC = 230 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 62 PROBLEM 2.61 For the cables of Problem 2.48, it is known that the maximum allowable tension is 600 N in cable AC and 750 N in cable BC. Determine (a) the maximum force P that can be applied at C, (b) the corresponding value of α. SOLUTION 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 sin β sin (25° + 45°) = 600 N 784.02 N β = 46.0° ∴ α = 46.0° + 25° α = 71.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 63 PROBLEM 2.62 A movable bin and its contents have a combined weight of 2.8 kN. Determine the shortest chain sling ACB that can be used to lift the loaded bin if the tension in the chain is not to exceed 5 kN. SOLUTION Free-Body Diagram tan α = h 0.6 m (1) Isosceles Force Triangle Law of sines: sin α = 1 2 (2.8 kN) TAC TAC = 5 kN sin α = 1 2 (2.8 kN) 5 kN α = 16.2602° From Eq. (1): tan16.2602° = h 0.6 m ∴ h = 0.175000 m Half length of chain = AC = (0.6 m) 2 + (0.175 m)2 = 0.625 m Total length: = 2 × 0.625 m 1.250 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 64 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 Force Triangle P 50 lb = 15 25 P = 30.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 65 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. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 2.65 Three forces are applied to a bracket as shown. The directions of the two 150-N forces may vary, but the angle between these forces is always 50°. Determine the range of values of α for which the magnitude of the resultant of the forces acting at A is less than 600 N. SOLUTION Combine the two 150-N forces into a resultant force Q: Q = 2(150 N) cos 25° = 271.89 N Equivalent loading at A: Using the law of cosines: (600 N) 2 = (500 N) 2 + (271.89 N)2 + 2(500 N)(271.89 N) cos(55° + α ) cos(55° + α ) = 0.132685 Two values for α : 55° + α = 82.375 α = 27.4° or 55° + α = −82.375° 55° + α = 360° − 82.375° α = 222.6° 27.4° < α < 222.6 For R < 600 lb: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 67 PROBLEM 2.66 A 200-kg crate is to be supported by the rope-and-pulley arrangement shown. Determine the magnitude and direction of the force P that must be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Ch. 4.) SOLUTION Free-Body Diagram: Pulley A 5 ΣFx = 0: − 2 P + P cos α = 0 281 cos α = 0.59655 α = ±53.377° For α = +53.377°: 16 ΣFy = 0: 2 P + P sin 53.377° − 1962 N = 0 281 P = 724 N 53.4° For α = −53.377°: 16 ΣFy = 0: 2 P + P sin(−53.377°) − 1962 N = 0 281 P = 1773 53.4° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 68 PROBLEM 2.67 A 600-lb crate is supported by several rope-andpulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.) SOLUTION Free-Body Diagram of Pulley (a) ΣFy = 0: 2T − (600 lb) = 0 T= 1 (600 lb) 2 T = 300 lb (b) ΣFy = 0: 2T − (600 lb) = 0 T= 1 (600 lb) 2 T = 300 lb (c) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb (d) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb (e) ΣFy = 0: 4T − (600 lb) = 0 T= 1 (600 lb) 4 T = 150.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 69 PROBLEM 2.68 Solve Parts b and d of Problem 2.67, assuming that the free end of the rope is attached to the crate. PROBLEM 2.67 A 600-lb crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (See the hint for Problem 2.66.) SOLUTION Free-Body Diagram of Pulley and Crate (b) ΣFy = 0: 3T − (600 lb) = 0 1 T = (600 lb) 3 T = 200 lb (d) ΣFy = 0: 4T − (600 lb) = 0 T= 1 (600 lb) 4 T = 150.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 70 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 ΣFy = 0: TACB (sin 25° + sin 55°) + (750 N) sin 55° − Q = 0 (b) (1292.88 N)(sin 25° + sin 55°) + (750 N) sin 55° − Q = 0 Q = 2219.8 N or Q = 2220 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 71 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 (b) P = 0.58010(1048.37 N) = 608.16 N Using (1), P = 608 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 72 PROBLEM 2.71 Determine (a) the x, y, and z components of the 900-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION Fh = F cos 65° = (900 N) cos 65° Fh = 380.36 N (a) Fx = Fh sin 20° = (380.36 N)sin 20° Fx = −130.091 N, Fx = −130.1 N Fy = F sin 65° = (900 N)sin 65° Fy = +815.68 N, Fy = +816 N Fz = Fh cos 20° = (380.36 N) cos 20° Fz = +357.42 N (b) cos θ x = cos θ y = cos θ z = Fx −130.091 N = 900 N F Fy F = +815.68 N 900 N Fz +357.42 N = 900 N F Fz = +357 N θ x = 98.3° θ y = 25.0° θ z = 66.6° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 73 PROBLEM 2.72 Determine (a) the x, y, and z components of the 750-N force, (b) the angles θx, θy, and θz that the force forms with the coordinate axes. SOLUTION Fh = F sin 35° = (750 N)sin 35° Fh = 430.18 N (a) Fx = Fh cos 25° = (430.18 N) cos 25° Fx = +389.88 N, Fx = +390 N Fy = F cos 35° = (750 N) cos 35° Fy = +614.36 N, Fy = +614 N Fz = Fh sin 25° = (430.18 N)sin 25° Fz = +181.802 N (b) cos θ x = cos θ y = cos θ z = Fx +389.88 N = 750 N F Fy F = +614.36 N 750 N Fz +181.802 N = 750 N F Fz = +181.8 N θ x = 58.7° θ y = 35.0° θ z = 76.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 74 PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) SOLUTION Recoil force F = 400 N ∴ FH = (400 N) cos 40° = 306.42 N (a) Fx = − FH sin 35° = −(306.42 N)sin 35° = −175.755 N Fx = −175.8 N Fy = − F sin 40° = −(400 N)sin 40° = −257.12 N Fy = −257 N Fz = + FH cos 35° = +(306.42 N) cos35° = +251.00 N (b) cos θ x = Fx −175.755 N = 400 N F −257.12 N 400 N cos θ y = Fy cos θ z = Fz 251.00 N = 400 N F F = Fz = +251 N θ x = 116.1° θ y = 130.0° θ z = 51.1° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 75 PROBLEM 2.74 Solve Problem 2.73, assuming that point A is located 15° north of west and that the barrel of the gun forms an angle of 25° with the horizontal. PROBLEM 2.73 A gun is aimed at a point A located 35° east of north. Knowing that the barrel of the gun forms an angle of 40° with the horizontal and that the maximum recoil force is 400 N, determine (a) the x, y, and z components of that force, (b) the values of the angles θx, θy, and θz defining the direction of the recoil force. (Assume that the x, y, and z axes are directed, respectively, east, up, and south.) SOLUTION Recoil force F = 400 N ∴ FH = (400 N) cos 25° = 362.52 N (a) Fx = + FH cos15° = + (362.52 N) cos15° = +350.17 N Fx = +350 N Fy = − F sin 25° = −(400 N)sin 25° = −169.047 N Fy = −169.0 N Fz = + FH sin15° = +(362.52 N)sin15° = +93.827 N (b) cos θ x = cos θ y = cos θ z = Fx +350.17 N = 400 N F Fy F = −169.047 N 400 N Fz +93.827 N = 400 N F Fz = +93.8 N θ x = 28.9° θ y = 115.0° θ z = 76.4° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 76 PROBLEM 2.75 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 Fy = + F cos θ y = (3900 lb)(0.86154) Fz = + (3900 lb)sin 30.51° sin 20° (b) cos θ x = From above: Fx 1861 lb =− = − 0.4771 3900 lb F θ y = 30.51° cos θ z = Fz 677 lb =+ = + 0.1736 3900 lb F Fy = +3360 lb Fz = +677 lb θ x = 118.5° θ y = 30.5° θ z = 80.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 77 PROBLEM 2.76 Cable AC is 70 ft long, and the tension in that cable is 5250 lb. Determine (a) the x, y, and z components of the force exerted by the cable on the anchor C, (b) the angles θx, θy, and θz defining the direction of that force. SOLUTION AC = 70 ft OA = 56 ft F = 5250 lb In triangle AOB: cos θ y = 56 ft 70 ft θ y = 36.870° FH = F sin θ y = (5250 lb) sin 36.870° = 3150.0 lb (a) (b) Fx = − FH sin 50° = −(3150.0 lb)sin 50° = −2413.04 lb Fx = −2413 lb Fy = + F cos θ y = + (5250 lb) cos 36.870° = +4200.0 lb Fy = +4200 lb Fz = − FH cos 50° = −3150cos50° = −2024.8 lb Fz = −2025 lb cos θ x = From above: Fx −2413.04 lb = 5250 lb F θ y = 36.870° θz = Fz −2024.8 lb = 5250 lb F θ x = 117.4° θ y = 36.9° θ z = 112.7° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 78 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 Fy = −(120 lb)sin 60° Fy = −103.923 lb Fy = −103.9 lb Fz = −(120 lb) cos 60° sin 20° Fz = −20.521 lb (b) Fz = −20.5 lb cos θ x = Fx 56.382 lb = F 120 lb cos θ y = Fy cos θ z = F = −103.923 lb 120 lb Fz −20.52 lb = F 120 lb θ x = 62.0° θ y = 150.0° θ z = 99.8° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 79 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 Fy = −(85 lb) cos 36° = −68.766 lb Fy = −68.8 lb Fz = (85 lb)sin 36° cos 48° Fz = 33.4 lb = 33.431 lb (b) cos θ x = Fx 37.129 lb = F 85 lb cos θ y = Fy cos θ z = Fz 33.431 lb = F 85 lb F = −68.766 lb 85 lb θ x = 64.1° θ y = 144.0° θ z = 66.8° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 80 PROBLEM 2.79 Determine the magnitude and direction of the force F = (690 lb)i + (300 lb)j – (580 lb)k. SOLUTION F = (690 N)i + (300 N) j − (580 N)k F = Fx2 + Fy2 + Fz2 = (690 N) 2 + (300 N)2 + (−580 N) 2 F = 950 N = 950 N cos θ x = Fx 690 N = F 950 N θ x = 43.4° cos θ y = Fy θ y = 71.6° cos θ z = F = 300 N 950 N Fz −580 N = F 950 N θ z = 127.6° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 81 PROBLEM 2.80 Determine the magnitude and direction of the force F = (650 N)i − (320 N)j + (760 N)k. SOLUTION F = (650 N)i − (320 N) j + (760 N)k F = Fx2 + Fy2 + Fz2 = (650 N) 2 + (−320 N)2 + (760 N)2 F = 1050 N cos θ x = Fx 650 N = F 1050 N θ x = 51.8° cos θ y = Fy −320 N 1050 N θ y = 107.7° cos θ z = Fz 760 N = F 1050 N θ z = 43.6° F = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 82 PROBLEM 2.81 A force acts at the origin of a coordinate system in a direction defined by the angles θx = 75° and θz = 130°. Knowing that the y component of the force is +300 lb, determine (a) the angle θy, (b) the other components and the magnitude of the force. SOLUTION cos 2 θ x + cos 2 θ y + cos 2 θ z = 1 cos 2 (75°) + cos 2 θ y + cos 2 (130°) = 1 cos θ y = ±0.72100 (a) (b) Since Fy ⬎ 0, we choose cos θ y ⫽⫹0.72100 ∴ θ y = 43.9° Fy = F cos θ y 300 lb = F (0.72100) F = 416.09 lb F = 416 lb Fx = F cos θ x = 416.09 lb cos 75° Fz = F cos θ z = 416.09 lb cos130° Fx = +107.7 lb Fz = −267 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 83 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 N, determine (a) the angle θx, (b) the other components and the magnitude of the force. SOLUTION cos 2 θ x + cos 2 θ y + cos 2 θ z = 1 cos 2 θ x + cos 2 55° + cos 2 45° = 1 cos θ x = ±0.41353 (a) (b) Since Fy ⬍ 0, we choose cos θ x ⫽⫺0.41353 ∴ θ x = 114.4° Fx = F cos θ x −500 N = F (−0.41353) F = 1209.10 N F = 1209.1 N Fy = F cos θ y = 1209.10 N cos55° Fy = +694 N Fz = F cos θ z = 1209.10 N cos 45° Fz = +855 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 84 PROBLEM 2.83 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 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 Fz = 108.036 N −60 N = = − 0.26087 F 230 N F 108.036 N cos θ z = z = = 0.46972 230 N F cos θ y = Fy θ y = 105.1° θ z = 62.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 85 PROBLEM 2.84 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 Fy = −62.0 lb cos θ x = Fx 80 N = = 0.38095 F 210 N θ x = 67.6° cos θ y = Fy θ y = 107.2° F = 61.929 N = −0.29490 210 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 86 PROBLEM 2.85 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AB is 2 kips, determine the components of the force exerted at A by the cable. SOLUTION Cable AB: λAB AB (−46.765 ft)i + (45 ft) j + (36 ft)k = = 74.216 ft AB TAB = TAB λAB = −46.765i + 45 j + 36k 74.216 (TAB ) x = −1.260 kips (TAB ) y = +1.213 kips (TAB ) z = +0.970 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 87 PROBLEM 2.86 In order to move a wrecked truck, two cables are attached at A and pulled by winches B and C as shown. Knowing that the tension in cable AC is 1.5 kips, determine the components of the force exerted at A by the cable. SOLUTION Cable AB: λAC AC (−46.765 ft)i + (55.8 ft) j + (−45 ft)k = = 85.590 ft AC TAC = TAC λAC = (1.5 kips) −46.765i + 55.8 j − 45k 85.590 (TAC ) x = −0.820 kips (TAC ) y = +0.978 kips (TAC ) z = −0.789 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 88 PROBLEM 2.87 Knowing that the tension in cable AB is 1425 N, determine the components of the force exerted on the plate at B. SOLUTION BA = −(900 mm)i + (600 mm) j + (360 mm)k BA = (900 mm) 2 + (600 mm) 2 + (360 mm) 2 = 1140 mm TBA = TBA λ BA BA = TBA BA 1425 N [ −(900 mm)i + (600 mm) j + (360 mm)k ] TBA = 1140 mm = −(1125 N)i + (750 N) j + (450 N)k (TBA ) x = −1125 N, (TBA ) y = 750 N, (TBA ) z = 450 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 89 PROBLEM 2.88 Knowing that the tension in cable AC is 2130 N, determine the components of the force exerted on the plate at C. SOLUTION CA = −(900 mm)i + (600 mm) j − (920 mm)k CA = (900 mm)2 + (600 mm)2 + (920 mm) 2 = 1420 mm TCA = TCA λ CA CA = TCA CA 2130 N TCA = [−(900 mm)i + (600 mm) j − (920 mm)k ] 1420 mm = −(1350 N)i + (900 N) j − (1380 N)k (TCA ) x = −1350 N, (TCA ) y = 900 N, (TCA ) z = −1380 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 90 PROBLEM 2.89 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 DB = (480 mm)i − (510 mm) j + (320 mm)k DB = (480 mm)2 + (510 mm 2 ) + (320 mm) 2 = 770 mm F = F λ DB DB =F DB 385 N = [(480 mm)i − (510 mm)j + (320 mm)k ] 770 mm = (240 N)i − (255 N) j + (160 N)k Fx = +240 N, Fy = −255 N, Fz = +160.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 91 PROBLEM 2.90 For the frame and cable of Problem 2.89, determine the components of the force exerted by the cable on the support at E. PROBLEM 2.89 A 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 EB = (270 mm)i − (400 mm) j + (600 mm)k EB = (270 mm)2 + (400 mm) 2 + (600 mm)2 = 770 mm F = F λ EB EB =F EB 385 N = [(270 mm)i − (400 mm)j + (600 mm)k ] 770 mm F = (135 N)i − (200 N) j + (300 N)k Fx = +135.0 N, Fy = −200 N, Fz = +300 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 92 PROBLEM 2.91 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 600 N and Q = 450 N. SOLUTION P = (600 N)[sin 40° sin 25°i + cos 40° j + sin 40° cos 25°k ] = (162.992 N)i + (459.63 N) j + (349.54 N)k Q = (450 N)[cos 55° cos 30°i + sin 55° j − cos 55° sin 30°k ] = (223.53 N)i + (368.62 N) j − (129.055 N)k R =P+Q = (386.52 N)i + (828.25 N) j + (220.49 N)k R = (386.52 N)2 + (828.25 N)2 + (220.49 N) 2 R = 940 N = 940.22 N cos θ x = Rx 386.52 N = R 940.22 N θ x = 65.7° cos θ y = Ry θ y = 28.2° cos θ z = Rz 220.49 N = R 940.22 N R = 828.25 N 940.22 N θ z = 76.4° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 93 PROBLEM 2.92 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 450 N and Q = 600 N. SOLUTION P = (450 N)[sin 40° sin 25°i + cos 40° j + sin 40° cos 25°k ] = (122.244 N)i + (344.72 N) j + (262.154 N)k Q = (600 N)[cos 55° cos 30°i + sin 55° j − cos 55° sin 30°k ] = (298.04 N)i + (491.49 N)j − (172.073 N)k R =P+Q = (420.28 N)i + (836.21 N) j + (90.081 N)k R = (420.28 N) 2 + (836.21 N) 2 + (90.081 N) 2 = 940.21 N R = 940 N cos θ x = Rx 420.28 = R 940.21 θ x = 63.4° cos θ y = Ry θ y = 27.2° cos θ z = Rz 90.081 = R 940.21 R = 836.21 940.21 θ z = 84.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 94 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 AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.)2 + (45 in.)2 + (60 in.) 2 = 85 in. AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.)2 + (45 in.)2 + (60 in.)2 = 125 in. (40 in.)i − (45 in.) j + (60 in.)k AB TAB = TAB λAB = TAB = (425 lb) AB 85 in. TAB = (200 lb)i − (225 lb) j + (300 lb)k (100 in.)i − (45 in.) j + (60 in.)k AC TAC = TAC λAC = TAC = (510 lb) 125 in. AC TAC = (408 lb)i − (183.6 lb) j + (244.8 lb)k R = TAB + TAC = (608)i − (408.6 lb) j + (544.8 lb)k Then: and R = 912.92 lb R = 913 lb cos θ x = 608 lb = 0.66599 912.92 lb cos θ y = 408.6 lb = −0.44757 912.92 lb cos θ z = 544.8 lb = 0.59677 912.92 lb θ x = 48.2° θ y = 116.6° θ z = 53.4° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 95 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 AB = (40 in.)i − (45 in.) j + (60 in.)k AB = (40 in.)2 + (45 in.)2 + (60 in.) 2 = 85 in. AC = (100 in.)i − (45 in.) j + (60 in.)k AC = (100 in.)2 + (45 in.)2 + (60 in.)2 = 125 in. (40 in.)i − (45 in.) j + (60 in.)k AB TAB = TAB λAB = TAB = (510 lb) AB 85 in. TAB = (240 lb)i − (270 lb) j + (360 lb)k (100 in.)i − (45 in.) j + (60 in.)k AC TAC = TAC λAC = TAC = (425 lb) 125 in. AC TAC = (340 lb)i − (153 lb) j + (204 lb)k R = TAB + TAC = (580 lb)i − (423 lb) j + (564 lb)k Then: and R = 912.92 lb R = 913 lb cos θ x = 580 lb = 0.63532 912.92 lb θ x = 50.6° cos θ y = −423 lb = −0.46335 912.92 lb θ y = 117.6° cos θ z = 564 lb = 0.61780 912.92 lb θ z = 51.8° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 96 PROBLEM 2.95 For the frame of Problem 2.89, determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N. PROBLEM 2.89 A 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 BD = −(480 mm)i + (510 mm) j − (320 mm)k BD = (480 mm) 2 + (510 mm) 2 + (320 mm) 2 = 770 mm BD FBD = TBD λ BD = TBD BD (385 N) [−(480 mm)i + (510 mm) j − (320 mm)k ] = (770 mm) = −(240 N)i + (255 N) j − (160 N)k BE = −(270 mm)i + (400 mm) j − (600 mm)k BE = (270 mm) 2 + (400 mm) 2 + (600 mm) 2 = 770 mm BE FBE = TBE λ BE = TBE BE (385 N) [−(270 mm)i + (400 mm) j − (600 mm)k ] = (770 mm) = −(135 N)i + (200 N) j − (300 N)k R = FBD + FBE = −(375 N)i + (455 N) j − (460 N)k R = (375 N)2 + (455 N)2 + (460 N)2 = 747.83 N R = 748 N cos θ x = −375 N 747.83 N θ x = 120.1° cos θ y = 455 N 747.83 N θ y = 52.5° cos θ z = −460 N 747.83 N θ z = 128.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 97 PROBLEM 2.96 For the cables of Problem 2.87, knowing that the tension is 1425 N in cable AB and 2130 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted at A by the two cables. SOLUTION TAB = −TBA (use results of Problem 2.87) (TAB ) x = +1125 N (TAB ) y = −750 N (TAB ) z = − 450 N TAC = −TCA (use results of Problem 2.88) (TAC ) x = +1350 N (TAC ) y = −900 N (TAC ) z = +1380 N Resultant: Rx = ΣFx = +1125 + 1350 = +2475 N Ry = ΣFy = −750 − 900 = −1650 N Rz = ΣFz = −450 + 1380 = + 930 N R = Rx2 + Ry2 + Rz2 = (+2475)2 + (−1650) 2 + (+930) 2 = 3116.6 N cos θ x = cos θ y = cos θ z = R = 3120 N Rx +2475 = R 3116.6 θ x = 37.4° −1650 3116.6 θ y = 122.0° Ry R = Rz + 930 = R 3116.6 θ z = 72.6° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 98 PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC. SOLUTION Cable AB: TAB = 183 lb TAB TAB Cable AC: TAC TAC Load P: AB (−48 in.)i + (29 in.) j + (24 in.)k = TAB λ AB = TAB = (183 lb) AB 61 in. = −(144 lb)i + (87 lb) j + (72 lb)k AC (−48 in.)i + (25 in.) j + (−36 in.)k = TAC λ AC = TAC = TAC AC 65 in. 48 25 36 = − TAC i + TAC j − TAC k 65 65 65 P = Pj For resultant to be directed along OA, i.e., x-axis Rz = 0: ΣFz = (72 lb) − 36 ′ =0 TAC 65 TAC = 130.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 99 PROBLEM 2.98 For the boom and loading of Problem. 2.97, determine the magnitude of the load P. PROBLEM 2.97 The boom OA carries a load P and is supported by two cables as shown. Knowing that the tension in cable AB is 183 lb and that the resultant of the load P and of the forces exerted at A by the two cables must be directed along OA, determine the tension in cable AC. SOLUTION See Problem 2.97. Since resultant must be directed along OA, i.e., the x-axis, we write Ry = 0: ΣFy = (87 lb) + 25 TAC − P = 0 65 TAC = 130.0 lb from Problem 2.97. Then (87 lb) + 25 (130.0 lb) − P = 0 65 P = 137.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 100 PROBLEM 2.99 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AB is 6 kN. SOLUTION Free-Body Diagram at A: The forces applied at A are: TAB , TAC , TAD , and W where W = W j. To express the other forces in terms of the unit vectors i, j, k, we write AB = − (450 mm)i + (600 mm) j AB = 750 mm AC = + (600 mm) j − (320 mm)k AC = 680 mm AD = + (500 mm)i + (600 mm) j + (360 mm)k AD = 860 mm AB (−450 mm)i + (600 mm) j and TAB = λ ABTAB = TAB = TAB AB 750 mm 45 60 j TAB = − i + 75 75 TAC = λ AC TAC = TAC TAD = λ ADTAD = TAD AC (600 mm)i − (320 mm) j = TAC AC 680 mm 32 60 = j − k TAC 68 68 AD (500 mm)i + (600 mm) j + (360 mm)k = TAD AD 860 mm 60 36 50 j + k TAD = i+ 86 86 86 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 101 PROBLEM 2.99 (Continued) Equilibrium condition: ΣF = 0: ∴ TAB + TAC + TAD + W = 0 Substituting the expressions obtained for TAB , TAC , and TAD ; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: From i: From j: From k: 45 50 TAB + TAD = 0 75 86 (1) 60 60 60 TAB + TAC + TAD − W = 0 75 68 86 (2) 32 36 TAC + TAD = 0 68 86 (3) − − Setting TAB = 6 kN in (1) and (2), and solving the resulting set of equations gives TAC = 6.1920 kN TAC = 5.5080 kN W = 13.98 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 102 PROBLEM 2.100 A container is supported by three cables that are attached to a ceiling as shown. Determine the weight W of the container, knowing that the tension in cable AD is 4.3 kN. SOLUTION See Problem 2.99 for the figure and analysis leading to the following set of linear algebraic equations: 45 50 TAB + TAD = 0 75 86 (1) 60 60 60 TAB + TAC + TAD − W = 0 75 68 86 (2) 32 36 TAC + TAD = 0 68 86 (3) − − Setting TAD = 4.3 kN into the above equations gives TAB = 4.1667 kN TAC = 3.8250 kN W = 9.71 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 103 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 The forces applied at A are: TAB , TAC , TAD , and P where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write AB = − (4.20 m)i − (5.60 m) j AB = 7.00 m AC = (2.40 m)i − (5.60 m) j + (4.20 m)k AC = 7.40 m AD = − (5.60 m)j − (3.30 m)k AD = 6.50 m AB and TAB = TAB λ AB = TAB = ( − 0.6i − 0.8 j)TAB AB AC TAC = TAC λ AC = TAC = (0.32432i − 0.75676 j + 0.56757k )TAC AC AD TAD = TAD λ AD = TAD = (− 0.86154 j − 0.50769k )TAD AD PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 104 PROBLEM 2.101 (Continued) Equilibrium condition: ΣF = 0: TAB + TAC + TAD + Pj = 0 Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (− 0.6TAB + 0.32432TAC )i + (−0.8TAB − 0.75676TAC − 0.86154TAD + P) j + (0.56757TAC − 0.50769TAD )k = 0 Equating to zero the coefficients of i, j, k: − 0.6TAB + 0.32432TAC = 0 (1) − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 (2) 0.56757TAC − 0.50769TAD = 0 (3) Setting TAD = 481 N in (2) and (3), and solving the resulting set of equations gives TAC = 430.26 N TAD = 232.57 N P = 926 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 105 PROBLEM 2.102 Three cables are used to tether a balloon as shown. Knowing that the balloon exerts an 800-N vertical force at A, determine the tension in each cable. SOLUTION See Problem 2.101 for the figure and analysis leading to the linear algebraic Equations (1), (2), and (3). − 0.6TAB + 0.32432TAC = 0 (1) − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 (2) 0.56757TAC − 0.50769TAD = 0 (3) From Eq. (1): TAB = 0.54053TAC From Eq. (3): TAD = 1.11795TAC Substituting for TAB and TAD in terms of TAC into Eq. (2) gives − 0.8(0.54053TAC ) − 0.75676TAC − 0.86154(1.11795TAC ) + P = 0 2.1523TAC = P ; P = 800 N 800 N 2.1523 = 371.69 N TAC = Substituting into expressions for TAB and TAD gives TAB = 0.54053(371.69 N) TAD = 1.11795(371.69 N) TAB = 201 N, TAC = 372 N, TAD = 416 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 106 PROBLEM 2.103 A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AB is 750 lb. SOLUTION The forces applied at A are: TAB , TAC , TAD and W where P = Pj. To express the other forces in terms of the unit vectors i, j, k, we write AB = − (36 in.)i + (60 in.) j − (27 in.)k AB = 75 in. AC = (60 in.) j + (32 in.)k AC = 68 in. AD = (40 in.)i + (60 in.) j − (27 in.)k AD = 77 in. AB and TAB = TAB λAB = TAB AB = (− 0.48i + 0.8 j − 0.36k )TAB AC TAC = TAC λAC = TAC AC = (0.88235 j + 0.47059k )TAC AD TAD = TAD λAD = TAD AD = (0.51948i + 0.77922 j − 0.35065k )TAD Equilibrium Condition with W = − Wj ΣF = 0: TAB + TAC + TAD − Wj = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 107 PROBLEM 2.103 (Continued) Substituting the expressions obtained for TAB , TAC , and TAD and factoring i, j, and k: (−0.48TAB + 0.51948TAD )i + (0.8TAB + 0.88235TAC + 0.77922TAD − W ) j + (−0.36TAB + 0.47059TAC − 0.35065TAD )k = 0 Equating to zero the coefficients of i, j, k: −0.48TAB + 0.51948TAD = 0 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 −0.36TAB + 0.47059TAC − 0.35065TAD = 0 Substituting TAB = 750 lb in Equations (1), (2), and (3) and solving the resulting set of equations, using conventional algorithms for solving linear algebraic equations, gives: TAC = 1090.1 lb TAD = 693 lb W = 2100 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 108 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 (1) 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 (2) − 0.36TAB + 0.47059TAC − 0.35065TAD = 0 (3) Substituting TAD = 616 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 667.67 lb TAC = 969.00 lb W = 1868 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 109 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 (1) 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 (2) − 0.36TAB + 0.47059TAC − 0.35065TAD = 0 (3) Substituting TAC = 544 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives: TAB = 374.27 lb TAD = 345.82 lb W = 1049 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 110 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 (1) 0.8TAB + 0.88235TAC + 0.77922TAD − W = 0 (2) −0.36TAB + 0.47059TAC − 0.35065TAD = 0 (3) Substituting W = 1600 lb in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms, gives TAB = 571 lb TAC = 830 lb TAD = 528 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 111 PROBLEM 2.107 Three cables are connected at A, where the forces P and Q are applied as shown. Knowing that Q = 0, find the value of P for which the tension in cable AD is 305 N. SOLUTION ΣFA = 0: TAB + TAC + TAD + P = 0 where P = Pi AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm AB 19 48 12 = TAB − i − j + k TAB = TAB λAB = TAB AB 53 53 53 AC 3 4 12 TAC = TAC λAC = TAC = TAC − i − j − k AC 13 13 13 305 N [( −960 mm)i + (720 mm) j − (220 mm)k ] TAD = TAD λAD = 1220 mm = −(240 N)i + (180 N) j − (55 N)k Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives: i: P = 48 12 TAB + TAC + 240 N 53 13 (1) j: 12 3 TAB + TAC = 180 N 53 13 (2) k: 19 4 TAB − TAC = 55 N 53 13 (3) Solving the system of linear equations using conventional algorithms gives: TAB = 446.71 N TAC = 341.71 N P = 960 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 112 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 AB = −(960 mm)i − (240 mm)j + (380 mm)k AB = 1060 mm AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm AB 48 12 19 TAB = TAB λAB = TAB = − i − j + k TAB AB 53 53 53 AC 12 3 4 TAC = TAC λAC = TAC = − i − j − k TAC AC 13 13 13 Substituting into ΣFA = 0, factoring i, j, k , and setting each coefficient equal to φ gives: i: − 48 12 TAB − TAC + 1200 N = 0 53 13 (1) j: − 12 3 TAB − TAC + Q = 0 53 13 (2) k: 19 4 TAB − TAC = 0 53 13 (3) Solving the resulting system of linear equations using conventional algorithms gives: TAB = 605.71 N TAC = 705.71 N Q = 300.00 N 0 ⱕ Q ⬍ 300 N Note: This solution assumes that Q is directed upward as shown (Q ⱖ 0), if negative values of Q are considered, cable AD remains taut, but AC becomes slack for Q = −460 N. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 113 PROBLEM 2.109 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AC is 60 N, determine the weight of the plate. SOLUTION We note that the weight of the plate is equal in magnitude to the force P exerted by the support on Point A. Free Body A: ΣF = 0: TAB + TAC + TAD + Pj = 0 We have: AB = −(320 mm)i − (480 mm)j + (360 mm)k AB = 680 mm AC = (450 mm)i − (480 mm) j + (360 mm)k AC = 750 mm AD = (250 mm)i − (480 mm) j − ( 360 mm ) k AD = 650 mm Thus: TAB TAC TAD AB 8 12 9 = TAB λ AB = TAB = − i − j + k TAB AB 17 17 17 AC = TAC λAC = TAC = ( 0.6i − 0.64 j + 0.48k ) TAC AC AD 5 9.6 7.2 = TAD λAD = TAD = i− j− k TAD AD 13 13 13 Dimensions in mm Substituting into the Eq. ΣF = 0 and factoring i, j, k : 5 8 − TAB + 0.6TAC + TAD i 13 17 9.6 12 TAD + P j + − TAB − 0.64TAC − 13 17 7.2 9 TAD k = 0 + TAB + 0.48TAC − 13 17 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 114 PROBLEM 2.109 (Continued) Setting the coefficient of i, j, k equal to zero: i: − 8 5 TAB + 0.6TAC + TAD = 0 17 13 (1) j: − 12 9.6 TAB − 0.64TAC − TAD + P = 0 7 13 (2) 9 7.2 TAB + 0.48TAC − TAD = 0 17 13 (3) 8 5 TAB + 36 N + TAD = 0 17 13 (1′) 9 7.2 TAB + 28.8 N − TAD = 0 17 13 (3′) k: Making TAC = 60 N in (1) and (3): − Multiply (1′) by 9, (3′) by 8, and add: 554.4 N − 12.6 TAD = 0 TAD = 572.0 N 13 Substitute into (1′) and solve for TAB : TAB = 17 5 36 + × 572 8 13 TAB = 544.0 N Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (544 N) + 0.64(60 N) + (572 N) 17 13 = 844.8 N P= Weight of plate = P = 845 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 115 PROBLEM 2.110 A rectangular plate is supported by three cables as shown. Knowing that the tension in cable AD is 520 N, determine the weight of the plate. SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 8 5 TAB + 0.6TAC + TAD = 0 17 13 (1) 12 9.6 TAB + 0.64 TAC − TAD + P = 0 17 13 (2) 9 7.2 TAB + 0.48TAC − TAD = 0 17 13 (3) − − Making TAD = 520 N in Eqs. (1) and (3): 8 TAB + 0.6TAC + 200 N = 0 17 (1′) 9 TAB + 0.48TAC − 288 N = 0 17 (3′) − Multiply (1′) by 9, (3′) by 8, and add: 9.24TAC − 504 N = 0 TAC = 54.5455 N Substitute into (1′) and solve for TAB : TAB = 17 (0.6 × 54.5455 + 200) TAB = 494.545 N 8 Substitute for the tensions in Eq. (2) and solve for P: 12 9.6 (494.545 N) + 0.64(54.5455 N) + (520 N) 17 13 Weight of plate = P = 768 N = 768.00 N P= PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 116 PROBLEM 2.111 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AB is 630 lb, determine the vertical force P exerted by the tower on the pin at A. SOLUTION Free Body A: We write ΣF = 0: TAB + TAC + TAD + Pj = 0 AB = −45i − 90 j + 30k AB = 105 ft AC = 30i − 90 j + 65k AC = 115 ft AD = 20i − 90 j − 60k AD = 110 ft AB TAB = TAB λ AB = TAB AB 6 2 3 = − i − j + k TAB 7 7 7 AC TAC = TAC λAC = TAC AC 18 13 6 = i− j + k TAC 23 23 23 AD TAD = TAD λAD = TAD AD 9 6 2 = i − j − k TAD 11 11 11 Substituting into the Eq. ΣF = 0 and factoring i, j, k : 6 2 3 − TAB + TAC + TAD i 23 11 7 6 18 9 + − TAB − TAC − TAD + P j 23 11 7 13 6 2 + TAB + TAC − TAD k = 0 23 11 7 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 117 PROBLEM 2.111 (Continued) Setting the coefficients of i, j, k , equal to zero: i: 3 6 2 − TAB + TAC + TAD = 0 7 23 11 (1) j: 6 18 9 − TAB − TAC − TAD + P = 0 7 23 11 (2) k: 2 13 6 TAB + TAC − TAD = 0 7 23 11 (3) Set TAB = 630 lb in Eqs. (1) – (3): 6 2 TAC + TAD = 0 23 11 (1′) 18 9 TAC − TAD + P = 0 23 11 (2′) 13 6 TAC − TAD = 0 23 11 (3′) −270 lb + −540 lb − 180 lb + Solving, TAC = 467.42 lb TAD = 814.35 lb P = 1572.10 lb P = 1572 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 118 PROBLEM 2.112 A transmission tower is held by three guy wires attached to a pin at A and anchored by bolts at B, C, and D. If the tension in wire AC is 920 lb, determine the vertical force P exerted by the tower on the pin at A. SOLUTION See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 3 6 2 − TAB + TAC + TAD = 0 7 23 11 (1) 6 18 9 − TAB − TAC − TAD + P = 0 7 23 11 (2) 2 13 6 TAB + TAC − TAD = 0 7 23 11 (3) Substituting for TAC = 920 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: Solving, 3 2 − TAB + 240 lb + TAD = 0 7 11 (1′) 6 9 − TAB − 720 lb − TAD + P = 0 7 11 (2′) 2 6 TAB + 520 lb − TAD = 0 7 11 (3′) TAB = 1240.00 lb TAD = 1602.86 lb P = 3094.3 lb P = 3090 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 119 PROBLEM 2.113 In trying to move across a slippery icy surface, a 180-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope. SOLUTION Free-Body Diagram at A 30 16 N= N i+ j 34 34 and W = W j = −(180 lb) j TAC = TAC λ AC = TAC AC ( −30 ft)i + (20 ft) j − (12 ft)k = TAC AC 38 ft 6 15 10 = TAC − i + j − k 19 19 19 TAB = TAB λ AB = TAB AB (−30 ft)i + (24 ft) j + (32 ft)k = TAB AB 50 ft 12 16 15 = TAB − i + j+ k 25 25 25 Equilibrium condition: ΣF = 0 TAB + TAC + N + W = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 120 PROBLEM 2.113 (Continued) Substituting the expressions obtained for TAB , TAC , N, and W; factoring i, j, and k; and equating each of the coefficients to zero gives the following equations: 15 15 16 TAB − TAC + N =0 25 19 34 (1) From j: 12 10 30 TAB + TAC + N − (180 lb) = 0 25 19 34 (2) From k: 16 6 TAB − TAC = 0 25 19 (3) From i: − Solving the resulting set of equations gives: TAB = 31.7 lb; TAC = 64.3 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 121 PROBLEM 2.114 Solve Problem 2.113, assuming that a friend is helping the man at A by pulling on him with a force P = −(60 lb)k. PROBLEM 2.113 In trying to move across a slippery icy surface, a 180-lb man uses two ropes AB and AC. Knowing that the force exerted on the man by the icy surface is perpendicular to that surface, determine the tension in each rope. SOLUTION Refer to Problem 2.113 for the figure and analysis leading to the following set of equations, Equation (3) being modified to include the additional force P = ( −60 lb)k. 15 15 16 TAB − TAC + N =0 25 19 34 (1) 12 10 30 TAB + TAC + N − (180 lb) = 0 25 19 34 (2) 16 6 TAB − TAC − (60 lb) = 0 25 19 (3) − Solving the resulting set of equations simultaneously gives: TAB = 99.0 lb TAC = 10.55 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 122 PROBLEM 2.115 For the rectangular plate of Problems 2.109 and 2.110, determine the tension in each of the three cables knowing that the weight of the plate is 792 N. SOLUTION See Problem 2.109 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below. Setting P = 792 N gives: 8 5 TAB + 0.6TAC + TAD = 0 17 13 (1) 12 9.6 TAB − 0.64TAC − TAD + 792 N = 0 17 13 (2) 9 7.2 TAB + 0.48TAC − TAD = 0 17 13 (3) − − Solving Equations (1), (2), and (3) by conventional algorithms gives TAB = 510.00 N TAB = 510 N TAC = 56.250 N TAC = 56.2 N TAD = 536.25 N TAD = 536 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 123 PROBLEM 2.116 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 0. SOLUTION ΣFA = 0: TAB + TAC + TAD + P + Q = 0 Where P = Pi and Q = Qj AB = −(960 mm)i − (240 mm) j + (380 mm)k AB = 1060 mm AC = −(960 mm)i − (240 mm) j − (320 mm)k AC = 1040 mm AD = −(960 mm)i + (720 mm) j − (220 mm)k AD = 1220 mm AB 19 48 12 TAB = TAB λAB = TAB = TAB − i − j + k AB 53 53 53 AC 3 4 12 TAC = TAC λAC = TAC = TAC − i − j − k AC 13 13 13 AD 36 11 48 = TAD − i + TAD = TAD λAD = TAD j− k AD 61 61 61 Substituting into ΣFA = 0, setting P = (2880 N)i and Q = 0, and setting the coefficients of i, j, k equal to 0, we obtain the following three equilibrium equations: i: − 48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61 (1) j: − 12 3 36 TAB − TAC + TAD = 0 53 13 61 (2) k: 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 124 PROBLEM 2.116 (Continued) Solving the system of linear equations using conventional algorithms gives: TAB = 1340.14 N TAC = 1025.12 N TAD = 915.03 N TAB = 1340 N TAC = 1025 N TAD = 915 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 125 PROBLEM 2.117 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = 576 N. SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 48 12 48 TAB − TAC − TAD + P = 0 53 13 61 (1) − 12 3 36 TAB − TAC + TAD + Q = 0 53 13 61 (2) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3) Setting P = 2880 N and Q = 576 N gives: − 48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61 (1′) 12 3 36 TAB − TAC + TAD + 576 N = 0 53 13 61 (2′) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3′) − Solving the resulting set of equations using conventional algorithms gives: TAB = 1431.00 N TAC = 1560.00 N TAD = 183.010 N TAB = 1431 N TAC = 1560 N TAD = 183.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 126 PROBLEM 2.118 For the cable system of Problems 2.107 and 2.108, determine the tension in each cable knowing that P = 2880 N and Q = −576 N. (Q is directed downward). SOLUTION See Problem 2.116 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 48 12 48 TAB − TAC − TAD + P = 0 53 13 61 (1) − 12 3 36 TAB − TAC + TAD + Q = 0 53 13 61 (2) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3) Setting P = 2880 N and Q = −576 N gives: − 48 12 48 TAB − TAC − TAD + 2880 N = 0 53 13 61 (1′) 12 3 36 TAB − TAC + TAD − 576 N = 0 53 13 61 (2′) 19 4 11 TAB − TAC − TAD = 0 53 13 61 (3′) − Solving the resulting set of equations using conventional algorithms gives: TAB = 1249.29 N TAC = 490.31 N TAD = 1646.97 N TAB = 1249 N TAC = 490 N TAD = 1647 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 127 PROBLEM 2.119 For the transmission tower of Problems 2.111 and 2.112, determine the tension in each guy wire knowing that the tower exerts on the pin at A an upward vertical force of 2100 lb. SOLUTION See Problem 2.111 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: 3 6 2 − TAB + TAC + TAD = 0 7 23 11 (1) 6 18 9 − TAB − TAC − TAD + P = 0 7 23 11 (2) 2 13 6 TAB + TAC − TAD = 0 7 23 11 (3) Substituting for P = 2100 lb in Equations (1), (2), and (3) above and solving the resulting set of equations using conventional algorithms gives: 3 6 2 − TAB + TAC + TAD = 0 7 23 11 (1′) 6 18 9 − TAB − TAC − TAD + 2100 lb = 0 7 23 11 (2′) 2 13 6 TAB + TAC − TAD = 0 7 23 11 (3′) TAB = 841.55 lb TAC = 624.38 lb TAD = 1087.81 lb TAB = 842 lb TAC = 624 lb TAD = 1088 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 128 PROBLEM 2.120 A horizontal circular plate weighing 60 lb is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Determine the tension in each wire. SOLUTION ΣFx = 0: −TAD (sin 30°)(sin 50°) + TBD (sin 30°)(cos 40°) + TCD (sin 30°)(cos 60°) = 0 Dividing through by sin 30° and evaluating: −0.76604TAD + 0.76604TBD + 0.5TCD = 0 (1) ΣFy = 0: −TAD (cos 30°) − TBD (cos 30°) − TCD (cos 30°) + 60 lb = 0 TAD + TBD + TCD = 69.282 lb or (2) ΣFz = 0: TAD sin 30° cos 50° + TBD sin 30° sin 40° − TCD sin 30° sin 60° = 0 or 0.64279TAD + 0.64279TBD − 0.86603TCD = 0 (3) Solving Equations (1), (2), and (3) simultaneously: TAD = 29.5 lb TBD = 10.25 lb TCD = 29.5 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 129 PROBLEM 2.121 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables. SOLUTION Free Body Diagram at A: Since TBAC = tension in cable BAC, it follows that TAB = TAC = TBAC TAB = TBAC λ AB = TBAC (−17.5 in.)i + (60 in.) j 60 −17.5 = TBAC i+ j 62.5 in. 62.5 62.5 TAC = TBAC λ AC = TBAC (60 in.)i + (25 in.)k 25 60 = TBAC j + k 65 in. 65 65 TAD = TAD λ AD = TAD (80 in.)i + (60 in.) j 3 4 = TAD i + j 100 in. 5 5 TAE = TAE λ AE = TAE (60 in.) j − (45 in.)k 3 4 = TAE j − k 75 in. 5 5 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 130 PROBLEM 2.121 (Continued) Substituting into ΣFA = 0, setting P = ( −200 lb) j, and setting the coefficients of i, j, k equal to φ , we obtain the following three equilibrium equations: 17.5 4 TBAC + TAD = 0 62.5 5 From i: − From 3 4 60 60 j: + TBAC + TAD + TAE − 200 lb = 0 62.5 65 5 5 From k: (1) 25 3 TBAC − TAE = 0 65 5 (2) (3) Solving the system of linear equations using convential acgorithms gives: TBAC = 76.7 lb; TAD = 26.9 lb; TAE = 49.2 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 131 PROBLEM 2.122 Knowing that the tension in cable AE of Prob. 2.121 is 75 lb, determine (a) the magnitude of the load P, (b) the tension in cables BAC and AD. PROBLEM 2.121 Cable BAC passes through a frictionless ring A and is attached to fixed supports at B and C, while cables AD and AE are both tied to the ring and are attached, respectively, to supports at D and E. Knowing that a 200-lb vertical load P is applied to ring A, determine the tension in each of the three cables. SOLUTION Refer to the solution to Problem 2.121 for the figure and analysis leading to the following set of equilibrium equations, Equation (2) being modified to include Pj as an unknown quantity: − 17.5 4 TBAC + TAD = 0 62.5 5 (1) 60 3 4 60 62.5 + 65 TBAC + 5 TAD + 5 TAE − P = 0 25 3 TBAC − TAE = 0 65 5 (2) (3) Substituting for TAE = 75 lb and solving simultaneously gives: P = 305 lb; TBAC = 117.0 lb; TAD = 40.9 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 132 PROBLEM 2.123 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 376 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.) SOLUTION TAB = T λ AB AB =T AB (−130 mm)i + (400 mm) j + (160 mm)k =T 450 mm 40 16 13 j+ k =T − i + 45 45 45 Free-Body A: TAC = T λ AC AC =T AC ( −150 mm)i + (400 mm) j + (−240 mm)k =T 490 mm 40 24 15 = T − i + j− k 49 49 49 ΣF = 0: TAB + TAC + Q + P + W = 0 Setting coefficients of i, j, k equal to zero: i: − 13 15 T − T +P=0 45 49 0.59501T = P (1) j: + 40 40 T + T −W = 0 45 49 1.70521T = W (2) k: + 16 24 T − T +Q =0 45 49 0.134240 T = Q (3) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 133 PROBLEM 2.123 (Continued) Data: W = 376 N 1.70521T = 376 N T = 220.50 N 0.59501(220.50 N) = P P = 131.2 N 0.134240(220.50 N) = Q Q = 29.6 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 134 PROBLEM 2.124 For the system of Problem 2.123, determine W and Q knowing that P = 164 N. PROBLEM 2.123 A container of weight W is suspended from ring A. Cable BAC passes through the ring and is attached to fixed supports at B and C. Two forces P = Pi and Q = Qk are applied to the ring to maintain the container in the position shown. Knowing that W = 376 N, determine P and Q. (Hint: The tension is the same in both portions of cable BAC.) SOLUTION Refer to Problem 2.123 for the figure and analysis resulting in Equations (1), (2), and (3) for P, W, and Q in terms of T below. Setting P = 164 N we have: Eq. (1): 0.59501T = 164 N Eq. (2): 1.70521(275.63 N) = W Eq. (3): 0.134240(275.63 N) = Q T = 275.63 N W = 470 N Q = 37.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 135 PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system. SOLUTION For both Problems 2.125 and 2.126: Free-Body Diagrams of Collars: ( AB) 2 = x 2 + y 2 + z 2 Here (0.525 m) 2 = (0.20 m) 2 + y 2 + z 2 y 2 + z 2 = 0.23563 m 2 or Thus, when y given, z is determined, Now λAB AB = AB 1 (0.20i − yj + zk )m 0.525 m = 0.38095i − 1.90476 yj + 1.90476 zk = Where y and z are in units of meters, m. From the F.B. Diagram of collar A: ΣF = 0: N x i + N z k + Pj + TAB λ AB = 0 Setting the j coefficient to zero gives P − (1.90476 y )TAB = 0 P = 341 N With TAB = 341 N 1.90476 y Now, from the free body diagram of collar B: ΣF = 0: N x i + N y j + Qk − TAB λAB = 0 Setting the k coefficient to zero gives Q − TAB (1.90476 z ) = 0 And using the above result for TAB , we have Q = TAB z = 341 N (341 N)( z ) (1.90476 z ) = (1.90476) y y PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 136 PROBLEM 2.125 (Continued) Then from the specifications of the problem, y = 155 mm = 0.155 m z 2 = 0.23563 m 2 − (0.155 m) 2 z = 0.46 m and 341 N 0.155(1.90476) = 1155.00 N TAB = (a) TAB = 1155 N or and 341 N(0.46 m)(0.866) (0.155 m) = (1012.00 N) Q= (b) Q = 1012 N or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 137 PROBLEM 2.126 Solve Problem 2.125 assuming that y = 275 mm. PROBLEM 2.125 Collars A and B are connected by a 525-mm-long wire and can slide freely on frictionless rods. If a force P = (341 N)j is applied to collar A, determine (a) the tension in the wire when y = 155 mm, (b) the magnitude of the force Q required to maintain the equilibrium of the system. SOLUTION From the analysis of Problem 2.125, particularly the results: y 2 + z 2 = 0.23563 m 2 341 N TAB = 1.90476 y 341 N Q= z y With y = 275 mm = 0.275 m, we obtain: z 2 = 0.23563 m 2 − (0.275 m) 2 z = 0.40 m and TAB = (a) 341 N = 651.00 (1.90476)(0.275 m) TAB = 651 N or and Q= (b) 341 N(0.40 m) (0.275 m) Q = 496 N or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 138 PROBLEM 2.127 Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 15 kN in member A and 10 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B. SOLUTION Using the force triangle and the laws of cosines and sines, we have γ = 180° − (40° + 20°) = 120° Then R 2 = (15 kN) 2 + (10 kN)2 − 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 = 0.39737 α = 23.414 φ = α + 50° = 73.414 R = 21.8 kN 73.4° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 139 PROBLEM 2.128 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 300-lb horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION P sin 35° = 300 lb (a) P= (b) Vertical component 300 lb sin 35° P = 523 lb Pv = P cos 35° = (523 lb) cos 35° Pv = 428 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 140 PROBLEM 2.129 Determine (a) the required tension in cable AC, knowing that the resultant of the three forces exerted at Point C of boom BC must be directed along BC, (b) the corresponding magnitude of the resultant. SOLUTION Using the x and y axes shown: Rx = ΣFx = TAC sin10° + (50 lb) cos 35° + (75 lb) cos 60° = TAC sin10° + 78.458 lb (1) Ry = ΣFy = (50 lb)sin 35° + (75 lb)sin 60° − TAC cos10° Ry = 93.631 lb − TAC cos10° (a) (2) Set Ry = 0 in Eq. (2): 93.631 lb − TAC cos10° = 0 TAC = 95.075 lb (b) TAC = 95.1 lb Substituting for TAC in Eq. (1): Rx = (95.075 lb)sin10° + 78.458 lb = 94.968 lb R = Rx R = 95.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 141 PROBLEM 2.130 Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC. SOLUTION 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 (b) TBC = (1962 N) sin 105° sin 60° TBC = 2190 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 142 PROBLEM 2.131 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 8 kN and FB = 16 kN, determine the magnitudes of the other two forces. SOLUTION 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 3 3 FB − FA = 0 5 5 3 3 FD = (16 kN) − (8 kN) 5 5 FD = 4.80 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 143 PROBLEM 2.132 Two cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 60 lb in either cable. SOLUTION ΣFx = 0: −TBC − Q cos 60° + 75 lb = 0 Free-Body Diagram TBC = 75 lb − Q cos 60° (1) ΣFy = 0: TAC − Q sin 60° = 0 TAC = Q sin 60° (2) TAC ⱕ 60 lb: Requirement: Q sin 60° ⱕ 60 lb From Eq. (2): Q ⱕ 69.3 lb TBC ⱕ 60 lb: Requirement: From Eq. (1): 75 lb − Q sin 60° ⱕ 60 lb Q ⱖ 30.0 lb 30.0 lb ⱕ Q ⱕ 69.3 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 144 PROBLEM 2.133 A horizontal circular plate is suspended as shown from three wires that are attached to a support at D and form 30° angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 110.3 N, determine (a) the tension in wire AD, (b) the angles θ x, θ y, and θ z that the force exerted at A forms with the coordinate axes. 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 Fx 110.3 N = = 0.38303 F 287.97 N θ x = 67.5° Fy = F cos 30° = 249.39 cos θ y = Fy F = 249.39 N = 0.86603 287.97 N θ y = 30.0° Fz = − F sin 30° cos 50° = −(287.97 N)sin 30°cos 50° = −92.552 N cos θ z = Fz −92.552 N = = −0.32139 F 287.97 N θ z = 108.7° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 145 PROBLEM 2.134 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 ⬍ 0, we must have cos θ x ⬍ 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° Then Fx 500 lb = = 1209.10 lb cos θ x 0.41353 F = 1209 lb Fy = F cos θ y = (1209.10 lb) cos 55° Fy = 694 lb Fz = F cos θ z = (1209.10 lb) cos 45° Fz = 855 lb F= and PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 146 PROBLEM 2.135 Find the magnitude and direction of the resultant of the two forces shown knowing that P = 300 N and Q = 400 N. SOLUTION 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 R = 515 N cos θ x = Rx 174.367 N = = 0.33853 515.07 N R θ x = 70.2° cos θ y = Ry θ y = 27.6° cos θ z = Rz 163.011 N = = 0.31648 R 515.07 N R = 456.42 N = 0.88613 515.07 N θ z = 71.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 147 PROBLEM 2.136 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 2.101 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) below: − 0.6TAB + 0.32432TAC = 0 (1) − 0.8TAB − 0.75676TAC − 0.86154TAD + P = 0 (2) 0.56757TAC − 0.50769TAD = 0 (3) Substituting TAC = 444 N in Equations (1), (2), and (3) above, and solving the resulting set of equations using conventional algorithms gives TAB = 240 N TAD = 496.36 N P = 956 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 148 PROBLEM 2.137 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. If a 60-lb force Q is applied to collar B as shown, determine (a) the tension in the wire when x = 9 in., (b) the corresponding magnitude of the force P required to maintain the equilibrium of the system. SOLUTION Free-Body Diagrams of Collars: A: B: λAB AB − xi − (20 in.) j + zk = = AB 25 in. ΣF = 0: Pi + N y j + N z k + TAB λ AB = 0 Collar A: Substitute for λAB and set coefficient of i equal to zero: P− Collar B: TAB x =0 25 in. (1) ΣF = 0: (60 lb)k + N x′ i + N y′ j − TAB λ AB = 0 Substitute for λAB and set coefficient of k equal to zero: 60 lb − x = 9 in. (a) From Eq. (2): (b) From Eq. (1): TAB z =0 25 in. (2) (9 in.)2 + (20 in.) 2 + z 2 = (25 in.) 2 z = 12 in. 60 lb − TAB (12 in.) 25 in. P= (125.0 lb)(9 in.) 25 in. TAB = 125.0 lb P = 45.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 149 PROBLEM 2.138 Collars A and B are connected by a 25-in.-long wire and can slide freely on frictionless rods. Determine the distances x and z for which the equilibrium of the system is maintained when P = 120 lb and Q = 60 lb. SOLUTION See Problem 2.137 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 = 6.7082 in. From Eq. (3): x = 2 z = 2(6.7082 in.) = 13.4164 in. x = 13.42 in., z = 6.71 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 150 PROBLEM 2F1 Two cables are tied together at C and loaded as shown. Draw the freebody diagram needed to determine the tension in AC and BC. SOLUTION Free-Body Diagram of Point C: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 151 PROBLEM 2.F2 A chairlift has been stopped in the position shown. Knowing that each chair weighs 250 N and that the skier in chair E weighs 765 N, draw the free-body diagrams needed to determine the weight of the skier in chair F. SOLUTION Free-Body Diagram of Point B: WE = 250 N + 765 N = 1015 N 8.25 = 30.510° 14 10 = tan −1 = 22.620° 24 θ AB = tan −1 θ BC Use this free body to determine TAB and TBC. Free-Body Diagram of Point C: θCD = tan −1 1.1 = 10.3889° 6 Use this free body to determine TCD and WF. Then weight of skier WS is found by WS = WF − 250 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 152 PROBLEM 2.F3 Two cables are tied together at A and loaded as shown. Draw the freebody diagram needed to determine the tension in each cable. SOLUTION Free-Body Diagram of Point A: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 153 PROBLEM 2.F4 The 60-lb collar A can slide on a frictionless vertical rod and is connected as shown to a 65-lb counterweight C. Draw the free-body diagram needed to determine the value of h for which the system is in equilibrium. SOLUTION Free-Body Diagram of Point A: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 154 PROBLEM 2.F5 A 36-lb triangular plate is supported by three cables as shown. Draw the free-body diagram needed to determine the tension in each wire. SOLUTION Free-Body Diagram of Point D: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 155 PROBLEM 2.F6 A 70-kg cylinder is supported by two cables AC and BC, which are attached to the top of vertical posts. A horizontal force P, perpendicular to the plane containing the posts, holds the cylinder in the position shown. Draw the free-body diagram needed to determine the magnitude of P and the force in each cable. SOLUTION Free-Body Diagram of Point C: W = (70 kg)(9.81 m/s 2 ) = 686.7 N W = −(686.7 N) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 156 PROBLEM 2.F7 Three cables are connected at point D, which is located 18 in. below the T-shaped pipe support ABC. The cables support a 180-lb cylinder as shown. Draw the free-body diagram needed to determine the tension in each cable. SOLUTION Free-Body Diagram of Point D: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 157 PROBLEM 2.F8 A 100-kg container is suspended from ring A, to which cables AC and AE are attached. A force P is applied to end F of a third cable that passes over a pulley at B and through ring A and then is attached to a support at D. Draw the free-body diagram needed to determine the magnitude of P. (Hint: The tension is the same in all portions of cable FBAD.) SOLUTION Free-Body Diagram of Ring A: W = (100 kg)(9.81 m/s 2 ) = 981 N W = −(681 N) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 158 CHAPTER 3 PROBLEM 3.1 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α = 25°, determine the moment of the force about Point B by resolving the force into horizontal and vertical components. SOLUTION Free-Body Diagram of Rod AB: x = (9 in.) cos 65° = 3.8036 in. y = (9 in.)sin 65° = 8.1568 in. F = Fx i + Fy j rA/B = (20 lb cos 25°)i + ( −20 lb sin 25°) j = (18.1262 lb)i − (8.4524 lb) j = BA = ( −3.8036 in.)i + (8.1568 in.)j M B = rA /B × F = (−3.8036i + 8.1568 j) × (18.1262i − 8.4524 j) = 32.150k − 147.852k = −115.702 lb-in. M B = 115.7 lb-in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 161 PROBLEM 3.2 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α = 25°, determine the moment of the force about Point B by resolving the force into components along AB and in a direction perpendicular to AB. SOLUTION Free-Body Diagram of Rod AB: θ = 90° − (65° − 25°) = 50° Q = (20 lb) cos 50° = 12.8558 lb M B = Q (9 in.) = (12.8558 lb)(9 in.) = 115.702 lb-in. M B = 115.7 lb-in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 162 PROBLEM 3.3 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that the moment of the force about B is 120 lb · in. clockwise, determine the value of α. SOLUTION Free-Body Diagram of Rod AB: α = θ − 25° Q = (20 lb) cos θ and Therefore, M B = (Q )(9 in.) 120 lb-in. = (20 lb)(cos θ )(9 in.) 120 lb-in. cos θ = 180 lb-in. or θ = 48.190° Therefore, α = 48.190° − 25° α = 23.2° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 163 PROBLEM 3.4 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B that creates a moment of equal magnitude and opposite sense about E. SOLUTION (a) By definition, We have W = mg = 80 kg(9.81 m/s 2 ) = 784.8 N ΣM E : M E = (784.8 N)(0.25 m) M E = 196.2 N ⋅ m (b) For the force at B to be the smallest, resulting in a moment (ME) about E, the line of action of force FB must be perpendicular to the line connecting E to B. The sense of FB must be such that the force produces a counterclockwise moment about E. Note: We have d = (0.85 m) 2 + (0.5 m) 2 = 0.98615 m ΣM E : 196.2 N ⋅ m = FB (0.98615 m) FB = 198.954 N and 0.85 m = 59.534° 0.5 m θ = tan −1 FB = 199.0 N or 59.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 164 PROBLEM 3.5 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at A that creates a moment of equal magnitude and opposite sense about E, (c) the magnitude, sense, and point of application on the bottom of the crate of the smallest vertical force that creates a moment of equal magnitude and opposite sense about E. SOLUTION First note. . . W = mg = (80 kg)(9.81 m/s 2 ) = 784.8 N (a) We have M E = rH /EW = (0.25 m)(784.8 N) = 196.2 N ⋅ m (b) (c) or M E = 196.2 N ⋅ m For FA to be minimum, it must be perpendicular to the line joining Points A and E. Then with FA directed as shown, we have (− M E ) = rA/E ( FA )min . Where rA /E = (0.35 m)2 + (0.5 m)2 = 0.61033 m then 196.2 N ⋅ m = (0.61033 m)( FA )min or ( FA ) min = 321 N Also tan φ = 0.35 m 0.5 m or φ = 35.0° (FA ) min = 321 N 35.0° For Fvertical to be minimum, the perpendicular distance from its line of action to Point E must be maximum. Thus, apply (Fvertical)min at Point D, and then (− M E ) = rD / E ( Fvertical ) min or (Fvertical )min = 231 N 196.2 N ⋅ m = (0.85 m)( Fvertical ) min at Point D PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 165 PROBLEM 3.6 A 300-N force P is applied at Point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part (a), determine the perpendicular distance from O to the line of action of P. SOLUTION x = (0.2 m) cos 40° = 0.153209 m y = (0.2 m)sin 40° = 0.128558 m ∴ rA /O = (0.153209 m)i + (0.128558 m) j (a) Fx = (300 N)sin 30° = 150 N Fy = (300 N) cos 30° = 259.81 N F = (150 N)i + (259.81 N) j M O = rA/ O × F = (0.153209i + 0.128558 j) m × (150i + 259.81j) N = (39.805k − 19.2837k ) N ⋅ m = (20.521 N ⋅ m)k (b) M O = 20.5 N ⋅ m M O = Fd 20.521 N ⋅ m = (300 N)(d ) d = 0.068403 m d = 68.4 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 166 PROBLEM 3.7 A 400-N force P is applied at Point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into components along line OA and in a direction perpendicular to that line. (b) Determine the magnitude and direction of the smallest force Q applied at B that has the same moment as P about O. SOLUTION (a) Portion OA of crank: θ = 90° − 30° − 40° θ = 20° S = P sin θ = (400 N) sin 20° = 136.81 N M O = rO /A S = (0.2 m)(136.81 N) = 27.362 N ⋅ m (b) M O = 27.4 N ⋅ m Smallest force Q must be perpendicular to OB. Portion OB of crank: M O = rO /B Q M O = (0.120 m)Q 27.362 N ⋅ m = (0.120 m)Q Q = 228 N 42.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 167 PROBLEM 3.8 It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if α = 10°, (c) the smallest force P that creates the same moment about B. SOLUTION (a) M B = rC/B FN We have = (4 in.)(200 lb) = 800 lb ⋅ in. or MB = 800 lb ⋅ in. (b) By definition, M B = rA/B P sin θ θ = 10° + (180° − 70°) = 120° Then 800 lb ⋅ in. = (18 in.) × P sin120° or P = 51.3 lb (c) For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, P must be directed as shown. Thus M B = dPmin d = rA/B or or 800 lb ⋅ in. = (18 in.)Pmin Pmin = 44.4 lb Pmin = 44.4 lb 20° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 168 PROBLEM 3.9 It is known that the connecting rod AB exerts on the crank BC a 500-lb force directed down and to the left along the centerline of AB. Determine the moment of the force about C. SOLUTION Using (a): M C = y1 ( FAB ) x + x1 ( FAB ) y 7 24 = (2.24 in.) × 500 lb + (1.68 in.) × 500 lb 25 25 = 1120 lb ⋅ in. (a) MC = 1.120 kip ⋅ in. Using (b): M C = y2 ( FAB ) x 7 = (8 in.) × 500 lb 25 = 1120 lb ⋅ in. (b) M C = 1.120 kip ⋅ in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 169 PROBLEM 3.10 It is known that the connecting rod AB exerts on the crank BC a 500-lb force directed down and to the left along the centerline of AB. Determine the moment of the force about C. SOLUTION Using (a): M C = − y1 ( FAB ) x + x1 ( FAB ) y 7 24 = −(2.24 in.) × 500 lb + (1.68 in.) × 500 lb 25 25 = +492.8 lb ⋅ in. (a) MC = 493 lb ⋅ in. Using (b): M C = y2 ( FAB ) x 7 = (3.52 in.) × 500 lb 25 = +492.8 lb ⋅ in. (b) M C = 493 lb ⋅ in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 170 PROBLEM 3.11 A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at Point C, (b) at Point E. SOLUTION (a) Slope of line: Then and Then EC = 0.875 m 5 = 1.90 m + 0.2 m 12 12 (TAB ) 13 12 = (1040 N) 13 = 960 N 5 TABy = (1040 N) 13 = 400 N TABx = (a) M D = TABx (0.875 m) − TABy (0.2 m) = (960 N)(0.875 m) − (400 N)(0.2 m) = 760 N ⋅ m (b) We have or M D = 760 N ⋅ m M D = TABx ( y ) + TABx ( x) = (960 N)(0) + (400 N)(1.90 m) = 760 N ⋅ m (b) or M D = 760 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 171 PROBLEM 3.12 It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If d = 2.80 m, determine the tension that must be developed in the cable of winch puller AB to create the required moment about Point D. SOLUTION Slope of line: EC = 0.875 m 7 = 2.80 m + 0.2 m 24 Then TABx = 24 TAB 25 and TABy = 7 TAB 25 We have M D = TABx ( y ) + TABy ( x) 24 7 TAB (0) + TAB (2.80 m) 25 25 = 1224 N 960 N ⋅ m = TAB or TAB = 1224 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 172 PROBLEM 3.13 It is known that a force with a moment of 960 N · m about D is required to straighten the fence post CD. If the capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified moment about Point D. 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 = 0.875 m (d + 0.20)2 + (0.875)2 m 0.875 960 N ⋅ m = 2400 N ( d + 0.20) 2 + (0.875) 2 or (d + 0.20) 2 + (0.875) 2 = 2.1875d or (d + 0.20) 2 + (0.875) 2 = 4.7852d 2 or (d ) 3.7852d 2 − 0.40d − 0.8056 = 0 Using the quadratic equation, the minimum values of d are 0.51719 m and − 0.41151 m. Since only the positive value applies here, d = 0.51719 m or d = 517 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 173 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 (65) + (72) 2 2 72 (65) + (72) 2 2 (485 N) = 325 N (485 N) = 360 N M C = (55 mm)(360 N) + (162)(325 N) = 72450 N ⋅ m = 72.450 N ⋅ m or M C = 72.5 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 174 PROBLEM 3.15 Form the vector products B × C and B′ × C, where B = B′, and use the results obtained to prove the identity sin α cos β = 1 1 sin (α + β ) + sin (α − β ). 2 2 SOLUTION Note: B = B(cos β i + sin β j) B′ = B(cos β i − sin β j) C = C (cos α i + sin α j) By definition, Now | B × C | = BC sin (α − β ) (1) | B′ × C | = BC sin (α + β ) (2) B × C = B(cos β i + sin β j) × C (cos α i + sin α j) = BC (cos β sin α − sin β cos α )k and (3) B′ × C = B(cos β i − sin β j) × C (cos α i + sin α j) = BC (cos β sin α + sin β cos α ) k (4) Equating the magnitudes of B × C from Equations (1) and (3) yields: BC sin(α − β ) = BC (cos β sin α − sin β cos α ) (5) Similarly, equating the magnitudes of B′ × C from Equations (2) and (4) yields: BC sin(α + β ) = BC (cos β sin α + sin β cos α ) (6) Adding Equations (5) and (6) gives: sin(α − β ) + sin(α + β ) = 2cos β sin α or sin α cos β = 1 1 sin(α + β ) + sin(α − β ) 2 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 175 PROBLEM 3.16 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = −7i + 3j − 3k and Q = 2i + 2j + 5k, (b) P = 6i − 5j − 2k and Q = −2i + 5j − k. SOLUTION (a) We have A = |P × Q| where P = −7i + 3j − 3k Q = 2i + 2 j + 5k 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 (b) We have A = |P × Q| where P = 6i − 5 j − 2k or A = 41.0 Q = −2i + 5 j − 1k Then i j k P × Q = 6 −5 −2 −2 5 −1 = [(5 + 10)i + (4 + 6) j + (30 − 10)k ] = (15)i + (10) j + (20)k A = (15) 2 + (10) 2 + (20) 2 or A = 26.9 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 176 PROBLEM 3.17 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) i + 2j − 5k and 4i − 7j − 5k, (b) 3i − 3j + 2k and −2i + 6j − 4k. SOLUTION (a) A×B |A × B| We have λ= where A = 1i + 2 j − 5k B = 4i − 7 j − 5k 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 λ= (b) 15(−3i − 1j − 1k ) 15 11 or λ = 1 11 (−3i − j − k ) A×B |A × B| We have λ= where A = 3i − 3 j + 2k B = −2i + 6 j − 4k Then i j k A × B = 3 −3 2 −2 6 −4 = (12 − 12)i + (−4 + 12) j + (18 − 6)k = (8 j + 12k ) and |A × B| = 4 (2) 2 + (3) 2 = 4 13 λ= 4(2 j + 3k ) 4 13 or λ = 1 13 (2 j + 3k ) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 177 PROBLEM 3.18 A line passes through the Points (20 m, 16 m) and (−1 m, −4 m). Determine the perpendicular distance d from the line to the 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 = Finally, 64 29 F = d ( F ) 64 d= m 29 d = 2.21 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 178 PROBLEM 3.19 Determine the moment about the origin O of the force F = 4i − 3j + 5k that acts at a Point A. Assume that the position vector of A is (a) r = 2i + 3j − 4k, (b) r = −8i + 6j − 10k, (c) r = 8i − 6j + 5k. SOLUTION MO = r × F (a) i j k M O = 2 3 −4 4 −3 5 = (15 − 12)i + (−16 − 10) j + (−6 − 12)k (b) i j k M O = −8 6 −10 4 −3 5 = (30 − 30)i + ( −40 + 40) j + (24 − 24)k (c) M O = 3i − 26 j − 18k MO = 0 i j k M O = 8 −6 5 4 −3 5 = (−30 + 15)i + (20 − 40) j + (−24 + 24)k M O = −15i − 20 j Note: The answer to Part (b) could have been anticipated since the elements of the last two rows of the determinant are proportional. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 179 PROBLEM 3.20 Determine the moment about the origin O of the force F = 2i + 3j − 4k that acts at a Point A. Assume that the position vector of A is (a) r = 3i − 6j + 5k, (b) r = i − 4j − 2k, (c) r = 4i + 6j − 8k. SOLUTION MO = r × F (a) i j k M O = 3 −6 5 2 3 −4 = (24 − 15)i + (10 + 12) j + (9 + 12)k (b) i j k M O = 1 −4 −2 2 3 −4 = (16 + 6)i + (−4 + 4) j + (3 + 8)k (c) M O = 9i + 22 j + 21k M O = 22i + 11k i j k M O = 4 6 −8 2 3 −4 = (−24 + 24)i + ( −16 + 16) j + (12 − 12)k MO = 0 Note: The answer to Part (c) could have been anticipated since the elements of the last two rows of the determinant are proportional. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 180 PROBLEM 3.21 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E. SOLUTION AE = (0.21 m)i − (0.16 m) j + (0.12 m)k (a) AE = (0.21 m) 2 + (−0.16 m) 2 + (0.12 m) 2 = 0.29 m AE FA = FA λ AE = F AE 0.21i − 0.16 j + 0.12k = (435 N) 0.29 = (315 N)i − (240 N) j + (180 N)k rA/O = −(0.09 m)i + (0.16 m) j i j k M O = −0.09 0.16 0 315 −240 180 = 28.8i + 16.20 j + (21.6 − 50.4)k (b) M O = (28.8 N ⋅ m)i + (16.20 N ⋅ m) j − (28.8 N ⋅ m)k FE = −FA = −(315 N)i + (240 N) j − (180 N)k rE / O = (0.12 m)i + (0.12 m)k i j k M O = 0.12 0 0.12 −315 240 −180 = −28.8i + (−37.8 + 21.6) j + 28.8k M O = −(28.8 N ⋅ m)i − (16.20 N ⋅ m) j + (28.8 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 181 PROBLEM 3.22 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A. SOLUTION We have R A = 2FAB + FAD where and FAB = −(82 lb) j AD 6i − 7.75 j − 3k FAD = FAD = (82 lb) AD 10.25 FAD = (48 lb)i − (62 lb) j − (24 lb)k Thus R A = 2FAB + FAD = (48 lb)i − (226 lb) j − (24 lb)k Also rA/C = (7.75 ft) j + (3 ft)k Using Eq. (3.21): i j k M C = 0 7.75 3 48 − 226 −24 = (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k M C = (492 lb ⋅ ft)i + (144.0 lb ⋅ ft) j − (372 lb ⋅ ft)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 182 PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. SOLUTION We have Then Txz = (6 lb) cos 8° = 5.9416 lb Tx = Txz sin 30° = 2.9708 lb Ty = TBC sin 8° = − 0.83504 lb Tz = Txz cos 30° = −5.1456 lb Now M A = rB/A × TBC where rB/A = (6sin 45°) j − (6cos 45°)k = Then or 6 ft 2 (j − k) i j k MA = 0 1 −1 2 2.9708 −0.83504 −5.1456 6 6 6 (−5.1456 − 0.83504)i − (2.9708) j − (2.9708)k = 2 2 2 6 M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 183 PROBLEM 3.24 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B. SOLUTION F=F BD BD where F = 900 N BD = −(1 m)i − (2 m) j + (2 m)k BD = (−1 m) 2 + ( −2 m) 2 + (2 m) 2 =3m −i − 2 j + 2k 3 = −(300 N)i − (600 N) j + (600 N)k = (2.5 m)i + (2 m) j F = (900 N) rB /O M O = rB /O × F i j k = 2.5 2 0 −300 −600 600 = 1200i − 1500 j + ( −1500 + 600)k MO = (1200 N ⋅ m)i − (1500 N ⋅ m) j − (900 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 184 PROBLEM 3.25 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 0 − cos 30° sin 30° = 200[(0.075sin 30°)i − (0.06sin 30°) j − (0.06 cos 30°)k ] or M A = (7.50 N ⋅ m)i − (6.00 N ⋅ m) j − (10.39 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 185 PROBLEM 3.26 The 6-m boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a Point C located on the vertical wall. If the tension in the cable is 2.5 kN, determine the moment about A of the force exerted by the cable at B. SOLUTION First note d BC = (−6)2 + (2.4) 2 + (−4) 2 = 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 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 3.27 In Prob. 3.21, determine the perpendicular distance from point O to wire AE. PROBLEM 3.21 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E. SOLUTION From the solution to Prob. 3.21 M O = (28.8 N ⋅ m)i + (16.20 N ⋅ m) j − (28.8 N ⋅ m)k M O = (28.8) 2 + (16.20) 2 + (28.8) 2 = 43.8329 N ⋅ m But M O = FA d or MO FA 43.8329 N ⋅ m d= 435 N = 0.100765 m d= d = 100.8 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 187 PROBLEM 3.28 In Prob. 3.21, determine the perpendicular distance from point B to wire AE. PROBLEM 3.21 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E. SOLUTION From the solution to Prob. 3.21 FA = (315 N)i − (240 N) j + (180 N)k rA/B = −(0.210 m)i M B = rA /B × FA = −0.21i × (315i − 240 j + 180k ) = 50.4k + 37.8 j M B = (50.4) 2 + (37.8) 2 = 63.0 N ⋅ m M B = FA d or MB FA 63.0 N ⋅ m d= 435 N = 0.144829 m d= d = 144.8 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 188 PROBLEM 3.29 In Problem 3.22, determine the perpendicular distance from point C to portion AD of the line ABAD. PROBLEM 3.22 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A. SOLUTION First compute the moment about C of the force FDA exerted by the line on D: From Problem 3.22: FDA = − FAD = −(48 lb) i + (62 lb) j + (24 lb)k M C = rD/C × FDA = + (6 ft)i × [−(48 lb)i + (62 lb) j + (24 lb)k ] = −(144 lb ⋅ ft) j + (372 lb ⋅ ft)k M C = (144) 2 + (372)2 = 398.90 lb ⋅ ft Then M C = FDA d Since FDA = 82 lb d= = MC FDA 398.90 lb ⋅ ft 82 lb d = 4.86 ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 189 PROBLEM 3.30 In Prob. 3.23, determine the perpendicular distance from point A to a line drawn through points B and C. PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. SOLUTION From the solution to Prob. 3.23: M A = −(25.4 lb ⋅ ft)i − (12.60 lb ⋅ ft) j − (12.60 lb ⋅ ft)k M A = (−25.4) 2 + (−12.60) 2 + (−12.60)2 = 31.027 lb ⋅ ft M A = TBC d MA TBC 31.027 lb ⋅ ft = 6 lb = 5.1712 ft d= or d = 5.17 ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 190 PROBLEM 3.31 In Prob. 3.23, determine the perpendicular distance from point D to a line drawn through points B and C. PROBLEM 3.23 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B. SOLUTION AB = 6 ft TBC = 6 lb We have | M D | = TBC d where d = perpendicular distance from D to line BC. M D = rB /D × TBC rB /D = (6sin 45° ft) j = (4.2426 ft) TBC : (TBC ) x = (6 lb) cos8° sin 30° = 2.9708 lb (TBC ) y = −(6 lb) sin 8° = −0.83504 lb (TBC ) z = −(6 lb) cos8° cos 30° = −5.1456 lb TBC = (2.9708 lb)i − (0.83504 lb) j − (5.1456 lb)k i j k MD = 0 4.2426 0 2.9708 −0.83504 −5.1456 = −(21.831 lb ⋅ ft)i − (12.6039 lb ⋅ ft) | M D | = ( −21.831) 2 + (−12.6039) 2 = 25.208 lb ⋅ ft 25.208 lb ⋅ ft = (6 lb)d d = 4.20 ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 191 PROBLEM 3.32 In Prob. 3.24, determine the perpendicular distance from point O to cable BD. PROBLEM 3.24 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B. SOLUTION From the solution to Prob. 3.24 we have M O = (1200 N ⋅ m)i − (1500 N ⋅ m) j − (900 N ⋅ m)k M O = (1200) 2 + (−1500) 2 + ( −900)2 = 2121.3 N ⋅ m M O = Fd MO F 2121.3 N ⋅ m = 900 N d= d = 2.36 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 192 PROBLEM 3.33 In Prob. 3.24, determine the perpendicular distance from point C to cable BD. PROBLEM 3.24 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B. SOLUTION From the solution to Prob. 3.24 we have F = −(300 N)i − (600 N) j + (600 N)k rB /C = (2 m) j M C = rB /C × F = (2 m) j × ( −300 Ni − 600 Nj + 600 Nk ) = (600 N ⋅ m)k + (1200 N ⋅ m)i M C = (600)2 + (1200) 2 = 1341.64 N ⋅ m M C = Fd MC F 1341.64 N ⋅ m = 900 N d= d = 1.491 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 193 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 = λ AB F = (24 ft) i + (24 ft) j − (28 ft) k (24) 2 + (24) 2 + (28) 2 ft F F (6) i + (6) j − (7) k 11 = (3 ft)i − (10 ft) j − (a − 10 ft)k = rA/C i j k F M C = 3 −10 10a 11 6 6 −7 = [(10 + 6a)i + (81 − 6a) j + 78 k ] |M C | = |rA/C × F 2 | Since or F 11 |rA/C × F 2 | = ( dF ) 2 1 (10 + 6a) 2 + (81 − 6a) 2 + (78)2 = d 2 121 Setting d da (d 2 ) = 0 to find a to minimize d: 1 [2(6)(10 + 6a) + 2(−6)(81 − 6a)] = 0 121 Solving a = 5.92 ft or a = 5.92 ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 194 PROBLEM 3.35 Given the vectors P = 3i − j + 2k, Q = 4i + 5j − 3k, and S = −2i + 3j − k, compute the scalar products P · Q, P · S, and Q · S. SOLUTION P ⋅ Q = (3i − j + 2k ) ⋅ (4i + 5 j − 3k ) = (3)(4) + (−1)(5) + (2)(−3) = 12 − 5 − 6 P ⋅ Q = +1 P ⋅ S = (3i − j + 2k ) ⋅ (−2i + 3j − k ) = (3)(−2) + (−1)(3) + (2)( −1) = −6 − 3 − 2 P ⋅ S = −11 Q ⋅ S = (4i + 5 j − 3k ) ⋅ (−2i + 3j − k ) = (4)(−2) + (5)(3) + ( −3)(−1) = −8 + 15 + 3 Q ⋅ S = +10 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 195 PROBLEM 3.36 Form the scalar product B · C and use the result obtained to prove the identity cos (α − β) = cos α cos β + sin α sin β . SOLUTION B = B cos α i + B sin α j (1) C = C cos β i + C sin β j (2) By definition: B ⋅ C = BC cos(α − β ) (3) From (1) and (2): B ⋅ C = ( B cos α i + B sin α j) ⋅ (C cos β i + C sin β j) = BC (cos α cos β + sin α sin β ) (4) Equating the right-hand members of (3) and (4), cos(α − β ) = cos α cos β + sin α sin β PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 196 PROBLEM 3.37 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 AB = −(6.5 ft)i − (8 ft) j + (2 ft)k AC = −(8 ft) j + (6 ft)k AB ⋅ AC = ( AB )( AC ) cos θ (−6.5i − 8 j + 2k ) ⋅ (−8 j + 6k ) = (10.5)(10) cos θ (−6.5)(0) + ( −8)( −8) + (2)(6) = 105cos θ cos θ = 0.72381 or θ = 43.6° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 197 PROBLEM 3.38 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 AC = −(8 ft)j + (6 ft)k AD = (4 ft)i − (8 ft) j + (1 ft)k AC ⋅ AD = ( AC )( AD ) cos θ (−8 j + 6k ) ⋅ (4i − 8 j + k ) = (10)(9) cos θ (0)(4) + ( −8)( −8) + (6)(1) = 90 cos θ or cos θ = 0.77778 or θ = 38.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 198 PROBLEM 3.39 Three cables are used to support a container as shown. Determine the angle formed by cables AB and AD. SOLUTION First note: AB = (450 mm)2 + (600 mm)2 = 750 mm AD = (−500 mm) 2 + (600 mm) 2 + (360 mm) 2 and By definition, = 860 mm AB = (450 mm)i + (600 mm) j AD = (−500 mm)i + (600 mm) j + (360 mm)k AB ⋅ AD = ( AB)( AD ) cos θ (450i + 600 j) ⋅ (−500i − 600 j + 360k ) = (750)(860) cos θ (450)(−500) + (600)(600) + (0)(360) = (750)(860) cos θ cos θ = 0.20930 or θ = 77.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 199 PROBLEM 3.40 Three cables are used to support a container as shown. Determine the angle formed by cables AC and AD. SOLUTION First note: AC = (600 mm) 2 + (−320 mm) 2 = 680 mm AD = (−500 mm) 2 + (600 mm) 2 + (360 mm) 2 and By definition, = 860 mm AC = (600 mm)j + (−320 mm)k AD = (−500 mm)i + (600 mm) j + (360 mm)k AC ⋅ AD = ( AC )( AD ) cos θ (600 j − 320k ) ⋅ (−500i + 600 j + 360k ) = (680)(860) cos θ 0(−500) + (600)(600) + (−320)(360) = (680)(860) cos θ cos θ = 0.41860 θ = 65.3° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 200 PROBLEM 3.41 The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x = 11 in., determine the angle formed by the two cords (a) using Eq. (3.32), (b) applying the law of cosines to triangle ABC. SOLUTION (a) Using Eq. (3.32): CA = 11i − 12 j + 24k CA = (11) 2 + ( −12) 2 + (24) 2 = 29 in. CB = 31i − 12 j + 24k CB = (31) 2 + ( −12) 2 + (24) 2 = 41 in. CA ⋅ CB cos θ = (CA)(CB) (11i − 12 j + 24k ) ⋅ (31i − 12 j + 24k ) = (29)(41) (11)(31) + (−12)(−12) + (24)(24) = (29)(41) = 0.89235 (b) θ = 26.8° Law of cosines: ( AB) 2 = (CA) 2 + (CB) 2 − 2(CA)(CB ) cos θ (20) 2 = (29)2 + (41) 2 − 2(29)(41) cos θ cos θ = 0.89235 θ = 26.8° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 201 PROBLEM 3.42 Solve Prob. 3.41 for the position corresponding to x = 4 in. PROBLEM 3.41 The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x = 11 in., determine the angle formed by the two cords (a) using Eq. (3.32), (b) applying the law of cosines to triangle ABC. SOLUTION (a) Using Eq. (3.32): CA = 4i − 12 j + 24k CA = (4) 2 + (−12) 2 + (24) 2 = 27.129 in. CB = 24i − 12 j + 24k CB = (24) 2 + (−12)2 + (24)2 = 36 in. CA ⋅ CB cos θ = (CA)(CB) (4i − 12 j + 24k ) ⋅ (24i − 12 j + 24k ) = (27.129)(36) = 0.83551 (b) θ = 33.3° Law of cosines: ( AB) 2 = (CA) 2 + (CB )2 − 2(CA)(CB ) cos θ (20) 2 = (27.129)2 + (36) 2 − 2(27.129)(36) cos θ cos θ = 0.83551 θ = 33.3° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 202 PROBLEM 3.43 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope AB is 540 N, determine (a) the angle between rope AB and the stake, (b) the projection on the stake of the force exerted by rope AB at Point B. SOLUTION First note: BA = (−3) 2 + (3)2 + (−1.5) 2 = 4.5 m BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m λ BD (a) TBA (−3i + 3j − 1.5k ) 4.5 T = BA (−2i + 2 j − k ) 3 1 BD ( −0.08i + 0.38 j + 0.16k ) = = BD 0.42 1 = (−4i + 19 j + 8k ) 21 TBA = Then We have or or TBA ⋅ λ BD = TBA cos θ TBA 1 ( −2i + 2 j − k ) ⋅ (−4i + 19 j + 8k ) = TBA cos θ 3 21 1 [(−2)( −4) + (2)(19) + (−1)(8)] 63 = 0.60317 cos θ = or θ = 52.9° (b) We have (TBA ) BD = TBA ⋅ λ BD = TBA cos θ = (540 N)(0.60317) or (TBA ) BD = 326 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 203 PROBLEM 3.44 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope BC is 490 N, determine (a) the angle between rope BC and the stake, (b) the projection on the stake of the force exerted by rope BC at Point B. SOLUTION First note: BC = (1) 2 + (3) 2 + (−1.5) 2 = 3.5 m BD = (−0.08) 2 + (0.38)2 + (0.16) 2 = 0.42 m TBC (i + 3j − 1.5k ) 3.5 T = BC (2i + 6 j − 3k ) 7 1 BD ( −0.08i + 0.38 j + 0.16k ) = = BD 0.42 1 = (−4i + 19 j + 8k ) 21 TBC = λBD (a) TBC ⋅ λBD = TBC cos θ TBC 1 (2i + 6 j − 3k ) ⋅ (−4i + 19 j + 8k ) = TBC cos θ 7 21 1 [(2)( −4) + (6)(19) + (−3)(8)] 147 = 0.55782 cos θ = θ = 56.1° (b) (TBC ) BD = TBC ⋅ λBD = TBC cos θ = (490 N)(0.55782) (TBC ) BD = 273 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 204 PROBLEM 3.45 Given the vectors P = 4i − 2 j + 3k , Q = 2i + 4 j − 5k , and S = S x i − j + 2k , determine the value of S x for which the three vectors are coplanar. SOLUTION If P, Q, and S are coplanar, then P must be perpendicular to (Q × S). P ⋅ (Q × S) = 0 (or, the volume of a parallelepiped defined by P, Q, and S is zero). Then or 4 2 Sx −2 3 4 −5 = 0 −1 2 32 + 10S x − 6 − 20 + 8 − 12S x = 0 Sx = 7 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 205 PROBLEM 3.46 Determine the volume of the parallelepiped of Fig. 3.25 when (a) P = 4i − 3j + 2k, Q = −2i − 5j + k, and S = 7i + j − k, (b) P = 5i − j + 6k, Q = 2i + 3j + k, and S = −3i − 2j + 4k. SOLUTION Volume of a parallelepiped is found using the mixed triple product. (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 (b) Vol. = P ⋅ (Q × S) 5 −1 6 = 2 3 1 in.3 −3 −2 4 = (60 + 3 − 24 + 54 + 8 + 10) = 111 or Volume = 111.0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 206 PROBLEM 3.47 Knowing that the tension in cable AB is 570 N, determine the moment about each of the coordinate axes of the force exerted on the plate at B. SOLUTION BA = (−900 mm)i + (600 mm) j + (360 mm)k BA = (−900) 2 + (600)2 + (360) 2 = 1140 mm BA FB = FB BA −900i + 600 j + 360k = (570 N) 1140 = −(450 N)i + (300 N) j + (180 N)k rB = (0.9 m)i M O = rB × FB = 0.9i × (−450i + 300 j + 180k ) = 270k − 162 j MO = M x i + M y j + M z k = −(162 N ⋅ m) j + (270 N ⋅ m)k M x = 0, M y = −162.0 N ⋅ m, M z = +270 N ⋅ m Therefore, PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 207 PROBLEM 3.48 Knowing that the tension in cable AC is 1065 N, determine the moment about each of the coordinate axes of the force exerted on the plate at C. SOLUTION CA = (−900 mm)i + (600 mm) j + (−920 mm)k CA = ( −900)2 + (600) 2 + (−920) 2 = 1420 mm CA FC = FC CA −900i + 600 j − 920k = (1065 N) 1420 = −(675 N)i + (450 N) j − (690 N)k rC = (0.9 m)i + (1.28 m)k Using Eq. (3.19): i j k 0.9 0 1.28 M O = rC × FC = −675 450 −690 M O = −(576 N ⋅ m)i − (243 N ⋅ m) j + (405 N ⋅ m)k But Therefore, MO = M x i + M y j + M z k M x = −576 N ⋅ m, M y = −243 N ⋅ m, M z = +405 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 208 PROBLEM 3.49 A small boat hangs from two davits, one of which is shown in the figure. It is known that the moment about the z-axis of the resultant force RA exerted on the davit at A must not exceed 279 lb⋅ft in absolute value. Determine the largest allowable tension in line ABAD when x = 6 ft. SOLUTION First note: R A = 2TAB + TAD Also note that only TAD will contribute to the moment about the z-axis. Now AD = (6) 2 + (−7.75) 2 + (−3) 2 = 10.25 ft AD =T AD T (6i − 7.75 j − 3k ) = 10.25 Then TAD Now M z = k ⋅ (rA/C × TAD ) where rA/C = (7.75 ft) j + (3 ft)k Then for Tmax , 0 0 1 Tmax 279 = 0 7.75 3 10.25 6 −7.75 −3 = Tmax | − (1)(7.75)(6)| 10.25 or Tmax = 61.5 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 209 PROBLEM 3.50 For the davit of Problem 3.49, determine the largest allowable distance x when the tension in line ABAD is 60 lb. SOLUTION From the solution of Problem 3.49, TAD is now AD TAD = T 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 x 2 + 69.0625 0 0 1 0 7.75 3 x −7.75 −3 | − (1)(7.75)( x) | 279 x 2 + 69.0625 = 465 x 0.6 x 2 + 69.0625 = x Squaring both sides: 0.36 x 2 + 24.8625 = x 2 x 2 = 38.848 x = 6.23 ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 210 PROBLEM 3.51 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x-axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb. SOLUTION The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at their intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. We have ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x where (TAB ) z = k ⋅ TAB = k ⋅ (TAB λ AB ) − i − 12 j + 12k = k ⋅ 255 lb 17 = 180 lb (TDE ) z = k ⋅ TDE = k ⋅ (TDE λDE ) 1.5i − 14 j + 12k = k ⋅ TDE 18.5 = 0.64865TDE y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft (180 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft and TDE = 282.79 lb or TDE = 283 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 211 PROBLEM 3.52 Solve Problem 3.51 when the tension in cable AB is 306 lb. PROBLEM 3.51 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x-axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ⋅ ft, determine the magnitude of TDE when TAB = 255 lb. SOLUTION The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. We have ΣM x : (TAB ) z ( y A ) + (TDE ) z ( yD ) = M x Where (TAB ) z = k ⋅ TAB = k ⋅ (TAB λAB ) − i − 12 j + 12k = k ⋅ 306 lb 17 = 216 lb (TDE ) z = k ⋅ TDE = k ⋅ (TDE λDE ) 1.5i − 14 j + 12k = k ⋅ TDE 18.5 = 0.64865TDE y A = 12 ft yD = 14 ft M x = 4728 lb ⋅ ft (216 lb)(12 ft) + (0.64865TDE )(14 ft) = 4728 lb ⋅ ft and or TDE = 235 lb TDE = 235.21 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 212 PROBLEM 3.53 A single force P acts at C in a direction perpendicular to the handle BC of the crank shown. Knowing that Mx = +20 N · m and My = −8.75 N · m, and Mz = −30 N · m, determine the magnitude of P and the values of φ and θ. SOLUTION rC = (0.25 m)i + (0.2 m) sin θ j + (0.2 m) cos θ k P = − P sin φ j + P cos φ k i j M O = rC × P = 0.25 0.2sin θ 0 − P sin φ k 0.2 cos θ P cos φ Expanding the determinant, we find M x = (0.2) P(sin θ cos φ + cos θ sin φ ) Dividing Eq. (3) by Eq. (2) gives: M x = (0.2) P sin(θ + φ ) (1) M y = −(0.25) P cos φ (2) M z = −(0.25) P sin φ (3) tan φ = Mz My (4) −30 N ⋅ m −8.75 N ⋅ m φ = 73.740 tan φ = φ = 73.7° Squaring Eqs. (2) and (3) and adding gives: M y2 + M z2 = (0.25) 2 P 2 or P = 4 M y2 + M z2 (5) P = 4 (8.75) 2 + (30) 2 = 125.0 N P = 125.0 N Substituting data into Eq. (1): (+20 N ⋅ m) = 0.2 m(125.0 N) sin(θ + φ ) (θ + φ ) = 53.130° and (θ + φ ) = 126.87° θ = −20.6° and θ = 53.1° Q = 53.1° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 213 PROBLEM 3.54 A single force P acts at C in a direction perpendicular to the handle BC of the crank shown. Determine the moment Mx of P about the x-axis when θ = 65°, knowing that My = −15 N · m and Mz = −36 N · m. SOLUTION See the solution to Prob. 3.53 for the derivation of the following equations: M x = (0.2) P sin(θ + φ ) M tan φ = z My P = 4 M y2 + M z2 (1) (4) (5) Substituting for known data gives: tan φ = −36 N ⋅ m −15 N ⋅ m φ = 67.380° P = 4 ( −15) 2 + (−36) 2 P = 156.0 N M x = 0.2 m(156.0 N) sin(65° + 67.380°) = 23.047 N ⋅ m M x = 23.0 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 214 PROBLEM 3.55 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B. SOLUTION First note: TAE = TAE AE AE AE = (0.9)2 + (−0.6) 2 + (0.2) 2 = 1.1 m Then Also, Then Now where Then 55 N (0.9i − 0.6 j + 0.2k ) 1.1 = 5[(9 N)i − (6 N) j + (2 N)k ] TAE = DB = (1.2) 2 + ( −0.35) 2 + (0)2 λ DB = 1.25 m DB = DB 1 (1.2i − 0.35 j) = 1.25 1 (24i − 7 j) = 25 M DB = λ DB ⋅ (rA/D × TAE ) rA/D = −(0.1 m) j + (0.2 m)k M DB 24 −7 0 1 (5) 0 −0.1 0.2 = 25 9 2 −6 1 = (−4.8 − 12.6 + 28.8) 5 or M DB = 2.28 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 215 PROBLEM 3.56 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B. SOLUTION First note: TCF = TCF CF CF CF = (0.6)2 + (−0.9) 2 + (−0.2) 2 = 1.1 m 33 N (0.6i − 0.9 j + 0.2k ) 1.1 = 3[(6 N)i − (9 N) j − (2 N)k ] Then TCF = Also, DB = (1.2) 2 + ( −0.35) 2 + (0)2 = 1.25 m DB = DB 1 (1.2i − 0.35 j) = 1.25 1 (24i − 7 j) = 25 Then λ DB Now M DB = λ DB ⋅ (rC/D × TCF ) where rC/D = (0.2 m) j − (0.4 m)k Then M DB 24 −7 0 1 (3) 0 0.2 −0.4 = 25 6 −9 −2 = 3 (−9.6 + 16.8 − 86.4) 25 or M DB = −9.50 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 216 PROBLEM 3.57 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 235-lb force P. SOLUTION AB = (32 in.)i − (30 in.) j − (24 in.)k AB = (32)2 + ( −30) 2 + ( −24) 2 = 50 in. AB = 0.64i − 0.60 − 0.48k λ AB = AB We shall apply the force P at Point G: rG /B = (5 in.)i + (30 in.)k DG = (21 in.)i − (38 in.) j + (18 in.)k DG = (21) 2 + (−38)2 + (18) 2 = 47 in. DG 21i − 38 j + 18k P=P = (235 lb) 47 DG P = (105 lb)i − (190 lb) j + (90 lb)k The moment of P about AB is given by Eq. (3.46): M AB 0.64 −0.60 −0.48 0 30 in. = λ AB ⋅ (rG /B × P) = 5 in. 105 lb −190 lb 90 lb M AB = 0.64[0 − (30 in.)(−190 lb)] − 0.60[(30 in.)(105 lb) − (5 in.)(90 lb)] − 0.48[(5 in.)( −190 lb) − 0] = +2484 lb ⋅ in. M AB = +207 lb ⋅ ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 217 PROBLEM 3.58 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q. SOLUTION AB = (32 in.)i − (30 in.) j − (24 in.)k AB = (32)2 + ( −30) 2 + ( −24) 2 = 50 in. AB = 0.64i − 0.60 j − 0.48k λ AB = AB We shall apply the force Q at Point H: rH /B = −(32 in.)i + (17 in.) j DH = −(16 in.)i − (21 in.) j − (12 in.)k DH = (16) 2 + (−21) 2 + (−12)2 = 29 in. DH −16i − 21j − 12k Q= = (174 lb) DH 29 Q = −(96 lb)i − (126 lb) j − (72 lb)k The moment of Q about AB is given by Eq. (3.46): M AB 0.64 −0.60 −0.48 = λ AB ⋅ (rH /B × Q) = −32 in. 17 in. 0 −96 lb −126 lb −72 lb M AB = 0.64[(17 in.)(−72 lb) − 0] − 0.60[(0 − (−32 in.)( −72 lb)] − 0.48[(−32 in.)(−126 lb) − (17 in.)(−96 lb)] = −2119.7 lb ⋅ in. M AB = 176.6 lb ⋅ ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 218 PROBLEM 3.59 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable. SOLUTION M AD = λ AD ⋅ (rB/A × TBH ) Where and 1 λ AD = (4i − 3k ) 5 rB/A = (0.5 m)i d BH = (0.375)2 + (0.75) 2 + (−0.75) 2 = 1.125 m Then Finally, 450 N (0.375i + 0.75 j − 0.75k ) 1.125 = (150 N)i + (300 N) j − (300 N)k TBH = MAD 4 0 −3 1 0 = 0.5 0 5 150 300 −300 1 = [(−3)(0.5)(300)] 5 or M AD = − 90.0 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 219 PROBLEM 3.60 In Problem 3.59, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable. PROBLEM 3.59 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable. SOLUTION M AD = λ AD ⋅ (rB/A × TBG ) Where and 1 λ AD = (4i − 3k ) 5 rB/A = (0.5 m) j BG = (−0.5) 2 + (0.925)2 + (−0.4)2 = 1.125 m Then Finally, 450 N (−0.5i + 0.925 j − 0.4k ) 1.125 = −(200 N)i + (370 N) j − (160 N)k TBG = MAD 4 0 −3 1 = 0.5 0 0 5 −200 370 −160 1 = [(−3)(0.5)(370)] 5 M AD = −111.0 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 220 PROBLEM 3.61 A regular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA. SOLUTION We have From triangle OBC: M OA = λOA ⋅ (rC /O × P) (OA) x = a 2 (OA) z = (OA) x tan 30° = Since a 1 a = 2 3 2 3 (OA) 2 = (OA) 2x + (OA) 2y + (OAz )2 2 or a a a = + (OA) 2y + 2 2 3 2 2 (OA) y = a 2 − 2 a2 a2 − =a 4 12 3 a 2 a i+a j+ k 2 3 2 3 Then rA/O = and λOA = i + 1 2 2 1 j+ k 3 2 3 P = λ BC P = (a sin 30°)i − (a cos 30°)k P ( P) = (i − 3k ) a 2 rC /O = ai M OA 1 2 = 1 2 3 0 2 3 P (a) 0 2 1 0 − 3 = 1 aP 2 aP − (1)(− 3) = 2 3 2 M OA = aP 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 221 PROBLEM 3.62 A regular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other. (b) Use this property and the result obtained in Problem 3.61 to determine the perpendicular distance between edges OA and BC. SOLUTION (a) For edge OA to be perpendicular to edge BC, OA ⋅ BC = 0 From triangle OBC: (OA) x = a 2 a 1 a = 2 3 2 3 a a OA = i + (OA) y j + k 2 2 3 BC = ( a sin 30°) i − (a cos 30°) k (OA) z = (OA) x tan 30° = and = Then or so that (b) We have M OA a a 3 a i− k = (i − 3 k ) 2 2 2 a a a i + (OA) y j + k ⋅ (i − 3k ) = 0 2 2 3 2 a2 a2 + (OA) y (0) − =0 4 4 OA ⋅ BC = 0 OA is perpendicular to BC. = Pd , with P acting along BC and d the perpendicular distance from OA to BC. From the results of Problem 3.57, M OA = Pa 2 Pa 2 = Pd or d = a 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 222 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 F1 and M 2 = moment of F1 about the line of action of F2 . Now, by definition, M1 = λ 1 ⋅ (rB /A × F2 ) = λ 1 ⋅ (rB /A × λ 2 ) F2 M 2 = λ 2 ⋅ (rA/B × F1 ) = λ 2 ⋅ (rA/B × λ 1 ) F1 Since F1 = F2 = F and rA /B = −rB /A M1 = λ 1 ⋅ (rB /A × λ 2 ) F M 2 = λ 2 ⋅ (−rB /A × λ 1 ) F Using Equation (3.39): so that λ 1 ⋅ (rB /A × λ 2 ) = λ 2 ⋅ (−rB /A × λ 1 ) M 2 = λ 1⋅ (rB /A × λ 2 ) F M12 = M 21 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 223 PROBLEM 3.64 In Problem 3.55, determine the perpendicular distance between cable AE and the line joining Points D and B. PROBLEM 3.55 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B. SOLUTION From the solution to Problem 3.55: ΤAE = 55 N TAE = 5[(9 N)i − (6 N) j + (2 N)k ] | M DB | = 2.28 N ⋅ m λ DB = 1 (24i − 7 j) 25 Based on the discussion of Section 3.11, it follows that only the perpendicular component of TAE will contribute to the moment of TAE about line DB. Now (TAE )parallel = TAE ⋅ λ DB = 5(9i − 6 j + 2k ) ⋅ 1 (24i − 7 j) 25 1 = [(9)(24) + (−6)(−7)] 5 = 51.6 N Also, so that TAE = (TAE ) parallel + (TAE ) perpendicular (TAE )perpendicular = (55) 2 + (51.6)2 = 19.0379 N Since λ DB and (TAE )perpendicular are perpendicular, it follows that M DB = d (TAE ) perpendicular or 2.28 N ⋅ m = d (19.0379 N) d = 0.119761 d = 0.1198 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 224 PROBLEM 3.65 In Problem 3.56, determine the perpendicular distance between cable CF and the line joining Points D and B. PROBLEM 3.56 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B. SOLUTION ΤCF = 33 N From the solution to Problem 3.56: TCF = 3[(6 N)i − (9 N) j − (2 N)k ] | M DB | = 9.50 N ⋅ m λ DB = 1 (24i − 7 j) 25 Based on the discussion of Section 3.11, it follows that only the perpendicular component of TCF will contribute to the moment of TCF about line DB. Now (TCF )parallel = TCF ⋅ λ DB = 3(6i − 9 j − 2k ) ⋅ 1 (24i − 7 j) 25 3 [(6)(24) + (−9)( −7)] 25 = 24.84 N = Also, so that TCF = (TCF ) parallel + (TCF ) perpendicular (TCF )perpendicular = (33) 2 − (24.84) 2 = 21.725 N Since λ DB and (TCF )perpendicular are perpendicular, it follows that | M DB | = d (TCF ) perpendicular or 9.50 N ⋅ m = d × 21.725 N or d = 0.437 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 225 PROBLEM 3.66 In Prob. 3.57, determine the perpendicular distance between rod AB and the line of action of P. PROBLEM 3.57 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 235-lb force P. SOLUTION AB = (32 in.)i − (30 in.) j − (24 in.)k AB = (32)2 + (−30) 2 + ( −24) 2 = 50 in. AB = 0.64i − 0.60 j − 0.48k λ AB = AB λP = P 105i − 190 j + 90k = P 235 Angle θ between AB and P: cos θ = λ AB ⋅ λ P = (0.64i − 0.60 j − 0.48k ) ⋅ 105i − 190 j + 90k 235 = 0.58723 ∴ θ = 54.039° The moment of P about AB may be obtained by multiplying the projection of P on a plane perpendicular to AB by the perpendicular distance d from AB to P: M AB = ( P sin θ )d From the solution to Prob. 3.57: M AB = 207 lb ⋅ ft = 2484 lb ⋅ in. We have 2484 lb ⋅ in. = (235 lb)(sin 54.039) d d = 13.06 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 226 PROBLEM 3.67 In Prob. 3.58, determine the perpendicular distance between rod AB and the line of action of Q. PROBLEM 3.58 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q. SOLUTION AB = (32 in.)i − (30 in.) j − (24 in.)k AB = (32)2 + (−30) 2 + ( −24) 2 = 50 in. AB λ AB = = 0.64i − 0.60 j − 0.48k AB λQ = Q −96i − 126 j − 72k = Q 174 Angle θ between AB and Q: cos θ = λ AB ⋅ λQ = (0.64i − 0.60 j − 0.48k ) ⋅ (−96i − 126 j − 72k ) 174 = 0.28000 ∴ θ = 73.740° The moment of Q about AB may be obtained by multiplying the projection of Q on a plane perpendicular to AB by the perpendicular distance d from AB to Q: M AB = (Q sin θ )d From the solution to Prob. 3.58: M AB = 176.6 lb ⋅ ft = 2119.2 lb ⋅ in. 2119.2 lb ⋅ in. = (174 lb)(sin 73.740°) d d = 12.69 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 227 PROBLEM 3.68 In Problem 3.59, determine the perpendicular distance between portion BH of the cable and the diagonal AD. PROBLEM 3.59 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable. SOLUTION From the solution to Problem 3.59: TBH = 450 N TBH = (150 N)i + (300 N) j − (300 N)k | M AD | = 90.0 N ⋅ m 1 λ AD = (4i − 3k ) 5 Based on the discussion of Section 3.11, it follows that only the perpendicular component of TBH will contribute to the moment of TBH about line AD. Now (TBH )parallel = TBH ⋅ λ AD 1 = (150i + 300 j − 300k ) ⋅ (4i − 3k ) 5 1 = [(150)(4) + (−300)(−3)] 5 = 300 N Also, so that TBH = (TBH ) parallel + (TBH )perpendicular (TBH )perpendicular = (450) 2 − (300) 2 = 335.41 N Since λ AD and (TBH )perpendicular are perpendicular, it follows that M AD = d (TBH ) perpendicular or 90.0 N ⋅ m = d (335.41 N) d = 0.26833 m d = 0.268 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 228 PROBLEM 3.69 In Problem 3.60, determine the perpendicular distance between portion BG of the cable and the diagonal AD. PROBLEM 3.60 In Problem 3.59, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable. SOLUTION From the solution to Problem 3.60: ΤBG = 450 N TBG = −(200 N)i + (370 N) j − (160 N)k | M AD | = 111 N ⋅ m 1 λ AD = (4i − 3k ) 5 Based on the discussion of Section 3.11, it follows that only the perpendicular component of TBG will contribute to the moment of TBG about line AD. Now (TBG ) parallel = TBG ⋅ λ AD 1 = ( −200i + 370 j − 160k ) ⋅ (4i − 3k ) 5 1 = [(−200)(4) + (−160)(−3)] 5 = −64 N Also, so that TBG = (TBG ) parallel + (TBG )perpendicular (TBG ) perpendicular = (450) 2 − ( −64) 2 = 445.43 N Since λ AD and (TBG ) perpendicular are perpendicular, it follows that M AD = d (TBG ) perpendicular or 111 N ⋅ m = d (445.43 N) d = 0.24920 m d = 0.249 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 229 PROBLEM 3.70 A plate in the shape of a parallelogram is acted upon by two couples. Determine (a) the moment of the couple formed by the two 21-lb forces, (b) the perpendicular distance between the 12-lb forces if the resultant of the two couples is zero, (c) the value of α if the resultant couple is 72 lb ⋅ in. clockwise and d is 42 in. SOLUTION (a) M1 = d1 F1 We have d1 = 16 in. where F1 = 21 lb M1 = (16 in.)(21 lb) = 336 lb ⋅ in. (b) (c) 336 lb ⋅ in. − d 2 (12 lb) = 0 d 2 = 28.0 in. M total = M1 + M 2 We have or M1 + M 2 = 0 We have or or M1 = 336 lb ⋅ in. −72 lb ⋅ in. = 336 lb ⋅ in. − (42 in.)(sin α )(12 lb) sin α = 0.80952 α = 54.049° and or α = 54.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 230 PROBLEM 3.71 Four 1-in.-diameter pegs are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. (a) Determine the resultant couple acting on the board. (b) If only one string is used, around which pegs should it pass and in what directions should it be pulled to create the same couple with the minimum tension in the string? (c) What is the value of that minimum tension? SOLUTION M = (35 lb)(7 in.) + (25 lb)(9 in.) = 245 lb ⋅ in. + 225 lb ⋅ in. (a) M = 470 lb ⋅ in. (b) With only one string, pegs A and D, or B and C should be used. We have 6 8 tan θ = θ = 36.9° 90° − θ = 53.1° Direction of forces: (c) With pegs A and D: θ = 53.1° With pegs B and C: θ = 53.1° The distance between the centers of the two pegs is 82 + 62 = 10 in. Therefore, the perpendicular distance d between the forces is 1 d = 10 in. + 2 in. 2 = 11 in. M = Fd We must have 470 lb ⋅ in. = F (11 in.) F = 42.7 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 231 PROBLEM 3.72 Four pegs of the same diameter are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. Determine the diameter of the pegs knowing that the resultant couple applied to the board is 485 lb·in. counterclockwise. SOLUTION M = d AD FAD + d BC FBC 485 lb ⋅ in. = [(6 + d ) in.](35 lb) + [(8 + d ) in.](25 lb) d = 1.250 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 232 PROBLEM 3.73 A piece of plywood in which several holes are being drilled successively has been secured to a workbench by means of two nails. Knowing that the drill exerts a 12-N·m couple on the piece of plywood, determine the magnitude of the resulting forces applied to the nails if they are located (a) at A and B, (b) at B and C, (c) at A and C. SOLUTION (a) M = Fd 12 N ⋅ m = F (0.45 m) F = 26.7 N (b) M = Fd 12 N ⋅ m = F (0.24 m) F = 50.0 N (c) M = Fd d = (0.45 m)2 + (0.24 m)2 = 0.510 m 12 N ⋅ m = F (0.510 m) F = 23.5 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 233 PROBLEM 3.74 Two parallel 40-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about Point A. SOLUTION (a) We have where ΣM B : − d1C x + d 2 C y = M d1 = (0.270 m) sin 55° = 0.22117 m d 2 = (0.270 m) cos 55° = 0.154866 m C x = (40 N) cos 20° = 37.588 N C y = (40 N) sin 20° = 13.6808 N M = −(0.22117 m)(37.588 N)k + (0.154866 m)(13.6808 N)k = −(6.1946 N ⋅ m)k (b) We have We have or M = 6.19 N ⋅ m or M = 6.19 N ⋅ m M = Fd (−k ) = 40 N[(0.270 m)sin(55° − 20°)](−k ) = −(6.1946 N ⋅ m)k (c) or M = 6.19 N ⋅ m ΣM A : Σ(rA × F ) = rB/A × FB + rC/A × FC = M i j k sin 55° 0 M = (0.390 m)(40 N) cos 55° − cos 20° − sin 20° 0 i j k + (0.660 m)(40 N) cos 55° sin 55° 0 cos 20° sin 20° 0 = (8.9478 N ⋅ m − 15.1424 N ⋅ m)k = −(6.1946 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 234 PROBLEM 3.75 The two shafts of a speed-reducer unit are subjected to couples of magnitude M1 = 15 lb·ft and M2 = 3 lb·ft, respectively. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION M1 = (15 lb ⋅ ft)k M 2 = (3 lb ⋅ ft)i M = M12 + M 22 = (15) 2 + (3) 2 = 15.30 lb ⋅ ft 15 tan θ x = =5 3 θ x = 78.7° θ y = 90° θ z = 90° − 78.7° = 11.30° M = 15.30 lb ⋅ ft; θ x = 78.7°, θ y = 90.0°, θ z = 11.30° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 235 PROBLEM 3.76 Replace the two couples shown with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION Replace the couple in the ABCD plane with two couples P and Q shown: P = (50 N) CD 160 mm = (50 N) = 40 N CG 200 mm Q = (50 N) CF 120 mm = (50 N) = 30 N CG 200 mm Couple vector M1 perpendicular to plane ABCD: M1 = (40 N)(0.24 m) − (30 N)(0.16 m) = 4.80 N ⋅ m Couple vector M2 in the xy plane: M 2 = −(12.5 N)(0.192 m) = −2.40 N ⋅ m 144 mm θ = 36.870° 192 mm M1 = (4.80 cos 36.870°) j + (4.80 sin 36.870°)k = 3.84 j + 2.88k tan θ = M 2 = −2.40 j M = M1 + M 2 = 1.44 j + 2.88k M = 3.22 N ⋅ m; θ x = 90.0°, θ y = 53.1°, θ z = 36.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 236 PROBLEM 3.77 Solve Prob. 3.76, assuming that two 10-N vertical forces have been added, one acting upward at C and the other downward at B. PROBLEM 3.76 Replace the two couples shown with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION Replace the couple in the ABCD plane with two couples P and Q shown. P = (50 N) CD 160 mm = (50 N) = 40 N CG 200 mm Q = (50 N) CF 120 mm = (50 N) = 30 N CG 200 mm Couple vector M1 perpendicular to plane ABCD. M1 = (40 N)(0.24 m) − (30 N)(0.16 m) = 4.80 N ⋅ m 144 mm θ = 36.870° 192 mm M1 = (4.80cos 36.870°) j + (4.80sin 36.870°)k = 3.84 j + 2.88k tan θ = M 2 = −(12.5 N)(0.192 m) = −2.40 N ⋅ m = −2.40 j M 3 = rB /C × M 3 ; rB /C = (0.16 m)i + (0.144 m) j − (0.192 m)k = (0.16 m)i + (0.144 m) j − (0.192 m)k × (−10 N) j = −1.92i − 1.6k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 237 PROBLEM 3.77 (Continued) M = M1 + M 2 + M 3 = (3.84 j + 2.88k ) − 2.40 j + (−1.92i − 1.6k ) = −(1.92 N ⋅ m)i + (1.44 N ⋅ m) j + (1.28 N ⋅ m)k M = ( −1.92) 2 + (1.44) 2 + (1.28) 2 = 2.72 N ⋅ m M = 2.72 N ⋅ m cos θ x = −1.92/2.72 cos θ y = 1.44/2.72 θ x = 134.9° θ y = 58.0° θ z = 61.9° cos θ z = 1.28/2.72 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 238 PROBLEM 3.78 If P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION M = M1 + M 2 ; F1 = 16 lb, F2 = 40 lb 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. M = 0.29617i + 0.88852 j − 0.35045k M cos θ x = 0.29617 λ axis = cos θ y = 0.88852 θ x = 72.8° θ y = 27.3° θ z = 110.5° cos θ z = −0.35045 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 239 PROBLEM 3.79 If P = 20 lb, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION From the solution to Problem. 3.78: 16-lb force: M1 = −(480 lb ⋅ in.)k 40-lb force: M 2 = 8 5[(10 lb ⋅ in.)i + (30 lb ⋅ in.) j + (15 lb ⋅ in.)k ] P = 20 lb M 3 = rC × P = (30 in.)i × (20 lb)k = (600 lb ⋅ in.) j M = M1 + M 2 + M 3 = −(480)k + 8 5 (10i + 30 j + 15k ) + 600 j = (178.885 lb ⋅ in.)i + (1136.66 lb ⋅ in.) j − (211.67 lb ⋅ in.)k M = (178.885) 2 + (113.66) 2 + (211.67)2 M = 1170 lb ⋅ in. = 1169.96 lb ⋅ in. M = 0.152898i + 0.97154 j − 0.180921k M cos θ x = 0.152898 λ axis = cos θ y = 0.97154 θ x = 81.2° θ y = 13.70° θ z = 100.4° cos θ z = −0.180921 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 240 PROBLEM 3.80 In a manufacturing operation, three holes are drilled simultaneously in a workpiece. If the holes are perpendicular to the surfaces of the workpiece, replace the couples applied to the drills with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION M = M1 + M 2 + M 3 = (1.5 N ⋅ m)(− cos 20° j + sin 20°k ) − (1.5 N ⋅ m) j + (1.75 N ⋅ m)(− cos 25° j + sin 25°k ) = −(4.4956 N ⋅ m) j + (0.22655 N ⋅ m)k M = (0) 2 + (−4.4956) 2 + (0.22655) 2 = 4.5013 N ⋅ m M = 4.50 N ⋅ m M = −(0.99873j + 0.050330k ) M cos θ x = 0 λaxis = cos θ y = −0.99873 θ x = 90.0°, θ y = 177.1°, θ z = 87.1° cos θ z = 0.050330 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 241 PROBLEM 3.81 A 260-lb force is applied at A to the rolled-steel section shown. Replace that force with an equivalent force-couple system at the center C of the section. SOLUTION AB = (2.5 in.) 2 + (6.0 in.)2 = 6.50 in. 2.5 in. 5 = 6.5 in. 13 6.0 in. 12 cos α = = 6.5 in. 13 sin α = α = 22.6° F = − F sin α i − F cos α j 5 12 = −(260 lb) i − (260 lb) j 13 13 = −(100.0 lb)i − (240 lb) j M C = rA /C × F = (2.5i + 4.0 j) × (−100.0i − 240 j) = 400k − 600k = −(200 lb ⋅ in.)k F = 260 lb 67.4°; M C = 200 lb ⋅ in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 242 PROBLEM 3.82 A 30-lb vertical force P is applied at A to the bracket shown, which is held by screws at B and C. (a) Replace P with an equivalent force-couple system at B. (b) Find the two horizontal forces at B and C that are equivalent to the couple obtained in part a. SOLUTION (a) M B = (30 lb)(5 in.) = 150.0 lb ⋅ in. F = 30.0 lb , M B = 150.0 lb ⋅ in. (b) B=C = 150 lb ⋅ in. = 50.0 lb 3.0 in. B = 50.0 lb ; C = 50.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 243 PROBLEM 3.83 The force P has a magnitude of 250 N and is applied at the end C of a 500-mm rod AC attached to a bracket at A and B. Assuming α = 30° and β = 60°, replace P with (a) an equivalent force-couple system at B, (b) an equivalent system formed by two parallel forces applied at A and B. SOLUTION (a) ΣF : F = P or F = 250 N Equivalence requires 60° ΣM B : M = −(0.3 m)(250 N) = −75 N ⋅ m The equivalent force-couple system at B is FB = 250 N (b) M B = 75.0 N ⋅ m 60° We 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 60.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 244 PROBLEM 3.84 Solve Problem 3.83, assuming α = β = 25°. PROBLEM 3.83 The force P has a magnitude of 250 N and is applied at the end C of a 500-mm rod AC attached to a bracket at A and B. Assuming α = 30° and β = 60°, replace P with (a) an equivalent forcecouple system at B, (b) an equivalent system formed by two parallel forces applied at A and B. 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 We require Equivalence requires M B = d AE Q (0.3 m)[(250 N) sin 50°] = [(0.2 m) sin 50°]Q Q = 375 N Adding the forces at B: FA = 375 N 25.0° FB = 625 N 25.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 245 PROBLEM 3.85 The 80-N horizontal force P acts on a bell crank as shown. (a) Replace P with an equivalent force-couple system at B. (b) Find the two vertical forces at C and D that are equivalent to the couple found in part a. SOLUTION (a) ΣF : FB = F = 80 N Based on or FB = 80.0 N M B = 4.00 N ⋅ m ΣM : M B = Fd B = 80 N (0.05 m) = 4.0000 N ⋅ m or (b) If the two vertical forces are to be equivalent to MB, they must be a couple. Further, the sense of the moment of this couple must be counterclockwise. Then with FC and FD acting as shown, ΣM : M D = FC d 4.0000 N ⋅ m = FC (0.04 m) FC = 100.000 N or FC = 100.0 N ΣFy : 0 = FD − FC FD = 100.000 N or FD = 100.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 246 PROBLEM 3.86 A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 1040 N, replace the force exerted by the cable at B with an equivalent system formed by two parallel forces applied at A and C. 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 60.0° Based on ΣM A : − [(1040 N) cos 30°](6.7 m) = ( FC cos 30°)(10.7 m) FC = 651.21 N FC = 651 N or 60.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 247 PROBLEM 3.87 Three control rods attached to a lever ABC exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at B. (b) Determine the single force that is equivalent to the force-couple system obtained in part a, and specify its point of application on the lever. SOLUTION (a) First note that the two 90-N forces form a couple. Then F = 216 N where and θ θ = 180° − (60° + 55°) = 65° M = ΣM B = (0.450 m)(216 N) cos 55° − (1.050 m)(90 N) cos 20° = −33.049 N ⋅ m The equivalent force-couple system at B is F = 216 N (b) 65.0°; M = 33.0 N ⋅ m The single equivalent force F′ is equal to F. Further, since the sense of M is clockwise, F′ must be applied between A and B. For equivalence, ΣM B : M = aF ′ cos 55° where a is the distance from B to the point of application of F′. Then −33.049 N ⋅ m = − a(216 N) cos 55° a = 0.26676 m or F ′ = 216 N 65.0° applied to the lever 267 mm to the left of B PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 248 PROBLEM 3.88 A hexagonal plate is acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E. SOLUTION From the statement of the problem, it follows that ΣM E = 0 for the given force-couple system. Further, for Pmin, we must require that P be perpendicular to rB/E . Then 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° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 3.89 A force and couple act as shown on a square plate of side a = 25 in. Knowing that P = 60 lb, Q = 40 lb, and α = 50°, replace the given force and couple by a single force applied at a point located (a) on line AB, (b) on line AC. In each case determine the distance from A to the point of application of the force. SOLUTION Replace the given force-couple system with an equivalent forcecouple system at A. Px = (60 lb)(cos 50°) = 38.567 lb Py = (60 lb)(sin 50°) = 45.963 lb M A = Py a − Qa = (45.963 lb)(25 in.) − (40 lb)(25 in.) = 149.075 lb ⋅ in. (a) Equating moments about A gives: 149.075 lb ⋅ in. = (45.963 lb) x x = 3.24 in. P = 60.0 lb (b) 50.0°; 3.24 in. from A 149.075 lb ⋅ in. = (38.567 lb) y y = 3.87 in. P = 60.0 lb 50.0°; 3.87 in. below A PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 250 PROBLEM 3.90 The force and couple shown are to be replaced by an equivalent single force. Knowing that P = 2Q, determine the required value of α if the line of action of the single equivalent force is to pass through (a) Point A, (b) Point C. SOLUTION (a) We must have M A = 0 ( P sin α )a − Q (a ) = 0 sin α = Q Q 1 = = P 2Q 2 α = 30.0° (b) MC = 0 We must have ( P sin α )a − ( P cos α ) a − Q (a) = 0 sin α − cos α = Q Q 1 = = P 2Q 2 sin α = cos α + 1 2 (1) 1 4 1 1 − cos 2 α = cos 2 α + cos α + 4 sin 2 α = cos 2 α + cos α + 2 cos 2 α + cos α − 0.75 = 0 (2) Solving the quadratic in cos α : cos α = −1 ± 7 4 α = 65.7° or 155.7° Only the first value of α satisfies Eq. (1), therefore α = 65.7° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 251 PROBLEM 3.91 The shearing forces exerted on the cross section of a steel channel can be represented by a 900-N vertical force and two 250-N horizontal forces as shown. Replace this force and couple with a single force F applied at Point C, and determine the distance x from C to line BD. (Point C is defined as the shear center of the section.) SOLUTION 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 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 3.92 A force and a couple are applied as shown to the end of a cantilever beam. (a) Replace this system with a single force F applied at Point C, and determine the distance d from C to a line drawn through Points D and E. (b) Solve part a if the directions of the two 360-N forces are reversed. SOLUTION (a) We have ΣF : F = (360 N) j − (360 N) j − (600 N)k or F = −(600 N)k ΣM D : (360 N)(0.15 m) = (600 N)(d ) and d = 0.09 m or d = 90.0 mm below ED (b) We have from part a: F = −(600 N)k ΣM D : − (360 N)(0.15 m) = −(600 N)(d ) and d = 0.09 m or d = 90.0 mm above ED PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 253 PROBLEM 3.93 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 M = (4.10 kip ⋅ ft)i + (16.38 kip ⋅ ft)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 254 PROBLEM 3.94 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 M = −(23.0 kip ⋅ ft)i + (11.52 kip ⋅ ft)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 255 PROBLEM 3.95 A 110-N force acting in a vertical plane parallel to the yz-plane is applied to the 220-mm-long horizontal handle AB of a socket wrench. Replace the force with an equivalent forcecouple system at the origin O of the coordinate system. SOLUTION We have ΣF : PB = F where PB = 110 N[− (sin15°) j + (cos15°)k ] = −(28.470 N) j + (106.252 N)k or F = −(28.5 N) j + (106.3 N)k We have where ΣM O : rB/O × PB = M O rB /O = [(0.22 cos 35°)i + (0.15) j − (0.22sin 35°)k ] m = (0.180213 m)i + (0.15 m) j − (0.126187 m)k i j k 0.180213 0.15 0.126187 N ⋅ m = M O 0 −28.5 106.3 M O = [(12.3487)i − (19.1566) j − (5.1361)k ] N ⋅ m or M O = (12.35 N ⋅ m)i − (19.16 N ⋅ m)j − (5.13 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 256 PROBLEM 3.96 An eccentric, compressive 1220-N force P is applied to the end of a cantilever beam. Replace P with an equivalent force-couple system at G. SOLUTION We have ΣF : − (1220 N)i = F F = − (1220 N)i Also, we have ΣM G : rA/G × P = M i j k 1220 0 −0.1 −0.06 N ⋅ m = M −1 0 0 M = (1220 N ⋅ m)[(−0.06)(−1) j − ( −0.1)( −1)k ] or M = (73.2 N ⋅ m) j − (122 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 257 PROBLEM 3.97 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 = λAB PAB = (33 mm)i + (990 mm) j − (594 mm)k (175 N) 1155.00 mm or FC = (5.00 N)i + (150.0 N) j − (90.0 N)k We have ΣM C : rB/C × PAB = M C i j k M C = 5 0.683 −0.860 0 N ⋅ m 1 30 −18 = (5){(− 0.860)(−18)i − (0.683)(−18) j + [(0.683)(30) − (0.860)(1)]k} or M C = (77.4 N ⋅ m)i + (61.5 N ⋅ m) j + (106.8 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 258 PROBLEM 3.98 A 46-lb force F and a 2120-lb⋅in. couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner H. 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 M ′ = (−1800i − 1120k ) + 45 14 0 36 −28 −6 = (−1800i − 1120k ) + {[(14)(−6)]i + [−(45)( −6)]j + [(45)(−28) − (14)(36)]k} = (−1800 − 84)i + (270) j + (−1120 − 1764)k = −(1884 lb ⋅ in.)i + (270 lb ⋅ in.)j − (2884 lb ⋅ in.)k = −(157 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k F′ = (36.0 lb)i − (28.0 lb) j − (6.00 lb)k The equivalent force-couple system at H is M ′ = −(157.0 lb ⋅ ft)i + (22.5 lb ⋅ ft) j − (240 lb ⋅ ft)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 259 PROBLEM 3.99 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 = λ EH F1 (0.06 m)i + (0.06 m) j − (0.07 m)k (77 N) 0.11 m = (42 N)i + (42 N) j − (49 N)k = M1z = k ⋅ M1 M1 = λEJ M 1 = − di + (0.03 m) j − (0.07 m)k d 2 + 0.0058 m (31 N ⋅ m) Then from Equation (1), 0 0 1 ( −0.07 m)(31 N ⋅ m) =0 0.31 −0.0233 0 + 2 + 0.0058 d −49 42 42 Solving for d, Equation (1) reduces to (13.0200 + 0.9786) − from which 2.17 N ⋅ m d 2 + 0.0058 d = 0.1350 m =0 or d = 135.0 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 260 PROBLEM 3.99 (Continued) (b) F2 = F1 = (42i + 42 j − 49k ) N or F2 = (42.0 N)i + (42.0 N) j − (49.0 N)k M 2 = rH /B × F1 + M1 i j k (0.1350)i + 0.03j − 0.07k (31 N ⋅ m) = 0.31 −0.0233 0 + 0.155000 42 42 −49 = (1.14170i + 15.1900 j + 13.9986k ) N ⋅ m + (−27.000i + 6.0000 j − 14.0000k ) N ⋅ m M 2 = − (25.858 N ⋅ m)i + (21.190 N ⋅ m) j or M 2 = − (25.9 N ⋅ m)i + (21.2 N ⋅ m) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 261 PROBLEM 3.100 A 2.6-kip force is applied at Point D of the cast iron post shown. Replace that force with an equivalent force-couple system at the center A of the base section. SOLUTION DE = −(12 in.) j − (5 in.)k; DE = 13.00 in. DE F = (2.6 kips) DE F = (2.6 kips) −12 j − 5k 13 F = −(2.40 kips) j − (1.000 kip)k M A = rD /A × F where rD /A = (6 in.)i + (12 in.) j i j k 12 in. 0 M A = 6 in. 0 −2.4 kips −1.0 kips M A = −(12.00 kip ⋅ in.)i + (6.00 kip ⋅ in.) j − (14.40 kip ⋅ in.)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 262 PROBLEM 3.101 A 3-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent forcecouple system at end A of the beam. (b) Which of the loadings are equivalent? SOLUTION (a) (a) We have ΣFY : − 300 N − 200 N = Ra or R a = 500 N and ΣM A : − 400 N ⋅ m − (200 N)(3 m) = M a or M a = 1000 N ⋅ m (b) We have ΣFY : 200 N + 300 N = Rb or R b = 500 N and ΣM A : − 400 N ⋅ m + (300 N)(3 m) = M b or M b = 500 N ⋅ m (c) We have ΣFY : − 200 N − 300 N = Rc or R c = 500 N and ΣM A : 400 N ⋅ m − (300 N)(3 m) = M c or M c = 500 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 263 PROBLEM 3.101 (Continued) (d) We have ΣFY : − 500 N = Rd or R d = 500 N and ΣM A : 400 N ⋅ m − (500 N)(3 m) = M d or M d = 1100 N ⋅ m (e) We have ΣFY : 300 N − 800 N = Re or R e = 500 N and ΣM A : 400 N ⋅ m + 1000 N ⋅ m − (800 N)(3 m) = M e or M e = 1000 N ⋅ m (f ) We have ΣFY : − 300 N − 200 N = R f or R f = 500 N and ΣM A : 400 N ⋅ m − (200 N)(3 m) = M f or M f = 200 N ⋅ m (g) We have ΣFY : − 800 N + 300 N = Rg or R g = 500 N and ΣM A : 1000 N ⋅ m + 400 N ⋅ m + (300 N)(3 m) = M g or M g = 2300 N ⋅ m (h) We have ΣFY : − 250 N − 250 N = Rh or R h = 500 N and ΣM A : 1000 N ⋅ m + 400 N ⋅ m − (250 N)(3 m) = M h or M h = 650 N ⋅ m (b) Therefore, loadings (a) and (e) are equivalent. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 264 PROBLEM 3.102 A 3-m-long beam is loaded as shown. Determine the loading of Prob. 3.101 that is equivalent to this loading. SOLUTION We have ΣFY : − 200 N − 300 N = R R = 500 N or and ΣM A : 500 N ⋅ m + 200 N ⋅ m − (300 N)(3 m) = M M = 200 N ⋅ m or Problem 3.101 equivalent force-couples at A: Case R M (a) 500 N 1000 N⋅m (b) 500 N 500 N⋅m (c) 500 N 500 N⋅m (d) 500 N 1100 N⋅m (e) 500 N 1000 N⋅m (f ) 500 N 200 N⋅m (g) 500 N 2300 N⋅m (h) 500 N 650 N⋅m Equivalent to case (f ) of Problem 3.101 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 265 PROBLEM 3.103 Determine the single equivalent force and the distance from Point A to its line of action for the beam and loading of (a) Prob. 3.101a, (b) Prob. 3.101b, (c) Prob. 3.102. SOLUTION For equivalent single force at distance d from A: (a) We have ΣFY : − 300 N − 200 N = R or R = 500 N ΣM C : − 400 N ⋅ m + (300 N)(d ) and − (200 N)(3 − d ) = 0 or d = 2.00 m (b) We have ΣFY : 200 N + 300 N = R or R = 500 N ΣM C : − 400 N ⋅ m − (200 N)(d ) and + (300 N)(3 − d ) = 0 or d = 1.000 m (c) We have ΣFY : − 200 N − 300 N = R or R = 500 N ΣM C : 500 N ⋅ m + 200 N ⋅ m and + (200 N)( d ) − (300 N)(3 − d ) = 0 or d = 0.400 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 266 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)i + (1 ft) j + (2 ft)k ] × 10 lb)i = (15 lb ⋅ ft)i + (15 lb ⋅ ft)k G : M G = ΣM O = (15 lb ⋅ ft)i + (15 lb ⋅ ft) j I : M I = ΣM I = (15 lb ⋅ ft) j − (5 lb ⋅ ft)k + [(4.5 ft)i + (1 ft) j] × (10 lb) j = (15 lb ⋅ ft) j − (15 lb ⋅ ft)k The equivalent force-couple system is the system at corner D. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 267 PROBLEM 3.105 Three horizontal forces are applied as shown to a vertical cast iron arm. Determine the resultant of the forces and the distance from the ground to its line of action when (a) P = 200 N, (b) P = 2400 N, (c) P = 1000 N. SOLUTION (a) RD = +200 N − 600 N − 400 N = −800 N M D = −(200 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m) = +150.0 N ⋅ m y= M D 150 N ⋅ m = = 0.1875 m R 800 N R = 800 N ; y = 187.5 mm (b) RD = +2400 N − 600 N − 400 N = +1400 N M D = −(2400 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m) = −840 N ⋅ m y= M D 840 N ⋅ m = = 0.600 m R 1400 N R = 1400 N ; y = 600 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 268 PROBLEM 3.105 (Continued) (c) RD = +1000 − 600 − 400 = 0 M D = −(1000 N)(0.450 m) + (600 N)(0.300 m) + (400 N)(0.1500 m) = −210 N ⋅ m ∴ y = ∞ System reduces to a couple. M D = 210 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 269 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. L = 42 in. (b) We have 375.4 + 3.5d = 11.7(42) or d = 33.1 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 270 PROBLEM 3.107 The weights of two children sitting at ends A and B of a seesaw are 84 lb and 64 lb, respectively. Where should a third child sit so that the resultant of the weights of the three children will pass through C if she weighs (a) 60 lb, (b) 52 lb. 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 (b) For the resultant weight to act at C, Then ΣM C = 0 WC = 52 lb (84 lb)(6 ft) − 52 lb(d ) − 64 lb(6 ft) = 0 d = 2.31 ft to the right of C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 271 PROBLEM 3.108 A couple of magnitude M = 54 lb ⋅ in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC. SOLUTION (a) We have ΣF : R = (−10 j) + (30 cos 60°)i + 30 sin 60° j + (−45i ) = −(30 lb)i + (15.9808 lb) j or R = 34.0 lb (b) 28.0° First reduce the given forces and couple to an equivalent force-couple system (R , M B ) at B. We have ΣM B : M B = (54 lb ⋅ in) + (12 in.)(10 lb) − (8 in.)(45 lb) = −186 lb ⋅ in. Then with R at D, or and with R at E, or ΣM B : −186 lb ⋅ in = a(15.9808 lb) a = 11.64 in. ΣM B : −186 lb ⋅ in = C (30 lb) C = 6.2 in. The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in. below B. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 272 PROBLEM 3.109 A couple M and the three forces shown are applied to an angle bracket. Find the moment of the couple if the line of action of the resultant of the force system is to pass through (a) Point A, (b) Point B, (c) Point C. SOLUTION In each case, we must have M1R = 0 (a) M AB = ΣM A = M + (12 in.)[(30 lb) sin 60°] − (8 in.)(45 lb) = 0 M = +48.231 lb ⋅ in. (b) M = 240 lb ⋅ in. M BR = ΣM B = M + (12 in.)(10 lb) − (8 in.)(45 lb) = 0 M = +240 lb ⋅ in. (c) M = 48.2 lb ⋅ in. M CR = ΣM C = M + (12 in.)(10 lb) − (8 in.)[(30 lb) cos 60°] = 0 M =0 M=0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 273 PROBLEM 3.110 A 32-lb motor is mounted on the floor. Find the resultant of the weight and the forces exerted on the belt, and determine where the line of action of the resultant intersects the floor. SOLUTION We have 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° ΣM O : ΣM O = xRy − [(140 lb) cos 30°][(4 + 2 cos 30°)in.] − [(140 lb) sin 30°][(2 in.)sin 30°] − (60 lb)(2 in.) = x(38.0 lb) x= and 1 (− 694.97 − 70.0 − 120) in. 38.0 x = −23.289 in. Or resultant intersects the base (x-axis) 23.3 in. to the left of the vertical centerline (y-axis) of the motor. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 274 PROBLEM 3.111 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH. SOLUTION We have First replace the applied forces and couples with an equivalent force-couple system at G. ΣFx : 200cos 15° − 120 cos 70° + P = Rx Thus, Rx = (152.142 + P) N or ΣFy : − 200sin 15° − 120sin 70° − 80 = Ry Ry = −244.53 N or ΣM G : − (0.47 m)(200 N) cos15° + (0.05 m)(200 N)sin15° + (0.47 m)(120 N) cos 70° − (0.19 m)(120 N)sin 70° − (0.13 m)( P N) − (0.59 m)(80 N) + 42 N ⋅ m + 40 N ⋅ m = M G M G = −(55.544 + 0.13P) N ⋅ m or (1) Setting P = 0 in Eq. (1): Now with R at I, ΣM G : − 55.544 N ⋅ m = − a(244.53 N) a = 0.227 m or and with R at J, ΣM G : − 55.544 N ⋅ m = −b(152.142 N) b = 0.365 m or (a) The rivet hole is 0.365 m above G. (b) The rivet hole is 0.227 m to the right of G. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 275 PROBLEM 3.112 Solve Problem 3.111, assuming that P = 60 N. PROBLEM 3.111 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P = 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH. SOLUTION See the solution to Problem 3.111 leading to the development of Equation (1): M G = −(55.544 + 0.13P) N ⋅ m Rx = (152.142 + P) N and P = 60 N For we have Rx = (152.142 + 60) = 212.14 N M G = −[55.544 + 0.13(60)] = −63.344 N ⋅ m Then with R at I, ΣM G : −63.344 N ⋅ m = −a(244.53 N) a = 0.259 m or and with R at J, ΣM G : −63.344 N ⋅ m = −b(212.14 N) b = 0.299 m or (a) The rivet hole is 0.299 m above G. (b) The rivet hole is 0.259 m to the right of G. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 276 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 R = Rx2 + Ry2 = (147.728) 2 + (758.36) 2 = 772.62 lb Ry Rx −758.36 = tan −1 −147.728 = 78.977° θ = tan −1 or We have ΣM A = dRy where ΣM A = −[240 lb cos 70°](6 ft) − [240 lbsin 70°](4 ft) R = 773 lb 79.0° − (160 lb)(12 ft) + [300 lb cos 40°](6 ft) − [300 lb sin 40°](20 ft) − (180 lb)(8 ft) = −7232.5 lb ⋅ ft −7232.5 lb ⋅ ft −758.36 lb = 9.5370 ft d= or d = 9.54 ft to the right of A PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 277 PROBLEM 3.114 Four ropes are attached to a crate and exert the forces shown. If the forces are to be replaced with a single equivalent force applied at a point on line AB, determine (a) the equivalent force and the distance from A to the point of application of the force when α = 30°, (b) the value of α so that the single equivalent force is applied at Point B. SOLUTION We have (a) For equivalence, ΣFx : −100 cos 30° + 400 cos 65° + 90 cos 65° = Rx or Rx = 120.480 lb ΣFy : 100 sin α + 160 + 400 sin 65° + 90 sin 65° = Ry or Ry = (604.09 + 100sin α ) lb With α = 30°, Ry = 654.09 lb Then R = (120.480) 2 + (654.09) 2 = 665 lb (1) 654.09 120.480 or θ = 79.6° tan θ = ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) sin 65° Also or + (26 in.)(400 lb) cos 65° + (66 in.)(90 lb)sin 65° + (36 in.)(90 lb) cos 65° = d (654.09 lb) ΣM A = 42, 435 lb ⋅ in. and d = 64.9 in. 79.6° and R is applied 64.9 in. to the right of A. (b) R = 665 lb We have d = 66 in. Then or Using Eq. (1): ΣM A : 42, 435 lb ⋅ in = (66 in.) Ry Ry = 642.95 lb 642.95 = 604.09 + 100sin α or α = 22.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 278 PROBLEM 3.115 Solve Prob. 3.114, assuming that the 90-lb force is removed. PROBLEM 3.114 Four ropes are attached to a crate and exert the forces shown. If the forces are to be replaced with a single equivalent force applied at a point on line AB, determine (a) the equivalent force and the distance from A to the point of application of the force when α = 30°, (b) the value of α so that the single equivalent force is applied at Point B. SOLUTION (a) For equivalence, ΣFx : − (100 lb) cos 30° + (400 lb)sin 25° = Rx or Rx = 82.445 lb ΣFy : 160 lb + (100 lb) sin 30° + (400 lb) cos 25° = Ry or Ry = 572.52 lb R = (82.445)2 + (572.52) = 578.43 lb tan θ = 572.52 82.445 or θ = 81.806° ΣM A : (46 in.)(160 lb) + (66 in.)(400 lb) cos 25° + (26 in.)(400 lb)sin 25° = d (527.52 lb) d = 62.3 in. R = 578 lb (b) 81.8° and is applied 62.3 in. to the right of A. We have d = 66.0 in. For R applied at B, ΣM A : Ry (66 in.) = (160 lb)(46 in.) + (66 in.)(400 lb) cos 25° + (26 in.)(400 lb)sin 25° Ry = 540.64 lb ΣFY : 160 lb + (100 lb)sin α + (400 lb) cos 25° = 540.64 lb α = 10.44° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 279 PROBLEM 3.116 Four forces act on a 700 × 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate. SOLUTION (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 = −1.20000 θ = −50.194° (b) R = 1562 N 50.2° M CR = Σr × F = (0.5 m)i × (300 N + 300 N) j = (300 N ⋅ m)k (300 N ⋅ m)k = xi × (1200 N) j x = 0.25000 m x = 250 mm (300 N ⋅ m) = yj × ( −1000 N)i y = 0.30000 m y = 300 mm Intersection 250 mm to right of C and 300 mm above C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 280 PROBLEM 3.117 Solve Problem 3.116, assuming that the 760-N force is directed to the right. PROBLEM 3.116 Four forces act on a 700 × 375-mm plate as shown. (a) Find the resultant of these forces. (b) Locate the two points where the line of action of the resultant intersects the edge of the plate. SOLUTION 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 θ = 66.5714° R = 1308 N 66.6° M CR = Σr × F (b) = (0.5 m)i × (300 N + 300 N) j = (300 N ⋅ m)k (300 N ⋅ m)k = xi × (1200 N) j x = 0.25000 m or x = 0.250 mm (300 N ⋅ m)k = [ x′i + (0.375 m) j] × [(520 N)i + (1200 N) j] = (1200 x′ − 195)k x′ = 0.41250 m or x′ = 412.5 mm Intersection 412 mm to the right of A and 250 mm to the right of C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 281 PROBLEM 3.118 As follower AB rolls along the surface of member C, it exerts a constant force F perpendicular to the surface. (a) Replace F with an equivalent force-couple system at Point D obtained by drawing the perpendicular from the point of contact to the x-axis. (b) For a = 1 m and b = 2 m, determine the value of x for which the moment of the equivalent force-couple system at D is maximum. SOLUTION (a) The slope of any tangent to the surface of member C is dy d x2 = b 1 − 2 dx dx a −2b = 2 x a Since the force F is perpendicular to the surface, dy tan α = − dx −1 = a2 1 2b x For equivalence, ΣF : F = R ΣM D : ( F cos α )( y A ) = M D where cos α = 2bx (a ) + (2bx)2 2 2 x2 y A = b 1 − 2 a x3 2 Fb 2 x − 2 a MD = 4 2 2 a + 4b x Therefore, the equivalent force-couple system at D is R=F a2 tan −1 2bx x3 2 Fb2 x − 2 a M= a 4 + 4b 2 x 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 282 PROBLEM 3.118 (Continued) (b) To maximize M, the value of x must satisfy dM =0 dx a = 1 m, b = 2 m where for M= 8F ( x − x3 ) 1 + 16 x 2 dM = 8F dx 1 1 + 16 x 2 (1 − 3x 2 ) − ( x − x3 ) (32 x)(1 + 16 x 2 ) −1/ 2 2 =0 (1 + 16 x 2 ) (1 + 16 x 2 )(1 − 3x 2 ) − 16 x( x − x3 ) = 0 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 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and 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 PROBLEM 3.119 As plastic bushings are inserted into a 60-mm-diameter cylindrical sheet metal enclosure, the insertion tools exert the forces shown on the enclosure. Each of the forces is parallel to one of the coordinate axes. Replace these forces with an equivalent force-couple system at C. SOLUTION For equivalence, ΣF : R = FA + FB + FC + FD = −(17 N) j − (12 N) j − (16 N)k − (21 N)i = −(21 N)i − (29 N) j − (16 N)k ΣM C : M = rA /C × FA + rB /C × FB + rD /C × FD M = [(0.11 m) j − (0.03 m)k ] × [−(17 N)] j + [(0.02 m)i + (0.11 m) j − (0.03 m)k ] × [ −(12 N)]j + [(0.03 m)i + (0.03 m) j − (0.03 m)k ] × [ −(21 N)]i = −(0.51 N ⋅ m)i + [−(0.24 N ⋅ m)k − (0.36 N ⋅ m)i] + [(0.63 N ⋅ m)k + (0.63 N ⋅ m) j] ∴ The equivalent force-couple system at C is R = −(21.0 N)i − (29.0 N) j − (16.00 N)k M = −(0.870 N ⋅ m)i + (0.630 N ⋅ m) j + (0.390 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 284 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 M A = M B + M C + rB/ A × R B + rC/ A × R C i j k 0 0 N⋅m = − (5.25 N ⋅ m)i + (6.3750 N ⋅ m)i + 0.225 0 −351.26 49.593 i j k + 0.45 0 0 N⋅m 0 −68.591 −389.00 = (1.12500 N ⋅ m)i + (163.892 N ⋅ m) j − (109.899 N ⋅ m)k or M A = (1.125 N ⋅ m)i + (163.9 N ⋅ m) j − (109.9 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 285 PROBLEM 3.121 Four forces are applied to the machine component ABDE as shown. Replace these forces with an equivalent force-couple system at A. 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 RA = rB × [−(300 N)i − (50 N) j] + rD × (−250 N)k + r × ( − 120 N)i i j k i j k = 0.2 m 0 0 + 0.2 m 0 0.16 m −300 N −50 N 0 0 0 −250 N i j k + 0.2 m −0.1 m 0.16 m −120 N 0 0 = −(10 N ⋅ m)k + (50 N ⋅ m) j − (19.2 N ⋅ m) j − (12 N ⋅ m)k Force-couple system at A is R = −(420 N)i − (50 N) j − (250 N)k M RA = (30.8 N ⋅ m) j − (220 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 286 PROBLEM 3.122 While using a pencil sharpener, a student applies the forces and couple shown. (a) Determine the forces exerted at B and C knowing that these forces and the couple are equivalent to a forcecouple system at A consisting of the force R = (2.6 lb)i + Ry j − (0.7 lb)k and the couple M RA = M x i + (1.0 lb · ft)j − (0.72 lb · ft)k. (b) Find the corresponding values of Ry and M x . SOLUTION (a) From the statement of the problem, equivalence requires ΣF : B + C = R or ΣFx : Bx + C x = 2.6 lb (1) ΣFy : − C y = R y (2) ΣFz : − C z = −0.7 lb or C z = 0.7 lb and ΣM A : (rB/A × B + M B ) + rC/A × C = M AR or 1.75 ΣM x : (1 lb ⋅ ft) + ft (C y ) = M x 12 (3) 3.75 1.75 3.5 ΣM y : ft ( Bx ) + ft (C x ) + ft (0.7 lb) = 1 lb ⋅ ft 12 12 12 or Using Eq. (1): 3.75Bx + 1.75C x = 9.55 3.75Bx + 1.75(2.6 Bx ) = 9.55 or Bx = 2.5 lb and C x = 0.1 lb 3.5 ΣM z : − ft (C y ) = −0.72 lb ⋅ ft 12 C y = 2.4686 lb or B = (2.50 lb)i C = (0.1000 lb)i − (2.47 lb) j − (0.700 lb)k (b) Eq. (2) Using Eq. (3): Ry = −2.47 lb 1.75 1+ (2.4686) = M x 12 or M x = 1.360 lb ⋅ ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 287 PROBLEM 3.123 A blade held in a brace is used to tighten a screw at A. (a) Determine the forces exerted at B and C, knowing that these forces are equivalent to a force-couple system at A consisting of R = −(30 N)i + Ry j + Rz k and M RA = − (12 N · m)i. (b) Find the corresponding values of Ry and Rz . (c) What is the orientation of the slot in the head of the screw for which the blade is least likely to slip when the brace is in the position shown? SOLUTION (a) Equivalence requires or Equating the i coefficients: Also, or Equating coefficients: ΣF : R = B + C −(30 N)i + Ry j + Rz k = − Bk + (−C x i + C y j + C z k ) i : − 30 N = −C x or C x = 30 N ΣM A : M RA = rB/A × B + rC/A × C −(12 N ⋅ m)i = [(0.2 m)i + (0.15 m)j] × (− B)k +(0.4 m)i × [−(30 N)i + C y j + C z k ] i : − 12 N ⋅ m = −(0.15 m) B k : 0 = (0.4 m)C y or or B = 80 N Cy = 0 j: 0 = (0.2 m)(80 N) − (0.4 m)C z or C z = 40 N B = −(80.0 N)k C = −(30.0 N)i + (40.0 N)k (b) Now we have for the equivalence of forces −(30 N)i + Ry j + Rz k = −(80 N)k + [(−30 N)i + (40 N)k ] Equating coefficients: j: R y = 0 Ry = 0 k : Rz = −80 + 40 (c) or Rz = −40.0 N First note that R = −(30 N)i − (40 N)k. Thus, the screw is best able to resist the lateral force Rz when the slot in the head of the screw is vertical. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 288 PROBLEM 3.124 In order to unscrew the tapped faucet A, a plumber uses two pipe wrenches as shown. By exerting a 40-lb force on each wrench, at a distance of 10 in. from the axis of the pipe and in a direction perpendicular to the pipe and to the wrench, he prevents the pipe from rotating, and thus avoids loosening or further tightening the joint between the pipe and the tapped elbow C. Determine (a) the angle θ that the wrench at A should form with the vertical if elbow C is not to rotate about the vertical, (b) the force-couple system at C equivalent to the two 40-lb forces when this condition is satisfied. SOLUTION We first reduce the given forces to force-couple systems at A and B, noting that | M A | = | M B | = (40 lb)(10 in.) = 400 lb ⋅ in. We now determine the equivalent force-couple system at C. R = (40 lb)(1 − cos θ )i − (40 lb) sin θ j (1) M CR = M A + M B + (15 in.)k × [−(40 lb) cos θ i − (40 lb)sin θ j] + (7.5 in.)k × (40 lb)i = + 400 − 400 − 600cos θ j + 600sin θ i + 300 j = (600 lb ⋅ in.)sin θ i + (300 lb ⋅ in.)(1 − 2 cos θ ) j (a) (2) For no rotation about vertical, y component of M CR must be zero. 1 − 2cos θ = 0 cos θ = 1/2 θ = 60.0° (b) For θ = 60.0° in Eqs. (1) and (2), R = (20.0 lb)i − (34.641 lb) j; M CR = (519.62 lb ⋅ in.)i R = (20.0 lb)i − (34.6 lb) j; M CR = (520 lb ⋅ in.)i PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 289 PROBLEM 3.125 Assuming θ = 60° in Prob. 3.124, replace the two 40-lb forces with an equivalent force-couple system at D and determine whether the plumber’s action tends to tighten or loosen the joint between (a) pipe CD and elbow D, (b) elbow D and pipe DE. Assume all threads to be right-handed. PROBLEM 3.124 In order to unscrew the tapped faucet A, a plumber uses two pipe wrenches as shown. By exerting a 40-lb force on each wrench, at a distance of 10 in. from the axis of the pipe and in a direction perpendicular to the pipe and to the wrench, he prevents the pipe from rotating, and thus avoids loosening or further tightening the joint between the pipe and the tapped elbow C. Determine (a) the angle θ that the wrench at A should form with the vertical if elbow C is not to rotate about the vertical, (b) the force-couple system at C equivalent to the two 40-lb forces when this condition is satisfied. SOLUTION The equivalent force-couple system at C for θ = 60° was obtained in the solution to Prob. 3.124: R = (20.0 lb)i − (34.641 lb) j M CR = (519.62 lb ⋅ in.)i The equivalent force-couple system at D is made of R and M RD where M RD = M CR + rC /D × R = (519.62 lb ⋅ in.)i + (25.0 in.) j × [(20.0 lb)i − (34.641 lb) j] = (519.62 lb ⋅ in.)i − (500 lb ⋅ in.)k Equivalent force-couple at D: R = (20.0 lb)i − (34.6 lb) j; M CR = (520 lb ⋅ in.)i − (500 lb ⋅ in.)k (a) (b) Since M RD has no component along the y-axis, the plumber’s action will neither loosen nor tighten the joint between pipe CD and elbow. , the plumber’s action will tend to tighten Since the x component of M RD is the joint between elbow and pipe DE. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 290 PROBLEM 3.126 As an adjustable brace BC is used to bring a wall into plumb, the force-couple system shown is exerted on the wall. Replace this force-couple system with an equivalent force-couple system at A if R = 21.2 lb and M = 13.25 lb · ft. SOLUTION We have where or We have where ΣF : R = R A = RλBC λBC = (42 in.)i − (96 in.) j − (16 in.)k 106 in. RA = 21.2 lb (42i − 96 j − 16k ) 106 R A = (8.40 lb)i − (19.20 lb) j − (3.20 lb)k ΣM A : rC/A × R + M = M A rC/A = (42 in.)i + (48 in.)k = 1 (42i + 48k )ft 12 = (3.5 ft)i + (4.0 ft)k R = (8.40 lb)i − (19.50 lb) j − (3.20 lb)k M = −λBC M −42i + 96 j + 16k (13.25 lb ⋅ ft) 106 = −(5.25 lb ⋅ ft)i + (12 lb ⋅ ft) j + (2 lb ⋅ ft)k = Then i j k 3.5 0 4.0 lb ⋅ ft + (−5.25i + 12 j + 2k ) lb ⋅ ft = M A 8.40 −19.20 −3.20 M A = (71.55 lb ⋅ ft)i + (56.80 lb ⋅ ft)j − (65.20 lb ⋅ ft)k or M A = (71.6 lb ⋅ ft)i + (56.8 lb ⋅ ft)j − (65.2 lb ⋅ ft)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 291 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 ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) = R ( z D ) (375 N)(3 m) + (260 N)(0.5 m) + (400 N)(4.75 m) = (1035 N)(z D ) z D = 3.0483 m We have or z D = 3.05 m ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) = R ( xD ) 375 N(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) = (1035 N)( xD ) xD = 2.5749 m or xD = 2.57 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 292 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 ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH ) (375 N)(1 m) + (260 N)(1.5 m) + (400 N)(4.75 m) + (425 N)(xD ) = (1460 N)(2.5 m) xD = 2.3235 m or xD = 2.32 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 293 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 ΣM x : (5 ft)(105 lb) − (1 ft)(90 lb) + (3 ft)(160 lb) + (5.5 ft)(50 lb) = − y (405 lb) or y = −2.94 ft ΣM y : (5.5 ft)(105 lb) + (12 ft)(90 lb) + (14.5 ft)(160 lb) + (22.5 ft)(50 lb) = − x(405 lb) x = 12.60 ft or R acts 12.60 ft to the right of member AB and 2.94 ft below member BC. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 294 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 ΣM y : − (9 ft)(105 ft) − (14.5 − b) ft × (90 lb) + (8 ft)(50 lb) = 0 or b = 20.6 ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 295 PROBLEM 3.131* A group of students loads a 2 × 3.3-m flatbed trailer with two 0.66 × 0.66 × 0.66-m boxes and one 0.66 × 0.66 × 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 × 0.66 × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.) SOLUTION For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the intersection with the center line of the trailer, the added 0.66 × 0.66 × 1.2-m box should be placed adjacent to one of the edges of the trailer with the 0.66 × 0.66-m side on the bottom. The edges to be considered are based on the location of the resultant for the three given weights. We have ΣF : − (224 N) j − (392 N) j − (176 N) j = R R = −(792 N) j We have ΣM z : − (224 N)(0.33 m) − (392 N)(1.67 m) − (176 N)(1.67 m) = ( −792 N)( x) xR = 1.29101 m We have ΣM x : (224 N)(0.33 m) + (392 N)(0.6 m) + (176 N)(2.0 m) = (792 N)( z ) z R = 0.83475 m From the statement of the problem, it is known that the resultant of R from the original loading and the lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0. Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G as possible without the box overhanging the trailer. These two requirements imply (0.33 m ≤ x ≤ 1 m) (1.5 m ≤ z ≤ 2.97 m) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 296 PROBLEM 3.131* (Continued) xL = 0.33 m With at G : ΣM z : (1 − 0.33) m × WL − (1.29101 − 1) m × (792 N) = 0 WL = 344.00 N or Now we must check if this is physically possible, at G : ΣM x : ( z L − 1.5) m × 344 N) − (1.5 − 0.83475) m × (792 N) = 0 z L = 3.032 m or which is not acceptable. z L = 2.97 m: With at G : ΣM x : (2.97 − 1.5) m × WL − (1.5 − 0.83475) m × (792 N) = 0 WL = 358.42 N or Now check if this is physically possible, at G : ΣM z : (1 − xL ) m × (358.42 N) − (1.29101 − 1) m × (792 N) = 0 or xL = 0.357 m ok! WL = 358 N The minimum weight of the fourth box is And it is placed on end A (0.66 × 0.66-m side down) along side AB with the center of the box 0.357 m from side AD. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 297 PROBLEM 3.132* Solve Problem 3.131 if the students want to place as much weight as possible in the fourth box and at least one side of the box must coincide with a side of the trailer. PROBLEM 3.131* A group of students loads a 2 × 3.3-m flatbed trailer with two 0.66 × 0.66 × 0.66-m boxes and one 0.66 × 0.66 × 1.2-m box. Each of the boxes at the rear of the trailer is positioned so that it is aligned with both the back and a side of the trailer. Determine the smallest load the students should place in a second 0.66 × 0.66 × 1.2-m box and where on the trailer they should secure it, without any part of the box overhanging the sides of the trailer, if each box is uniformly loaded and the line of action of the resultant of the weights of the four boxes is to pass through the point of intersection of the centerlines of the trailer and the axle. (Hint: Keep in mind that the box may be placed either on its side or on its end.) SOLUTION First replace the three known loads with a single equivalent force R applied at coordinate ( xR , 0, z R ). Equivalence requires ΣFy : − 224 − 392 − 176 = − R or R = 792 N ΣM x : (0.33 m)(224 N) + (0.6 m)(392 N) + (2 m)(176 N) = z R (792 N) or z R = 0.83475 m ΣM z : − (0.33 m)(224 N) − (1.67 m)(392 N) − (1.67 m)(176 N) = xR (792 N) or xR = 1.29101 m From the statement of the problem, it is known that the resultant of R and the heaviest loads WH passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0 Further, since WH is to be as large as possible, the fourth box should be placed as close to G as possible while keeping one of the sides of the box coincident with a side of the trailer. Thus, the two limiting cases are xH = 0.6 m or z H = 2.7 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 298 PROBLEM 3.132* (Continued) Now consider these two possibilities. With xH = 0.6 m at G : ΣM z : (1 − 0.6) m × WH − (1.29101 − 1) m × (792 N) = 0 WH = 576.20 N or Checking if this is physically possible at or G : ΣM x : ( z H − 1.5) m × (576.20 N) − (1.5 − 0.83475) m × (792 N) = 0 z H = 2.414 m which is acceptable. With z H = 2.7 m at or G : ΣM x : (2.7 − 1.5) m × WH − (1.5 − 0.83475) m × (792 N) = 0 WH = 439 N Since this is less than the first case, the maximum weight of the fourth box is WH = 576 N and it is placed with a 0.66 × 1.2-m side down, a 0.66-m edge along side AD, and the center 2.41 m from side DC. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 299 PROBLEM 3.133* A piece of sheet metal is bent into the shape shown and is acted upon by three forces. If the forces have the same magnitude P, replace them with an equivalent wrench and determine (a) the magnitude and the direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench. SOLUTION ( ) First reduce the given forces to an equivalent force-couple system R, M OR at the origin. We have ΣF : − Pj + Pj + Pk = R R = Pk or 5 ΣM O : − (aP ) j + −( aP )i + aP k = M OR 2 5 M OR = aP −i − j + k 2 or (a) Then for the wrench, R=P and λ axis = R =k R cos θ x = 0 cos θ y = 0 cos θ z = 1 or (b) θ x = 90° θ y = 90° θ z = 0° Now M1 = λ axis ⋅ M OR 5 = k ⋅ aP −i − j + k 2 5 = aP 2 Then P= M1 25 aP = R P or P = 5 a 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 300 PROBLEM 3.133* (Continued) (c) The components of the wrench are (R , M1 ), where M1 = M1λ axis , and the axis of the wrench is assumed to intersect the xy-plane at Point Q, whose coordinates are (x, y, 0). Thus, we require M z = rQ × R R where M z = M O × M1 Then 5 5 aP −i − j + k − aPk = ( xi + yj) + Pk 2 2 Equating coefficients: i : − aP = yP or j: − aP = − xP or y = −a x=a The axis of the wrench is parallel to the z-axis and intersects the xy-plane at x = a, y = −a. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 301 PROBLEM 3.134* Three forces of the same magnitude P act on a cube of side a as shown. Replace the three forces by an equivalent wrench and determine (a) the magnitude and direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench. SOLUTION Force-couple system at O: R = Pi + Pj + Pk = P(i + j + k ) M OR = aj × Pi + ak × Pj + ai × Pk = − Pak − Pai − Paj M OR = − Pa (i + j + k ) Since R and M OR have the same direction, they form a wrench with M1 = M OR . Thus, the axis of the wrench is the diagonal OA. We note that cos θ x = cos θ y = cos θ z = a a 3 = 1 3 R = P 3 θ x = θ y = θ z = 54.7° M1 = M OR = − Pa 3 Pitch = p = M 1 − Pa 3 = = −a R P 3 (a) R = P 3 θ x = θ y = θ z = 54.7° (b) – a (c) Axis of the wrench is diagonal OA. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 302 PROBLEM 3.135* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane. SOLUTION First, reduce the given force system to a force-couple at the origin. We have ΣF : − (10 lb) j − (11 lb) j = R R = − (21 lb) j We have ΣM O : Σ(rO × F ) + ΣM C = M OR M OR i j k i j k = 0 0 20 lb ⋅ in. + 0 0 −15 lb ⋅ in. − (12 lb ⋅ in) j 0 −10 0 0 −11 0 = (35 lb ⋅ in.)i − (12 lb ⋅ in.) j R = − (21 lb) j (a) (b) We have or R = − (21.0 lb) j R R = (− j) ⋅ [(35 lb ⋅ in.)i − (12 lb ⋅ in.) j] M1 = λ R ⋅ M OR λR = = 12 lb ⋅ in. and M1 = −(12 lb ⋅ in.) j and pitch p= M 1 12 lb ⋅ in. = = 0.57143 in. R 21 lb or p = 0.571 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 303 PROBLEM 3.135* (Continued) (c) We have M OR = M1 + M 2 M 2 = M OR − M1 = (35 lb ⋅ in.)i We require M 2 = rQ/O × R (35 lb ⋅ in.)i = ( xi + zk ) × [ −(21 lb) j] 35i = −(21x)k + (21z )i From i: From k: 35 = 21z z = 1.66667 in. 0 = − 21x z=0 The axis of the wrench is parallel to the y-axis and intersects the xz-plane at x = 0, z = 1.667 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 304 PROBLEM 3.136* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane. SOLUTION First, reduce the given force system to a force-couple system. We have ΣF : − (20 N)i − (15 N) j = R We have ΣM O : Σ(rO × F ) + ΣM C = M OR R = 25 N M OR = −20 N(0.1 m)j − (4 N ⋅ m)i − (1 N ⋅ m)j = −(4 N ⋅ m)i − (3 N ⋅ m) j R = −(20.0 N)i − (15.00 N)j (a) (b) We have R R = (−0.8i − 0.6 j) ⋅ [−(4 N ⋅ m)i − (3 N ⋅ m)j] M1 = λR ⋅ M OR λ= = 5 N⋅m Pitch: p= M1 5 N ⋅ m = = 0.200 m R 25 N or p = 0.200 m (c) From above, note that M1 = M OR Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in the xy-plane with a slope of y= 3 x 4 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 305 PROBLEM 3.137* Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz-plane. SOLUTION First, reduce the given force system to a force-couple at the origin. We have ΣF : − (84 N) j − (80 N)k = R and ΣM O : Σ(rO × F ) + ΣM C = M OR R = 116 N i j k i j k 0.6 0 0.1 + 0.4 0.3 0 + (−30 j − 32k ) N ⋅ m = M OR 0 84 0 0 0 80 M OR = − (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k R = − (84.0 N) j − (80.0 N)k (a) (b) We have M1 = λ R ⋅ M OR λR = R R −84 j − 80k ⋅ [− (15.6 N ⋅ m)i + (2 N ⋅ m) j − (82.4 N ⋅ m)k ] 116 = 55.379 N ⋅ m =− and Then pitch M1 = M1λR = − (40.102 N ⋅ m) j − (38.192 N ⋅ m)k p= M 1 55.379 N ⋅ m = = 0.47741 m R 116 N or p = 0.477 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 306 PROBLEM 3.137* (Continued) (c) We have M OR = M1 + M 2 M 2 = M OR − M1 = [(−15.6i + 2 j − 82.4k ) − (40.102 j − 38.192k )] N ⋅ m = − (15.6 N ⋅ m)i + (42.102 N ⋅ m) j − (44.208 N ⋅ m)k We require M 2 = rQ/O × R (−15.6i + 42.102 j − 44.208k ) = ( xi + zk ) × (84 j − 80k ) = (84 z )i + (80 x) j − (84 x)k From i: or From k: or −15.6 = 84 z z = − 0.185714 m z = − 0.1857 m −44.208 = −84 x x = 0.52629 m x = 0.526 m The axis of the wrench intersects the xz-plane at x = 0.526 m y = 0 z = − 0.1857 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 307 PROBLEM 3.138* Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz-plane. SOLUTION First, reduce the given force system to a force-couple at the origin at B. (a) We have 15 8 ΣF : − (26.4 lb)k − (17 lb) i + j = R 17 17 R = − (8.00 lb)i − (15.00 lb) j − (26.4 lb)k and We have R = 31.4 lb ΣM B : rA/B × FA + M A + M B = M BR M RB i j k 15 8 0 − 220k − 238 i + j = 264i − 220k − 14(8i + 15 j) = 0 −10 17 17 0 0 − 26.4 M RB = (152 lb ⋅ in.)i − (210 lb ⋅ in.)j − (220 lb ⋅ in.)k (b) We have R R −8.00i − 15.00 j − 26.4k = ⋅ [(152 lb ⋅ in.)i − (210 lb ⋅ in.) j − (220 lb ⋅ in.)k ] 31.4 = 246.56 lb ⋅ in. M1 = λ R ⋅ M OR λR = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 308 PROBLEM 3.138* (Continued) and Then pitch (c) We have M1 = M1λR = − (62.818 lb ⋅ in.)i − (117.783 lb ⋅ in.) j − (207.30 lb ⋅ in.)k p= M 1 246.56 lb ⋅ in. = = 7.8522 in. 31.4 lb R or p = 7.85 in. M RB = M1 + M 2 M 2 = M RB − M1 = (152i − 210 j − 220k ) − ( − 62.818i − 117.783j − 207.30k ) = (214.82 lb ⋅ in.)i − (92.217 lb ⋅ in.) j − (12.7000 lb ⋅ in.)k We require M 2 = rQ/B × R i j k 214.82i − 92.217 j − 12.7000k = x 0 z −8 −15 −26.4 = (15 z )i − (8 z ) j + (26.4 x) j − (15 x)k From i: 214.82 = 15 z z = 14.3213 in. From k: −12.7000 = −15 x x = 0.84667 in. The axis of the wrench intersects the xz-plane at x = 0.847 in. y = 0 z = 14.32 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 309 PROBLEM 3.139* A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane. SOLUTION (a) First reduce the given force system to a force-couple at the origin. We have ΣF : Pλ BA + P λDC + P λDE = R 4 3 3 4 −9 4 12 R = P j − k + i − j + i − j + k 5 5 5 5 25 5 25 R= R= We have 3P (2i − 20 j − k ) 25 3P 27 5 P (2) 2 + (20) 2 + (1)2 = 25 25 ΣM : Σ(rO × P) = M OR 3P 4P 4P 12 P −4 P 3P −9 P (24a) j × j− k + (20a) j × i− j + (20a) j × i− j+ k = M OR 5 5 5 5 25 5 25 M OR = 24 Pa ( −i − k ) 5 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 310 PROBLEM 3.139* (Continued) (b) We have M1 = λ R ⋅ M OR where λR = Then M1 = 25 1 R 3P (2i − 20 j − k ) (2i − 20 j − k ) = = R 25 27 5 P 9 5 p= and pitch 1 9 5 (2i − 20 j − k ) ⋅ M 1 −8Pa 25 −8a = = R 15 5 27 5 P 81 M1 = M1λ R = (c) Then M 2 = M OR − M1 = 24 Pa −8 Pa (−i − k ) = 5 15 5 or p = − 0.0988a −8Pa 1 8 Pa ( −2i + 20 j + k ) (2i − 20 j − k ) = 675 15 5 9 5 24 Pa 8Pa 8 Pa ( −i − k ) − (−2i + 20 j + k ) = (−430i − 20 j − 406k ) 5 675 675 M 2 = rQ/O × R We require 8Pa 3P 675 (−403i − 20 j − 406k ) = ( xi + zk ) × 25 (2i − 20 j − k ) 3P = [20 zi + ( x + 2 z ) j − 20 xk ] 25 From i: 8(− 403) Pa 3P = 20 z 675 25 From k: 8(−406) Pa 3P = −20 x x = 2.0049a 675 25 z = −1.99012a The axis of the wrench intersects the xz-plane at x = 2.00a, z = −1.990a PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 311 PROBLEM 3.140* Two ropes attached at A and B are used to move the trunk of a fallen tree. Replace the forces exerted by the ropes with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the yz-plane. SOLUTION (a) ( ) First replace the given forces with an equivalent force-couple system R, M OR at the origin. We have d AC = (6) 2 + (2) 2 + (9) 2 = 11 m d BD = (14) 2 + (2)2 + (5) 2 = 15 m Then 1650 N = (6i + 2 j + 9k ) 11 = (900 N)i + (300 N) j + (1350 N)k TAC = and 1500 N = (14i + 2 j + 5k ) 15 = (1400 N)i + (200 N) j + (500 N)k TBD = Equivalence then requires ΣF : R = TAC + TBD = (900i + 300 j + 1350k ) +(1400i + 200 j + 500k ) = (2300 N)i + (500 N) j + (1850 N)k ΣM O : M OR = rA × TAC + rB × TBD = (12 m)k × [(900 N)i + (300 N)j + (1350 N)k ] + (9 m)i × [(1400 N)i + (200 N)j + (500 N)k ] = −(3600)i + (10,800 − 4500) j + (1800)k = −(3600 N ⋅ m)i + (6300 N ⋅ m)j + (1800 N ⋅ m)k The components of the wrench are (R , M1 ), where R = (2300 N)i + (500 N) j + (1850 N)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 312 PROBLEM 3.140* (Continued) (b) We have R = 100 (23)2 + (5) 2 + (18.5) 2 = 2993.7 N Let λ axis = Then 1 R (23i + 5 j + 18.5k ) = R 29.937 M1 = λ axis ⋅ M OR 1 (23i + 5 j + 18.5k ) ⋅ (−3600i + 6300 j + 1800k ) 29.937 1 [(23)( −36) + (5)(63) + (18.5)(18)] = 0.29937 = −601.26 N ⋅ m = Finally, P= M1 −601.26 N ⋅ m = R 2993.7 N or P = − 0.201 m (c) We have M1 = M 1 λ axis = (−601.26 N ⋅ m) × 1 (23i + 5 j + 18.5k ) 29.937 or M1 = −(461.93 N ⋅ m)i − (100.421 N ⋅ m) j − (371.56 N ⋅ m)k Now M 2 = M OR − M1 = (−3600i + 6300 j + 1800k ) − ( −461.93i − 100.421j − 371.56k ) = − (3138.1 N ⋅ m)i + (6400.4 N ⋅ m)j + (2171.6 N ⋅ m)k For equivalence: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 313 PROBLEM 3.140* (Continued) Thus, we require M 2 = rP × R r = ( yj + zk ) Substituting: −3138.1i + 6400.4 j + 2171.6k = i j k 0 y z 2300 500 1850 Equating coefficients: j : 6400.4 = 2300 z or k : 2171.6 = −2300 y or The axis of the wrench intersects the yz-plane at z = 2.78 m y = − 0.944 m y = −0.944 m z = 2.78 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 314 PROBLEM 3.141* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz-plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz-plane. SOLUTION First determine the resultant of the forces at D. We have d DA = (−12) 2 + (9) 2 + (8) 2 = 17 in. d ED = (−6) 2 + (0)2 + (−8)2 = 10 in. Then 34 lb = (−12i + 9 j + 8k ) 17 = −(24 lb)i + (18 lb) j + (16 lb)k FDA = and 30 lb = (−6i − 8k ) 10 = −(18 lb)i − (24 lb)k FED = Then ΣF : R = FDA + FED = (−24i + 18 j + 16k + ( −18i − 24k ) = −(42 lb)i + (18 lb)j − (8 lb)k For the applied couple d AK = ( −6) 2 + (−6) 2 + (18)2 = 6 11 in. Then M= 160 lb ⋅ in. ( −6i − 6 j + 18k ) 6 11 160 = [−(1 lb ⋅ in.)i − (1 lb ⋅ in.)j + (3 lb ⋅ in.)k ] 11 To be able to reduce the original forces and couple to a single equivalent force, R and M must be perpendicular. Thus ? R ⋅ M =0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 315 PROBLEM 3.141* (Continued) Substituting (−42i + 18 j − 8k ) ⋅ 160 or 11 160 11 ? ( −i − j + 3k ) = 0 ? [(−42)(−1) + (18)(−1) + (−8)(3)] = 0 0 =0 or R and M are perpendicular so that the given system can be reduced to the single equivalent force. R = −(42.0 lb)i + (18.00 lb) j − (8.00 lb)k Then for equivalence, M = rP/D × R Thus, we require where rP/D = −(12 in.)i + [( y − 3)in.] j + ( z in.)k Substituting: i j k ( −i − j + 3k ) = −12 ( y − 3) z 11 18 −42 −8 = [( y − 3)( −8) − ( z )(18)]i 160 + [( z )(−42) − (−12)(−8)]j + [( −12)(18) − ( y − 3)(−42)]k Equating coefficients: j: − k: 160 = − 42 z − 96 or 11 480 = −216 + 42( y − 3) or 11 The line of action of R intersects the yz-plane at x=0 z = −1.137 in. y = 11.59 in. y = 11.59 in. z = −1.137 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 316 PROBLEM 3.142* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz-plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz-plane. SOLUTION First, reduce the given force system to a force-couple at the origin. We have ΣF : FA + FG = R (40 mm)i + (60 mm) j − (120 mm)k R = (50 N)k + 70 N 140 mm = (20 N)i + (30 N) j − (10 N)k and We have R = 37.417 N ΣM O : Σ(rO × F) + ΣM C = M OR M OR = [(0.12 m) j × (50 N)k ] + {(0.16 m)i × [(20 N)i + (30 N) j − (60 N)k ]} (160 mm)i − (120 mm) j + (10 N ⋅ m) 200 mm (40 mm)i − (120 mm) j + (60 mm)k + (14 N ⋅ m) 140 mm M 0R = (18 N ⋅ m)i − (8.4 N ⋅ m) j + (10.8 N ⋅ m)k To be able to reduce the original forces and couples to a single equivalent force, R and M must be perpendicular. Thus, R ⋅ M = 0. Substituting ? (20i + 30 j − 10k ) ⋅ (18i − 8.4 j + 10.8k ) = 0 ? or (20)(18) + (30)(−8.4) + (−10)(10.8) = 0 or 0=0 R and M are perpendicular so that the given system can be reduced to the single equivalent force. R = (20.0 N)i + (30.0 N) j − (10.00 N)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 317 PROBLEM 3.142* (Continued) Then for equivalence, Thus, we require M OR = rp × R rp = yj + zk Substituting: i j k 18i − 8.4 j + 10.8k = 0 y z 20 30 −10 Equating coefficients: j: − 8.4 = 20 z k: or z = −0.42 m 10.8 = −20 y or y = −0.54 m The line of action of R intersects the yz-plane at x=0 y = −0.540 m z = −0.420 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 318 PROBLEM 3.143* Replace the wrench shown with an equivalent system consisting of two forces perpendicular to the y-axis and applied respectively at A and B. SOLUTION Express the forces at A and B as A = Ax i + Az k B = Bx i + Bz k Then, for equivalence to the given force system, ΣFx : Ax + Bx = 0 (1) ΣFz : Az + Bz = R (2) ΣM x : Az ( a) + Bz ( a + b) = 0 (3) ΣM z : − Ax (a) − Bx (a + b) = M (4) Bx = − Ax From Equation (1), Substitute into Equation (4): − Ax ( a) + Ax ( a + b) = M M M and Bx = − Ax = b b From Equation (2), Bz = R − Az and Equation (3), Az a + ( R − Az )(a + b) = 0 a Az = R 1 + b and a Bz = R − R 1 + b a Bz = − R b M A= b Then a i + R 1 + b k M B = − b a i − b Rk PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 319 PROBLEM 3.144* Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes through a given point while the other force lies in a given plane. SOLUTION First, choose a coordinate system so that the xy-plane coincides with the given plane. Also, position the coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a. Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the scalar components of R and M are known relative to the shown coordinate system. A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the given Point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B. The known components of the wrench can be expressed as R = Rx i + Ry j + Rz k and M = M x i + M y j + M z k while the unknown forces A and B can be expressed as A = Ax i + Ay j + Az k and B = Bx i + Bz k Since the position vector of Point P is given, it follows that the scalar components (x, y, z) of the position vector rP are also known. Then, for equivalence of the two systems, ΣFx : Rx = Ax + Bx (1) ΣFy : Ry = Ay (2) ΣFz : Rz = Az + Bz (3) ΣM x : M x = yAz − zAy (4) ΣM y : M y = zAx − xAz − bBz (5) ΣM z : M z = xAy − yAx (6) Based on the above six independent equations for the six unknowns ( Ax , Ay , Az , Bx , Bz , b), there exists a unique solution for A and B. Ay = Ry From Equation (2), PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 320 PROBLEM 3.144* (Continued) 1 Ax = ( xRy − M z ) y Equation (6): 1 Bx = Rx − ( xRy − M z ) y Equation (1): Equation (4): 1 Az = ( M x + zRy ) y Equation (3): 1 Bz = Rz − ( M x + zRy ) y b= Equation (5): ( xM x + yM y + zM z ) ( M x − yRz + zRy ) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 321 PROBLEM 3.145* Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point. SOLUTION First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular coordinate system with the axis of the wrench while one of the other axes passes through the given point. See Figures a and b. We have R = Rj and M = Mj and are known. The unknown forces A and B can be expressed as A = Ax i + Ay j + Az k and B = Bx i + By j + Bz k The distance a is known. It is assumed that force B intersects the xz-plane at (x, 0, z). Then for equivalence, ΣFx : 0 = Ax + Bx (1) ΣFy : R = Ay + By (2) ΣFz : 0 = Az + Bz (3) 0 = − zBy (4) ΣM x : ΣM y : M = − aAz − xBz + zBx (5) ΣM z : (6) 0 = aAy + xBy Since A and B are made perpendicular, A ⋅ B = 0 or There are eight unknowns: Ax Bx + Ay B y + Az Bz = 0 (7) Ax , Ay , Az , Bx , By , Bz , x, z But only seven independent equations. Therefore, there exists an infinite number of solutions. 0 = − zBy Next, consider Equation (4): If By = 0, Equation (7) becomes Ax Bx + Az Bz = 0 Ax2 + Az2 = 0 Using Equations (1) and (3), this equation becomes PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 322 PROBLEM 3.145* (Continued) Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By ≠ 0, so that from Equation (4), z = 0. To obtain one possible solution, arbitrarily let Ax = 0. (Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.) The defining equations then become 0 = Bx (1)′ R = Ay + By (2) 0 = Az + Bz (3) M = − aAz − xBz (5)′ 0 = aAy + xBy (6) Ay By + Az Bz = 0 (7)′ Then Equation (2) can be written Ay = R − By Equation (3) can be written Bz = − Az Equation (6) can be written x=− aAy By Substituting into Equation (5)′, R − By M = − aAz − − a By M Az = − By aR or ( − Az ) (8) Substituting into Equation (7)′, M M ( R − By ) By + − By By = 0 aR aR or By = a 2 R3 a2 R2 + M 2 Then from Equations (2), (8), and (3), a2 R2 RM 2 = 2 2 2 2 a R +M a R + M2 M a 2 R3 aR 2 M Az = − = − aR a 2 R 2 + M 2 a2 R2 + M 2 Ay = R − Bz = 2 aR 2 M a2 R2 + M 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 323 PROBLEM 3.145* (Continued) In summary, A= RM ( Mj − aRk ) a R2 + M 2 B= aR 2 (aRj + Mk ) a2 R2 + M 2 2 Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a given point. Lastly, if R > 0 and M > 0, it follows from the equations found for A and B that Ay > 0 and B y > 0. From Equation (6), x < 0 (assuming a > 0). Then, as a consequence of letting Ax = 0, force A lies in a plane parallel to the yz-plane and to the right of the origin, while force B lies in a plane parallel to the yz-plane but to the left to the origin, as shown in the figure below. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 324 PROBLEM 3.146* Show that a wrench can be replaced with two forces, one of which has a prescribed line of action. SOLUTION First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and another axis intersects the prescribed line of action (AA′). Note that it has been assumed that the line of action of force B intersects the xz-plane at Point P(x, 0, z). Denoting the known direction of line AA′ by λ A = λx i + λ y j + λz k it follows that force A can be expressed as A = Aλ A = A(λx i + λ y j + λz k ) Force B can be expressed as B = Bx i + B y j + Bz k Next, observe that since the axis of the wrench and the prescribed line of action AA′ are known, it follows that the distance a can be determined. In the following solution, it is assumed that a is known. Then for equivalence, ΣFx : 0 = Aλx + Bx (1) ΣFy : R = Aλ y + By (2) ΣFz : 0 = Aλz + Bz (3) ΣM x : 0 = − zBy (4) ΣM y : M = − aAλz + zBx − xBz (5) ΣM x : 0 = − aAλ y + xB y (6) Since there are six unknowns (A, Bx, By, Bz, x, z) and six independent equations, it will be possible to obtain a solution. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 325 PROBLEM 3.146* (Continued) Case 1: Let z = 0 to satisfy Equation (4). Now Equation (2): Equation (3): Equation (6): Aλ y = R − B y Bz = − Aλz x=− aAλ y By a = − By ( R − By ) Substitution into Equation (5): a M = − aAλz − − By 1 M A=− B λz aR y ( R − By )(− Aλz ) Substitution into Equation (2): R=− By = Then 1 M B λ + By λz aR y y λz aR 2 λz aR − λ y M MR R = λz aR − λ y M λ − aR λ y z M λx MR Bx = − Aλx = λz aR − λ y M A=− Bz = − Aλz = λz MR λz aR − λ y M A= In summary, B= and R x = a 1 − By λz aR − λ y M = a 1 − R λ aR 2 z P λA aR λy − λz M R (λ M i + λz aRj + λz M k ) λz aR − λ y M x or x = λy M λz R Note that for this case, the lines of action of both A and B intersect the x-axis. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 326 PROBLEM 3.146* (Continued) Case 2: Let By = 0 to satisfy Equation (4). Now Equation (2): A= R λy Equation (1): λ Bx = − R x λy Equation (3): λ Bz = − R z λy Equation (6): aAλ y = 0 which requires a = 0 Substitution into Equation (5): λ M = z − R x λ y λ − x − R z λ y M or λz x − λx z = R λy This last expression is the equation for the line of action of force B. In summary, R A= λy λA R B= λy ( −λ x i − λx k ) Assuming that λx , λ y , λz ⬎ 0, the equivalent force system is as shown below. Note that the component of A in the xz-plane is parallel to B. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 327 PROBLEM 3.147 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D. SOLUTION (a) Fx = (300 N) cos 25° = 271.89 N Fy = (300 N) sin 25° = 126.785 N F = (271.89 N)i + (126.785 N) j r = DA = −(0.1 m)i − (0.2 m) j MD = r × F M D = [−(0.1 m)i − (0.2 m) j] × [(271.89 N)i + (126.785 N) j] = −(12.6785 N ⋅ m)k + (54.378 N ⋅ m)k = (41.700 N ⋅ m)k M D = 41.7 N ⋅ m (b) The smallest force Q at B must be perpendicular to DB at 45° M D = Q ( DB) 41.700 N ⋅ m = Q (0.28284 m) Q = 147.4 N 45.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 328 PROBLEM 3.148 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A. SOLUTION First note dCB = (12.0 in.) 2 + (2.33 in.) 2 = 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 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 329 PROBLEM 3.149 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C. SOLUTION (a) We have M A = rE/A × TDE where rE/A = (2.3 m) j TDE = λ DE TDE = (0.6 m)i + (3.3 m) j − (3 m)k (0.6) 2 + (3.3)2 + (3) 2 m (810 N) = (108 N)i + (594 N) j − (540 N)k i j k MA = 0 2.3 0 N⋅m 108 594 −540 = −(1242 N ⋅ m)i − (248.4 N ⋅ m)k or M A = −(1242 N ⋅ m)i − (248 N ⋅ m)k (b) We have M A = rG/A × TCG where rG/A = (2.7 m)i + (2.3 m) j TCG = λ CGTCG = −(.6 m)i + (3.3 m) j − (3 m)k (.6) 2 + (3.3) 2 + (3) 2 m (810 N) = −(108 N)i + (594 N) j − (540 N)k i j k M A = 2.7 2.3 0 N⋅m −108 594 −540 = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k or M A = −(1242 N ⋅ m)i + (1458 N ⋅ m) j + (1852 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 330 PROBLEM 3.150 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 AB = AB(sin 37° j − cos 37°k ) CD = CD(− cos 40° cos 55° j + sin 40° j − cos 40° sin 55°k ) Now AB ⋅ CD = ( AB)(CD ) cos θ or AB(sin 37° j − cos 37°k ) ⋅ CD (− cos 40° cos 55°i + sin 40° j − cos 40° sin 55°k ) = (AB)(CD) cos θ or cos θ = (sin 37°)(sin 40°) + (− cos 37°)(− cos 40° sin 55°) = 0.88799 or θ = 27.4° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 331 PROBLEM 3.151 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the moments about the y and the z axes of the force exerted at B by portion AB of the rope are, respectively, 120 N ⋅ m and −460 N ⋅ m, determine the distance a. SOLUTION First note BA = (2.2 m)i − (3.2 m) j − ( a m)k Now M D = rA/D × TBA where rA/D = (2.2 m)i + (1.6 m) j TBA = Then i j k TBA MD = 2.2 1.6 0 d BA 2.2 −3.2 − a = Thus TBA (2.2i − 3.2 j − ak ) (N) d BA TBA {−1.6a i + 2.2a j + [(2.2)(−3.2) − (1.6)(2.2)]k} d BA M y = 2.2 TBA a d BA M z = −10.56 Then forming the ratio TBA d BA (N ⋅ m) (N ⋅ m) My Mz T 2.2 dBA (N ⋅ m) 120 N ⋅ m BA = −460 N ⋅ m −10.56 TdBA (N ⋅ m) or a = 1.252 m BA PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 332 PROBLEM 3.152 To loosen a frozen valve, a force F of magnitude 70 lb is applied to the handle of the valve. Knowing that θ = 25°, Mx = −61 lb ⋅ ft, and M z = − 43 lb ⋅ ft, determine φ and d. 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. and M x = (70 lb)(9.9694sin φ − 0.42262d ) in. = −(61 lb ⋅ ft)(12 in./ft) (1) M y = (70 lb)(−0.90631d cos φ + 3.6252sin φ ) in. (2) M z = (70 lb)(1.69048 − 9.9694 cos φ ) in. = − 43 lb ⋅ ft(12 in./ft) (3) 634.33 = 24.636° 697.86 From Equation (3): φ = cos −1 or From Equation (1): 1022.90 d = = 34.577 in. 29.583 or d = 34.6 in. φ = 24.6° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 333 PROBLEM 3.153 The tension in the cable attached to the end C of an adjustable boom ABC is 560 lb. Replace the force exerted by the cable at C with an equivalent forcecouple system (a) at A, (b) at B. SOLUTION (a) Based on ΣF : FA = T = 560 lb FA = 560 lb or 20.0° ΣM A : M A = (T sin 50°)(d A ) = (560 lb)sin 50°(18 ft) = 7721.7 lb ⋅ ft M A = 7720 lb ⋅ ft or (b) Based on ΣF : FB = T = 560 lb FB = 560 lb or 20.0° ΣM B : M B = (T sin 50°)(d B ) = (560 lb) sin 50°(10 ft) = 4289.8 lb ⋅ ft M B = 4290 lb ⋅ ft or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 334 PROBLEM 3.154 While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle. 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. We have FB = 2.9 lb − 2.65 lb = 0.25 lb, where the 2.65-lb force is part of the couple. Combining the two parallel forces, M couple = (2.65 lb)[(3.2 in. + 2.8 in.) cos 25°] = 14.4103 lb ⋅ in. and M couple = 14.4103 lb ⋅ in. A single equivalent force will be located in the negative z direction. Based on ΣM B : −14.4103 lb ⋅ in. = [(0.25 lb) cos 25°](a ) a = 63.600 in. F′ = (0.25 lb)(cos 25°i + sin 25°k ) F′ = (0.227 lb)i + (0.1057 lb)k and is applied on an extension of handle BD at a distance of 63.6 in. to the right of B. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 335 PROBLEM 3.155 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 −0.12 0.1 N ⋅ m M = 0.18 0 −122.873 −86.036 = [( −0.12)(−86.036) − (0.1)(−122.873)]i + [−(0.18)(−86.036)]j + [(0.18)(−122.873)]k = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k The equivalent force-couple system at A is F = −(122.9 N) j − (86.0 N)k M = (22.6 N ⋅ m)i + (15.49 N ⋅ m) j − (22.1 N ⋅ m)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 336 PROBLEM 3.156 A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position. If b = 1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that the distance from support A to the line of action of the equivalent force is maximum, (b) the magnitude of the equivalent force and its point of application on the beam. SOLUTION For equivalence, ΣFy : −1300 + 400 a − 400 − 600 = − R b a R = 2300 − 400 N b or ΣM A : a a 400 − a (400) − (a + b)(600) = − LR 2 b L= or Then with (1) 1000a + 600b − 200 2300 − 400 b = 1.5 m L= a2 b a b 10a + 9 − 4 2 a 3 8 23 − a 3 (2) where a, L are in m. (a) Find value of a to maximize L. 8 8 4 8 10 − a 23 − a − 10a + 9 − a 2 − dL 3 3 3 3 = 2 da 8 23 − 3 a PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 337 PROBLEM 3.156 (Continued) 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 (b) 184 80 64 2 80 32 a− a+ a + a + 24 − a 2 = 0 3 3 9 3 9 or AB = 9 m, a must be less than 9 m a = 6.91 m 6.9065 1.5 or R = 458 N Using Eq. (1), R = 2300 − 400 and using Eq. (2), 4 10(6.9065) + 9 − (6.9065)2 3 L= = 3.16 m 8 23 − (6.9065) 3 R is applied 3.16 m to the right of A. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 338 PROBLEM 3.157 A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic holds the socket wrench handle at Points A and B and applies forces at these points. Knowing that these forces are equivalent to a force-couple system at C consisting of the force C = (8 lb)i + (4 lb)k and the couple M C = (360 lb · in.)i, determine the forces applied at A and at B when Az = 2 lb. SOLUTION We have ΣF : A+B=C or Fx : Ax + Bx = 8 lb Bx = −( Ax + 8 lb) (1) ΣFy : Ay + By = 0 Ay = − By or (2) ΣFz : 2 lb + Bz = 4 lb Bz = 2 lb or (3) ΣM C : rB/C × B + rA/C × A = M C We have i 8 Bx j 0 By k i 2 + 8 2 Ax j 0 Ay k 8 2 lb ⋅ in. = (360 lb ⋅ in.)i (2 By − 8 Ay )i + (2 Bx − 16 + 8 Ax − 16) j or + (8By + 8 Ay )k = (360 lb ⋅ in.)i From i-coefficient: j-coefficient: k-coefficient: 2 By − 8 Ay = 360 lb ⋅ in. −2 Bx + 8 Ax = 32 lb ⋅ in. 8 By + 8 Ay = 0 (4) (5) (6) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 339 PROBLEM 3.157 (Continued) From Equations (2) and (4): 2 By − 8(− By ) = 360 By = 36 lb From Equations (1) and (5): Ay = 36 lb 2(− Ax − 8) + 8 Ax = 32 Ax = 1.6 lb From Equation (1): Bx = −(1.6 + 8) = −9.6 lb A = (1.600 lb)i − (36.0 lb) j + (2.00 lb)k B = −(9.60 lb)i + (36.0 lb) j + (2.00 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 340 PROBLEM 3.158 A concrete foundation mat in the shape of a regular hexagon of side 12 ft supports four column loads as shown. Determine the magnitudes of the additional loads that must be applied at B and F if the resultant of all six loads is to pass through the center of the mat. SOLUTION From the statement of the problem, it can be concluded that the six applied loads are equivalent to the resultant R at O. It then follows that ΣM O = 0 or ΣM x = 0 ΣM z = 0 For the applied loads: Then ΣM x = 0: (6 3 ft) FB + (6 3 ft)(10 kips) − (6 3 ft)(20 kips) − (6 3 ft) FF = 0 FB − FF = 10 or (1) ΣM z = 0: (12 ft)(15 kips) + (6 ft) FB − (6 ft)(10 kips) − (12 ft)(30 kips) − (6 ft)(20 kips) + (6 ft) FF = 0 FB + FF = 60 or (2) Then Eqs. (1) + (2) FB = 35.0 kips and FF = 25.0 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 341 CHAPTER 4 PROBLEM 4.1 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B. SOLUTION Free-Body Diagram: W = (350 kg)(9.81 m/s 2 ) = 3.4335 kN Wt = (1400 kg)(9.81 m/s 2 ) = 13.7340 kN (a) Rear wheels: ΣM B = 0: W (1.7 m + 2.05 m) + W (2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0 (3.4335 kN)(3.75 m) + (3.4335 kN)(2.05 m) + (13.7340 kN)(1.2 m) − 2 A(3 m) = 0 A = +6.0659 kN (b) Front wheels: A = 6.07 kN ΣFy = 0: − W − W − Wt + 2 A + 2 B = 0 −3.4335 kN − 3.4335 kN − 13.7340 kN + 2(6.0659 kN) + 2B = 0 B = +4.2346 kN B = 4.23 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 345 PROBLEM 4.2 Solve Problem 4.1, assuming that crate D is removed and that the position of crate C is unchanged. PROBLEM 4.1 Two crates, each of mass 350 kg, are placed as shown in the bed of a 1400-kg pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B. SOLUTION Free-Body Diagram: W = (350 kg)(9.81 m/s 2 ) = 3.4335 kN Wt = (1400 kg)(9.81 m/s 2 ) = 13.7340 kN (a) Rear wheels: ΣM B = 0: W (1.7 m + 2.05 m) + Wt (1.2 m) − 2 A(3 m) = 0 (3.4335 kN)(3.75 m) + (13.7340 kN)(1.2 m) − 2 A(3 m) = 0 A = + 4.8927 kN (b) Front wheels: A = 4.89 kN ΣM y = 0: − W − Wt + 2 A + 2 B = 0 −3.4335 kN − 13.7340 kN + 2(4.8927 kN) + 2B = 0 B = +3.6911 kN B = 3.69 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 346 PROBLEM 4.3 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in. SOLUTION Free-Body Diagram: ΣFx = 0: Bx = 0 ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0 (40a − 160) 12 A= (1) ΣM A = 0: − (40 lb)(6 in.) − (50 lb)(12 in.) − (30 lb)(a + 12 in.) − (10 lb)(a + 20 in.) + (12 in.) B y = 0 By = Bx = 0, B = Since (a) (b) (1400 + 40a) 12 (1400 + 40a ) 12 (2) For a = 10 in., Eq. (1): A= (40 × 10 − 160) = +20.0 lb 12 Eq. (2): B= (1400 + 40 × 10) = +150.0 lb 12 B = 150.0 lb Eq. (1): A= (40 × 7 − 160) = +10.00 lb 12 A = 10.00 lb Eq. (2): B= (1400 + 40 × 7) = +140.0 lb 12 B = 140.0 lb A = 20.0 lb For a = 7 in., PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 347 PROBLEM 4.4 For the bracket and loading of Problem 4.3, determine the smallest distance a if the bracket is not to move. PROBLEM 4.3 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if a = 10 in., (b) if a = 7 in. SOLUTION Free-Body Diagram: For no motion, reaction at A must be downward or zero; smallest distance a for no motion corresponds to A = 0. ΣM B = 0: (40 lb)(6 in.) − (30 lb)a − (10 lb)(a + 8 in.) + (12 in.) A = 0 A= (40a − 160) 12 A = 0: (40a − 160) = 0 a = 4.00 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 348 PROBLEM 4.5 A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when α = 35°, (b) the corresponding reaction at each of the two wheels. SOLUTION Free-Body Diagram: W = mg = (40 kg)(9.81 m/s 2 ) = 392.40 N a1 = (300 mm)sinα − (80 mm)cosα a2 = (430 mm)cosα − (300 mm)sinα b = (930 mm)cosα From free-body diagram of hand truck, Dimensions in mm ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0 (1) ΣFy = 0: P − 2W + 2 B = 0 (2) α = 35° For a1 = 300sin 35° − 80 cos 35° = 106.541 mm a2 = 430 cos 35° − 300sin 35° = 180.162 mm b = 930cos 35° = 761.81 mm (a) From Equation (1): P(761.81 mm) − 392.40 N(180.162 mm) + 392.40 N(106.54 mm) = 0 P = 37.921 N (b) or P = 37.9 N From Equation (2): 37.921 N − 2(392.40 N) + 2 B = 0 or B = 373 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 349 PROBLEM 4.6 Solve Problem 4.5 when α = 40°. PROBLEM 4.5 A hand truck is used to move two kegs, each of mass 40 kg. Neglecting the mass of the hand truck, determine (a) the vertical force P that should be applied to the handle to maintain equilibrium when α = 35°, (b) the corresponding reaction at each of the two wheels. SOLUTION Free-Body Diagram: W = mg = (40 kg)(9.81 m/s 2 ) W = 392.40 N a1 = (300 mm)sinα − (80 mm)cosα a2 = (430 mm)cosα − (300 mm)sinα b = (930 mm)cosα From F.B.D.: ΣM B = 0: P(b) − W ( a2 ) + W (a1 ) = 0 P = W ( a2 − a1 )/b (1) ΣFy = 0: − W − W + P + 2 B = 0 B =W − For 1 P 2 (2) α = 40°: a1 = 300sin 40° − 80 cos 40° = 131.553 mm a2 = 430 cos 40° − 300sin 40° = 136.563 mm b = 930cos 40° = 712.42 mm (a) From Equation (1): P= 392.40 N (0.136563 m − 0.131553 m) 0.71242 m P = 2.7595 N (b) From Equation (2): B = 392.40 N − P = 2.76 N 1 (2.7595 N) 2 B = 391 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 350 PROBLEM 4.7 A 3200-lb forklift truck is used to lift a 1700-lb crate. Determine the reaction at each of the two (a) front wheels A, (b) rear wheels B. SOLUTION Free-Body Diagram: (a) Front wheels: ΣM B = 0: (1700 lb)(52 in.) + (3200 lb)(12 in.) − 2 A(36 in.) = 0 A = +1761.11 lb (b) Rear wheels: A = 1761 lb ΣFy = 0: − 1700 lb − 3200 lb + 2(1761.11 lb) + 2 B = 0 B = +688.89 lb B = 689 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 351 PROBLEM 4.8 For the beam and loading shown, determine (a) the reaction at A, (b) the tension in cable BC. SOLUTION Free-Body Diagram: (a) Reaction at A: ΣFx = 0: Ax = 0 ΣMB = 0: (15 lb)(28 in.) + (20 lb)(22 in.) + (35 lb)(14 in.) + (20 lb)(6 in.) − Ay (6 in.) = 0 Ay = +245 lb (b) Tension in BC: A = 245 lb ΣM A = 0: (15 lb)(22 in.) + (20 lb)(16 in.) + (35 lb)(8 in.) − (15 lb)(6 in.) − FBC (6 in.) = 0 FBC = +140.0 lb Check: FBC = 140.0 lb ΣFy = 0: − 15 lb − 20 lb = 35 lb − 20 lb + A − FBC = 0 −105 lb + 245 lb − 140.0 = 0 0 = 0 (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 352 PROBLEM 4.9 For the beam and loading shown, determine the range of the distance a for which the reaction at B does not exceed 100 lb downward or 200 lb upward. SOLUTION Assume B is positive when directed . Sketch showing distance from D to forces. ΣM D = 0: (300 lb)(8 in. − a ) − (300 lb)(a − 2 in.) − (50 lb)(4 in.) + 16 B = 0 −600a + 2800 + 16B = 0 (2800 + 16B) 600 (1) [2800 + 16( −100)] 1200 = = 2 in. 600 600 a ≥ 2.00 in. a= For B = 100 lb = −100 lb, Eq. (1) yields: a≥ For B = 200 = +200 lb, Eq. (1) yields: a≤ Required range: [2800 + 16(200)] 6000 = = 10 in. 600 600 2.00 in. ≤ a ≤ 10.00 in. a ≤ 10.00 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 353 PROBLEM 4.10 The maximum allowable value of each of the reactions is 180 N. Neglecting the weight of the beam, determine the range of the distance d for which the beam is safe. SOLUTION ΣFx = 0: Bx = 0 B = By ΣM A = 0: (50 N) d − (100 N)(0.45 m − d ) − (150 N)(0.9 m − d ) + B(0.9 m − d ) = 0 50d − 45 + 100d − 135 + 150d + 0.9 B − Bd = 0 d= 180 N ⋅ m − (0.9 m) B 300 A − B (1) ΣM B = 0: (50 N)(0.9 m) − A(0.9 m − d ) + (100 N)(0.45 m) = 0 45 − 0.9 A + Ad + 45 = 0 (0.9 m) A − 90 N ⋅ m A (2) d≥ 180 − (0.9)180 18 = = 0.15 m 300 − 180 120 d ≥ 150.0 mm d≤ (0.9)180 − 90 72 = = 0.40 m 180 180 d= Since B ≤ 180 N, Eq. (1) yields Since A ≤ 180 N, Eq. (2) yields Range: 150.0 mm ≤ d ≤ 400 mm d ≤ 400 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 354 PROBLEM 4.11 Three loads are applied as shown to a light beam supported by cables attached at B and D. Neglecting the weight of the beam, determine the range of values of Q for which neither cable becomes slack when P = 0. SOLUTION ΣM B = 0: (3.00 kN)(0.500 m) + TD (2.25 m) − Q (3.00 m) = 0 Q = 0.500 kN + (0.750) TD (1) ΣM D = 0: (3.00 kN)(2.75 m) − TB (2.25 m) − Q(0.750 m) = 0 Q = 11.00 kN − (3.00) TB (2) For cable B not to be slack, TB ≥ 0, and from Eq. (2), Q ≤ 11.00 kN For cable D not to be slack, TD ≥ 0, and from Eq. (1), Q ≥ 0.500 kN For neither cable to be slack, 0.500 kN ≤ Q ≤ 11.00 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 355 PROBLEM 4.12 Three loads are applied as shown to a light beam supported by cables attached at B and D. Knowing that the maximum allowable tension in each cable is 4 kN and neglecting the weight of the beam, determine the range of values of Q for which the loading is safe when P = 0. SOLUTION ΣM B = 0: (3.00 kN)(0.500 m) + TD (2.25 m) − Q (3.00 m) = 0 Q = 0.500 kN + (0.750) TD (1) ΣM D = 0: (3.00 kN)(2.75 m) − TB (2.25 m) − Q(0.750 m) = 0 Q = 11.00 kN − (3.00) TB (2) For TB ≤ 4.00 kN, Eq. (2) yields Q ≥ 11.00 kN − 3.00(4.00 kN) Q ≥ −1.000 kN For TD ≤ 4.00 kN, Eq. (1) yields Q ≤ 0.500 kN + 0.750(4.00 kN) Q ≤ 3.50 kN For loading to be safe, cables must also not be slack. Combining with the conditions obtained in Problem 4.11, 0.500 kN ≤ Q ≤ 3.50 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 356 PROBLEM 4.13 For the beam of Problem 4.12, determine the range of values of Q for which the loading is safe when P = 1 kN. PROBLEM 4.12 Three loads are applied as shown to a light beam supported by cables attached at B and D. Knowing that the maximum allowable tension in each cable is 4 kN and neglecting the weight of the beam, determine the range of values of Q for which the loading is safe when P = 0. SOLUTION ΣM B = 0: (3.00 kN)(0.500 m) − (1.000 kN)(0.750 m) + TD (2.25 m) − Q(3.00 m) = 0 Q = 0.250 kN + 0.75 TD (1) ΣM D = 0: (3.00 kN)(2.75 m) + (1.000 kN)(1.50 m) − TB (2.25 m) − Q (0.750 m) = 0 Q = 13.00 kN − 3.00 TB (2) For the loading to be safe, cables must not be slack and tension must not exceed 4.00 kN. Making 0 ≤ TB ≤ 4.00 kN in Eq. (2), we have 13.00 kN − 3.00(4.00 kN) ≤ Q ≤ 13.00 kN − 3.00(0) 1.000 kN ≤ Q ≤ 13.00 kN (3) Making 0 ≤ TD ≤ 4.00 kN in Eq. (1), we have 0.250 kN + 0.750(0) ≤ Q ≤ 0.250 kN + 0.750(4.00 kN) 0.250 kN ≤ Q ≤ 3.25 kN (4) 1.000 kN ≤ Q ≤ 3.25 kN Combining Eqs. (3) and (4), PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 357 PROBLEM 4.14 For the beam of Sample Problem 4.2, determine the range of values of P for which the beam will be safe, knowing that the maximum allowable value of each of the reactions is 30 kips and that the reaction at A must be directed upward. SOLUTION ΣFx = 0: Bx = 0 B = By ΣM A = 0: − P(3 ft) + B(9 ft) − (6 kips)(11 ft) − (6 kips)(13 ft) = 0 P = 3B − 48 kips (1) ΣM B = 0: − A(9 ft) + P (6 ft) − (6 kips)(2 ft) − (6 kips)(4 ft) = 0 P = 1.5 A + 6 kips (2) Since B ≤ 30 kips, Eq. (1) yields P ≤ (3)(30 kips) − 48 kips P ≤ 42.0 kips Since 0 ≤ A ≤ 30 kips, Eq. (2) yields 0 + 6 kips ≤ P ≤ (1.5)(30 kips)1.6 kips 6.00 kips ≤ P ≤ 51.0 kips Range of values of P for which beam will be safe: 6.00 kips ≤ P ≤ 42.0 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 358 PROBLEM 4.15 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION At B: Ty Tx = 0.18 m 0.24 m 3 Ty = Tx 4 (a) (1) ΣM C = 0: Tx (0.18 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0 Tx = +1600 N From Eq. (1): Ty = 3 (1600 N) = 1200 N 4 T = Tx2 + Ty2 = 16002 + 12002 = 2000 N (b) T = 2.00 kN ΣFx = 0: Cx − Tx = 0 Cx − 1600 N = 0 C x = +1600 N C x = 1600 N ΣFy = 0: C y − Ty − 240 N − 240 N = 0 C y − 1200 N − 480 N = 0 C y = +1680 N C y = 1680 N α = 46.4° C = 2320 N C = 2.32 kN 46.4° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 359 PROBLEM 4.16 Solve Problem 4.15, assuming that a = 0.32 m. PROBLEM 4.15 The bracket BCD is hinged at C and attached to a control cable at B. For the loading shown, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION At B: Ty Tx = 0.32 m 0.24 m 4 Ty = Tx 3 ΣM C = 0: Tx (0.32 m) − (240 N)(0.4 m) − (240 N)(0.8 m) = 0 Tx = 900 N From Eq. (1): Ty = 4 (900 N) = 1200 N 3 T = Tx2 + Ty2 = 9002 + 12002 = 1500 N T = 1.500 kN ΣFx = 0: C x − Tx = 0 C x − 900 N = 0 C x = +900 N C x = 900 N ΣFy = 0: C y − Ty − 240 N − 240 N = 0 C y − 1200 N − 480 N = 0 C y = +1680 N C y = 1680 N α = 61.8° C = 1906 N C = 1.906 kN 61.8° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 360 PROBLEM 4.17 The lever BCD is hinged at C and attached to a control rod at B. If P = 100 lb, determine (a) the tension in rod AB, (b) the reaction at C. SOLUTION Free-Body Diagram: (a) ΣM C = 0: T (5 in.) − (100 lb)(7.5 in.) = 0 T = 150.0 lb (b) 3 ΣFx = 0: C x + 100 lb + (150.0 lb) = 0 5 C x = −190 lb C x = 190 lb 4 ΣFy = 0: C y + (150.0 lb) = 0 5 C y = −120 lb C y = 120 lb α = 32.3° C = 225 lb C = 225 lb 32.3° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 361 PROBLEM 4.18 The lever BCD is hinged at C and attached to a control rod at B. Determine the maximum force P that can be safely applied at D if the maximum allowable value of the reaction at C is 250 lb. SOLUTION Free-Body Diagram: ΣM C = 0: T (5 in.) − P (7.5 in.) = 0 T = 1.5P 3 ΣFx = 0: P + C x + (1.5P) = 0 5 C x = −1.9 P ΣFy = 0: C y + C x = 1.9 P 4 (1.5P) = 0 5 C y = −1.2 P C y = 1.2 P C = C x2 + C y2 = (1.9 P) 2 + (1.2 P) 2 C = 2.2472 P For C = 250 lb, 250 lb = 2.2472P P = 111.2 lb P = 111.2 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 362 PROBLEM 4.19 Two links AB and DE are connected by a bell crank as shown. Knowing that the tension in link AB is 720 N, determine (a) the tension in link DE, (b) the reaction at C. SOLUTION Free-Body Diagram: ΣM C = 0: FAB (100 mm) − FDE (120 mm) = 0 FDE = (a) For (1) FAB = 720 N FDE = (b) 5 FAB 6 5 (720 N) 6 FDE = 600 N 3 ΣFx = 0: − (720 N) + C x = 0 5 C x = +432 N 4 ΣFy = 0: − (720 N) + C y − 600 N = 0 5 C y = +1176 N C = 1252.84 N α = 69.829° C = 1253 N 69.8° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 363 PROBLEM 4.20 Two links AB and DE are connected by a bell crank as shown. Determine the maximum force that may be safely exerted by link AB on the bell crank if the maximum allowable value for the reaction at C is 1600 N. SOLUTION See solution to Problem 4.15 for F.B.D. and derivation of Eq. (1). FDE = 5 FAB 6 (1) 3 ΣFx = 0: − FAB + C x = 0 5 ΣFy = 0: − Cx = 3 FAB 5 4 FAB + C y − FDE = 0 5 4 5 − FAB + C y − FAB = 0 5 6 49 Cy = FAB 30 C = C x2 + C y2 1 (49) 2 + (18) 2 FAB 30 C = 1.74005FAB = For C = 1600 N, 1600 N = 1.74005FAB FAB = 920 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 364 PROBLEM 4.21 Determine the reactions at A and C when (a) α = 0, (b) α = 30°. SOLUTION (a) α =0 From F.B.D. of member ABC: ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − A(0.8 m) = 0 A = 225 N or A = 225 N ΣFy = 0: C y + 225 N = 0 C y = −225 N or C y = 225 N ΣFx = 0: 300 N + 300 N + C x = 0 C x = −600 N or C x = 600 N Then C = C x2 + C y2 = (600) 2 + (225) 2 = 640.80 N and θ = tan −1 Cy −1 −225 = tan = 20.556° −600 Cx or (b) C = 641 N 20.6° α = 30° From F.B.D. of member ABC: ΣM C = 0: (300 N)(0.2 m) + (300 N)(0.4 m) − ( A cos 30°)(0.8 m) + ( A sin 30°)(20 in.) = 0 A = 365.24 N or A = 365 N 60.0° ΣFx = 0: 300 N + 300 N + (365.24 N) sin 30° + C x = 0 C x = −782.62 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 365 PROBLEM 4.21 (Continued) ΣFy = 0: C y + (365.24 N) cos 30° = 0 C y = −316.31 N or C y = 316 N Then C = C x2 + C y2 = (782.62) 2 + (316.31) 2 = 884.12 N and θ = tan −1 Cy −1 −316.31 = tan = 22.007° −782.62 Cx C = 884 N or 22.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 366 PROBLEM 4.22 Determine the reactions at A and B when (a) α = 0, (b) α = 90°, (c) α = 30°. SOLUTION (a) α =0 ΣM A = 0: B(20 in.) − 75 lb(10 in.) = 0 B = 37.5 lb ΣFx = 0: Ax = 0 + ΣFy = 0: Ay − 75 lb + 37.5 lb = 0 Ay = 37.5 lb A = B = 37.5 lb (b) α = 90° ΣM A = 0: B(12 in.) − 75 lb(10 in.) = 0 B = 62.5 lb ΣFx = 0: Ax − B = 0 Ax = 62.5 lb ΣFy = 0: Ay − 75 lb = 0 Ay = 75 lb A = Ax2 + Ay2 = (62.5 lb) 2 + (75 lb) 2 = 97.6 lb 75 62.5 θ = 50.2° tan θ = A = 97.6 lb 50.2°; B = 62.51 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 367 PROBLEM 4.22 (Continued) (c) α = 30° ΣM A = 0: ( B cos 30°)(20 in.) + ( B sin 30°)(12 in.) − (75 lb)(10 in.) = 0 B = 32.161 lb ΣFx = 0: Ax − (32.161) sin 30° = 0 Ax = 16.0805 lb ΣFy = 0: Ay + (32.161) cos 30° − 75 = 0 Ay = 47.148 lb A = Ax2 + Ay2 = (16.0805) 2 + (47.148) 2 = 49.8 lb 47.148 16.0805 θ = 71.2° tan θ = A = 49.8 lb 71.2°; B = 32.2 lb 60.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 368 PROBLEM 4.23 Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm. SOLUTION Free-Body Diagram: ΣM A = 0: ( B cos 60°)(0.5 m) − ( B sin 60°)h − (150 N)(0.25 m) = 0 37.5 B= 0.25 − 0.866h (a) (1) When h = 0, B= From Eq. (1): 37.5 = 150 N 0.25 B = 150.0 N 30.0° ΣFy = 0: Ax − B sin 60° = 0 Ax = (150)sin 60° = 129.9 N A x = 129.9 N ΣFy = 0: Ay − 150 + B cos 60° = 0 Ay = 150 − (150) cos 60° = 75 N A y = 75 N α = 30° A = 150.0 N (b) A = 150.0 N 30.0° When h = 200 mm = 0.2 m, From Eq. (1): B= 37.5 = 488.3 N 0.25 − 0.866(0.2) B = 488 N 30.0° ΣFx = 0: Ax − B sin 60° = 0 Ax = (488.3) sin 60° = 422.88 N A x = 422.88 N ΣFy = 0: Ay − 150 + B cos 60° = 0 Ay = 150 − (488.3) cos 60° = −94.15 N A y = 94.15 N α = 12.55° A = 433.2 N A = 433 N 12.55° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 369 PROBLEM 4.24 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 75-lb vertical force at B, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION Free-Body Diagram: Geometry: x AC = (10 in.) cos 20° = 9.3969 in. y AC = (10 in.)sin 20° = 3.4202 in. yDA = 12 in. − 3.4202 in. = 8.5798 in. yDA −1 8.5798 = tan = 42.397° 9.3969 x AC α = tan −1 β = 90° − 20° − 42.397° = 27.603° Equilibrium for lever: ΣM C = 0: TAD cos 27.603°(10 in.) − (75 lb)[(15 in.)cos 20°] = 0 (a) TAD = 119.293 lb TAD = 119.3 lb ΣFx = 0: C x + (119.293 lb) cos 42.397° = 0 (b) C x = −88.097 lb ΣFy = 0: C y − 75 lb − (119.293 lb) sin 42.397° = 0 C y = 155.435 Thus, C = C x2 + C y2 = (−88.097) 2 + (155.435) 2 = 178.665 lb and θ = tan −1 Cy Cx = tan −1 155.435 = 60.456° 88.097 C = 178.7 lb 60.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 370 PROBLEM 4.25 For each of the plates and loadings shown, determine the reactions at A and B. SOLUTION (a) Free-Body Diagram: ΣM A = 0: B(20 in.) − (50 lb)(4 in.) − (40 lb)(10 in.) = 0 B = +30 lb B = 30.0 lb ΣFx = 0: Ax + 40 lb = 0 Ax = −40 lb A x = 40.0 lb ΣFy = 0: Ay + B − 50 lb = 0 Ay + 30 lb − 50 lb = 0 Ay = +20 lb A y = 20.0 lb α = 26.56° A = 44.72 lb A = 44.7 lb 26.6° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 371 PROBLEM 4.25 (Continued) (b) Free-Body Diagram: ΣM A = 0: ( B cos 30°)(20 in.) − (40 lb)(10 in.) − (50 lb)(4 in.) = 0 B = 34.64 lb B = 34.6 lb 60.0° ΣFx = 0: Ax − B sin 30° + 40 lb Ax − (34.64 lb) sin 30° + 40 lb = 0 Ax = −22.68 lb A x = 22.68 lb ΣFy = 0: Ay + B cos 30° − 50 lb = 0 Ay + (34.64 lb) cos 30° − 50 lb = 0 Ay = +20 lb A y = 20.0 lb α = 41.4° A = 30.2 lb A = 30.24 lb 41.4° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 372 PROBLEM 4.26 For each of the plates and loadings shown, determine the reactions at A and B. SOLUTION (a) Free-Body Diagram: ΣM B = 0: A(20 in.) + (50 lb)(16 in.) − (40 lb)(10 in.) = 0 A = +20 lb A = 20.0 lb ΣFx = 0: 40 lb + Bx = 0 Bx = −40 lb B x = 40 lb ΣFy = 0: A + By − 50 lb = 0 20 lb + By − 50 lb = 0 By = +30 lb α = 36.87° B = 50 lb B y = 30 lb B = 50.0 lb 36.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 373 PROBLEM 4.26 (Continued) (b) ΣM A = 0: − ( A cos 30°)(20 in.) − (40 lb)(10 in.) + (50 lb)(16 in.) = 0 A = 23.09 lb A = 23.1 lb 60.0° ΣFx = 0: A sin 30° + 40 lb + Bx = 0 (23.09 lb) sin 30° + 40 lb + 8 x = 0 Bx = −51.55 lb B x = 51.55 lb ΣFy = 0: A cos 30° + By − 50 lb = 0 (23.09 lb) cos 30° + By − 50 lb = 0 By = +30 lb B y = 30 lb α = 30.2° B = 59.64 lb B = 59.6 lb 30.2° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 374 PROBLEM 4.27 A rod AB hinged at A and attached at B to cable BD supports the loads shown. Knowing that d = 200 mm, determine (a) the tension in cable BD, (b) the reaction at A. SOLUTION Free-Body Diagram: (a) Move T along BD until it acts at Point D. ΣM A = 0: (T sin 45°)(0.2 m) + (90 N)(0.1 m) + (90 N)(0.2 m) = 0 T = 190.919 N (b) T = 190.9 N ΣFx = 0: Ax − (190.919 N) cos 45° = 0 Ax = +135.0 N A x = 135.0 N ΣFy = 0: Ay − 90 N − 90 N + (190.919 N) sin 45° = 0 Ay = +45.0 N A y = 45.0 N A = 142.3 N 18.43° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 375 PROBLEM 4.28 A rod AB, hinged at A and attached at B to cable BD, supports the loads shown. Knowing that d = 150 mm, determine (a) the tension in cable BD, (b) the reaction at A. SOLUTION Free-Body Diagram: tan α = (a) 10 ; α = 33.690° 15 Move T along BD until it acts at Point D. ΣM A = 0: (T sin 33.690°)(0.15 m) − (90 N)(0.1 m) − (90 N)(0.2 m) = 0 T = 324.50 N (b) T = 324 N ΣFx = 0: Ax − (324.50 N) cos 33.690° = 0 Ax = +270 N A x = 270 N ΣFy = 0: Ay − 90 N − 90 N + (324.50 N) sin 33.690° = 0 Ay = 0 A = 270 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 376 PROBLEM 4.29 A force P of magnitude 90 lb is applied to member ACDE, which is supported by a frictionless pin at D and by the cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at D. SOLUTION Free-Body Diagram: (a) ΣM D = 0: (90 lb)(9 in.) − 5 12 T (9 in.) − T (7 in.) + T (3 in.) = 0 13 13 T = 117 lb (b) ΣFx = 0: Dx − 117 lb − T = 117.0 lb 5 (117 lb) + 90 = 0 13 Dx = +72 lb ΣFy = 0: D y + 12 (117 lb) = 0 13 Dy = −108 lb D = 129.8 lb 56.3° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 377 PROBLEM 4.30 Solve Problem 4.29 for a = 6 in. PROBLEM 4.29 A force P of magnitude 90 lb is applied to member ACDE, which is supported by a frictionless pin at D and by the cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at D. SOLUTION Free-Body Diagram: (a) ΣM D = 0: (90 lb)(6 in.) − 5 12 T (6 in.) − T (7 in.) + T (6 in.) = 0 13 13 T = 195 lb (b) T = 195.0 lb ΣFx = 0: Dx − 195 lb − 5 (195 lb) + 90 = 0 13 Dx = +180 lb ΣFy = 0: D y + 12 (195 lb) = 0 13 Dy = −180 lb D = 255 lb 45.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 378 PROBLEM 4.31 Neglecting friction, determine the tension in cable ABD and the reaction at support C. SOLUTION Free-Body Diagram: ΣM C = 0: T (0.25 m) − T (0.1 m) − (120 N)(0.1 m) = 0 ΣFx = 0: C x − 80 N = 0 C x = +80 N ΣFy = 0: C y − 120 N + 80 N = 0 C y = +40 N T = 80.0 N C x = 80.0 N C y = 40.0 N C = 89.4 N 26.6° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 379 PROBLEM 4.32 Neglecting friction and the radius of the pulley, determine (a) the tension in cable ADB, (b) the reaction at C. SOLUTION Free-Body Diagram: Dimensions in mm Geometry: Distance: AD = (0.36) 2 + (0.150) 2 = 0.39 m Distance: BD = (0.2)2 + (0.15)2 = 0.25 m Equilibrium for beam: ΣM C = 0: (a) 0.15 0.15 (120 N)(0.28 m) − T (0.36 m) − T (0.2 m) = 0 0.39 0.25 T = 130.000 N or T = 130.0 N 0.36 0.2 ΣFx = 0: C x + (130.000 N) + 0.25 (130.000 N) = 0 0.39 (b) C x = − 224.00 N 0.15 0.15 (130.00 N) + ΣFy = 0: C y + (130.00 N) − 120 N = 0 0.39 0.25 C y = − 8.0000 N Thus, C = C x2 + C y2 = (−224) 2 + (− 8) 2 = 224.14 N and θ = tan −1 Cy Cx = tan −1 8 = 2.0454° 224 C = 224 N 2.05° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 380 PROBLEM 4.33 Rod ABC is bent in the shape of an arc of circle of radius R. Knowing the θ = 30°, determine the reaction (a) at B, (b) at C. SOLUTION Free-Body Diagram: ΣM D = 0: C x ( R) − P( R) = 0 Cx = + P ΣFx = 0: C x − B sin θ = 0 P − B sin θ = 0 B = P/sin θ B= P sin θ θ ΣFy = 0: C y + B cos θ − P = 0 C y + ( P/sin θ ) cos θ − P = 0 1 C y = P 1 − tan θ For θ = 30°, (a) (b) B = P/sin 30° = 2 P Cx = + P B = 2P 60.0° Cx = P C y = P(1 − 1/tan 30°) = − 0.732/P C y = 0.7321P C = 1.239P 36.2° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 381 PROBLEM 4.34 Rod ABC is bent in the shape of an arc of circle of radius R. Knowing that θ = 60°, determine the reaction (a) at B, (b) at C. SOLUTION See the solution to Problem 4.33 for the free-body diagram and analysis leading to the following expressions: Cx = + P 1 C y = P 1 − tan θ P B= sin θ For θ = 60°, (a) (b) B = P/sin 60° = 1.1547 P Cx = + P B = 1.155P 30.0° Cx = P C y = P(1 − 1/tan 60°) = + 0.4226 P C y = 0.4226 P C = 1.086P 22.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 382 PROBLEM 4.35 A movable bracket is held at rest by a cable attached at C and by frictionless rollers at A and B. For the loading shown, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION Free-Body Diagram: (a) ΣFy = 0: T − 600 N = 0 T = 600 N (b) ΣFx = 0: B − A = 0 ∴ B=A Note that the forces shown form two couples. ΣM = 0: (600 N)(600 mm) − A(90 mm) = 0 A = 4000 N ∴ B = 4000 N A = 4.00 kN ; B = 4.00 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 383 PROBLEM 4.36 A light bar AB supports a 15-kg block at its midpoint C. Rollers at A and B rest against frictionless surfaces, and a horizontal cable AD is attached at A. Determine (a) the tension in cable AD, (b) the reactions at A and B. SOLUTION Free-Body Diagram: W = (15 kg)(9.81 m/s 2 ) = 147.150 N (a) ΣFx = 0: TAD − 105.107 N = 0 TAD = 105.1 N (b) ΣFy = 0: A − W = 0 A − 147.150 N = 0 A = 147.2 N ΣM A = 0: B(350 mm) − (147.150 N) (250 mm) = 0 B = 105.107 N B = 105.1 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 384 PROBLEM 4.37 A light bar AD is suspended from a cable BE and supports a 50-lb block at C. The ends A and D of the bar are in contact with frictionless vertical walls. Determine the tension in cable BE and the reactions at A and D. SOLUTION Free-Body Diagram: ΣFx = 0: A= D ΣFy = 0: TBE = 50.0 lb We note that the forces shown form two couples. ΣM = 0: A(8 in.) − (50 lb)(3 in.) = 0 A = 18.75 lb A = 18.75 lb D = 18.75 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 385 PROBLEM 4.38 A light rod AD is supported by frictionless pegs at B and C and rests against a frictionless wall at A. A vertical 120-lb force is applied at D. Determine the reactions at A, B, and C. SOLUTION Free-Body Diagram: ΣFx = 0: A cos 30° − (120 lb) cos 60° = 0 A = 69.28 lb A = 69.3 lb ΣM B = 0: C (8 in.) − (120 lb)(16 in.) cos 30° + (69.28 lb)(8 in.)sin 30° = 0 C = 173.2 lb C = 173.2 lb 60.0° B = 34.6 lb 60.0° ΣM C = 0: B(8 in.) − (120 lb)(8 in.) cos 30° + (69.28 lb)(16 in.) sin 30° = 0 B = 34.6 lb Check: ΣFy = 0: 173.2 − 34.6 − (69.28)sin 30° − (120)sin 60° = 0 0 = 0 (check) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 386 PROBLEM 4.39 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α = 0), determine the tension in the cord and the reactions at A and C. SOLUTION Free-Body Diagram: ΣFy = 0: − T cos 30° + (80 N) cos 30° = 0 T = 80 N T = 80.0 N ΣM C = 0: ( A sin 30°)(0.4 m) − (80 N)(0.2 m) − (80 N)(0.2 m) = 0 A = + 160 N A = 160.0 N 30.0° C = 160.0 N 30.0° ΣM A = 0: (80 N)(0.2 m) − (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0 C = + 160 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 387 PROBLEM 4.40 Solve Problem 4.39 if the cord BE is parallel to the rods (α = 30°). PROBLEM 4.39 Bar AD is attached at A and C to collars that can move freely on the rods shown. If the cord BE is vertical (α = 0), determine the tension in the cord and the reactions at A and C. SOLUTION Free-Body Diagram: ΣFy = 0: − T + (80 N) cos 30° = 0 T = 69.282 N T = 69.3 N ΣM C = 0: − (69.282 N) cos 30°(0.2 m) − (80 N)(0.2 m) + ( A sin 30°)(0.4 m) = 0 A = + 140.000 N A = 140.0 N 30.0° C = 180.0 N 30.0° ΣM A = 0: + (69.282 N) cos 30°(0.2 m) − (80 N)(0.6 m) + (C sin 30°)(0.4 m) = 0 C = + 180.000 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 388 PROBLEM 4.41 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C, D, and E when θ = 30°. SOLUTION Free-Body Diagram: ΣFy = 0: E cos 30° − 20 − 40 = 0 E= 60 lb = 69.282 lb cos 30° E = 69.3 lb 60.0° ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.) + E sin 30°(3 in.) = 0 −80 − 3C + 69.282(0.5)(3) = 0 C = 7.9743 lb C = 7.97 lb D = 42.6 lb ΣFx = 0: E sin 30° + C − D = 0 (69.282 lb)(0.5) + 7.9743 lb − D = 0 D = 42.615 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 389 PROBLEM 4.42 The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine (a) the smallest value of θ for which the equilibrium of the bracket is maintained, (b) the corresponding reactions at C, D, and E. SOLUTION Free-Body Diagram: ΣFy = 0: E cos θ − 20 − 40 = 0 E= 60 cos θ (1) ΣM D = 0: (20 lb)(4 in.) − (40 lb)(4 in.) − C (3 in.) 60 + sin θ 3 in. = 0 cos θ 1 C = (180 tan θ − 80) 3 (a) For C = 0, 180 tan θ = 80 tan θ = From Eq. (1): E= 4 θ = 23.962° 9 θ = 24.0° 60 = 65.659 cos 23.962° ΣFx = 0: −D + C + E sin θ = 0 D = (65.659) sin 23.962 = 26.666 lb (b) C = 0 D = 26.7 lb E = 65.7 lb 66.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 390 PROBLEM 4.43 Beam AD carries the two 40-lb loads shown. The beam is held by a fixed support at D and by the cable BE that is attached to the counterweight W. Determine the reaction at D when (a) W = 100 lb, (b) W = 90 lb. SOLUTION W = 100 lb (a) From F.B.D. of beam AD: ΣFx = 0: Dx = 0 ΣFy = 0: D y − 40 lb − 40 lb + 100 lb = 0 Dy = −20.0 lb or D = 20.0 lb ΣM D = 0: M D − (100 lb)(5 ft) + (40 lb)(8 ft) + (40 lb)(4 ft) = 0 M D = 20.0 lb ⋅ ft or M D = 20.0 lb ⋅ ft W = 90 lb (b) From F.B.D. of beam AD: ΣFx = 0: Dx = 0 ΣFy = 0: D y + 90 lb − 40 lb − 40 lb = 0 Dy = −10.00 lb or D = 10.00 lb ΣM D = 0: M D − (90 lb)(5 ft) + (40 lb)(8 ft) + (40 lb)(4 ft) = 0 M D = −30.0 lb ⋅ ft or M D = 30.0 lb ⋅ ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 391 PROBLEM 4.44 For the beam and loading shown, determine the range of values of W for which the magnitude of the couple at D does not exceed 40 lb ⋅ ft. SOLUTION For Wmin , From F.B.D. of beam AD: M D = − 40 lb ⋅ ft ΣM D = 0: (40 lb)(8 ft) − Wmin (5 ft) + (40 lb)(4 ft) − 40 lb ⋅ ft = 0 Wmin = 88.0 lb For Wmax , From F.B.D. of beam AD: M D = 40 lb ⋅ ft ΣM D = 0: (40 lb)(8 ft) − Wmax (5 ft) + (40 lb)(4 ft) + 40 lb ⋅ ft = 0 Wmax = 104.0 lb or 88.0 lb ≤ W ≤ 104.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 392 PROBLEM 4.45 An 8-kg mass can be supported in the three different ways shown. Knowing that the pulleys have a 100-mm radius, determine the reaction at A in each case. SOLUTION W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N (a) ΣFx = 0: Ax = 0 ΣFy = 0: Ay − W = 0 A y = 78.480 N ΣM A = 0: M A − W (1.6 m) = 0 M A = + (78.480 N)(1.6 m) M A = 125.568 N ⋅ m A = 78.5 N (b) M A = 125.6 N ⋅ m ΣFx = 0: Ax − W = 0 A x = 78.480 ΣFy = 0: Ay − W = 0 A y = 78.480 A = (78.480 N) 2 = 110.987 N 45° ΣM A = 0: M A − W (1.6 m) = 0 M A = + (78.480 N)(1.6 m) A = 111.0 N (c) M A = 125.568 N ⋅ m M A = 125.6 N ⋅ m 45° ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 2W = 0 Ay = 2W = 2(78.480 N) = 156.960 N ΣM A = 0: M A − 2W (1.6 m) = 0 M A = + 2(78.480 N)(1.6 m) A = 157.0 N M A = 251.14 N ⋅ m M A = 251 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 393 PROBLEM 4.46 A tension of 20 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C. SOLUTION Free-Body Diagram: ΣFx = 0: C x + (20 N) = 0 C x = −20 N ΣFy = 0: C y − (20 N) = 0 C y = +20 N C = 28.3 N 45.0° ΣM C = 0: M C + (20 N)(0.160 m) + (20 N) (0.055 m) = 0 M C = −4.30 N ⋅ m M C = 4.30 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 394 PROBLEM 4.47 Solve Problem 4.46, assuming that 15-mm-radius pulleys are used. PROBLEM 4.46 A tension of 20 N is maintained in a tape as it passes through the support system shown. Knowing that the radius of each pulley is 10 mm, determine the reaction at C. SOLUTION Free-Body Diagram: ΣFx = 0: C x + (20 N) = 0 C x = −20 N ΣFy = 0: C y − (20 N) = 0 C y = +20 N C = 28.3 N 45.0° ΣM C = 0: M C + (20 N) (0.165 m) + (20 N) (0.060 m) = 0 M C = −4.50 N ⋅ m M C = 4.50 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 395 PROBLEM 4.48 The rig shown consists of a 1200-lb horizontal member ABC and a vertical member DBE welded together at B. The rig is being used to raise a 3600-lb crate at a distance x = 12 ft from the vertical member DBE. If the tension in the cable is 4 kips, determine the reaction at E, assuming that the cable is (a) anchored at F as shown in the figure, (b) attached to the vertical member at a point located 1 ft above E. SOLUTION Free-Body Diagram: M E = 0: M E + (3600 lb) x + (1200 lb) (6.5 ft) − T (3.75 ft) = 0 M E = 3.75T − 3600 x − 7800 (a) (1) For x = 12 ft and T = 4000 lbs, M E = 3.75(4000) − 3600(12) − 7800 = 36, 000 lb ⋅ ft ΣFx = 0 ∴ Ex = 0 ΣFy = 0: E y − 3600 lb − 1200 lb − 4000 = 0 E y = 8800 lb E = 8.80 kips ; M E = 36.0 kip ⋅ ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 396 PROBLEM 4.48 (Continued) ΣM E = 0: M E + (3600 lb)(12 ft) + (1200 lb)(6.5 ft) = 0 (b) M E = −51, 000 lb ⋅ ft ΣFx = 0 ∴ Ex = 0 ΣFy = 0: E y − 3600 lb − 1200 lb = 0 E y = 4800 lb E = 4.80 kips ; M E = 51.0 kip ⋅ ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 397 PROBLEM 4.49 For the rig and crate of Prob. 4.48, and assuming that cable is anchored at F as shown, determine (a) the required tension in cable ADCF if the maximum value of the couple at E as x varies from 1.5 to 17.5 ft is to be as small as possible, (b) the corresponding maximum value of the couple. SOLUTION Free-Body Diagram: M E = 0: M E + (3600 lb) x + (1200 lb)(6.5 ft) − T (3.75 ft) = 0 M E = 3.75T − 3600 x − 7800 (1) For x = 1.5 ft, Eq. (1) becomes ( M E )1 = 3.75T − 3600(1.5) − 7800 (2) For x = 17.5 ft, Eq. (1) becomes ( M E ) 2 = 3.75T − 3600(17.5) − 7800 (a) For smallest max value of |M E |, we set ( M E )1 = − ( M E )2 3.75T − 13, 200 = −3.75T + 70,800 (b) T = 11.20 kips From Equation (2), then M E = 3.75(11.20) − 13.20 |M E | = 28.8 kip ⋅ ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 398 PROBLEM 4.50 A 6-m telephone pole weighing 1600 N is used to support the ends of two wires. The wires form the angles shown with the horizontal axis and the tensions in the wires are, respectively, T1 = 600 N and T2 = 375 N. Determine the reaction at the fixed end A. SOLUTION Free-Body Diagram: ΣFx = 0: Ax + (375 N) cos 20° − (600 N) cos10° = 0 Ax = +238.50 N ΣFy = 0: Ay − 1600 N − (600 N)sin10° − (375 N) sin 20° = 0 Ay = +1832.45 N A = 238.502 + 1832.452 1832.45 θ = tan −1 238.50 A = 1848 N 82.6° ΣM A = 0: M A + (600 N) cos10°(6 m) − (375 N) cos 20°(6 m) = 0 M A = −1431.00 N ⋅ m M A = 1431 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 399 PROBLEM 4.51 A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of θ corresponding to equilibrium if P = 2W. SOLUTION Free-Body Diagram: (a) Triangle ABC is isosceles. We have θ θ CD = ( BC ) cos = l cos 2 2 θ ΣM C = 0: P(l cos θ ) − W l cos = 0 2 Setting cos θ = 2 cos 2 θ 2 − 1: θ θ Pl 2 cos 2 − 1 − Wl cos = 0 2 2 cos 2 θ θ 1 W − cos − = 0 2 2P 2 2 cos θ 2 = 1 W W2 ± 8 + 4 P P2 1 W θ = 2cos −1 4 P ± W2 + 8 2 P PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 400 PROBLEM 4.51 (Continued) (b) For P = 2W , cos θ cos θ 2 2 θ 2 = 1 11 1 + 8 = 1 ± 33 ± 8 42 4 ( = 0.84307 and cos = 32.534° θ = 65.1° θ 2 θ 2 ) = −0.59307 = 126.375° θ = 252.75° (discard) θ = 65.1° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 401 PROBLEM 4.52 A vertical load P is applied at end B of rod BC. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position in terms of P, l, and the counterweight W. (b) Determine the value of θ corresponding to equilibrium if P = 2W. SOLUTION (a) Triangle ABC is isosceles. We have CD = ( BC ) cos θ 2 = l cos θ 2 θ ΣM C = 0: W l cos − P(l sin θ ) = 0 2 Setting sin θ = 2sin θ θ θ θ W − 2 P sin (b) For P = 2W , sin θ 2 θ 2 or θ cos : Wl cos − 2 Pl sin cos = 0 2 2 2 2 2 θ 2 = θ 2 =0 θ = 2sin −1 W 2P W W = = 0.25 2 P 4W θ = 29.0° = 14.5° = 165.5° θ = 331° (discard) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 402 PROBLEM 4.53 A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by an elastic cord that passes over a pulley at C. (a) Express the tension in the cord in terms of W and θ. (b) Determine the value of θ for which the tension in the cord is equal to 3W. SOLUTION (a) From F.B.D. of rod AB: 1 ΣM C = 0: T (l sin θ ) + W cos θ − T (l cos θ ) = 0 2 T= W cos θ 2(cosθ − sin θ ) Dividing both numerator and denominator by cos θ, T= (b) For T = 3W , or 3W = W 2 1 1 − tan θ or T = ( W2 ) (1 − tan θ ) ( W2 ) (1 − tan θ ) 1 1 − tan θ = 6 5 θ = tan −1 = 39.806° 6 or θ = 39.8° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 403 PROBLEM 4.54 Rod AB is acted upon by a couple M and two forces, each of magnitude P. (a) Derive an equation in θ, P, M, and l that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ corresponding to equilibrium when M = 150 N · m, P = 200 N, and l = 600 mm. SOLUTION Free-Body Diagram: (a) From free-body diagram of rod AB: ΣM C = 0: P(l cos θ ) + P(l sin θ ) − M = 0 or sinθ + cosθ = (b) M Pl For M = 150 lb ⋅ in., P = 20 lb, and l = 6 in., sin θ + cos θ = 150 lb ⋅ in. 5 = = 1.25 (20 lb)(6 in.) 4 sin 2 θ + cos 2 θ = 1 Using identity sin θ + (1 − sin 2 θ )1/2 = 1.25 (1 − sin 2 θ )1/2 = 1.25 − sin θ 1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ 2sin 2 θ − 2.5sin θ + 0.5625 = 0 Using quadratic formula sin θ = = or −( −2.5) ± (625) − 4(2)(0.5625) 2(2) 2.5 ± 1.75 4 sin θ = 0.95572 and sin θ = 0.29428 θ = 72.886° and θ = 17.1144° or θ = 17.11° and θ = 72.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 404 PROBLEM 4.55 Solve Sample Problem 4.5, assuming that the spring is unstretched when θ = 90°. SOLUTION First note: T = tension in spring = ks where s = deformation of spring = rβ F = kr β From F.B.D. of assembly: or ΣM 0 = 0: W (l cos β ) − F (r ) = 0 Wl cos β − kr 2 β = 0 cos β = For kr 2 β Wl k = 250 lb/in. r = 3 in. l = 8 in. W = 400 lb cos β = or (250 lb/in.)(3 in.)2 β (400 lb)(8 in.) cos β = 0.703125β Solving numerically, β = 0.89245 rad or β = 51.134° Then θ = 90° + 51.134° = 141.134° or θ = 141.1° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 405 PROBLEM 4.56 A slender rod AB, of weight W, is attached to blocks A and B that move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Neglecting the weight of the blocks, derive an equation in W, k, l, and θ that must be satisfied when the rod is in equilibrium. (b) Determine the value of θ when W = 75 lb, l = 30 in., and k = 3 lb/in. SOLUTION Free-Body Diagram: Spring force: Fs = ks = k (l − l cos θ ) = kl (1 − cos θ ) l ΣM D = 0: Fs (l sin θ ) − W cos θ = 0 2 (a) kl (1 − cos θ )l sin θ − kl (1 − cos θ ) tan θ − (b) For given values of W l cos θ = 0 2 W =0 2 or (1 − cos θ ) tan θ = W 2kl W = 75 lb l = 30 in. k = 3 lb/in. (1 − cos θ ) tan θ = tan θ − sin θ 75 lb = 2(3 lb/in.)(30 in.) = 0.41667 Solving numerically, θ = 49.710° or θ = 49.7° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 406 PROBLEM 4.57 A vertical load P is applied at end B of rod BC. The constant of the spring is k, and the spring is unstretched when θ = 60°. (a) Neglecting the weight of the rod, express the angle θ corresponding to the equilibrium position terms of P, k, and l. (b) Determine the value of θ 1 corresponding to equilibrium if P = 4 kl. SOLUTION Free-Body Diagram: (a) Triangle ABC is isosceles. We have θ θ AB = 2( AD ) = 2l sin ; CD = l cos 2 2 Elongation of spring: x = ( AB)θ − ( AB )θ = 60° θ = 2l sin − 2l sin 30° 2 θ 1 T = k x = 2kl sin − 2 2 θ ΣM C = 0: T l cos − P(l sin θ ) = 0 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 407 PROBLEM 4.57 (Continued) θ θ θ θ 1 2kl sin − l cos − Pl 2sin cos = 0 2 2 2 2 2 cos θ 2 =0 2(kl − P ) sin or θ = 180° (trivial) sin θ 2 θ 2 − kl = 0 = 1 2 kl kl − P 1 2 θ = 2sin −1 kl /( kl − P) (b) For P = 1 kl , 4 sin θ 2 θ 2 = 1 2 3 4 kl kl = 2 3 θ = 83.6° = 41.8° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 408 PROBLEM 4.58 A collar B of weight W can move freely along the vertical rod shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Derive an equation in θ, W, k, and l that must be satisfied when the collar is in equilibrium. (b) Knowing that W = 300 N, l = 500 mm, and k = 800 N/m, determine the value of θ corresponding to equilibrium. SOLUTION First note: T = ks where k = spring constant s = elongation of spring l = −l cos θ l (1 − cos θ ) = cos θ kl T= (1 − cos θ ) cos θ (a) From F.B.D. of collar B: or (b) For ΣFy = 0: T sin θ − W = 0 kl (1 − cos θ )sin θ − W = 0 cos θ or tan θ − sin θ = W kl W = 3 lb l = 6 in. k = 8 lb/ft 6 in. l= = 0.5 ft 12 in./ft tan θ − sin θ = Solving numerically, 3 lb = 0.75 (8 lb/ft)(0.5 ft) θ = 57.957° or θ = 58.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 409 PROBLEM 4.59 Eight identical 500 × 750-mm rectangular plates, each of mass m = 40 kg, are held in a vertical plane as shown. All connections consist of frictionless pins, rollers, or short links. In each case, determine whether (a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible, compute the reactions. SOLUTION 1. Three non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: determinate (c) Equilibrium maintained A = C = 196.2 N 2. Three non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: determinate (c) Equilibrium maintained B = 0, C = D = 196.2 N 3. Four non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: indeterminate (c) Equilibrium maintained A x = 294 N , D x = 294 N ( A y + D y = 392 N ) 4. Three concurrent reactions (through D): (a) Plate: improperly constrained (b) Reactions: indeterminate (c) No equilibrium (ΣM D ≠ 0) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 410 PROBLEM 4.59 (Continued) 5. Two reactions: (a) Plate: partial constraint (b) Reactions: determinate (c) Equilibrium maintained C = D = 196.2 N 6. Three non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: determinate (c) Equilibrium maintained B = 294 N 7. 8. , D = 491 N 53.1° Two reactions: (a) Plate: improperly constrained (b) Reactions determined by dynamics (c) No equilibrium (ΣFy ≠ 0) Four non-concurrent, non-parallel reactions: (a) Plate: completely constrained (b) Reactions: indeterminate (c) Equilibrium maintained B = D y = 196.2 N (C + D x = 0) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 411 PROBLEM 4.60 The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins, rollers, or short links. For each case, answer the questions listed in Problem 4.59, and, wherever possible, compute the reactions, assuming that the magnitude of the force P is 100 lb. SOLUTION 1. Three non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: determinate (c) Equilibrium maintained A = 120.2 lb 2. 3. 56.3°, B = 66.7 lb Four concurrent, reactions (through A): (a) Bracket: improper constraint (b) Reactions: indeterminate (c) No equilibrium (ΣM A ≠ 0) Two reactions: (a) Bracket: partial constraint (b) Reactions: indeterminate (c) Equilibrium maintained A = 50 lb , C = 50 lb 4. Three non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: determinate (c) Equilibrium maintained A = 50 lb , B = 83.3 lb 36.9°, C = 66.7 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 412 PROBLEM 4.60 (Continued) 5. 6. Four non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: indeterminate (c) Equilibrium maintained (ΣM C = 0) A y = 50 lb Four non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: indeterminate (c) Equilibrium maintained A x = 66.7 lb B x = 66.7 lb ( A y + B y = 100 lb ) 7. Three non-concurrent, non-parallel reactions: (a) Bracket: complete constraint (b) Reactions: determinate (c) Equilibrium maintained A = C = 50 lb 8. Three concurrent, reactions (through A) (a) Bracket: improper constraint (b) Reactions: indeterminate (c) No equilibrium (ΣM A ≠ 0) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 413 PROBLEM 4.61 Determine the reactions at A and B when a = 150 mm. SOLUTION Free-Body Diagram: Force triangle 80 mm 80 mm = a 150 mm β = 28.072° tan β = A= 320 N sin 28.072° B= 320 N tan 28.072° A = 680 N 28.1° B = 600 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 414 PROBLEM 4.62 Determine the value of a for which the magnitude of the reaction at B is equal to 800 N. SOLUTION Free-Body Diagram: Force triangle tan β = 80 mm a a= 80 mm tan β (1) From force triangle: tan β = From Eq. (1): a= 320 N = 0.4 800 N 80 mm 0.4 a = 200 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 415 PROBLEM 4.63 Using the method of Sec. 4.7, solve Problem 4.22b. PROBLEM 4.22 Determine the reactions at A and B when (a) α = 0, (b) α = 90°, (c) α = 30°. SOLUTION Free-Body Diagram: (Three-force body) The line of action at A must pass through C, where B and the 75-lb load intersect. In triangle ACE: Force triangle tan θ = 10 in. 12 in. θ = 39.806° B = (75 lb) tan 39.806° = 62.5 lb 75 lb A= = 97.6° cos 39.806° A = 97.6 lb 50.2°; B = 62.5 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 416 PROBLEM 4.64 A 500-lb cylindrical tank, 8 ft in diameter, is to be raised over a 2-ft obstruction. A cable is wrapped around the tank and pulled horizontally as shown. Knowing that the corner of the obstruction at A is rough, find the required tension in the cable and the reaction at A. SOLUTION Free-Body Diagram: Force triangle cos α = GD 2 ft = = 0.5 AG 4 ft α = 60° 1 ( β = 60°) 2 T = (500 lb) tan 30° T = 289 lb θ = α = 30° A= 500 lb cos 30° A = 577 lb 60.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 417 PROBLEM 4.65 For the frame and loading shown, determine the reactions at A and C. SOLUTION Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with member forces intersecting at E. The F.B.D.’s of members AB and BCD illustrate the above conditions. The force triangle for member BCD is also shown. The angle β is found from the member dimensions: 6 in. = 30.964° 10 in. β = tan −1 Applying the law of sines to the force triangle for member BCD, 30 lb B C = = sin(45° − β ) sin β sin135° or 30 lb B C = = sin14.036° sin 30.964° sin135° A= B= (30 lb)sin 30.964° = 63.641 lb sin14.036° or and C= A = 63.6 lb 45.0° C = 87.5 lb 59.0° (30 lb) sin135° = 87.466 lb sin14.036° or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 418 PROBLEM 4.66 For the frame and loading shown, determine the reactions at C and D. SOLUTION Since BD is a two-force member, the reaction at D must pass through Points B and D. Free-Body Diagram: (Three-force body) Reaction at C must pass through E, where the reaction at D and the 150-lb load intersect. Triangle CEF: tan β = 4.5 ft 3 ft β = 56.310° Triangle ABE: tan γ = 1 2 γ = 26.565° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 419 PROBLEM 4.66 (Continued) Force Triangle Law of sines: 150 lb C D = = sin 29.745° sin116.565° sin 33.690° C = 270.42 lb, D = 167.704 lb C = 270 lb 56.3°; D = 167.7 lb 26.6° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 420 PROBLEM 4.67 Determine the reactions at B and D when b = 60 mm. SOLUTION Since CD is a two-force member, the line of action of reaction at D must pass through Points C and D. Free-Body Diagram: (Three-force body) Reaction at B must pass through E, where the reaction at D and the 80-N force intersect. 220 mm 250 mm β = 41.348° tan β = Force triangle Law of sines: 80 N B D = = sin 3.652° sin 45° sin131.348° B = 888.0 N D = 942.8 N B = 888 N 41.3° D = 943 N 45.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 421 PROBLEM 4.68 Determine the reactions at B and D when b = 120 mm. SOLUTION Since CD is a two-force member, line of action of reaction at D must pass through C and D . Free-Body Diagram: (Three-force body) Reaction at B must pass through E, where the reaction at D and the 80-N force intersect. 280 mm 250 mm β = 48.24° tan β = Force triangle Law of sines: 80 N B D = = sin 3.24° sin135° sin 41.76° B = 1000.9 N D = 942.8 N B = 1001 N 48.2° D = 943 N 45.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 422 PROBLEM 4.69 A T-shaped bracket supports a 300-N load as shown. Determine the reactions at A and C when α = 45°. SOLUTION Free-Body Diagram: (Three-force body) The line of action of C must pass through E, where A and the 300-N force intersect. Triangle ABE is isosceles: EA = AB = 400 mm In triangle CEF: tan θ = CF CF 150 mm = = EF EA + AF 700 mm θ = 12.0948° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 423 PROBLEM 4.69 (Continued) Force Triangle Law of sines: A C 300 N = = sin 32.905° sin135° sin12.0948° A = 778 N ; C = 1012 N 77.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 424 PROBLEM 4.70 A T-shaped bracket supports a 300-N load as shown. Determine the reactions at A and C when α = 60°. SOLUTION Free-Body Diagram: EA = (400 mm) tan 30° = 230.94 mm In triangle CEF: tan θ = CF CF = EF EA + AF 150 230.94 + 300 θ = 15.7759° tan θ = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 425 PROBLEM 4.70 (Continued) Force Triangle Law of sines: A C 300 N = = sin 44.224° sin120° sin15.7759° A = 770 N C = 956 N A = 770 N ; C = 956 N 74.2° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 426 PROBLEM 4.71 A 40-lb roller, of diameter 8 in., which is to be used on a tile floor, is resting directly on the subflooring as shown. Knowing that the thickness of each tile is 0.3 in., determine the force P required to move the roller onto the tiles if the roller is (a) pushed to the left, (b) pulled to the right. SOLUTION Geometry: For each case as roller comes into contact with tile, 3.7 in. 4 in. α = 22.332° α = cos −1 (a) Roller pushed to left (three-force body): Forces must pass through O. Law of sines: Force Triangle 40 lb P = ; P = 24.87 lb sin 37.668° sin 22.332° P = 24.9 lb (b) 30.0° Roller pulled to right (three-force body): Forces must pass through O. Law of sines: 40 lb P = ; P = 15.3361 lb sin 97.668° sin 22.332° P = 15.34 lb 30.0° Force Triangle PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 427 PROBLEM 4.72 One end of rod AB rests in the corner A and the other end is attached to cord BD. If the rod supports a 40-lb load at its midpoint C, find the reaction at A and the tension in the cord. SOLUTION Free-Body Diagram: (Three-force body) The line of action of reaction at A must pass through E, where T and the 40-lb load intersect. Force triangle EF 23 = AF 12 α = 62.447° 5 EH tan β = = DH 12 β = 22.620° tan α = A T 40 lb = = sin 67.380° sin 27.553° sin 85.067° A = 37.1 lb 62.4° T = 18.57 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 428 PROBLEM 4.73 A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1.5 m, determine (a) the tension in cable CD, (b) the reaction at B. SOLUTION Three-force body: W and TCD intersect at E. 0.7497 m 1.5 m β = 26.556° tan β = Three forces intersect at E. W = (50 kg) 9.81 m/s 2 = 490.50 N Law of sines: Force triangle TCD 490.50 N B = = sin 61.556° sin 63.444° sin 55° TCD = 498.99 N B = 456.96 N TCD = 499 N (a) B = 457 N (b) 26.6° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 429 PROBLEM 4.74 Solve Problem 4.73, assuming that a = 3 m. PROBLEM 4.73 A 50-kg crate is attached to the trolley-beam system shown. Knowing that a = 1.5 m, determine (a) the tension in cable CD, (b) the reaction at B. SOLUTION W and TCD intersect at E. Free-Body Diagram: Three-Force Body AE 0.301 m = AB 3m β = 5.7295° tan β = Three forces intersect at E. Force Triangle W = (50 kg) 9.81 m/s 2 = 490.50 N Law of sines: TCD 490.50 N B = = sin 29.271° sin 95.730° sin 55° TCD = 998.18 N B = 821.76 N TCD = 998 N (a) B = 822 N (b) 5.73° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 430 PROBLEM 4.75 Determine the reactions at A and B when β = 50°. SOLUTION Free-Body Diagram: (Three-force body) Reaction A must pass through Point D where the 100-N force and B intersect. In right Δ BCD: α = 90° − 75° = 15° BD = 250 tan 75° = 933.01 mm In right Δ ABD: Dimensions in mm AB 150 mm = BD 933.01 mm γ = 9.1333° tan γ = Force Triangle Law of sines: 100 N A B = = sin 9.1333° sin15° sin155.867° A = 163.1 N; B = 257.6 N A = 163.1 N 74.1° B = 258 N 65.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 431 PROBLEM 4.76 Determine the reactions at A and B when β = 80°. SOLUTION Free-Body Diagram: (Three-force body) Reaction A must pass through D where the 100-N force and B intersect. In right triangle BCD: α = 90° − 75° = 15° BD = BC tan 75° = 250 tan75° BD = 933.01 mm In right triangle ABD: tan γ = AB 150 mm = BD 933.01 mm γ = 9.1333° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 432 PROBLEM 4.76 (Continued) Force Triangle Law of sines: 100 N A B = = sin 9.1333° sin15° sin155.867° A = 163.1 N 55.9° B = 258 N 65.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 433 PROBLEM 4.77 Knowing that θ = 30°, determine the reaction (a) at B, (b) at C. SOLUTION Free-Body Diagram: (Three-force body) Reaction at C must pass through D where force P and reaction at B intersect. In Δ CDE: ( 3 − 1) R R = 3 −1 β = 36.2° tan β = Force Triangle Law of sines: P B C = = sin 23.8° sin126.2° sin 30° B = 2.00 P C = 1.239 P (a) B = 2P (b) C = 1.239P 60.0° 36.2° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 434 PROBLEM 4.78 Knowing that θ = 60°, determine the reaction (a) at B, (b) at C. SOLUTION Reaction at C must pass through D where force P and reaction at B intersect. In ΔCDE: tan β = R− =1− R 3 Free-Body Diagram: (Three-force body) R 1 3 β = 22.9° Force Triangle Law of sines: P B C = = sin 52.9° sin 67.1° sin 60° B = 1.155P C = 1.086 P (a) B = 1.155P 30.0° (b) C = 1.086P 22.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 435 PROBLEM 4.79 Using the method of Section 4.7, solve Problem 4.23. PROBLEM 4.23 Determine the reactions at A and B when (a) h = 0, (b) h = 200 mm. SOLUTION Free-Body Diagram: (a) h=0 Reaction A must pass through C where the 150-N weight and B interect. Force triangle is equilateral. (b) A = 150.0 N 30.0° B = 150.0 N 30.0° h = 200 mm 55.662 250 β = 12.5521° tan β = Law of sines: A B 150 N = = sin17.4480° sin 60° sin102.552° A = 433.24 N B = 488.31 N Force Triangle A = 433 N B = 488 N 12.55° 30.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 436 PROBLEM 4.80 Using the method of Section 4.7, solve Problem 4.24. PROBLEM 4.24 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 75-lb vertical force at B, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION Reaction at C must pass through E, where the 75-lb force and T intersect. 9.3969 in. 8.5798 in. α = 47.602° tan α = 14.0954 in. 24.870 in. β = 29.543° tan β = Force Triangle Law of sines: 75 lb T C = = sin18.0590° sin 29.543° sin132.398° Free-Body Diagram: Dimensions in in. (a) (b) T = 119.3 lb C = 178.7 lb 60.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 437 PROBLEM 4.81 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B. SOLUTION Free-Body Diagram: Reaction at B must pass through D. 7 in. 12 in. α = 30.256° 7 in. tan β = 24 in. β = 16.26° tan α = Force Triangle Law of sines: T T − 72 lb B = = sin 59.744° sin13.996° sin106.26 T (sin13.996°) = (T − 72 lb)(sin 59.744°) T (0.24185) = (T − 72)(0.86378) T = 100.00 lb sin 106.26° sin 59.744° = 111.14 lb T = 100.0 lb B = (100 lb) B = 111.1 lb 30.3° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 438 PROBLEM 4.82 Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portions AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B. SOLUTION Free-Body Diagram: Force Triangle Reaction at B must pass through D. tan α = 120 ; α = 36.9° 160 T T − 75 N B = = 4 3 5 3T = 4T − 300; T = 300 N 5 5 B = T = (300 N) = 375 N 4 4 B = 375 N 36.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 439 PROBLEM 4.83 A thin ring of mass 2 kg and radius r = 140 mm is held against a frictionless wall by a 125-mm string AB. Determine (a) the distance d, (b) the tension in the string, (c) the reaction at C. SOLUTION Free-Body Diagram: (Three-force body) The force T exerted at B must pass through the center G of the ring, since C and W intersect at that point. Thus, points A, B, and G are in a straight line. (a) From triangle ACG: d = ( AG ) 2 − (CG )2 = (265 mm)2 − (140 mm) 2 = 225.00 mm d = 225 mm Force Triangle W = (2 kg)(9.81 m/s 2 ) = 19.6200 N Law of sines: T C 19.6200 N = = 265 mm 140 mm 225.00 mm T = 23.1 N (b) C = 12.21 N (c) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 440 PROBLEM 4.84 A uniform rod AB of length 2R rests inside a hemispherical bowl of radius R as shown. Neglecting friction, determine the angle θ corresponding to equilibrium. SOLUTION Based on the F.B.D., the uniform rod AB is a three-force body. Point E is the point of intersection of the three forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to the rod, triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle. Note that the angle α of triangle DOA is the central angle corresponding to the inscribed angle θ of triangle DCA. α = 2θ The horizontal projections of AE , ( x AE ), and AG , ( x AG ), are equal. x AE = x AG = x A or ( AE ) cos 2θ = ( AG ) cos θ and (2 R) cos 2θ = R cos θ Now then or cos 2θ = 2 cos 2 θ − 1 4 cos 2 θ − 2 = cos θ 4 cos 2 θ − cos θ − 2 = 0 Applying the quadratic equation, cos θ = 0.84307 and cos θ = − 0.59307 θ = 32.534° and θ = 126.375° (Discard) or θ = 32.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 441 PROBLEM 4.85 A slender rod BC of length L and weight W is held by two cables as shown. Knowing that cable AB is horizontal and that the rod forms an angle of 40° with the horizontal, determine (a) the angle θ that cable CD forms with the horizontal, (b) the tension in each cable. SOLUTION Free-Body Diagram: (Three-force body) (a) The line of action of TCD must pass through E, where TAB and W intersect. CF EF L sin 40° = 1 L cos 40° 2 tan θ = = 2 tan 40° = 59.210° θ = 59.2° (b) Force Triangle TAB = W tan 30.790° = 0.59588W TAB = 0.596W W cos 30.790° = 1.16408W TCD = TCD = 1.164W PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 442 PROBLEM 4.86 A slender rod of length L and weight W is attached to a collar at A and is fitted with a small wheel at B. Knowing that the wheel rolls freely along a cylindrical surface of radius R, and neglecting friction, derive an equation in θ, L, and R that must be satisfied when the rod is in equilibrium. SOLUTION Free-Body Diagram (Three-force body) Reaction B must pass through D where B and W intersect. Note that ΔABC and ΔBGD are similar. AC = AE = L cos θ In Δ ABC: (CE ) 2 + ( BE )2 = ( BC )2 (2 L cos θ ) 2 + ( L sin θ )2 = R 2 2 R 2 2 = 4cos θ + sin θ L 2 R 2 2 = 4cos θ + 1 − cos θ L 2 R 2 = 3cos θ + 1 L 2 1 cos 2 θ = R − 1 3 L PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 443 PROBLEM 4.87 Knowing that for the rod of Problem 4.86, L = 15 in., R = 20 in., and W = 10 lb, determine (a) the angle θ corresponding to equilibrium, (b) the reactions at A and B. SOLUTION See the solution to Problem 4.86 for the free-body diagram and analysis leading to the following equation: 2 1 cos 2 θ = R − 1 3 L For L = 15 in., R = 20 in., and W = 10 lb, 2 1 20 in. cos θ = − 1 ; θ = 59.39° 3 15 in. 2 (a) In Δ ABC: θ = 59.4° BE L sin θ 1 = = tan θ CE 2 L cos θ 2 1 tan α = tan 59.39° = 0.8452 2 α = 40.2° tan α = Force Triangle A = W tan α = (10 lb) tan 40.2° = 8.45 lb W (10 lb) B= = = 13.09 lb cos α cos 40.2° A = 8.45 lb (b) B = 13.09 lb 49.8° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 444 PROBLEM 4.88 Rod AB is bent into the shape of an arc of circle and is lodged between two pegs D and E. It supports a load P at end B. Neglecting friction and the weight of the rod, determine the distance c corresponding to equilibrium when a = 20 mm and R = 100 mm. SOLUTION Free-Body Diagram: Since yED = xED = a, slope of ED is 45°; slope of HC is 45°. Also DE = 2 a and a 1 DH = HE = DE = 2 2 For triangles DHC and EHC, sin β = a 2 R = a 2R Now c = R sin(45° − β ) For a = 20 mm and sin β = R = 100 mm 20 mm 2(100 mm) = 0.141421 β = 8.1301° and c = (100 mm) sin(45° − 8.1301°) = 60.00 mm or c = 60.0 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 445 PROBLEM 4.89 A slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium, derive an expression for the angle θ in terms of the angle β. SOLUTION As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the force geometry: Free-Body Diagram: tan β = xGB y AB where y AB = L cos θ and xGB = tan β = 1 L sin θ 2 1 2 L sin θ L cos θ 1 = tan θ 2 or tan θ = 2 tan β PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 446 PROBLEM 4.90 An 8-kg slender rod of length L is attached to collars that can slide freely along the guides shown. Knowing that the rod is in equilibrium and that β = 30°, determine (a) the angle θ that the rod forms with the vertical, (b) the reactions at A and B. SOLUTION (a) As shown in the free-body diagram of the slender rod AB, the three forces intersect at C. From the geometry of the forces: Free-Body Diagram: tan β = xCB yBC where xCB = 1 L sin θ 2 and yBC = L cos θ tan β = 1 tan θ 2 or tan θ = 2 tan β For β = 30° tan θ = 2 tan 30° = 1.15470 θ = 49.107° or W = mg = (8 kg)(9.81 m/s2 ) = 78.480 N (b) From force triangle: A = W tan β = (78.480 N) tan 30° = 45.310 N and θ = 49.1° B= or W 78.480 N = = 90.621 N cos β cos 30° A = 45.3 N or B = 90.6 N 60.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 447 PROBLEM 4.91 A 200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Dimensions in mm We have six unknowns and six equations of equilibrium. —OK ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × T i + (80k − 200i ) × (−720 j) = 0 −120 Dx j + 120 D y i − 120T k − 160Tj + 57.6 × 103 i + 144 × 103 k = 0 Equating to zero the coefficients of the unit vectors: k: −120T + 144 × 103 = 0 i: 120 Dy + 57.6 × 103 = 0 j: − 120 Dx − 160(1200 N) = 0 (b) ΣFx = 0: C x + Dx + T = 0 ΣFy = 0: C y + Dy − 720 = 0 ΣFz = 0: Cz = 0 (a) T = 1200 N Dy = −480 N Dx = −1600 N C x = 1600 − 1200 = 400 N C y = 480 + 720 = 1200 N C = (400 N)i + (1200 N) j; D = −(1600 N)i − (480 N) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 448 PROBLEM 4.92 Solve Problem 4.91, assuming that the axle has been rotated clockwise in its bearings by 30° and that the 720-N load remains vertical. PROBLEM 4.91 A 200-mm lever and a 240-mmdiameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Dimensions in mm We have six unknowns and six equations of equilibrium. ΣM C = 0: (−120k ) × ( Dx i + Dy j) + (120 j − 160k ) × T i + (80k − 173.21i ) × (−720 j) = 0 −120 Dx j + 120 D y i −120T k −160T j + 57.6 × 103 i + 124.71 × 103 k = 0 Equating to zero the coefficients of the unit vectors, k : − 120T + 124.71 × 103 = 0 i: T = 1039 N 120 Dy + 57.6 × 103 = 0 Dy = −480 N j: − 120 Dx − 160(1039.2) (b) T = 1039.2 N ΣFx = 0: C x + Dx + T = 0 ΣFy = 0: C y + Dy − 720 = 0 ΣFz = 0: Cz = 0 Dx = −1385.6 N C x = 1385.6 − 1039.2 = 346.4 C y = 480 + 720 = 1200 N C = (346 N)i + (1200 N) j D = −(1386 N)i − (480 N) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 449 PROBLEM 4.93 A 4 × 8-ft sheet of plywood weighing 40 lb has been temporarily propped against column CD. It rests at A and B on small wooden blocks and against protruding nails. Neglecting friction at all surfaces of contact, determine the reactions at A, B, and C. SOLUTION Free-Body Diagram: We have five unknowns and six equations of equilibrium. Plywood sheet is free to move in x direction, but equilibrium is maintained (ΣFx = 0). ΣM A = 0: rB /A × ( B y j + Bz k ) + rC /A × C k + rG /A × ( −40 lb) j = 0 i j 5 0 0 By k i j k i j k 0 + 4 4sin 60° −4cos 60° + 2 2sin 60° −2 cos 60° = 0 Bz 0 C −40 0 0 0 (4C sin 60° − 80 cos 60°) i + (−5Bz − 4C ) j + (5B y − 80)k = 0 Equating the coefficients of the unit vectors to zero, i: 4C sin 60° − 80 cos 60° = 0 j: −5Bz − 4C = 0 k: 5B y − 80 = 0 ΣFy = 0: Ay + B y − 40 = 0 ΣFz = 0: Az + Bz + C = 0 C = 11.5470 lb Bz = 9.2376 lb B y = 16.0000 lb Ay = 40 − 16.0000 = 24.000 lb Az = 9.2376 − 11.5470 = −2.3094 lb A = (24.0 lb)j − (2.31 lb)k ; B = (16.00 lb) j − (9.24 lb)k ; C = (11.55 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 450 PROBLEM 4.94 Two tape spools are attached to an axle supported by bearings at A and D. The radius of spool B is 1.5 in. and the radius of spool C is 2 in. Knowing that TB = 20 lb and that the system rotates at a constant rate, determine the reactions at A and D. Assume that the bearing at A does not exert any axial thrust and neglect the weights of the spools and axle. SOLUTION Free-Body Diagram: We have six unknowns and six equations of equilibrium. ΣM A = 0: (4.5i + 1.5k ) × (−20 j) + (10.5i + 2 j) × (−TC k ) + (15i) × ( Dx i + Dy j + Dz k ) = 0 −90k + 30i + 10.5TC j − 2TC i + 15D y k − 15Dz j = 0 Equate coefficients of unit vectors to zero: i: 30 − 2TC = 0 j: 10.5TC − 15Dz = 0 10.5(15) − 15Dz = 0 k: −90 + 15 Dy = 0 ΣFx = 0: Dx = 0 ΣFy = 0: Ay + D y − 20 lb = 0 Ay = 20 − 6 = 14 lb ΣFz = 0: Az + Dz − 15 lb = 0 Az = 15 − 10.5 = 4.5 lb TC = 15 lb Dz = 10.5 lb Dy = 6 lb A = (14.00 lb) j + (4.50 lb)k ; D = (6.00 lb) j + (10.50 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 451 PROBLEM 4.95 Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radius of the inner sheave is 125 mm and the radius of the outer sheave is 250 mm. Knowing that when the system is at rest, the tension is 90 N in both portions of belt B and 150 N in both portions of belt C, determine the reactions at A and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION We replace TB and TB′ by their resultant (−180 N)j and TC and TC′ by their resultant (−300 N)k. Dimensions in mm We have five unknowns and six equations of equilibrium. Axle AD is free to rotate about the x-axis, but equilibrium is maintained (ΣMx = 0). ΣM A = 0: (150i ) × (−180 j) + (250i) × ( −300k ) + (450i ) × ( Dy j + Dz k ) = 0 −27 × 103 k + 75 × 103 j + 450 Dy k − 450 Dz j = 0 Equating coefficients of j and k to zero, j: 75 × 103 − 450 Dz = 0 Dz = 166.7 N k: − 27 × 103 + 450 Dy = 0 Dy = 60.0 N ΣFx = 0: Ax = 0 ΣFy = 0: Ay + Dy − 180 N = 0 ΣFz = 0: Az + Dz − 300 N = 0 Ay = 180 − 60 = 120.0 N Az = 300 − 166.7 = 133.3 N A = (120.0 N) j + (133.3 N)k ; D = (60.0 N) j + (166.7 N)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 452 PROBLEM 4.96 Solve Problem 4.95, assuming that the pulley rotates at a constant rate and that TB = 104 N, T′B = 84 N, TC = 175 N. PROBLEM 4.95 Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radius of the inner sheave is 125 mm and the radius of the outer sheave is 250 mm. Knowing that when the system is at rest, the tension is 90 N in both portions of belt B and 150 N in both portions of belt C, determine the reactions at A and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Dimensions in mm We have six unknowns and six equations of equilibrium. —OK ΣM A = 0: (150i + 250k ) × (−104 j) + (150i − 250k ) × ( −84 j) + (250i + 125 j) × (−175k ) + (250i − 125 j) × (−TC ) + 450i × ( D y j + Dz k ) = 0 −150(104 + 84)k + 250(104 − 84)i + 250(175 + TC′ ) j − 125(175 − TC′ ) + 450 D y k − 450 Dz j = 0 Equating the coefficients of the unit vectors to zero, i : 250(104 − 84) − 125(175 − TC′ ) = 0 175 = TC′ = 40 j: 250(175 + 135) − 450 Dz = 0 Dz = 172.2 N k : − 150(104 + 84) + 450 D y = 0 Dy = 62.7 N TC′ = 135; PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 453 PROBLEM 4.96 (Continued) ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 104 − 84 + 62.7 = 0 Ay = 125.3 N ΣFz = 0: Az − 175 − 135 + 172.2 = 0 Az = 137.8 N A = (125.3 N) j + (137.8 N)k; D = (62.7 N) j + (172.2 N)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 454 PROBLEM 4.97 Two steel pipes AB and BC, each having a mass per unit length of 8 kg/m, are welded together at B and supported by three wires. Knowing that a = 0.4 m, determine the tension in each wire. SOLUTION W1 = 0.6m′g W2 = 1.2m′g ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0 (−0.4i + 0.6k ) × TA j + (−0.4i + 0.3k ) × (−W1 j) + 0.2i × (−W2 j) + 0.8i × TC j = 0 −0.4TAk − 0.6TA i + 0.4W1k + 0.3W1i − 0.2W2 k + 0.8TC k = 0 Equate coefficients of unit vectors to zero: 1 1 i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g 2 2 k : − 0.4TA + 0.4W1 − 0.2W2 + 0.8TC = 0 −0.4(0.3m′g ) + 0.4(0.6m′g ) − 0.2(1.2m′g ) + 0.8TC = 0 TC = (0.12 − 0.24 − 0.24)m′g = 0.15m′g 0.8 ΣFy = 0: TA + TC + TD − W1 − W2 = 0 0.3m′g + 0.15m′g + TD − 0.6m′g − 1.2m′g = 0 TD = 1.35m′g m′g = (8 kg/m)(9.81m/s 2 ) = 78.48 N/m TA = 0.3m′g = 0.3 × 78.45 TA = 23.5 N TB = 0.15m′g = 0.15 × 78.45 TB = 11.77 N TC = 1.35m′g = 1.35 × 78.45 TC = 105.9 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 455 PROBLEM 4.98 For the pipe assembly of Problem 4.97, determine (a) the largest permissible value of a if the assembly is not to tip, (b) the corresponding tension in each wire. SOLUTION W1 = 0.6m′g W2 = 1.2m′g ΣM D = 0: rA/D × TA j + rE/D × (−W1 j) + rF/D × (−W2 j) + rC/D × TC j = 0 (− ai + 0.6k ) × TA j + (− ai + 0.3k ) × (−W1 j) + (0.6 − a)i × (−W2 j) + (1.2 − a)i × TC j = 0 −TA ak − 0.6TA i + W1ak + 0.3W1i − W2 (0.6 − a)k + TC (1.2 − a )k = 0 Equate coefficients of unit vectors to zero: 1 1 i : − 0.6TA + 0.3W1 = 0; TA = W1 = 0.6m′g = 0.3m′g 2 2 k : − TA a + W1a − W2 (0.6 − a ) + TC (1.2 − a) = 0 −0.3m′ga + 0.6m′ga − 1.2m′g (0.6 − a ) + TC (1.2 − a) = 0 TC = (a) 0.3a − 0.6a + 1.2(0.6 − a) 1.2 − a For maximum a and no tipping, TC = 0. −0.3a + 1.2(0.6 − a) = 0 −0.3a + 0.72 − 1.2a = 0 1.5a = 0.72 a = 0.480 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 456 PROBLEM 4.98 (Continued) (b) Reactions: m′g = (8 kg/m) 9.81 m/s 2 = 78.48 N/m TA = 0.3m′g = 0.3 × 78.48 = 23.544 N TA = 23.5 N ΣFy = 0: TA + TC + TD − W1 − W2 = 0 TA + 0 + TD − 0.6m′g − 1.2m′g = 0 TD = 1.8m′g − TA = 1.8 × 78.48 − 23.544 = 117.72 TD = 117.7 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 457 PROBLEM 4.99 The 45-lb square plate shown is supported by three vertical wires. Determine the tension in each wire. SOLUTION Free-Body Diagram: ΣM B = 0: rC /B × TC j + rA/B × TA j + rG /B × (−45 lb) j = 0 [−(20 in.)i + (15 in.)k ] × TC j + (20 in.)k × TA j + [−(10 in.)i + (10 in.)k ] × [−(45 lb)j] = 0 −20TC k − 15TC i − 20TAi + 450k + 450i = 0 Equating to zero the coefficients of the unit vectors, k: −20TC + 450 = 0 TC = 22.5 lb i: −15(22.5) − 20TA + 450 = 0 TA = 5.625 lb ΣFy = 0: TA + TB + TC − 45 lb = 0 5.625 lb + TB + 22.5 lb − 45 lb = 0 TB = 16.875 lb TA = 5.63 lb; TB = 16.88 lb; TC = 22.5 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 458 PROBLEM 4.100 The table shown weighs 30 lb and has a diameter of 4 ft. It is supported by three legs equally spaced around the edge. A vertical load P of magnitude 100 lb is applied to the top of the table at D. Determine the maximum value of a if the table is not to tip over. Show, on a sketch, the area of the table over which P can act without tipping the table. SOLUTION r = 2 ft b = r sin 30° = 1 ft We shall sum moments about AB. (b + r )C + (a − b) P − bW = 0 (1 + 2)C + (a − 1)100 − (1)30 = 0 1 C = [30 − (a − 1)100] 3 If table is not to tip, C ≥ 0. [30 − ( a − 1)100] ≥ 0 30 ≥ (a − 1)100 a − 1 ≤ 0.3 a ≤ 1.3 ft a = 1.300 ft Only ⊥ distance from P to AB matters. Same condition must be satisfied for each leg. P must be located in shaded area for no tipping. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 459 PROBLEM 4.101 An opening in a floor is covered by a 1 × 1.2-m sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C. SOLUTION rB/A = 0.6i rC/A = 0.8i + 1.05k rG/A = 0.3i + 0.6k W = mg = (18 kg)9.81 W = 176.58 N ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0 (0.6i ) × Bj + (0.8i + 1.05k ) × Cj + (0.3i + 0.6k ) × ( −Wj) = 0 0.6 Bk + 0.8Ck − 1.05Ci − 0.3Wk + 0.6Wi = 0 Equate coefficients of unit vectors to zero: 0.6 i : 1.05C + 0.6W = 0 C = 176.58 N = 100.90 N 1.05 k : 0.6 B + 0.8C − 0.3W = 0 0.6 B + 0.8(100.90 N) − 0.3(176.58 N) = 0 B = −46.24 N ΣFy = 0: A + B + C − W = 0 A − 46.24 N + 100.90 N + 176.58 N = 0 A = 121.92 N (a ) A = 121.9 N (b) B = −46.2 N (c) C = 100.9 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 460 PROBLEM 4.102 Solve Problem 4.101, assuming that the small block C is moved and placed under edge DE at a point 0.15 m from corner E. PROBLEM 4.101 An opening in a floor is covered by a 1 × 1.2-m sheet of plywood of mass 18 kg. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C. SOLUTION rB/A = 0.6i rC/A = 0.65i + 1.2k rG/A = 0.3i + 0.6k W = mg = (18 kg) 9.81 m/s 2 W = 176.58 N ΣM A = 0: rB/A × Bj + rC/A × Cj + rG/A × (−Wj) = 0 0.6i × Bj + (0.65i + 1.2k ) × Cj + (0.3i + 0.6k ) × (−Wj) = 0 0.6 Bk + 0.65Ck − 1.2Ci − 0.3Wk + 0.6Wi = 0 Equate coefficients of unit vectors to zero: 0.6 C = 176.58 N = 88.29 N 1.2 i : −1.2C + 0.6W = 0 k : 0.6 B + 0.65C − 0.3W = 0 0.6 B + 0.65(88.29 N) − 0.3(176.58 N) = 0 B = −7.36 N ΣFy = 0: A + B + C − W = 0 A − 7.36 N + 88.29 N − 176.58 N = 0 A = 95.648 N (a ) A = 95.6 N (b) B = − 7.36 N (c) C = 88.3 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 461 PROBLEM 4.103 The rectangular plate shown weighs 80 lb and is supported by three vertical wires. Determine the tension in each wire. SOLUTION Free-Body Diagram: ΣM B = 0: rA/B × TA j + rC/B × TC j + rG/B × (−80 lb) j = 0 (60 in.)k × TA j + [(60 in.)i + (15 in.)k ] × TC j + [(30 in.)i + (30 in.)k ] × ( −80 lb) j = 0 −60TAi + 60TC k − 15TC i − 2400k + 2400i = 0 Equating to zero the coefficients of the unit vectors, i: 60TA − 15(40) + 2400 = 0 TA = 30.0 lb k: 60TC − 2400 = 0 TC = 40.0 lb ΣFy = 0: TA + TB + TC − 80 lb = 0 30 lb + TB + 40 lb − 80 lb = 0 TB = 10.00 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 462 PROBLEM 4.104 The rectangular plate shown weighs 80 lb and is supported by three vertical wires. Determine the weight and location of the lightest block that should be placed on the plate if the tensions in the three wires are to be equal. SOLUTION Free-Body Diagram: Let −Wb j be the weight of the block and x and z the block’s coordinates. Since tensions in wires are equal, let TA = TB = TC = T ΣM 0 = 0: (rA × Tj) + (rB × Tj) + (rC × Tj) + rG × (−Wj) + ( xi + zk ) × ( −Wb j) = 0 or (75 k ) × Tj + (15 k ) × Tj + (60i + 30k ) × Tj + (30i + 45k ) × (−Wj) + ( xi + zk ) × (−Wb j) = 0 or −75T i − 15T i + 60T k − 30T i − 30W k + 45W i − Wb × k + Wb z i = 0 Equate coefficients of unit vectors to zero: i: −120T + 45W + Wb z = 0 (1) k: 60T − 30W − Wb x = 0 (2) ΣFy = 0: 3T − W − Wb = 0 (3) Eq. (1) + 40 Eq. (3): 5W + ( z − 40)Wb = 0 (4) Eq. (2) – 20 Eq. (3): −10W − ( x − 20)Wb = 0 (5) Also, PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 463 PROBLEM 4.104 (Continued) Solving Eqs. (4) and (5) for Wb /W and recalling that 0 ≤ x ≤ 60 in., 0 ≤ z ≤ 90 in., Eq. (4): Wb 5 5 = ≥ = 0.125 W 40 − z 40 − 0 Eq. (5): Wb 10 10 = ≥ = 0.5 W 20 − x 20 − 0 Thus, (Wb ) min = 0.5W = 0.5(80) = 40 lb (Wb ) min = 40.0 lb Making Wb = 0.5W in Eqs. (4) and (5): 5W + ( z − 40)(0.5W ) = 0 z = 30.0 in. −10W − ( x − 20)(0.5W ) = 0 x = 0 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 464 PROBLEM 4.105 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: (ΣMAC = 0). Five unknowns and six equations of equilibrium, but equilibrium is maintained rB = 1.2k TAD TAE rA = 2.4k AD = −0.8i + 0.6 j − 2.4k AD = 2.6 m AE = 0.8i + 1.2 j − 2.4k AE = 2.8 m AD TAD = = (−0.8i + 0.6 j − 2.4k ) AD 2.6 AE TAE = = (0.8i + 1.2 j − 2.4k ) AE 2.8 ΣM C = 0: rA × TAD + rA × TAE + rB × (−3 kN) j = 0 i j k i j k TAD T 0 0 2.4 0 2.4 AE + 1.2k × (−3.6 kN) j = 0 + 0 2.6 2.8 0.8 1.2 −2.4 −0.8 0.6 −2.4 Equate coefficients of unit vectors to zero: i : − 0.55385TAD − 1.02857TAE + 4.32 = 0 (1) j : − 0.73846TAD + 0.68671TAE = 0 TAD = 0.92857TAE From Eq. (1): (2) −0.55385(0.92857)TAE − 1.02857TAE + 4.32 = 0 1.54286TAE = 4.32 TAE = 2.800 kN TAE = 2.80 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 465 PROBLEM 4.105 (Continued) From Eq. (2): TAD = 0.92857(2.80) = 2.600 kN TAD = 2.60 kN 0.8 0.8 (2.6 kN) + (2.8 kN) = 0 2.6 2.8 0.6 1.2 (2.6 kN) + (2.8 kN) − (3.6 kN) = 0 ΣFy = 0: C y + 2.6 2.8 2.4 2.4 (2.6 kN) − (2.8 kN) = 0 ΣFz = 0: C z − 2.6 2.8 ΣFx = 0: C x − Cx = 0 C y = 1.800 kN C z = 4.80 kN C = (1.800 kN) j + (4.80 kN)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 466 PROBLEM 4.106 Solve Problem 4.105, assuming that the 3.6-kN load is applied at Point A. PROBLEM 4.105 A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE. Determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: (ΣMAC = 0). Five unknowns and six equations of equilibrium, but equilibrium is maintained AD = −0.8i + 0.6 j − 2.4k AD = 2.6 m AE = 0.8i + 1.2 j − 2.4k AE = 2.8 m AD TAD (−0.8i + 0.6 j − 2.4k ) TAD = = AD 2.6 AE TAE (0.8i + 1.2 j − 2.4k ) TAE = = AE 2.8 ΣM C = 0: rA × TAD + rA × TAE + rA × (−3.6 kN) j Factor rA : or Coefficient of i: rA × (TAD + TAE − (3.6 kN) j) TAD + TAE − (3 kN) j = 0 − (Forces concurrent at A) TAD T (0.8) + AE (0.8) = 0 2.6 2.8 TAD = Coefficient of j: 2.6 TAE 2.8 (1) TAD T (0.6) + AE (1.2) − 3.6 kN = 0 2.6 2.8 2.6 0.6 1.2 TAE TAE − 3.6 kN = 0 + 2.8 2.6 2.8 0.6 + 1.2 TAE = 3.6 kN 2.8 TAE = 5.600 kN TAE = 5.60 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 467 PROBLEM 4.106 (Continued) From Eq. (1): TAD = 2.6 (5.6) = 5.200 kN 2.8 TAD = 5.20 kN 0.8 0.8 (5.2 kN) + (5.6 kN) = 0 2.6 2.8 0.6 1.2 ΣFy = 0: C y + (5.2 kN) + (5.6 kN) − 3.6 kN = 0 2.6 2.8 2.4 2.4 ΣFz = 0: Cz − (5.2 kN) − (5.6 kN) = 0 2.6 2.8 ΣFx = 0: C x − Cx = 0 Cy = 0 Cz = 9.60 kN C = (9.60 kN)k Note: Since the forces and reaction are concurrent at A, we could have used the methods of Chapter 2. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 468 PROBLEM 4.107 A 10-ft boom is acted upon by the 840-lb force shown. Determine the tension in each cable and the reaction at the ball-and-socket joint at A. SOLUTION We have five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣM x = 0). Free-Body Diagram: BD = (−6 ft)i + (7 ft) j + (6 ft)k BD = 11 ft BE = (−6 ft)i + (7 ft) j − (6 ft)k BE = 11 ft BD TBD TBD = TBD = (−6i + 7 j + 6k ) BD 11 BE TBE TBE = TBE = (−6i + 7 j − 6k ) BE 11 ΣM A = 0: rB × TBD + rB × TBE + rC × ( −840 j) = 0 6i × TBD T (−6i + 7 j + 6k ) + 6i × BE (−6i + 7 j − 6k ) + 10i × (−840 j) = 0 11 11 42 36 42 36 TBD k − TBD j + TBE k + TBE j − 8400k = 0 11 11 11 11 Equate coefficients of unit vectors to zero: i: − k: 36 36 TBD + TBE = 0 TBE = TBD 11 11 42 42 TBD + TBE − 8400 = 0 11 11 42 2 TBD = 8400 11 TBD = 1100 lb TBE = 1100 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 469 PROBLEM 4.107 (Continued) ΣFx = 0: Ax − 6 6 (1100 lb) − (1100 lb) = 0 11 11 Ax = 1200 lb ΣFy = 0: Ay + 7 7 (1100 lb) + (1100 lb) − 840 lb = 0 11 11 Ay = −560 lb ΣFz = 0: Az + 6 6 (1100 lb) − (1100 lb) = 0 11 11 Az = 0 A = (1200 lb)i − (560 lb) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 470 PROBLEM 4.108 A 12-m pole supports a horizontal cable CD and is held by a ball and socket at A and two cables BE and BF. Knowing that the tension in cable CD is 14 kN and assuming that CD is parallel to the x-axis (φ = 0), determine the tension in cables BE and BF and the reaction at A. SOLUTION Free-Body Diagram: There are five unknowns and six equations of equilibrium. The pole is free to rotate about the y-axis, but equilibrium is maintained under the given loading (ΣM y = 0). Resolve BE and BF into components: BE = (7.5 m)i − (8 m) j + (6 m)k BF = (7.5 m)i − (8 m) j − (6 m)k BE = 12.5 m BF = 12.5 m Express TBE and TBF in terms of components: TBE = TBE TBF = TBF BE = TBE (0.60i − 0.64 j + 0.48k ) BE BF = TBF (0.60i − 0.64 j − 0.48k ) BF (1) (2) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 471 PROBLEM 4.108 (Continued) ΣM A = 0: rB/A × TBE + rB/A × TBF + rC/A × (−14 kN)i = 0 8 j × TBE (0.60i − 0.64 j + 0.48k ) + 8 j × TBF (0.60i − 0.64 j − 0.48k ) + 12 j × (−14i ) = 0 −4.8 TBE k + 3.84 TBE i − 4.8TBF k − 3.84TBF i + 168k = 0 Equating the coefficients of the unit vectors to zero, i: 3.84TBE − 3.84TBF = 0 TBE = TBF k: −4.8TBE − 4.8TBF + 168 = 0 ΣFx = 0: Ax + 2(0.60)(17.50 kN) − 14 kN = 0 Ax = 7.00 kN ΣFy = 0: Ay − z (0.64) (17.50 kN) = 0 Ay = 22.4 kN ΣFz = 0: Az + 0 = 0 TBE = TBF = 17.50 kN Az = 0 A = −(7.00 kN)i + (22.4 kN) j Because of the symmetry, we could have noted at the outset that TBF = TBE and eliminated one unknown. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 472 PROBLEM 4.109 Solve Problem 4.108, assuming that cable CD forms an angle φ = 25° with the vertical xy plane. PROBLEM 4.108 A 12-m pole supports a horizontal cable CD and is held by a ball and socket at A and two cables BE and BF. Knowing that the tension in cable CD is 14 kN and assuming that CD is parallel to the x-axis (φ = 0), determine the tension in cables BE and BF and the reaction at A. SOLUTION Free-Body Diagram: BE = (7.5 m)i − (8 m) j + (6 m)k BE = 12.5 m BF = (7.5 m)i − (8 m)j − (6 m)k BF = 12.5 m BE TBE = TBE = TBE (0.60i − 0.64 j + 0.48k ) BE BF TBF = TBF = TBF (0.60i − 0.64 j − 0.48k ) BF ΣM A = 0: rB/A × TBE + rB/A × TBF + rC/A × TCD = 0 8 j × TBE (0.60i − 0.64 j + 0.48k ) + 8 j × TBF (0.60i − 0.64 j − 0.48k ) + 12 j × (19 kN)(− cos 25° i + sin 25°k ) = 0 −4.8TBE k + 3.84TBE i − 4.8TBF k − 3.84TBF i + 152.6 k − 71.00 i = 0 Equating the coefficients of the unit vectors to zero, Solving simultaneously, i: 3.84TBE − 3.84TBF + 71.00 = 0; TBF − TBE = 18.4896 k: − 4.8TBE − 4.8 TBF + 152.26 = 0; TBF + TBE = 31.721 TBE = 6.6157 kN; TBF = 25.105 kN TBE = 6.62 kN; TBF = 25.1 kN ΣFx = 0: Ax + (0.60) (TBF + TBE ) − 14 cos 25° = 0 Ax = 12.6883 − 0.60(31.7207) Ax = −6.34 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 473 PROBLEM 4.109 (Continued) ΣFy = 0: Ay − (0.64) (TBF + TBE ) = 0 Ay = 0.64(31.721) Ay = 20.3 kN ΣFz = 0: Az − 0.48(TBF − TBE ) + 14sin 25° = 0 Az = 0.48(18.4893) − 5.9167 Az = 2.96 kN A = −(6.34 kN)i + (20.3 kN) j + (2.96 kN) k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 474 PROBLEM 4.110 A 48-in. boom is held by a ball-and-socket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣMAC = 0). T = Tension in both parts of cable DAE. rB = 30k rA = 48k AD = −20i − 48k AD = 52 in. AE = 20 j − 48k AE = 52 in. BF = 16i − 30k BF = 34 in. AD T T TAD = T = (−20i − 48k ) = ( −5i − 12k ) AD 52 13 AE T T TAE = T = (20 j − 48k ) = (5 j − 12k ) AE 52 13 T BF TBF TBF = TBF = (16i − 30k ) = BF (8i − 15k ) BF 34 17 ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rB × (−320 lb) j = 0 i j k i j k i j k T T T + 0 0 48 + 0 0 30 BF + (30k ) × (−320 j) = 0 0 0 48 13 13 17 −5 0 −12 0 5 −12 8 0 −15 Coefficient of i: − 240 T + 9600 = 0 13 T = 520 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 475 PROBLEM 4.110 (Continued) Coefficient of j: − 240 240 T+ TBD = 0 13 17 TBD = 17 17 T = (520) TBD = 680 lb 13 13 ΣF = 0: TAD + TAE + TBF − 320 j + C = 0 Coefficient of i: − 20 8 (520) + (680) + Cx = 0 52 17 −200 + 320 + Cx = 0 Coefficient of j: 20 (520) − 320 + C y = 0 52 200 − 320 + C y = 0 Coefficient of k: Cx = −120 lb − C y = 120 lb 48 48 30 (520) − (520) − (680) + Cz = 0 52 52 34 −480 − 480 − 600 + Cz = 0 Cz = 1560 lb Answers: TDAE = T TDAE = 520 lb TBD = 680 lb C = −(120.0 lb)i + (120.0 lb) j + (1560 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 476 PROBLEM 4.111 Solve Problem 4.110, assuming that the 320-lb load is applied at A. PROBLEM 4.110 A 48-in. boom is held by a ball-andsocket joint at C and by two cables BF and DAE; cable DAE passes around a frictionless pulley at A. For the loading shown, determine the tension in each cable and the reaction at C. SOLUTION Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣMAC = 0). T = tension in both parts of cable DAE. rB = 30k rA = 48k AD = −20i − 48k AD = 52 in. AE = 20 j − 48k AE = 52 in. BF = 16i − 30k BF = 34 in. AD T T TAD = T = (−20i − 48k ) = ( −5i − 12k ) AD 52 13 AE T T TAE = T = (20 j − 48k ) = (5 j − 12k ) AE 52 13 T BF TBF TBF = TBF = (16i − 30k ) = BF (8i − 15k ) BF 34 17 ΣM C = 0: rA × TAD + rA × TAE + rB × TBF + rA × ( −320 lb) j = 0 i j k i j k i j k T T T + 0 0 48 + 0 0 30 BF + 48k × (−320 j) = 0 0 0 48 13 13 17 −5 0 −12 0 5 −12 8 0 −15 Coefficient of i: − 240 T + 15,360 = 0 13 T = 832 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 477 PROBLEM 4.111 (Continued) Coefficient of j: − 240 240 T+ TBD = 0 13 17 TBD = 17 17 T = (832) 13 13 TBD = 1088 lb ΣF = 0: TAD + TAE + TBF − 320 j + C = 0 − Coefficient of i: 20 8 (832) + (1088) + C x = 0 52 17 −320 + 512 + Cx = 0 20 (832) − 320 + C y = 0 52 Coefficient of j: 320 − 320 + C y = 0 Coefficient of k: − Cy = 0 48 48 30 (832) − (852) − (1088) + Cz = 0 52 52 34 −768 − 768 − 960 + Cz = 0 Answers: Cx = −192 lb TDAE = T Cz = 2496 lb TDAE = 832 lb TBD = 1088 lb C = −(192.0 lb)i + (2496 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 478 PROBLEM 4.112 A 600-lb crate hangs from a cable that passes over a pulley B and is attached to a support at H. The 200-lb boom AB is supported by a ball-and-socket joint at A and by two cables DE and DF. The center of gravity of the boom is located at G. Determine (a) the tension in cables DE and DF, (b) the reaction at A. SOLUTION Free-Body Diagram: WC = 600 lb WG = 200 lb We have five unknowns (TDE , TDF , Ax , Ay , Az ) and five equilibrium equations. The boom is free to spin about the AB axis, but equilibrium is maintained, since ΣM AB = 0. BH = (30 ft)i − (22.5 ft) j BH = 37.5 ft We have 8.8 DE = (13.8 ft)i − (22.5 ft) j + (6.6 ft)k 12 = (13.8 ft)i − (16.5 ft) j + (6.6 ft)k DE = 22.5 ft DF = (13.8 ft)i − (16.5 ft) j − (6.6 ft)k DF = 22.5 ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 479 PROBLEM 4.112 (Continued) TBH = TBH Thus: TDE = TDE TDF = TDF (a) BH 30i − 22.5 j = (600 lb) = (480 lb)i − (360 lb) j BH 37.5 DE TDE = (13.8i − 16.5 j + 6.6k ) DE 22.5 DF TDE = (13.8i − 16.5 j − 6.6k ) DF 22.5 ΣM A = 0: (rJ × WC ) + (rK × WG ) + (rH × TBH ) + (rE × TDE ) + (rF × TDF ) = 0 − (12i ) × (−600 j) − (6i ) × (−200 j) + (18i ) × (480i − 360 j) i j k i j k TDE TDF 5 0 6.6 + 5 0 + −6.6 = 0 22.5 22.5 13.8 −16.5 6.6 13.8 −16.5 −6.6 7200k + 1200k − 6480k + 4.84(TDE − TDF )i or + 58.08 82.5 (TDE − TDF ) j − (TDE + TDF )k = 0 22.5 22.5 Equating to zero the coefficients of the unit vectors, i or j: TDE − TDF = 0 k : 7200 + 1200 − 6480 − TDE = TDF * 82.5 (2TDE ) = 0 22.5 TDE = 261.82 lb TDE = TDF = 262 lb (b) 13.8 ΣFx = 0: Ax + 480 + 2 (261.82) = 0 22.5 16.5 ΣFy = 0: Ay − 600 − 200 − 360 − 2 (261.82) = 0 22.5 ΣFz = 0: Az = 0 Ax = −801.17 lb Ay = 1544.00 lb A = −(801 lb)i + (1544 lb) j *Remark: The fact is that TDE = TDF could have been noted at the outset from the symmetry of structure with respect to xy plane. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 480 PROBLEM 4.113 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust. SOLUTION rB/A (960 − 180)i = 780i Dimensions in mm 960 450 rG/A = − 90 i + k 2 2 = 390i + 225k rC/A = 600i + 450k T = Tension in cable DCE CD = −690i + 675 j − 450k CE = 270i + 675 j − 450k CD = 1065 mm CE = 855 mm T (−690i + 675 j − 450k ) 1065 T TCE = (270i + 675 j − 450k ) 855 W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N) j TCD = ΣM A = 0: rC/A × TCD + rC/A × TCE + rG/A × (−Wj) + rB/A × B = 0 i j k i j k T T 600 0 450 450 + 600 0 1065 855 270 675 −450 −690 675 −450 i + 390 0 j 0 −981 k i 225 + 780 0 0 j 0 k 0 =0 By Bz PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 481 PROBLEM 4.113 (Continued) Coefficient of i: −(450)(675) T T − (450)(675) + 220.73 × 103 = 0 1065 855 T = 344.64 N Coefficient of j: (−690 × 450 + 600 × 450) T = 345 N 344.64 344.64 + (270 × 450 + 600 × 450) − 780 Bz = 0 1065 855 Bz = 185.516 N Coefficient of k: (600)(675) 344.64 344.64 + (600)(675) − 382.59 × 103 + 780 By = 0 By = 113.178 N 1065 855 B = (113.2 N) j + (185.5 N)k ΣF = 0: A + B + TCD + TCE + W = 0 690 270 (344.64) + (344.64) = 0 1065 855 Coefficient of i: Ax − Ax = 114.5 N Coefficient of j: Ay + 113.178 + 675 675 (344.64) + (344.64) − 981 = 0 1065 855 Ay = 377 N Coefficient of k: Az + 185.516 − 450 450 (344.64) − (344.64) = 0 1065 855 Az = 141.5 N A = (114.5 N)i + (377 N) j + (144.5 N)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 482 PROBLEM 4.114 Solve Problem 4.113, assuming that cable DCE is replaced by a cable attached to Point E and hook C. PROBLEM 4.113 A 100-kg uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE that passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust. SOLUTION See solution to Problem 4.113 for free-body diagram and analysis leading to the following: CD = 1065 mm CE = 855 mm T (−690i + 675 j − 450k ) 1065 T (270i + 675 j − 450k ) TCE = 855 W = −mgi = −(100 kg)(9.81 m/s 2 ) j = −(981 N)j TCD = Now, ΣM A = 0: rC/A × TCE + rG/A × (−W j) + rB/A × B = 0 i j k i j k i j T 600 0 450 0 225 + 780 0 + 390 855 270 675 −450 0 −981 0 0 By Coefficient of i: −(450)(675) k 0 =0 Bz T + 220.73 × 103 = 0 855 T = 621.31 N Coefficient of j: Coefficient of k: (270 × 450 + 600 × 450) (600)(675) T = 621 N 621.31 − 780 Bz = 0 Bz = 364.74 N 855 621.31 − 382.59 × 103 + 780 By = 0 By = 113.186 N 855 B = (113.2 N)j + (365 N)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 483 PROBLEM 4.114 (Continued) ΣF = 0: A + B + TCE + W = 0 270 (621.31) = 0 855 Coefficient of i: Ax + Coefficient of j: Ay + 113.186 + Coefficient of k: Az + 364.74 − Ax = −196.2 N 675 (621.31) − 981 = 0 855 450 (621.31) = 0 855 Ay = 377.3 N Az = −37.7 N A = −(196.2 N)i + (377 N)j − (37.7 N)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 484 PROBLEM 4.115 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION rB/A = (38 − 8)i = 30i rE/A = (30 − 4)i + 20k rG/A = 26i + 20k 38 = i + 10k 2 = 19i + 10k EF = 8i + 25 j − 20k EF = 33 in. AE T T =T = (8i + 25 j − 20k ) AE 33 ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0 i j k i j k i j T + 19 0 10 + 30 0 26 0 20 33 8 25 −20 0 −75 0 0 By −(25)(20) Coefficient of i: Coefficient of j: Coefficient of k: (160 + 520) (26)(25) T + 750 = 0: 33 k 0 =0 Bz T = 49.5 lb 49.5 − 30 Bz = 0: Bz = 34 lb 33 49.5 − 1425 + 30 By = 0: By = 15 lb 33 B = (15.00 lb)j + (34.0 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 485 PROBLEM 4.115 (Continued) ΣF = 0: A + B + T − (75 lb)j = 0 Ax + Coefficient of i: Coefficient of j: Coefficient of k: Ay + 15 + 8 (49.5) = 0 33 25 (49.5) − 75 = 0 33 Az + 34 − 20 (49.5) = 0 33 Ax = −12.00 lb Ay = 22.5 lb Az = −4.00 lb A = −(12.00 lb)i + (22.5 lb)j − (4.00 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 486 PROBLEM 4.116 Solve Problem 4.115, assuming that cable EF is replaced by a cable attached at points E and H. PROBLEM 4.115 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION rB/A = (38 − 8)i = 30i rE/A = (30 − 4)i + 20k = 26i + 20k 38 i + 10k 2 = 19i + 10k rG/A = EH = −30i + 12 j − 20k EH = 38 in. EH T T=T = (−30i + 12 j − 20k ) EH 38 ΣM A = 0: rE/A × T + rG/A × (−75 j) + rB/A × B = 0 i j k i j k i j T + 19 0 10 + 30 0 26 0 20 38 −30 12 −20 0 −75 0 0 By −(12)(20) Coefficient of i: Coefficient of j: Coefficient of k: (−600 + 520) (26)(12) T + 750 = 0 38 T = 118.75 k 0 =0 Bz T = 118.8lb 118.75 − 30 Bz = 0 Bz = −8.33lb 38 118.75 − 1425 + 30 By = 0 B y = 15.00 lb 38 B = (15.00 lb)j − (8.33 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 487 PROBLEM 4.116 (Continued) ΣF = 0: Ax − Coefficient of i: Coefficient of j: Coefficient of k: Ay + 15 + A + B + T − (75 lb)j = 0 30 (118.75) = 0 38 12 (118.75) − 75 = 0 38 Az − 8.33 − 20 (118.75) = 0 38 Ax = 93.75 lb Ay = 22.5 lb Az = 70.83 lb A = (93.8 lb)i + (22.5 lb)j + (70.8 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 488 PROBLEM 4.117 A 20-kg cover for a roof opening is hinged at corners A and B. The roof forms an angle of 30° with the horizontal, and the cover is maintained in a horizontal position by the brace CE. Determine (a) the magnitude of the force exerted by the brace, (b) the reactions at the hinges. Assume that the hinge at A does not exert any axial thrust. SOLUTION Force exerted by CE: F = F (cos 75°)i + F (sin 75°) j F = F (0.25882i + 0.96593j) W = mg = 20 kg(9.81 m/s 2 ) = 196.2 N rA /B = 0.6k rC /B = 0.9i + 0.6k rG /B = 0.45i + 0.3k F = F (0.25882i + 0.96593j) ΣM B = 0: rG/B × (−196.2 j) + rC/B × F + rA/B × A = 0 (a) i j k i j k i 0.45 0 0.3 + 0.9 0 0.6 F + 0 0 −196.2 0 0.25882 +0.96593 0 Ax j 0 Ay k 0.6 = 0 0 Coefficient of i : +58.86 − 0.57956 F − 0.6 Ay = 0 (1) Coefficient of j: +0.155292 F + 0.6 Ax = 0 (2) Coefficient of k: −88.29 + 0.86934 F = 0: From Eq. (2): +58.86 − 0.57956(101.56) − 0.6 Ay = 0 From Eq. (3): +0.155292(101.56) + 0.6 Ax = 0 F = 101.56 N Ay = 0 Ax = −26.286 N F = (101.6 N) (b) Coefficient of i: Coefficient of j: ΣF : A + B + F − Wj = 0 26.286 + Bx + 0.25882(101.56) = 0 Bx = 0 B y + 0.96593(101.56) − 196.2 = 0 B y = 98.1 N Bz = 0 Coefficient of k: A = −(26.3 N)i; B = (98.1 N) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 489 PROBLEM 4.118 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION ΔABH is equilateral. Free-Body Diagram: Dimensions in mm rH/C = −50i + 250 j rD/C = 300i rF/C = 350i + 250k T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k ) ΣM C = 0: rH/C × T + rD × D + rF/C × (−400 j) = 0 i j k i j −50 250 0 T + 300 0 0 0.5 −0.866 0 Dy Coefficient i: k i j k 0 + 350 0 250 = 0 Dz 0 −400 0 −216.5T + 100 × 103 = 0 T = 461.9 N Coefficient of j: T = 462 N −43.3T − 300 Dz = 0 −43.3(461.9) − 300 Dz = 0 Dz = −66.67 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 490 PROBLEM 4.118 (Continued) Coefficient of k: −25T + 300 D y − 140 × 103 = 0 −25(461.9) + 300 Dy − 140 × 103 = 0 D y = 505.1 N D = (505 N) j − (66.7 N)k ΣF = 0: C + D + T − 400 j = 0 Coefficient i: Cx = 0 Cx = 0 Coefficient j: C y + (461.9)0.5 + 505.1 − 400 = 0 C y = −336 N Coefficient k: Cz − (461.9)0.866 − 66.67 = 0 Cz = 467 N C = −(336 N) j + (467 N)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 491 PROBLEM 4.119 Solve Problem 4.115, assuming that the hinge at B is removed and that the hinge at A can exert couples about axes parallel to the y and z axes. PROBLEM 4.115 The rectangular plate shown weighs 75 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B. SOLUTION rE/A = (30 − 4)i + 20k = 26i + 20k rG/A = (0.5 × 38)i + 10k = 19i + 10k AE = 8i + 25 j − 20k AE = 33 in. AE T T =T = (8i + 25 j − 20k ) AE 33 ΣM A = 0: rE/A × T + rG/A × (−75 j) + ( M A ) y j + ( M A ) z k = 0 i j k i j k T + 19 0 10 + ( M A ) y j + ( M A ) z k = 0 26 0 20 33 8 25 −20 0 −75 0 −(20)(25) Coefficient of i: Coefficient of j: Coefficient of k: (160 + 520) (26)(25) T + 750 = 0 33 T = 49.5 lb 49.5 + ( M A ) y = 0 ( M A ) y = −1020 lb ⋅ in. 33 49.5 − 1425 + ( M A ) z = 0 33 ( M A ) z = 450 lb ⋅ in. ΣF = 0: A + T − 75 j = 0 Ax + Coefficient of i: 8 (49.5) = 0 33 M A = −(1020 lb ⋅ in.)j + (450 lb ⋅ in.)k Ax = 12.00 lb Coefficient of j: Ay + 25 (49.5) − 75 = 0 33 Ay = 37.5 lb Coefficient of k: Az − 20 (49.5) = 0 33 Az = 30.0 lb A = −(12.00 lb)i + (37.5 lb)j + (30.0 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 492 PROBLEM 4.120 Solve Problem 4.118, assuming that the bearing at D is removed and that the bearing at C can exert couples about axes parallel to the y and z axes. PROBLEM 4.118 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust. SOLUTION Free-Body Diagram: ΔABH is equilateral. Dimensions in mm rH/C = −50i + 250 j rF/C = 350i + 250k T = T (sin 30°) j − T (cos 30°)k = T (0.5 j − 0.866k ) ΣM C = 0: rF/C × (−400 j) + rH/C × T + ( M C ) y j + ( M C ) z k = 0 i j k i j k 350 0 250 + −50 250 0 T + (M C ) y j + (M C ) z k = 0 0 −400 0 0 0.5 −0.866 Coefficient of i: Coefficient of j: +100 × 103 − 216.5T = 0 T = 461.9 N T = 462 N −43.3(461.9) + ( M C ) y = 0 ( M C ) y = 20 × 103 N ⋅ mm (M C ) y = 20.0 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 493 PROBLEM 4.120 (Continued) Coefficient of k: −140 × 103 − 25(461.9) + ( M C ) z = 0 ( M C ) z = 151.54 × 103 N ⋅ mm (M C ) z = 151.5 N ⋅ m ΣF = 0: C + T − 400 j = 0 M C = (20.0 N ⋅ m)j + (151.5 N ⋅ m)k Coefficient of i: Coefficient of j: Coefficient of k: Cx = 0 C y + 0.5(461.9) − 400 = 0 C y = 169.1 N C z − 0.866(461.9) = 0 C z = 400 N C = (169.1 N)j + (400 N)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 494 PROBLEM 4.121 The assembly shown is welded to collar A that fits on the vertical pin shown. The pin can exert couples about the x and z axes but does not prevent motion about or along the y-axis. For the loading shown, determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram: First note: −(0.08 m)i + (0.06 m) j TCF = λ CF TCF = (0.08) 2 + (0.06)2 m TCF = TCF (−0.8i + 0.6 j) TDE = λ DE TDE = (0.12 m)j − (0.09 m)k (0.12) 2 + (0.09)2 m TDE = TDE (0.8 j − 0.6k ) (a) From F.B.D. of assembly: ΣFy = 0: 0.6TCF + 0.8TDE − 480 N = 0 0.6TCF + 0.8TDE = 480 N or (1) ΣM y = 0: − (0.8TCF )(0.135 m) + (0.6TDE )(0.08 m) = 0 TDE = 2.25TCF or (2) Substituting Equation (2) into Equation (1), 0.6TCF + 0.8[(2.25)TCF ] = 480 N TCF = 200.00 N TCF = 200 N or and from Equation (2): TDE = 2.25(200.00 N) = 450.00 TDE = 450 N or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 495 PROBLEM 4.121 (Continued) (b) From F.B.D. of assembly: ΣFz = 0: Az − (0.6)(450.00 N) = 0 Az = 270.00 N ΣFx = 0: Ax − (0.8)(200.00 N) = 0 Ax = 160.000 N or A = (160.0 N)i + (270 N)k ΣM x = 0: MAx + (480 N)(0.135 m) − [(200.00 N)(0.6)](0.135 m) − [(450 N)(0.8)](0.09 m) = 0 M Ax = −16.2000 N ⋅ m ΣM z = 0: MAz − (480 N)(0.08 m) + [(200.00 N)(0.6)](0.08 m) + [(450 N)(0.8)](0.08 m) = 0 M Az = 0 or M A = −(16.20 N ⋅ m)i PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 496 PROBLEM 4.122 The assembly shown is used to control the tension T in a tape that passes around a frictionless spool at E. Collar C is welded to rods ABC and CDE. It can rotate about shaft FG but its motion along the shaft is prevented by a washer S. For the loading shown, determine (a) the tension T in the tape, (b) the reaction at C. SOLUTION Free-Body Diagram: rA/C = 4.2 j + 2k rE/C = 1.6i − 2.4 j ΣM C = 0: rA/C × (−6 j) + rE/C × T (i + k ) + ( M C ) y j + ( M C ) z k = 0 (4.2 j + 2k ) × (−6 j) + (1.6i − 2.4 j) × T (i + k ) + ( M C ) y j + ( M C ) z k = 0 Coefficient of i: Coefficient of j: Coefficient of k: 12 − 2.4T = 0 T = 5.00 lb −1.6(5 lb) + (M C ) y = 0 ( M C ) y = 8 lb ⋅ in. 2.4(5 lb) + ( M C ) z = 0 ( M C ) z = −12 lb ⋅ in. M C = (8.00 lb ⋅ in.)j − (12.00 lb ⋅ in.)k ΣF = 0: Cx i + C y j + C z k − (6 lb)j + (5 lb)i + (5 lb)k = 0 Equate coefficients of unit vectors to zero. Cx = −5 lb C y = 6 lb Cz = −5 lb C = −(5.00 lb)i + (6.00 lb)j − (5.00 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 497 PROBLEM 4.123 The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram: rB/A = 12i rF/A = 12 j − 8k rD/A = 12i − 16k rE/A = 12i − 24k rF/A = 12i − 32k BG = −12i + 9k BG = 15 in. λ BG = −0.8i + 0.6k DH = −12i + 16 j; DH = 20 in.; λDH = −0.6i + 0.8 j FJ = −12i + 16 j; FJ = 20 in.; λFJ = −0.6i + 0.8 j ΣM A = 0: rB/A × TBG λBG + rDH × TDH λDH + rF/A × TFJ λFJ +rF/A × (−24 j) + rE/A × ( −24 j) = 0 i j k i j k i j k 12 0 0 TBG + 12 0 −16 TDH + 12 0 −32 TFJ −0.8 0 0.6 −0.6 0.8 0 −0.6 0.8 0 i j k i j k + 12 0 −8 + 12 0 −24 = 0 0 −24 0 0 −24 0 Coefficient of i: +12.8TDH + 25.6TFJ − 192 − 576 = 0 (1) Coefficient of k: +9.6TDH + 9.6TFJ − 288 − 288 = 0 (2) 9.6TFJ = 0 TFJ = 0 3 4 Eq. (1) − Eq. (2): PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 498 PROBLEM 4.123 (Continued) 12.8TDH − 268 = 0 TDH = 60 lb −7.2TBG + (16 × 0.6)(60.0 lb) = 0 TBG = 80.0 lb From Eq. (1): Coefficient of j: ΣF = 0: A + TBG λ BG + TDH λ DH + TFJ − 24 j − 24 j = 0 Coefficient of i: Ax + (80)( −0.8) + (60.0)(−0.6) = 0 Coefficient of j: Ay + (60.0)(0.8) − 24 − 24 = 0 Coefficient of k: Az + (80.0)(+0.6) = 0 Ax = 100.0 lb Ay = 0 Az = −48.0 lb A = (100.0 lb)i − (48.0 lb) j Note: The value Ay = 0 can be confirmed by considering ΣM BF = 0. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 499 PROBLEM 4.124 Solve Problem 4.123, assuming that the load at C has been removed. PROBLEM 4.123 The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram: rB/A = 12i rB/A = 12i − 16k rE/A = 12i − 24k rF /A = 12i − 32k BG = −12i + 9k ; BG = 15 in.; λ BG = −0.8i + 0.6k DH = −12i + 16 j; DH = 20 in.; λ DH = −0.6i + 0.8 j FJ = −12i + 16 j; FJ = 20 in.; λ FJ = −0.6i + 0.8 j ΣM A = 0: rB/A × TBG λ BG + rD/A × TDH λ DH + rF /A × TFJ λ FJ + rE/A × (−24 j) = 0 i j k i j k i j k i j k 12 0 0 TBG + 12 0 −16 TDH + 12 0 −32 TFJ + 12 0 −24 = 0 −0.8 0 0.6 −0.6 0.8 0 −0.6 0.8 0 0 −24 0 i: + 12.8TDH + 25.6TFJ − 576 = 0 (1) k: +9.6TDH + 9.6TFJ − 288 = 0 (2) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 500 PROBLEM 4.124 (Continued) Multiply Eq. (1) by 3 4 and subtract Eq. (2): From Eq. (1): 9.6TFJ − 144 = 0 TFJ = 15.00 lb 12.8TDH + 25.6(15.00) − 576 = 0 TDH = 15.00 lb j: −7.2TBG + (16)(0.6)(15) + (32)(0.6)(15) = 0 − 7.2TBG + 432 = 0 TBG = 60.0 lb ΣF = 0: A + TBG λ BG + TDAλ DH + TFJ λ FJ − 24 j = 0 i : Ax + (60)(−0.8) + (15)( −0.6) + (15)(−0.6) = 0 j: Ay + (15)(0.8) + (15)(0.8) − 24 = 0 k: Az + (60)(0.6) = 0 Ax = 66.0 lb Ay = 0 Az = −36.0 lb A = (66.0 lb)i − (36.0 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 501 PROBLEM 4.125 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 1.8-kN load is applied at F, determine the tension in each cable. SOLUTION Free-Body Diagram: In this problem: We have Thus, Dimensions in mm a = 210 mm CD = (240 mm)j − (320 mm)k CD = 400 mm BD = −(420 mm)i + (240 mm)j − (320 mm)k BD = 580 mm BE = (420 mm)i − (320 mm)k BE = 528.02 mm TCD TBD TBE CD = TCD = TCD (0.6 j − 0.8k ) CD BD = TBD = TBD (−0.72414i + 0.41379 j − 0.55172k ) BD BE = TBE = TBE (0.79542i − 0.60604k ) BE ΣM A = 0: (rC × TCD ) + (rB × TBD ) + (rB × TBE ) + (rW × W) = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 502 PROBLEM 4.125 (Continued) rC = −(420 mm)i + (320 mm)k Noting that rB = (320 mm)k rW = − ai + (320 mm)k and using determinants, we write i j k i j k −420 0 320 TCD + 0 0 320 TBD 0 0.6 −0.8 −0.72414 0.41379 −0.55172 i j k i j k + 0 0 320 TBE + −a 0 320 = 0 0.79542 0 −0.60604 0 −1.8 0 Equating to zero the coefficients of the unit vectors, i: −192TCD − 132.413TBD + 576 = 0 (1) j: −336TCD − 231.72TBD + 254.53TBE = 0 (2) k: −252TCD + 1.8a = 0 (3) Recalling that a = 210 mm, Eq. (3) yields TCD = From Eq. (1): 1.8(210) = 1.500 kN 252 −192(1.5) − 132.413TBD + 576 = 0 TBD = 2.1751 kN From Eq. (2): TCD = 1.500 kN TBD = 2.18 kN −336(1.5) − 231.72(2.1751) + 254.53TBE = 0 TBE = 3.9603 kN TBE = 3.96 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 503 PROBLEM 4.126 Solve Problem 4.125, assuming that the 1.8-kN load is applied at C. PROBLEM 4.125 The rigid L-shaped member ABC is supported by a ball-and-socket joint at A and by three cables. If a 1.8-kN load is applied at F, determine the tension in each cable. SOLUTION See solution of Problem 4.125 for free-body diagram and derivation of Eqs. (1), (2), and (3): −192TCD − 132.413TBD + 576 = 0 (1) −336TCD − 231.72TBD + 254.53TBE = 0 (2) −252TCD + 1.8a = 0 (3) In this problem, the 1.8-kN load is applied at C and we have a = 420 mm. Carrying into Eq. (3) and solving for TCD , TCD = 3.00 From Eq. (1): From Eq. (2): −(192)(3) − 132.413TBD + 576 = 0 −336(3) − 0 + 254.53TBE = 0 TCD = 3.00 kN TBD = 0 TBE = 3.96 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 504 PROBLEM 4.127 The assembly shown consists of an 80-mm rod AF that is welded to a cross consisting of four 200-mm arms. The assembly is supported by a ball-and-socket joint at F and by three short links, each of which forms an angle of 45° with the vertical. For the loading shown, determine (a) the tension in each link, (b) the reaction at F. SOLUTION rE/F = −200 i + 80 j TB = TB (i − j) / 2 rB/F = 80 j − 200k TC = TC (− j + k ) / 2 rC/F = 200i + 80 j TD = TD (− i + j) / 2 rD/E = 80 j + 200k ΣM F = 0: rB/F × TB + rC/F × TC + rD/F × TD + rE/F × (− Pj) = 0 i j k i j k i j k i j k TC TB TD 0 80 −200 + 200 80 0 + 0 80 200 + −200 80 0 = 0 2 2 2 1 −1 0 0 −1 1 0 −1 −1 0 −P 0 Equate coefficients of unit vectors to zero and multiply each equation by 2. i: −200 TB + 80 TC + 200 TD = 0 (1) j: −200 TB − 200 TC − 200 TD = 0 (2) k: −80 TB − 200 TC + 80 TD + 200 2 P = 0 (3) −80 TB − 80 TC − 80 TD = 0 (4) 80 (2): 200 Eqs. (3) + (4): −160TB − 280TC + 200 2 P = 0 Eqs. (1) + (2): −400TB − 120TC = 0 TB = − (5) 120 TC − 0.3TC 400 (6) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 505 PROBLEM 4.127 (Continued) Eqs. (6) −160( −0.3TC ) − 280TC + 200 2 P = 0 (5): −232TC + 200 2 P = 0 TC = 1.2191P TC = 1.219 P From Eq. (6): TB = −0.3(1.2191P) = − 0.36574 = P From Eq. (2): − 200(− 0.3657 P) − 200(1.2191P) − 200Tθ D = 0 TD = − 0.8534 P ΣF = 0: TB = −0.366 P TD = − 0.853P F + TB + TC + TD − Pj = 0 i : Fx + (− 0.36574 P) 2 − Fx = − 0.3448P j: Fy − k : Fz + =0 2 Fx = − 0.345P (− 0.36574 P) 2 Fy = P ( − 0.8534 P) − (1.2191P) 2 − (− 0.8534 P) 2 − 200 = 0 Fy = P (1.2191P) 2 =0 Fz = − 0.8620 P Fz = − 0.862 P F = − 0.345P i + Pj − 0.862Pk PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 506 PROBLEM 4.128 The uniform 10-kg rod AB is supported by a ball-and-socket joint at A and by the cord CG that is attached to the midpoint G of the rod. Knowing that the rod leans against a frictionless vertical wall at B, determine (a) the tension in the cord, (b) the reactions at A and B. SOLUTION Free-Body Diagram: Five unknowns and six equations of equilibrium, but equilibrium is maintained (ΣMAB = 0). W = mg = (10 kg) 9.81m/s 2 W = 98.1 N GC = − 300i + 200 j − 225k GC = 425 mm GC T T =T = (− 300i + 200 j − 225k ) GC 425 rB/ A = − 600i + 400 j + 150 mm rG/ A = − 300i + 200 j + 75 mm ΣMA = 0: rB/ A × B + rG/ A × T + rG/ A × (− W j) = 0 i j k i j k i j k T − 600 400 150 + − 300 200 75 + − 300 200 75 425 B 0 0 − 300 200 − 225 0 − 98.1 0 Coefficient of i : (−105.88 − 35.29)T + 7357.5 = 0 T = 52.12 N T = 52.1 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 507 PROBLEM 4.128 (Continued) Coefficient of j : 150 B − (300 × 75 + 300 × 225) 52.12 =0 425 B = 73.58 N B = (73.6 N)i ΣF = 0: A + B + T − W j = 0 Coefficient of i : Coefficient of j: Coefficient of k : 300 =0 425 Ax = −36.8 N 200 − 98.1 = 0 425 Ay = 73.6 N Ax + 73.58 − 52.15 Ay + 52.15 Az − 52.15 225 =0 425 Az = 27.6 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 508 PROBLEM 4.129 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in. SOLUTION From F.B.D. of weldment: ΣM O = 0: rA/O × A + rB/O × B + rC/O × C = 0 i 12 0 j 0 Ay k i 0 + 0 Az Bx j k i 8 0 + 0 0 Bz Cx j k 0 10 = 0 Cy 0 (−12 Az j + 12 Ay k ) + (8 Bz i − 8 Bx k ) + (−10 C y i + 10 Cx j) = 0 From i-coefficient: Bz = 1.25C y or j-coefficient: k-coefficient: or (3) ΣF = 0: A + B + C − P = 0 ( Bx + Cx )i + ( Ay + C y − 240 lb) j + ( Az + Bz )k = 0 From i-coefficient: Bx + Cx = 0 C x = − Bx or j-coefficient: or (2) 12 Ay − 8 Bx = 0 Bx = 1.5 Ay or (1) −12 Az + 10 C x = 0 Cx = 1.2 Az or 8 Bz − 10 C y = 0 (4) Ay + C y − 240 lb = 0 Ay + C y = 240 lb (5) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 509 PROBLEM 4.129 (Continued) k-coefficient: Az + Bz = 0 Az = − Bz or (6) Substituting Cx from Equation (4) into Equation (2), − Bz = 1.2 Az (7) Using Equations (1), (6), and (7), Cy = Bz − Az 1 Bx Bx = = = 1.25 1.25 1.25 1.2 1.5 (8) From Equations (3) and (8): Cy = 1.5 Ay 1.5 or C y = Ay and substituting into Equation (5), 2 Ay = 240 lb Ay = C y = 120 lb (9) Using Equation (1) and Equation (9), Bz = 1.25(120 lb) = 150.0 lb Using Equation (3) and Equation (9), Bx = 1.5(120 lb) = 180.0 lb From Equation (4): Cx = −180.0 lb From Equation (6): Az = −150.0 lb Therefore, A = (120.0 lb) j − (150.0 lb)k B = (180.0 lb)i + (150.0 lb)k C = −(180.0 lb)i + (120.0 lb) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 510 PROBLEM 4.130 Solve Problem 4.129, assuming that the force P is removed and is replaced by a couple M = +(600 lb ⋅ in.)j acting at B. PROBLEM 4.129 Three rods are welded together to form a “corner” that is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 240 lb, a = 12 in., b = 8 in., and c = 10 in. SOLUTION From F.B.D. of weldment: ΣM O = 0: rA/O × A + rB/O × B + rC/O × C + M = 0 i 12 0 j 0 Ay k i 0 + 0 Az Bx j k i 8 0 + 0 0 Bz Cx j k 0 10 + (600 lb ⋅ in.) j = 0 Cy 0 (−12 Az j + 12 Ay k ) + (8 Bz j − 8 Bx k ) + ( −10C y i + 10C x j) + (600 lb ⋅ in.) j = 0 From i-coefficient: 8 Bz − 10 C y = 0 C y = 0.8Bz or j-coefficient: (1) −12 Az + 10 Cx + 600 = 0 Cx = 1.2 Az − 60 or k-coefficient: (2) 12 Ay − 8 Bx = 0 Bx = 1.5 Ay or (3) ΣF = 0: A + B + C = 0 ( Bx + C x )i + ( Ay + C y ) j + ( Az + Bz )k = 0 From i-coefficient: C x = − Bx (4) j-coefficient: C y = − Ay (5) k-coefficient: Az = − Bz (6) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 511 PROBLEM 4.130 (Continued) Substituting Cx from Equation (4) into Equation (2), B Az = 50 − x 1.2 (7) 2 C y = 0.8 Bz = − 0.8 Az = Bx − 40 3 (8) Using Equations (1), (6), and (7), From Equations (3) and (8): C y = Ay − 40 Substituting into Equation (5), 2 Ay = 40 Ay = 20.0 lb From Equation (5): C y = −20.0 lb Equation (1): Bz = −25.0 lb Equation (3): Bx = 30.0 lb Equation (4): Cx = −30.0 lb Equation (6): Az = 25.0 lb A = (20.0 lb) j + (25.0 lb)k Therefore, B = (30.0 lb)i − (25.0 lb)k C = − (30.0 lb)i − (20.0 lb) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 512 PROBLEM 4.131 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe. SOLUTION From F.B.D. of pipe assembly ABCD: ΣFx = 0: Bx = 0 ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0 Bz = 60.0 N and B = (60.0 N)k ΣM D ( z -axis) = 0: C y (3 m) − 90 N ⋅ m = 0 C y = 30.0 N ΣM D ( y -axis) = 0: − C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0 Cz = −16.00 N and C = (30.0 N) j − (16.00 N)k ΣFy = 0: Dy + 30.0 = 0 Dy = −30.0 N ΣFz = 0: Dz − 16.00 N + 60.0 N − 48 N = 0 Dz = 4.00 N and D = − (30.0 N) j + (4.00 N)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 513 PROBLEM 4.132 Solve Problem 4.131, assuming that the plumber exerts a force F = −(48 N)k and that the motor is turned off (M = 0). PROBLEM 4.131 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor that rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F = −(48 N)k , M = −(90 N ⋅ m)k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe. SOLUTION From F.B.D. of pipe assembly ABCD: ΣFx = 0: Bx = 0 ΣM D ( x -axis) = 0: (48 N)(2.5 m) − Bz (2 m) = 0 Bz = 60.0 N and B = (60.0 N)k ΣM D ( z -axis) = 0: C y (3 m) − Bx (2 m) = 0 Cy = 0 ΣM D ( y -axis) = 0: C z (3 m) − (60.0 N)(4 m) + (48 N)(4 m) = 0 Cz = −16.00 N and C = − (16.00 N)k ΣFy = 0: Dy + C y = 0 Dy = 0 ΣFz = 0: Dz + Bz + C z − F = 0 Dz + 60.0 N − 16.00 N − 48 N = 0 Dz = 4.00 N and D = (4.00 N)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 514 PROBLEM 4.133 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire. SOLUTION Free-Body Diagram: W = mg = (50 kg)(9.81 m/s 2 ) W = 490.50 N CE = − 240i + 600 j − 400k CE = 760 mm CE T = (− 240i + 600 j − 400k ) T =T CE 760 AB 480i − 200 j 1 λ AB = = = (12i − 5 j) AB 520 13 ΣMAB = 0: λ AB ⋅ (rE/ A × T ) + λ AB ⋅ (rG/ A × − W j) = 0 rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k 12 0 12 0 −5 −5 1 T 240 400 0 + 240 −100 200 =0 13 × 20 13 − 240 600 − 400 0 −W 0 (−12 × 400 × 400 − 5 × 240 × 400) T + 12 × 200W = 0 760 T = 0.76W = 0.76(490.50 N) T = 373 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 515 PROBLEM 4.134 Solve Problem 4.133, assuming that wire CE is replaced by a wire connecting E and D. PROBLEM 4.133 The 50-kg plate ABCD is supported by hinges along edge AB and by wire CE. Knowing that the plate is uniform, determine the tension in the wire. SOLUTION Free-Body Diagram: Dimensions in mm W = mg = (50 kg)(9.81 m/s 2 ) W = 490.50 N DE = − 240i + 400 j − 400k DE = 614.5 mm DE T = (240i + 400 j − 400k ) T =T DE 614.5 AB 480i − 200 j 1 λ AB = = = (12i − 5 j) AB 520 13 rE/ A = 240i + 400 j; rG/ A = 240i − 100 j + 200k 12 −5 0 12 5 0 1 T 240 400 0 + 240 −100 200 =0 13 × 614.5 13 240 400 − 400 0 −W 0 (−12 × 400 × 400 − 5 × 240 × 400) T + 12 × 200 × W = 0 614.5 T = 0.6145W = 0.6145(490.50 N) T = 301 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 516 PROBLEM 4.135 Two rectangular plates are welded together to form the assembly shown. The assembly is supported by ball-and-socket joints at B and D and by a ball on a horizontal surface at C. For the loading shown, determine the reaction at C. SOLUTION λ BD = First note: −(6 in.)i − (9 in.) j + (12 in.)k (6) 2 + (9) 2 + (12)2 in. 1 (− 6i − 9 j + 12k ) 16.1555 = − (6 in.)i = rA/B P = (80 lb)k rC/D = (8 in.)i C = (C ) j From the F.B.D. of the plates: ΣM BD = 0: λ BD ⋅ (rA/B × P ) + λ BD ⋅ ( rC/D × C ) = 0 −6 −9 12 −6 −9 12 6(80) C (8) −1 0 0 + 1 0 0 =0 16.1555 16.1555 0 0 1 0 1 0 ( −9)(6)(80) + (12)(8)C = 0 C = 45.0 lb or C = (45.0 lb) j PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 517 PROBLEM 4.136 Two 2 × 4-ft plywood panels, each of weight 12 lb, are nailed together as shown. The panels are supported by ball-and-socket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension. SOLUTION Free-Body Diagram: AF = 4i − 2 j − 4k AF = 6 ft 1 λ AF = (2i − j − 2k ) 3 rG1/A = 2i − j rG2 /A = 4i − j − 2k rB/A = 4i ΣM AF = 0: λ AF ⋅ (rG1 /A × ( −12 j) + λ AF ⋅ (rG2 /A × (−12 j)) + λ AF ⋅ (rB /A × T ) = 0 2 −1 −2 2 −1 −2 1 1 + 4 −1 −2 + λ AF ⋅ (rB/A × T) = 0 2 −1 0 3 3 0 −12 0 0 −12 0 1 1 (2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) + λ AF ⋅ (rB/A × T) = 0 3 3 λ AF ⋅ (rB/A × T) = −32 or T ⋅ (λ A/F × rB/A ) = −32 (1) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 518 PROBLEM 4.136 (Continued) Projection of T on (λ AF × rB/A ) is constant. Thus, Tmin is parallel to 1 1 λ AF × rB/A = (2i − j − 2k ) × 4i = (−8 j + 4k ) 3 3 Corresponding unit vector is 1 5 (−2 j + k ). Tmin = T ( −2 j + k ) From Eq. (1): 1 (2) 5 T 1 (−2 j + k ) ⋅ (2i − j − 2k ) × 4i = −32 5 3 T 1 (−2 j + k ) ⋅ ( −8 j + 4k ) = −32 3 5 T (16 + 4) = −32 3 5 T =− 3 5(32) = 4.8 5 20 T = 10.7331 lb From Eq. (2): Tmin = T ( −2 j + k ) 1 5 = 4.8 5( −2 j + k ) Tmin 1 5 = −(9.6 lb)j + (4.8 lb k ) Since Tmin has no i component, wire BH is parallel to the yz plane, and x = 4 ft. (a) (b) x = 4.00 ft; y = 8.00 ft Tmin = 10.73 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 519 PROBLEM 4.137 Solve Problem 4.136, subject to the restriction that H must lie on the y-axis. PROBLEM 4.136 Two 2 × 4-ft plywood panels, each of weight 12 lb, are nailed together as shown. The panels are supported by ball-andsocket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension. SOLUTION Free-Body Diagram: AF = 4i − 2 j − 4k 1 λ AF = (2i − j − 2k ) 3 rG1/A = 2i − j rG2 /A = 4i − j − 2k rB/A = 4i ΣMAF = 0: λ AF ⋅ (rG/A × (−12 j) + λ AF ⋅ (rG2 /A × (−12 j)) + λ AF ⋅ (rB/A × T ) = 0 2 −1 2 2 −1 −2 1 1 2 −1 0 + 4 −1 −2 + λ AF ⋅ (rB/A × T) = 0 3 3 0 −12 0 0 −12 0 1 1 (2 × 2 × 12) + (−2 × 2 × 12 + 2 × 4 × 12) + λ AF ⋅ (rB/A × T) = 0 3 3 λ AF ⋅ (rB/A × T) = −32 BH = −4i + yj − 4k BH = (32 + y 2 )1/2 BH −4i + yj − 4k =T T=T BH (32 + y 2 )1/ 2 (1) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 520 PROBLEM 4.137 (Continued) From Eq. (1): λ AF 2 −1 −2 T ⋅ (rB/A × T ) = 4 0 0 = −32 3(32 + y 2 )1/2 −4 y −4 (−16 − 8 y )T = −3 × 32(32 + y 2 )1/2 T = 96 (32 + y 2 )1/2 8 y + 16 (2) (8y +16) 12 (32 + y 2 ) −1/ 2 (2 y ) + (32 + y 2 )1/2 (8) dT = 0: 96 dy (8 y + 16) 2 Numerator = 0: (8 y + 16) y = (32 + y 2 )8 8 y 2 + 16 y = 32 × 8 + 8 y 2 From Eq. (2): T = 96 (32 + 162 )1/ 2 = 11.3137 lb 8 × 16 + 16 x = 0 ft; y = 16.00 ft Tmin = 11.31 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 521 PROBLEM 4.138 The frame ACD is supported by ball-and-socket joints at A and D and by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the frame supports at Point C a load of magnitude P = 268 N, determine the tension in the cable. SOLUTION Free-Body Diagram: λ AD λ AD TBG TBH AD (1 m)i − (0.75 m)k = = AD 1.25 m = 0.8i − 0.6k BG = TBG BG −0.5i + 0.925 j − 0.4k = TBG 1.125 BH = TBH BH 0.375i + 0.75 j − 0.75k = TBH 1.125 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 522 PROBLEM 4.138 (Continued) rB /A = (0.5 m)i; rC/A = (1 m)i; P = −(268 N)j To eliminate the reactions at A and D, we shall write ΣM AD = 0: λ AD ⋅ (rB /A × TBG ) + λ AD ⋅ (rB /A × TBH ) + λ AD ⋅ (rC /A × P) = 0 (1) Substituting for terms in Eq. (1) and using determinants, −0.6 −0.6 −0.6 0.8 0 0.8 0 0.8 0 TBG TBH + 0.5 + 1 0.5 0 0 0 0 0 0 =0 1.125 1.125 −0.5 0.925 −0.4 0.375 0.75 −0.75 0 −268 0 Multiplying all terms by (–1.125), 0.27750TBG + 0.22500TBH = 180.900 For this problem, (2) TBG = TBH = T (0.27750 + 0.22500)T = 180.900 T = 360 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 523 PROBLEM 4.139 Solve Prob. 4.138, assuming that cable GBH is replaced by a cable GB attached at G and B. PROBLEM 4.138 The frame ACD is supported by ball-andsocket joints at A and D and by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the frame supports at Point C a load of magnitude P = 268 N, determine the tension in the cable. SOLUTION Free-Body Diagram: λ AD λ AD TBG TBH AD (1 m)i − (0.75 m)k = = AD 1.25 m = 0.8i − 0.6k BG = TBG BG −0.5i + 0.925 j − 0.4k = TBG 1.125 BH = TBH BH 0.375i + 0.75 j − 0.75k = TBH 1.125 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 524 PROBLEM 4.139 (Continued) rB /A = (0.5 m)i; rC/A = (1 m)i; P = −(268 N)j To eliminate the reactions at A and D, we shall write ΣM AD = 0: λ AD ⋅ (rB /A × TBG ) + λ AD ⋅ (rB /A × TBH ) + λ AD ⋅ (rC /A × P) = 0 (1) Substituting for terms in Eq. (1) and using determinants, −0.6 −0.6 −0.6 0.8 0 0.8 0 0.8 0 TBG TBH + 0.5 + 1 0.5 0 0 0 0 0 0 =0 1.125 1.125 −0.5 0.925 −0.4 0.375 0.75 −0.75 0 −268 0 Multiplying all terms by (–1.125), 0.27750TBG + 0.22500TBH = 180.900 (2) For this problem, TBH = 0. Thus, Eq. (2) reduces to 0.27750TBG = 180.900 TBG = 652 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 525 PROBLEM 4.140 The bent rod ABDE is supported by ball-and-socket joints at A and E and by the cable DF. If a 60-lb load is applied at C as shown, determine the tension in the cable. SOLUTION Free-Body Diagram: DF = −16i + 11j − 8k DF = 21 in. DE T T=T = (−16i + 11j − 8k ) DF 21 rD/E = 16i rC/E = 16i − 14k EA 7i − 24k = λ EA = EA 25 ΣM EA = 0: λ EA ⋅ (rB/E × T) + λ EA ⋅ (rC/E ⋅ (− 60 j)) = 0 7 0 −24 7 0 −24 1 T 16 0 0 + 16 0 −14 =0 21 × 25 25 0 −60 0 −16 11 −8 − 24 × 16 × 11 −7 × 14 × 60 + 24 × 16 × 60 T+ =0 21 × 25 25 201.14 T + 17,160 = 0 T = 85.314 lb T = 85.3 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 526 PROBLEM 4.141 Solve Problem 4.140, assuming that cable DF is replaced by a cable connecting B and F. SOLUTION Free-Body Diagram: rB/ A = 9i rC/ A = 9i + 10k BF = −16i + 11j + 16k BF = 25.16 in. BF T = (−16i + 11j + 16k ) T =T BF 25.16 AE 7i − 24k λ AE = = AE 25 ΣMAE = 0: λ AF ⋅ (rB/ A × T) + λ AE ⋅ (rC/ A ⋅ (− 60 j)) = 0 7 0 −24 7 0 −24 1 T 9 0 0 +9 0 10 =0 25 × 25.16 25 −16 11 16 0 −60 0 − 24 × 9 × 11 24 × 9 × 60 + 7 × 10 × 60 T+ =0 25 × 25.16 25 94.436 T − 17,160 = 0 T = 181.7 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 527 PROBLEM 4.142 A gardener uses a 60-N wheelbarrow to transport a 250-N bag of fertilizer. What force must she exert on each handle? SOLUTION Free-Body Diagram: ΣM A = 0: (2 F )(1 m) − (60 N)(0.15 m) − (250 N)(0.3 m) = 0 F = 42.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 528 PROBLEM 4.143 The required tension in cable AB is 200 lb. Determine (a) the vertical force P that must be applied to the pedal, (b) the corresponding reaction at C. SOLUTION Free-Body Diagram: BC = 7 in. (a) ΣM C = 0: P(15 in.) − (200 lb)(6.062 in.) = 0 P = 80.83 lb (b) P = 80.8 lb ΣFy = 0: C x − 200 lb = 0 C x = 200 lb ΣFy = 0: C y − P = 0 C y − 80.83 lb = 0 C y = 80.83 lb α = 22.0° C = 215.7 lb C = 216 lb 22.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 529 PROBLEM 4.144 A lever AB is hinged at C and attached to a control cable at A. If the lever is subjected to a 500-N horizontal force at B, determine (a) the tension in the cable, (b) the reaction at C. SOLUTION Triangle ACD is isosceles with C = 90° + 30° = 120° A = D = 1 (180° − 120°) = 30°. 2 Thus, DA forms angle of 60° with the horizontal axis. (a) We resolve FAD into components along AB and perpendicular to AB. ΣM C = 0: ( FAD sin 30°)(250 mm) − (500 N)(100 mm) = 0 (b) FAD = 400 N ΣFx = 0: − (400 N) cos 60° + C x − 500 N = 0 C x = +300 N ΣFy = 0: − (400 N) sin 60° + C y = 0 C y = +346.4 N C = 458 N 49.1° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 530 PROBLEM 4.145 A force P of magnitude 280 lb is applied to member ABCD, which is supported by a frictionless pin at A and by the cable CED. Since the cable passes over a small pulley at E, the tension may be assumed to be the same in portions CE and ED of the cable. For the case when a = 3 in., determine (a) the tension in the cable, (b) the reaction at A. SOLUTION Free-Body Diagram: (a) ΣM A = 0: − (280 lb)(8 in.) 7 T (12 in.) 25 24 − T (8 in.) = 0 25 T (12 in.) − (12 − 11.04)T = 840 (b) ΣFx = 0: T = 875 lb 7 (875 lb) + 875 lb + Ax = 0 25 Ax = −1120 ΣFy = 0: Ay − 280 lb − Ay = +1120 A x = 1120 lb 24 (875 lb) = 0 25 A y = 1120 lb A = 1584 lb 45.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 531 PROBLEM 4.146 Two slots have been cut in plate DEF, and the plate has been placed so that the slots fit two fixed, frictionless pins A and B. Knowing that P = 15 lb, determine (a) the force each pin exerts on the plate, (b) the reaction at F. SOLUTION Free-Body Diagram: (a) (b) ΣFx = 0: 15 lb − B sin 30° = 0 B = 30.0 lb 60.0° ΣM A = 0: − (30 lb)(4 in.) + B sin 30°(3 in.) + B cos 30°(11 in.) − F (13 in.) = 0 −120 lb ⋅ in. + (30 lb) sin 30°(3 in.) + (30 lb) cos 30°(11 in.) − F (13 in.) = 0 F = + 16.2145 lb (a) F = 16.21 lb ΣFy = 0: A − 30 lb + B cos 30° − F = 0 A − 30 lb + (30 lb) cos 30° − 16.2145 lb = 0 A = + 20.23 lb A = 20.2 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 532 PROBLEM 4.147 Knowing that the tension in wire BD is 1300 N, determine the reaction at the fixed support C of the frame shown. SOLUTION T = 1300 N 5 Tx = T 13 = 500 N 12 Ty = T 13 = 1200 N ΣM x = 0: C x − 450 N + 500 N = 0 C x = −50 N ΣFy = 0: C y − 750 N − 1200 N = 0 C y = +1950 N C x = 50 N C y = 1950 N C = 1951 N 88.5° ΣM C = 0: M C + (750 N)(0.5 m) + (4.50 N)(0.4 m) − (1200 N)(0.4 m) = 0 M C = −75.0 N ⋅ m M C = 75.0 N ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 533 PROBLEM 4.148 The spanner shown is used to rotate a shaft. A pin fits in a hole at A, while a flat, frictionless surface rests against the shaft at B. If a 60-lb force P is exerted on the spanner at D, find the reactions at A and B. SOLUTION Free-Body Diagram: (Three-force body) The line of action of A must pass through D, where B and P intersect. 3sin 50° 3cos 50° + 15 = 0.135756 α = 7.7310° tan α = 60 lb sin 7.7310° = 446.02 lb 60 lb B= tan 7.7310° = 441.97 lb A= Force triangle A = 446 lb 7.73° B = 442 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 534 PROBLEM 4.149 Rod AB is supported by a pin and bracket at A and rests against a frictionless peg at C. Determine the reactions at A and C when a 170-N vertical force is applied at B. SOLUTION Free-Body Diagram: (Three-force body) The reaction at A must pass through D where C and the 170-N force intersect. 160 mm 300 mm α = 28.07° tan α = We note that triangle ABD is isosceles (since AC = BC) and, therefore, CAD = α = 28.07° Also, since CD ⊥ CB, reaction C forms angle α = 28.07° with the horizontal axis. Force triangle We note that A forms angle 2α with the vertical axis. Thus, A and C form angle 180° − (90° − α ) − 2α = 90° − α Force triangle is isosceles, and we have A = 170 N C = 2(170 N)sin α = 160.0 N A = 170.0 N 33.9°; C = 160.0 N 28.1° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 535 PROBLEM 4.150 The 24-lb square plate shown is supported by three vertical wires. Determine (a) the tension in each wire when a = 10 in., (b) the value of a for which the tension in each wire is 8 lb. SOLUTION rB/A = ai + 30k rC/A = 30i + ak rG/A = 15i + 15k By symmetry, B = C. ΣM A = 0: rB/A × Bj + rC × Cj + rG/A × (−W j) = 0 (ai + 30k ) × Bj + (30i + ak ) × Bj + (15i + 15k ) × (−W j) = 0 Bak − 30 Bi + 30 Bk − Bai − 15Wk + 15W i = 0 Equate coefficient of unit vector i to zero: i : − 30 B − Ba + 15W = 0 B= 15W 30 + a C=B= 15W 30 + a (1) ΣFy = 0: A + B + C − W = 0 15W A+ 2 − W = 0; 30 + a (a) For a = 10 in. From Eq. (1): C=B= From Eq. (2): A= A= aW 30 + a (2) 15(24 lb) = 9.00 lb 30 + 10 10(24 lb) = 6.00 lb 30 + 10 A = 6.00 lb; B = C = 9.00 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 536 PROBLEM 4.150 (Continued) (b) For tension in each wire = 8 lb, From Eq. (1): 8 lb = 15(24 lb) 30 + a 30 in. + a = 45 a = 15.00 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 537 PROBLEM 4.151 Frame ABCD is supported by a ball-and-socket joint at A and by three cables. For a = 150 mm, determine the tension in each cable and the reaction at A. SOLUTION First note: TDG = λ DG TDG = −(0.48 m)i + (0.14 m)j (0.48) 2 + (0.14) 2 m TDG −0.48i + 0.14 j TDG 0.50 T = DG (24i + 7 j) 25 = TBE = λ BE TBE = −(0.48 m)i + (0.2 m)k (0.48)2 + (0.2)2 m TBE −0.48i + 0.2k TBE 0.52 T = BE (−12 j + 5k ) 13 = From F.B.D. of frame ABCD: 7 ΣM x = 0: TDG (0.3 m) − (350 N)(0.15 m) = 0 25 TDG = 625 N or 24 5 ΣM y = 0: × 625 N (0.3 m) − TBE (0.48 m) = 0 13 25 TBE = 975 N or 7 ΣM z = 0: TCF (0.14 m) + × 625 N (0.48 m) − (350 N)(0.48 m) = 0 25 TCF = 600 N or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 538 PROBLEM 4.151 (Continued) ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0 12 24 Ax − 600 N − × 975 N − × 625 N = 0 13 25 Ax = 2100 N ΣFy = 0: Ay + (TDG ) y − 350 N = 0 7 Ay + × 625 N − 350 N = 0 25 Ay = 175.0 N ΣFz = 0: Az + (TBE ) z = 0 5 Az + × 975 N = 0 13 Az = −375 N A = (2100 N)i + (175.0 N) j − (375 N)k Therefore, PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 539 PROBLEM 4.152 The pipe ACDE is supported by ball-and-socket joints at A and E and by the wire DF. Determine the tension in the wire when a 640-N load is applied at B as shown. SOLUTION Free-Body Diagram: Dimensions in mm AE = 480i + 160 j − 240k AE = 560 mm AE 480i + 160 j − 240k λ AE = = AE 560 6i + 2 j − 3k λ AE = 7 rB/A = 200i rD/A = 480i + 160 j DF = −480i + 330 j − 240k ; DF = 630 mm DF −480i + 330 j − 240k −16i + 11j − 8k = TDF = TDF TDF = TDF DF 630 21 ΣM AE = λ AE ⋅ (rD/A × TDF ) + λ AE ⋅ (rB/A × (−600 j)) = 0 6 2 −3 6 2 −3 TDF 1 480 160 0 0 0 =0 + 200 21 × 7 7 0 −640 0 −16 11 −8 3 × 200 × 640 −6 × 160 × 8 + 2 × 480 × 8 − 3 × 480 × 11 − 3 × 160 × 16 TDF + =0 21 × 7 7 −1120TDF + 384 × 103 = 0 TDF = 342.86 N TDF = 343 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 540 PROBLEM 4.153 A force P is applied to a bent rod ABC, which may be supported in four different ways as shown. In each case, if possible, determine the reactions at the supports. SOLUTION (a) ΣM A = 0: − Pa + (C sin 45°)2a + (cos 45°)a = 0 3 C 2 =P C= 2 P 3 2 1 ΣFx = 0: Ax − P 3 2 Ax = P 3 2 1 P ΣFy = 0: Ay − P + 3 2 Ay = 2P 3 C = 0.471P A = 0.745P (b) 45° 63.4° ΣM C = 0: +Pa − ( A cos 30°)2a + ( A sin 30°)a = 0 A(1.732 − 0.5) = P A = 0.812 P A = 0.812 P ΣFx = 0: (0.812 P )sin 30° + C x = 0 60.0° Cx = −0.406 P ΣFy = 0: (0.812 P) cos 30° − P + C y = 0 C y = −0.297 P C = 0.503P 36.2° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 541 PROBLEM 4.153 (Continued) (c) ΣM C = 0: + Pa − ( A cos 30°)2a + ( A sin 30°)a = 0 A(1.732 + 0.5) = P A = 0.448P A = 0.448P ΣFx = 0: − (0.448P) sin 30° + Cx = 0 60.0° Cx = 0.224 P ΣFy = 0: (0.448 P) cos 30° − P + C y = 0 C y = 0.612 P C = 0.652 P 69.9° (d) Force T exerted by wire and reactions A and C all intersect at Point D. ΣM D = 0: Pa = 0 Equilibrium is not maintained. Rod is improperly constrained. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 542 PROBLEM 4.F1 For the frame and loading shown, draw the free-body diagram needed to determine the reactions at A and E when α = 30°. SOLUTION Free-Body Diagram of Frame: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 543 PROBLEM 4.F2 Neglecting friction, draw the free-body diagram needed to determine the tension in cable ABD and the reaction at C when θ = 60°. SOLUTION Free-Body Diagram of Member ACD: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 544 PROBLEM 4.F3 Bar AC supports two 400-N loads as shown. Rollers at A and C rest against frictionless surfaces and a cable BD is attached at B. Draw the free-body diagram needed to determine the tension in cable BD and the reactions at A and C. SOLUTION Free-Body Diagram of Bar AC: Note: By similar triangles yB 0.15 m = 0.25 m 0.5 m yB = 0.075 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 545 PROBLEM 4.F4 Draw the free-body diagram needed to determine the tension in each cable and the reaction at D. SOLUTION Free-Body Diagram of Member ABCD: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 546 PROBLEM 4.F5 A 4 × 8-ft sheet of plywood weighing 34 lb has been temporarily placed among three pipe supports. The lower edge of the sheet rests on small collars at A and B and its upper edge leans against pipe C. Neglecting friction on all surfaces, draw the free-body diagram needed to determine the reactions at A, B, and C. SOLUTION Free-Body Diagram of Plywood sheet: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 547 PROBLEM 4.F6 Two transmission belts pass over sheaves welded to an axle supported by bearings at B and D. The sheave at A has a radius of 2.5 in. and the sheave at C has a radius of 2 in. Knowing that the system rotates at a constant rate, draw the free-body diagram needed to determine the tension T and the reactions at B and D. Assume that the bearing at D does not exert any axial thrust and neglect the weights of the sheaves and axle. SOLUTION Free-Body Diagram of axle-sheave system: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 548 PROBLEM 4.F7 The 6-m pole ABC is acted upon by a 455-N force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. Draw the free-body diagram needed to determine the tension in each cable and the reaction at A. SOLUTION Free-Body Diagram of Pole: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 549 CHAPTER 5 PROBLEM 5.1 Locate the centroid of the plane area shown. SOLUTION A, in 2 x , in y , in x A,in 3 y A,in 3 1 8 0.5 4 4 32 2 3 2.5 2.5 7.5 7.5 Σ 11 11.5 39.5 X ΣA= xA X (11 in 2 ) = 11.5 in 3 X = 1.045 in. Y ΣA=ΣyA Y (11) = 39.5 Y = 3.59 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 553 PROBLEM 5.2 Locate the centroid of the plane area shown. SOLUTION For the area as a whole, it can be concluded by observation that Y = 2 (72 mm) 3 or Y = 48.0 mm Dimensions in mm A, mm 2 x , mm x A, mm3 1 1 × 30 × 72 = 1080 2 20 21,600 2 1 × 48 × 72 = 1728 2 46 79,488 Σ Then X A = Σ x A 2808 101,088 X (2808) = 101, 088 or X = 36.0 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 554 PROBLEM 5.3 Locate the centroid of the plane area shown. SOLUTION Then A, mm 2 x , mm y , mm xA, mm3 yA, mm3 1 126 × 54 = 6804 9 27 61,236 183,708 2 1 × 126 × 30 = 1890 2 30 64 56,700 120,960 3 1 × 72 × 48 = 1728 2 48 −16 82,944 −27,648 Σ 10,422 200,880 277,020 X ΣA = Σ xA X (10, 422 m 2 ) = 200,880 mm 2 and or X = 19.27 mm Y Σ A = Σ yA Y (10, 422 m 2 ) = 270, 020 mm3 or Y = 26.6 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 555 PROBLEM 5.4 Locate the centroid of the plane area shown. SOLUTION Then A, in 2 x , in y , in x A, in 3 y A, in 3 1 1 (12)(6) = 36 2 4 4 144 144 2 (6)(3) = 18 9 7.5 162 135 Σ 54 306 279 XA = Σ xA X (54) = 306 X = 5.67 in. YA = Σ yA Y (54) = 279 Y = 5.17 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 556 PROBLEM 5.5 Locate the centroid of the plane area shown. SOLUTION By symmetry, X = Y A, in 2 Component I Quarter circle II Square Σ π 4 x , in. x A, in 3 (10) 2 = 78.54 4.2441 333.33 −(5)2 = −25 2.5 −62.5 53.54 270.83 X Σ A = Σ x A: X (53.54 in 2 ) = 270.83 in 3 X = 5.0585 in. X = Y = 5.06 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 557 PROBLEM 5.6 Locate the centroid of the plane area shown. SOLUTION Then A, in 2 x , in. y , in. x A, in 3 y A, in 3 1 14 × 20 = 280 7 10 1960 2800 2 −π (4) 2 = −16π 6 12 –301.59 –603.19 Σ 229.73 1658.41 2196.8 X= Σ xA 1658.41 = ΣA 229.73 X = 7.22 in. Y = Σ y A 2196.8 = Σ A 229.73 Y = 9.56 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 558 PROBLEM 5.7 Locate the centroid of the plane area shown. SOLUTION By symmetry, X = 0 Component A, in 2 y , in. y A, in 3 I Rectangle (3)(6) = 18 1.5 27.0 II Semicircle 2.151 −13.51 Σ − π 2 (2) 2 = −6.28 13.49 11.72 Y ΣA=ΣyA Y (11.72 in.2 ) = 13.49 in 3 Y = 1.151 in. X =0 Y = 1.151 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 559 PROBLEM 5.8 Locate the centroid of the plane area shown. SOLUTION 1 Σ Then x , mm y , mm x A, mm3 y A, mm3 (60)(120) = 7200 –30 60 −216 × 103 432 × 103 (60) 2 = 2827.4 25.465 95.435 72.000 × 103 269.83 × 103 (60) 2 = −2827.4 –25.465 25.465 72.000 × 103 −72.000 × 103 −72.000 × 103 629.83 × 103 π 2 3 A, mm 2 4 − π 4 7200 XA = Σ x A X (7200) = −72.000 × 103 YA = Σ y A Y (7200) = 629.83 × 103 X = −10.00 mm Y = 87.5 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 560 PROBLEM 5.9 Locate the centroid of the plane area shown. SOLUTION A, mm 2 x , mm y , mm x A, mm3 y A, mm3 1 1 (120)(75) = 4500 2 80 25 360 × 103 112.5 × 103 2 (75)(75) = 5625 157.5 37.5 885.94 × 103 210.94 × 103 163.169 43.169 −720.86 × 103 −190.716 × 103 525.08 × 103 132.724 × 103 3 Σ Then − π 4 (75) 2 = −4417.9 5707.1 XA = Σx A X (5707.1) = 525.08 × 103 X = 92.0 mm YA = Σ y A Y (5707.1) = 132.724 × 103 Y = 23.3 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 561 PROBLEM 5.10 Locate the centroid of the plane area shown. SOLUTION X = 0 First note that symmetry implies A, in 2 − 1 2 π (12) 2 2 2 Σ Then π (8) 2 y , in. yA, in 3 = −100.531 3.3953 –341.33 = 226.19 5.0930 1151.99 125.659 Y = 810.66 Σ y A 810.66 in 3 = Σ A 125.66 in 2 or Y = 6.45 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 562 PROBLEM 5.11 Locate the centroid of the plane area shown. SOLUTION X =0 First note that symmetry implies A, m 2 4 × 4.5 × 3 = 18 3 1 2 y, m − π Σ 2 (1.8) 2 = −5.0894 12.9106 yA, m3 1.2 21.6 0.76394 −3.8880 17.7120 Then Y = Σ y A 17.7120 m3 = Σ A 12.9106 m 2 or Y = 1.372 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 563 PROBLEM 5.12 Locate the centroid of the plane area shown. SOLUTION Area mm2 x , mm y , mm xA, mm3 yA, mm3 1 1 (200)(480) = 32 × 103 3 360 60 11.52 × 106 1.92 × 106 2 1 − (50)(240) = 4 × 103 3 180 15 −0.72 × 106 −0.06 × 106 Σ 28 × 103 10.80 × 106 1.86 × 106 X Σ A = Σ x A: X (28 × 103 mm 2 ) = 10.80 × 106 mm3 X = 385.7 mm Y Σ A = Σ y A: X = 386 mm Y (28 × 103 mm 2 ) = 1.86 × 106 mm3 Y = 66.43 mm Y = 66.4 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 564 PROBLEM 5.13 Locate the centroid of the plane area shown. SOLUTION Then A, mm 2 x , mm y , mm xA, mm3 yA, mm3 1 (15)(80) = 1200 40 7.5 48 × 103 9 × 103 2 1 (50)(80) = 1333.33 3 60 30 80 × 103 40 × 103 Σ 2533.3 128 × 103 49 × 103 X A = Σ xA X (2533.3) = 128 × 103 X = 50.5 mm YA = Σ yA Y (2533.3) = 49 × 103 Y = 19.34 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 565 PROBLEM 5.14 Locate the centroid of the plane area shown. SOLUTION Dimensions in in. Then A, in 2 x , in. y , in. xA, in 3 yA, in 3 1 2 (4)(8) = 21.333 3 4.8 1.5 102.398 32.000 2 1 − (4)(8) = −16.0000 2 5.3333 1.33333 85.333 −21.333 Σ 5.3333 17.0650 10.6670 X ΣA = Σ xA X (5.3333 in 2 ) = 17.0650 in 3 and or X = 3.20 in. Y Σ A = Σ yA Y (5.3333 in 2 ) = 10.6670 in 3 or Y = 2.00 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 566 PROBLEM 5.15 Locate the centroid of the plane area shown. SOLUTION Dimensions in mm A, mm 2 1 Then π 2 × 47 × 26 = 1919.51 2 1 × 94 × 70 = 3290 2 Σ 5209.5 x , mm y , mm x A, mm3 y A, mm3 0 11.0347 0 21,181 −15.6667 −23.333 −51,543 −76,766 −51,543 −55,584 X= Σ x A −51,543 = ΣA 5209.5 X = −9.89 mm Y = Σ y A −55,584 = ΣA 5209.5 Y = −10.67 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 567 PROBLEM 5.16 Determine the x coordinate of the centroid of the trapezoid shown in terms of h1, h2, and a. SOLUTION A x xA 1 1 h1a 2 1 a 3 1 h1a 2 6 2 1 h2 a 2 2 a 3 2 h2 a2 6 Σ 1 a(h1 + h 2 ) 2 X= 1 2 a ( h1 + 2h 2 ) 6 1 2 Σ x A 6 a ( h1 + 2h 2 ) = 1 a (h1 + h 2 ) ΣA 2 1 h1 + 2h 2 X= a 3 h1 + h 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 568 PROBLEM 5.17 For the plane area of Problem 5.5, determine the ratio a/r so that the centroid of the area is located at point B. SOLUTION By symmetry, X = Y . For centroid to be at B, X = a. Area x xA I Quarter circle 1 2 πr 4 4r 3π 1 3 r 3 II Square −a 2 1 a 2 1 − a3 2 π Σ 4 r 2 − a2 X Σ A = Σ x A: 1 π 1 X r 2 − a 2 = r 3 − a3 2 4 3 Set X = a : 1 π 1 a r 2 − a 2 = r 3 − a3 2 4 3 1 3 1 3 r − a 3 2 1 3 π 2 1 a − r a + r3 = 0 2 4 3 Divide by 1 3 r : 2 3 a π a 2 r − 2 r + 3 =0 a = 0.508 r PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 569 PROBLEM 5.18 Determine the y coordinate of the centroid of the shaded area in terms of r1, r2, and α. SOLUTION First, determine the location of the centroid. y2 = From Figure 5.8A: Similarly, = 2 cos α r2 π 3 ( 2 −α ) y1 = 2 cos α r1 3 ( π2 − α ) Σ yA = Then π 2 sin ( 2 − α ) r2 π 3 ( 2 −α ) = π A2 = − α r22 2 π A1 = − α r12 2 2 cos α π 2 cos α π 2 − α r22 − r1 π r2 − α r1 3 ( π2 − α ) 2 3 2 α ( 2 − ) ( ) 2 3 3 r2 − r1 cos α 3 π π ΣA = − α r22 − − α r12 2 2 and π = − α r22 − r12 2 ( ) Y ΣA = Σ yA Now π 2 Y − α r22 − r12 = r23 − r13 cos α 2 3 ( ) ( ) Y = 2 r23 − r13 3 r22 − r12 2 cos α π − 2α PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 570 PROBLEM 5.19 Show that as r1 approaches r2, the location of the centroid approaches that for an arc of circle of radius (r1 + r2 )/2. SOLUTION First, determine the location of the centroid. y2 = From Figure 5.8A: Similarly, = 2 cos α r2 3 ( π2 − α ) y1 = 2 cos α r1 3 ( π2 − α ) Σ yA = Then π 2 sin ( 2 − α ) r2 π 3 ( 2 −α ) = π A2 = − α r22 2 π A1 = − α r12 2 2 cos α π 2 cos α π 2 r2 − α r22 − r1 π − α r1 3 ( π2 − α ) 2 3 2 ( 2 − α ) ( ) 2 3 3 r2 − r1 cos α 3 π π ΣA = − α r22 − − α r12 2 2 and π = − α r22 − r12 2 ( ) Y ΣA = Σ yA Now π 2 Y − α r22 − r12 = r23 − r13 2 3 2 r3 − r3 Y = 22 12 3 r2 − r1 ( ) ( ) cos α 2cos α π − 2α PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 571 PROBLEM 5.19 (Continued) Using Figure 5.8B, Y of an arc of radius Y = = Now r23 − r13 r22 − r12 = = 1 (r1 + r2 ) is 2 sin( π − α ) 1 (r1 + r2 ) π 2 2 ( 2 −α) 1 cos α (r1 + r2 ) π 2 ( 2 −α) ( (r2 − r1 ) r22 + r1r2 + r12 (r2 − r1 )(r2 + r1 ) (1) ) r22 + r1r2 + r12 r2 + r1 r2 = r + Δ Let r1 = r − Δ r= Then and In the limit as Δ r23 − r13 r22 − r12 1 (r1 + r2 ) 2 = (r + Δ) 2 + (r + Δ)(r − Δ)( r − Δ) 2 (r + Δ ) + (r − Δ ) = 3r 2 + Δ 2 2r 0 (i.e., r1 = r2 ), then r23 − r13 r22 So that − r12 = 3 r 2 = 3 1 × ( r1 + r2 ) 2 2 Y = 2 3 cos α × (r1 + r2 ) π 3 4 −α 2 or Y = ( r1 + r2 ) cos α π − 2α which agrees with Equation (1). PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 572 PROBLEM 5.20 The horizontal x-axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x-axis, and explain the results obtained. SOLUTION Length of BD: BD = 0.48 in. + (1.44 in. − 0.48 in.) 0.84in. = 0.48 + 0.56 = 1.04 in. 0.84 in. × 0.60 in. Area above x-axis (consider two triangular areas): 1 1 Q1 = Σ y A = (0.28 in.) (0.84 in.)(1.04 in.) + (0.56 in.) (0.84 in.)(0.48 in.) 2 2 = 0.122304 in 3 + 0.112896 in 3 Q1 = 0.2352 in 3 Area below x-axis: 1 1 Q2 = Σ yA = −(0.40 in.) (0.60 in.)(1.44 in.) − (0.20 in.) (0.60 in.) 2 2 = −0.1728 in 3 − 0.0624 in 3 Q2 = −0.2352 in 3 |Q| = |Q2 |, since C is centroid and thus, Q = Σ y A = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 573 PROBLEM 5.21 The horizontal x-axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x-axis, and explain the results obtained. SOLUTION Dimensions in mm Area above x-axis (Area A1): Q1 = Σ y A = (25)(20 × 80) + (7.5)(15 × 20) = 40 × 103 + 2.25 × 103 Q1 = 42.3 × 103 mm3 Area below x-axis (Area A2): Q2 = Σ y A = (−32.5)(65 × 20) Q2 = −42.3 × 103 mm3 |Q1| = |Q2 |, since C is centroid and thus, Q = Σ y A = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 574 PROBLEM 5.22 A composite beam is constructed by bolting four plates to four 60 × 60 × 12-mm angles as shown. The bolts are equally spaced along the beam, and the beam supports a vertical load. As proved in mechanics of materials, the shearing forces exerted on the bolts at A and B are proportional to the first moments with respect to the centroidal x-axis of the red-shaded areas shown, respectively, in parts a and b of the figure. Knowing that the force exerted on the bolt at A is 280 N, determine the force exerted on the bolt at B. SOLUTION From the problem statement, F is proportional to Qx . (Qx ) B FA (Qx ) A Therefore, FA FB = , or (Qx ) A (Qx ) B For the first moments, 12 (Qx ) A = 225 + (300 × 12) 2 FB = = 831, 600 mm3 12 (Qx ) B = (Qx ) A + 2 225 − (48 × 12) + 2(225 − 30)(12 × 60) 2 = 1,364,688 mm3 Then FB = 1,364, 688 (280 N) 831, 600 or FB = 459 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 575 PROBLEM 5.23 The first moment of the shaded area with respect to the x-axis is denoted by Qx. (a) Express Qx in terms of b, c, and the distance y from the base of the shaded area to the x-axis. (b) For what value of y is Qx maximum, and what is that maximum value? SOLUTION Shaded area: A = b (c − y ) Qx = yA = Qx = (a) (b) For Qmax , For y = 0, 1 (c + y )[b(c − y )] 2 1 b (c 2 − y 2 ) 2 dQ = 0 or dy (Qx ) = 1 b(−2 y ) = 0 2 1 2 bc 2 y=0 (Qx ) = 1 2 bc 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 576 PROBLEM 5.24 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. L, mm Then x , mm y , mm yL, mm 2 yL, mm 2 1 302 + 722 = 78 15 36 1170.0 2808.0 2 482 + 722 = 86.533 54 36 4672.8 3115.2 39 72 3042.0 5616.0 8884.8 11,539.2 3 78 Σ 242.53 X ΣL = Σx L X (242.53) = 8884.8 or X = 36.6 mm and Y ΣL = Σ yL Y (242.53) = 11,539.2 or Y = 47.6 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 577 PROBLEM 5.25 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION L, mm 1 Then 722 + 482 = 86.533 x , mm y , mm xL, mm 2 yL, mm 2 36 −24 3115.2 −2076.8 2 132 72 18 9504.0 2376.0 3 1262 + 302 = 129.522 9 69 1165.70 8937.0 4 54 −54 27 −2916.0 5 54 −27 0 −1458.0 Σ 456.06 9410.9 1458.0 0 10,694.2 X ΣL = Σx L X (456.06) = 9410.9 or X = 20.6 mm Y ΣL = Σ y L Y (456.06) = 10, 694.2 or Y = 23.4 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 578 PROBLEM 5.26 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION Then y , in. xL, in 2 yL, in 2 3 80.498 40.249 7.5 36 22.5 9 9 54 54.0 3 6 7.5 18 22.5 5 6 3 6 18 36.0 6 6 0 3 0 18.0 Σ 37.416 206.50 193.249 L, in. x , in. 1 122 + 62 = 13.4164 6 2 3 12 3 6 4 X ΣL = Σx L Y ΣL =Σ y L X (37.416) = 206.50 Y (37.416) = 193.249 X = 5.52 in. Y = 5.16 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 579 PROBLEM 5.27 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION By symmetry, X = Y . x , in. L, in. 1 Then 1 π (10) = 15.7080 2 2(10) π = 6.3662 yL, in 2 100 2 5 0 0 3 5 2.5 12.5 4 5 5 25 5 5 7.5 37.5 Σ 35.708 X ΣL = Σx L 175 X (35.708) = 175 X = Y = 4.90 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 580 PROBLEM 5.28 The homogeneous wire ABCD is bent as shown and is attached to a hinge at C. Determine the length L for which portion BCD of the wire is horizontal. SOLUTION First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through C. Further, because the wire is homogeneous, the center of gravity of the wire will coincide with the centroid of the corresponding line. Thus, X = 0 so that Σ X L = 0 Then L + (−40 mm)(80 mm) + (−40 mm)(100 mm) = 0 2 L2 = 14, 400 mm 2 L = 120.0 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 581 PROBLEM 5.29 The homogeneous wire ABCD is bent as shown and is attached to a hinge at C. Determine the length L for which portion AB of the wire is horizontal. SOLUTION WI = 80w WII = 100w WIII = Lw Σ M C = 0: (80 w)(32) + (100 w)(14) − ( Lw)(0.4 L) = 0 L2 = 9900 L = 99.5 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 582 PROBLEM 5.30 The homogeneous wire ABC is bent into a semicircular arc and a straight section as shown and is attached to a hinge at A. Determine the value of θ for which the wire is in equilibrium for the indicated position. SOLUTION First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through A. Further, because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. Thus, X =0 so that Σx L = 0 Then 1 2r − 2 r cos θ (r ) + π − r cos θ (π r ) = 0 or cos θ = 4 1 + 2π = 0.54921 or θ = 56.7° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 583 PROBLEM 5.31 A uniform circular rod of weight 8 lb and radius 10 in. is attached to a pin at C and to the cable AB. Determine (a) the tension in the cable, (b) the reaction at C. SOLUTION For quarter circle, (a) r = 2r π 2r ΣM C = 0: W − Tr = 0 π 2 2 T = W = (8 lb) π π (b) T = 5.09 lb ΣFx = 0: T − C x = 0 5.09 lb − C x = 0 C x = 5.09 lb ΣFy = 0: C y − W = 0 C y = 8 lb C y − 8 lb = 0 C = 9.48 lb 57.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 584 PROBLEM 5.32 Determine the distance h for which the centroid of the shaded area is as far above line BB′ as possible when (a) k = 0.10, (b) k = 0.80. SOLUTION A y yA 1 1 ba 2 1 a 3 1 2 a b 6 2 1 − (kb)h 2 1 h 3 1 − kbh 2 6 Σ b (a − kh) 2 b 2 ( a − kh 2 ) 6 Y Σ A = Σ yA Then b b Y ( a − kh) = (a 2 − kh 2 ) 2 6 Y = or (1) dY 1 −2kh(a − kh) − (a 2 − kh 2 )(− k ) = =0 dh 3 ( a − kh) 2 and or a 2 − kh 2 3(a − kh) 2h(a − kh) − a 2 + kh 2 = 0 (2) Simplifying Eq. (2) yields kh 2 − 2ah + a 2 = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 585 PROBLEM 5.32 (Continued) Then 2a ± (−2a) 2 − 4( k )(a 2 ) 2k a = 1 ± 1 − k k h= Note that only the negative root is acceptable since h < a. Then (a) k = 0.10 h= (b) a 1 − 1 − 0.10 0.10 or h = 0.513a k = 0.80 h= a 1 − 1 − 0.80 0.80 or h = 0.691a PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 586 PROBLEM 5.33 Knowing that the distance h has been selected to maximize the distance y from line BB′ to the centroid of the shaded area, show that y = 2h/3. SOLUTION See solution to Problem 5.32 for analysis leading to the following equations: Y = a 2 − kh 2 3(a − kh) (1) 2h(a − kh) − a 2 + kh 2 = 0 (2) Rearranging Eq. (2) (which defines the value of h which maximizes Y ) yields a 2 − kh 2 = 2h(a − kh) Then substituting into Eq. (1) (which defines Y ), Y = 1 × 2h(a − kh) 3( a − kh) or Y = 2 h 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 587 PROBLEM 5.34 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION h x a y= We have dA = (h − y )dx and x = h 1 − dx a xEL = x 1 (h + y ) 2 h x = 1 + 2 a yEL = A = dA = Then and x y a = EL dA = EL dA = 0 a 0 h2 2 a 0 a x x2 1 h 1 − dx = h x − = ah a 2 a 0 2 a x 2 x3 x 1 x h 1 − dx = h − = a 2 h a 2 3a 0 6 h x x 1 + h 1 − dx a a 2 a 0 x2 − 1 2 a a h2 x3 1 = − = ah2 dx x 2 2 3a 0 3 1 1 xA = xEL dA: x ah = a 2 h 2 6 x= 2 a 3 1 1 y A = yEL dA: y ah = ah2 2 3 y= 2 h 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 588 PROBLEM 5.35 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION y = h(1 − kx3 ) For x = a, y = 0. 0 = h(1 − k a3 ) ∴ k= 1 a3 x3 y = h 1 − 3 a xEL = x, yEL = a A = dA = x EL dA = yEL dA = = a 0 0 xydx = 1 y dA = ydx 2 ydx = a 0 h2 2 0 a x3 x4 3 h 1 − 3 dx = h x − 3 = ah 4a 0 4 a a x2 x4 x5 3 h x − 3 dx = h − 3 = a 2 h a 2 5a 0 10 a 0 a 1 1 y ydx = 2 2 a 0 x3 h2 h 2 1 − 3 dx = 2 a a 0 2 x3 x 6 1 − 3 + 6 a a dx a x4 x7 9 2 ah x − 3 + 6 = 28 2a 7a 0 3 3 xA = xEL dA: x ah = a 2 h 4 10 x= 2 a 5 3 9 2 yA = yEL dA: y ah = ah 4 28 y= 3 h 7 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 589 PROBLEM 5.36 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION y1 : h = ka 2 At (a, h), k= or h a2 y2 : h = ma m= or xEL = x Now yEL = 1 ( y1 + y2 ) 2 h h dA = ( y2 − y1 )dx = x − 2 x 2 dx a a h = 2 (ax − x 2 ) dx a and A = dA = Then and h a a 0 a h h a 1 1 (ax − x 2 )dx = 2 x 2 − x3 = ah 2 3 0 6 a a 2 a h a 1 1 h xEL dA = x 2 (ax − x 2 ) dx = 2 x3 − x 4 = a 2 h 0 a 4 0 12 a 3 1 1 2 yEL dA = y2 − y12 dx ( y1 + y2 )[( y2 − y1 ) dx] = 2 2 a ( = 1 2 ) a 0 h 2 2 h2 4 2 x − 4 x dx a a a 1 h2 a 2 3 1 5 = x − x 2 a4 3 5 0 1 = ah 2 15 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 590 PROBLEM 5.36 (Continued) 1 1 xA = xEL dA: x ah = a 2 h 6 12 x= 1 a 2 1 1 yA = yEL dA: y ah = ah 2 6 15 y= 2 h 5 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 591 PROBLEM 5.37 Determine by direct integration the centroid of the area shown. SOLUTION x , k y2 = y1 = k x 2 a = k a 2 , thus, k = But y2 = a x , y1 = 1 a2 x2 a xEL = x x2 dA = ( y2 − y1 ) dx = ax − dx a A = dA = a x2 ax − dx 0 a a 2 1 x3 = a x3/2 − = a 2 3a 0 3 3 x EL dA = a 0 x2 x a x − dx = a a 0 a x 1 3 3 xA = xEL dA: x a 2 = a 3 20 a 3/2 2 x3 x4 3 3 − dx = a x5/2 − = a a 4a 0 20 5 x= 9a 20 y=x= By symmetry, 9a 20 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 592 PROBLEM 5.38 Determine by direct integration the centroid of the area shown. SOLUTION y= For the element (EL) shown, b 2 a − x2 a dA = (b − y ) dx and xEL ) ( b a − a 2 − x 2 dx a =x = 1 ( y + b) 2 b = a + a 2 − x2 2a yEL = ( A = dA = Then To integrate, let x = a sin θ : Then A= π /2 0 a 0 ) ) ( b a − a 2 − x 2 dx a a 2 − x 2 = a cos θ , dx = a cos θ dθ b (a − a cos θ )(a cos θ dθ ) a π /2 = b 2 2θ 2 θ a sin θ − a + sin a 4 0 2 π = ab 1 − 4 and x EL dA a ) ( b x a − a 2 − x 2 dx a = = b a 2 1 2 2 3/2 x + ( a − x ) a 2 3 0 = 1 3 ab 6 0 π /2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 593 PROBLEM 5.38 (Continued) y EL dA = a 0 b2 = 2 2a = ) ( ( ) b b a + a 2 − x 2 a − a 2 − x 2 dx 2a a a a 0 b 2 x3 ( x ) dx = 2 2a 3 0 2 1 2 ab 6 π 1 x ab 1 − = a 2 b 4 6 or x = 2a 3(4 − π ) π 1 y ab 1 − = ab 2 4 6 or y = 2b 3(4 − π ) xA = xEL dA: yA = yEL dA: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 594 PROBLEM 5.39 Determine by direct integration the centroid of the area shown. SOLUTION x =0 First note that symmetry implies For the element (EL) shown, yEL = 2r (Figure 5.8B) π dA = π rdr A = dA = Then and So yEL dA = r2 r1 r2 r1 r2 π rdr = π 2 r2 π 2 2 r2 − r1 = r1 2 r2 ( 2 1 (π rdr ) = 2 r 3 = r23 − r13 π 3 r1 3 2r ( ) π 2 yA = yEL dA: y r22 − r12 = r23 − r13 2 3 ( ) ( ) ) or y = 4 r23 − r13 3π r22 − r12 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 595 PROBLEM 5.40 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION b 2 x a2 b y2 = k2 x 4 but b = k2 a 4 y2 = 4 x 4 a 4 b x dA = ( y2 − y1 )dx = 2 x 2 − 2 dx a a xEL = x y1 = k1 x 2 but b = k1a 2 1 ( y1 + y2 ) 2 b x4 = 2 x 2 + 2 2a a y1 = yEL = A = dA = b a2 a 0 x4 2 − x a2 dx a b x3 x5 = 2 − 2 a 3 5a 0 2 = ba 15 a b x4 xEL dA = x 2 x 2 − 2 dx 0 a a b a x5 = 2 x3 − 2 dx a 0 a a x4 x6 − 2 4 6a 0 = b a2 = 1 2 a b 12 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 596 PROBLEM 5.40 (Continued) yEL dA = = a 0 b 2a 2 b2 2a 4 2 x4 b x + 2 2 a a 2 x4 x − 2 dx a a 0 x8 4 x − 4 dx a a b 2 x5 x9 2 2 = 4 − 4 = ab 2a 5 9a 0 45 2 1 xA = xEL dA: x ba = a 2b 15 12 5 x= a 8 2 2 2 yA = yEL dA: y ba = ab 15 45 1 y = b 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 597 PROBLEM 5.41 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION y =b First note that symmetry implies x = a, At y=b y1 : b = ka 2 y1 = Then k= or b a2 b 2 x a2 y2 : b = 2b − ca 2 c= or x2 y2 = b 2 − 2 a Then x2 dA = ( y2 − y1 )dx2 = b 2 − 2 a x2 = 2b 1 − 2 dx a Now b 2 − 2 x dx a xEL = x and A = dA Then and b a2 a 0 x2 2b 1 − 2 a a x3 4 dx = 2b x − 2 = ab 3a 0 3 a x2 x2 x4 1 xEL dA = x 2b 1 − 2 dx = 2b − 2 = a 2b 0 a 2 4a 0 2 4 1 xA = xEL dA: x ab = a 2b 3 2 a 3 x= a 8 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 598 PROBLEM 5.42 Determine by direct integration the centroid of the area shown. SOLUTION xEL = x yEL = L A = dA = xEL dA = L 0 0 1 y dA = y dx 2 L x x2 x 2 2 x3 5 h 1 + − 2 2 dx = h x + − = hL 2 L 2L 3 L 0 6 L x x2 xh 1 + − 2 2 dx = h L L L 0 x2 x3 − 2 2 dx x + L L L x 2 1 x3 2 x 4 1 = h + − = hL2 2 2 3 L 4 L 0 3 5 1 xA = xEL dA: x hL = hL2 6 3 A= 5 hL 6 yEL = = h2 2 h2 = 2 L 0 2 L 5 x x2 y = h 1 + − 2 2 L L 1 y 2 1 2 h2 yEL dA = y dx = 2 2 x= 2 x x2 1 + − 2 2 dx L L L 0 x2 x4 x x2 x3 1 + 2 + 4 4 + 2 − 4 2 − 4 3 dx L L L L L L 4 x3 4 x5 x 2 4 x3 x 4 x + + 4 + − 2 − 3 = h2 L 2 10 L 3 5 3 L L L L 0 5 4 yA = yEL dA: y hL = h 2 L 6 10 y= 12 h 25 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 599 PROBLEM 5.43 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION For y1 at x = a, y = 2b, 2b = ka 2 , or k = Then y1 = By observation, b x y2 = − ( x + 2b) = b 2 − a a 2b a2 2b 2 x a2 Now xEL = x and for 0 ≤ x ≤ a, yEL = 1 b y1 = 2 x 2 2 a For a ≤ x ≤ 2a, yEL = 1 b x x y2 = 2 − and dA = y2 dx = b 2 − dx a a 2 2 A = dA = Then a 0 and dA = y1 dx = 2b 2 x dx + a2 2a a 2b 2 x dx a2 x b 2 − dx a 2a 2 a 2b x3 x 7 = 2 + b − 2 − = ab a 6 a 3 0 2 0 a and xEL dA = a 0 2b x 2 x 2 dx + a 2a a a x x b 2 − dx a 2a 2b x 4 x3 = 2 + b x2 − 3a 0 a 4 0 1 2 1 2 (2a ) − (a )3 a b + b (2a) 2 − (a) 2 + 2 3 a 7 2 = a b 6 = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 600 PROBLEM 5.43 (Continued) yEL dA = a 0 b 2 2b 2 x x dx + a 2 a 2 2a 0 b x x 2 − b 2 − dx 2 a a 2a 3 2b 2 x5 b2 a x = 4 + − 2 − a a 5 0 2 3 a 17 2 ab = 30 a Hence, 7 7 xA = xEL dA: x ab = a 2b 6 6 7 17 2 yA = yEL dA: y ab = ab 6 30 x =a y= 17 b 35 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 601 PROBLEM 5.44 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION For y2 at x = a, y = b, a = kb 2 , or k = Then y2 = Now xEL = x and for 0 ≤ x ≤ a , 2 yEL = b a x1/2 y2 b x1/ 2 = 2 2 a x1/2 dA = y2 dx = b For a ≤ x ≤ a, 2 yEL = a b2 a dx 1 b x 1 x1/ 2 ( y1 + y2 ) = − + 2 2a 2 a x1/2 x 1 dA = ( y2 − y1 )dx = b − + dx a a 2 Then A = dA = a /2 b 0 x1/ 2 a dx + x1/2 x 1 b − + dx a/ 2 a a 2 a a a/ 2 2 x3/2 x 2 1 b 2 3/2 = x + b − + x a 3 0 3 a 2a 2 a/2 3/ 2 3/ 2 2 b a a 3/ 2 = + ( a ) − 3 a 2 2 2 1 2 a 1 a + b − (a ) − + (a) − 2a 2 2 2 13 ab = 24 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 602 PROBLEM 5.44 (Continued) and xEL dA = a/2 0 x1/ 2 x b dx + a x1/ 2 x 1 x b − + dx a/2 a a 2 a a a/ 2 2 x5/2 x3 x 4 b 2 5/2 = + − + x b a 5 0 5 a 3a 4 a/2 5/2 5/2 2 b a a 5/2 = + ( a ) − 5 a 2 2 3 2 1 a 1 a + b − ( a )3 − + ( a ) 2 − 2 4 2 3a 71 2 a b = 240 yEL dA = a/2 0 + b x1/ 2 2 a x1/ 2 dx b a b x 1 x1/2 x1/ 2 x 1 − + dx − + b a/ 2 2 a 2 a a a 2 a a a/ 2 3 b2 1 2 b 2 x 2 1 x 1 x = + − − 2a 2 0 2 2a 3a a 2 a/2 Hence, = 3 2 2 2 b a a b a 1 − + ( a ) 2 − − 4a 2 2 6a 2 2 = 11 2 ab 48 13 71 2 xA = xEL dA : x ab = a b 24 240 13 11 2 yA = yEL dA: y ab = ab 24 48 x= 17 a = 0.546a 130 y= 11 b = 0.423b 26 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 603 PROBLEM 5.45 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line. Now xEL = r cos θ Then L = dL = and xEL dL = 7π /4 π /4 and dL = rdθ 7π / 4 π /4 3 r dθ = r[θ ]π7π/ 4/ 4 = π r 2 r cos θ (rdθ ) = r 2 [sin θ ] π7π/4/ 4 1 1 = r2 − − 2 2 = −r 2 2 Thus 3 xL = x dL : x π r = − r 2 2 2 x =− 2 2 r 3π PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 604 PROBLEM 5.46 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line. xEL = a cos3 θ Now and dL = dx 2 + dy 2 x = a cos3 θ : dx = −3a cos 2 θ sin θ dθ where y = a sin 3 θ : dy = 3a sin 2 θ cos θ dθ dL = [(−3a cos 2 θ sin θ dθ ) 2 + (3a sin 2 θ cos θ dθ )2 ]1/ 2 Then = 3a cos θ sin θ (cos 2 θ + sin 2 θ )1/ 2 dθ = 3a cos θ sin θ dθ L = dL = and x EL dL = 3 a 2 = π /2 0 π /2 0 π /2 1 3a cos θ sin θ dθ = 3a sin 2 θ 2 0 a cos3θ (3a cos θ sin θ dθ ) π /2 1 = 3a 2 − cos5 θ 5 0 Hence, = 3 2 a 5 3 3 xL = xEL dL : x a = a 2 2 5 x= 2 a 5 Alternative Solution: x x = a cos3 θ cos 2 θ = a y y = a sin θ sin θ = a 3 x a 2/3 2/3 2/3 2 y + a 2/3 = 1 or y = (a 2/3 − x 2/3 )3/2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 605 PROBLEM 5.46 (Continued) dy = (a 2/3 − x 2/3 )1/ 2 (− x −1/3 ) dx Then Now xEL = x and dy dL = 1 + dx 2 { dx = 1 + ( a 2/3 − x 2/3 )1/2 (− x −1/3 ) L = dL = Then and Hence xEL dL = a 0 a 0 2 } 1/2 dx a a1/3 3 3 dx = a1/3 x 2/3 = a 1/3 x 2 0 2 a a1/3 3 3 x 1/3 dx = a1/3 x5/3 = a 2 5 0 5 x 3 3 xL = xEL dL : x a = a 2 2 5 x= 2 a 5 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 606 PROBLEM 5.47* A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. Express your answer in terms of a. SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. We have at x = a, y = a, a = ka3/2 , or k = Then y= 1 a 1 a x3/ 2 dy 3 1/2 = x dx 2 a and Now xEL = x and dy dL = 1 + dx dx 2 1/2 2 3 x1/2 = 1 + 2 a 1 = 4a + 9 x dx 2 a L = dL = Then a 1 0 2 a dx 4a + 9 x dx a 1 2 1 × (4a + 9 x)3/ 2 3 9 2 a 0 a = [(13)3/2 − 8] 27 = 1.43971a = and x EL dL = a 0 1 x 2 a 4a + 9 x dx PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 607 PROBLEM 5.47* (Continued) Use integration by parts with u=x du = dx Then dv = 4a + 9 x dx 2 v= (4a + 9 x)3/ 2 27 a 1 2 3/2 xEL dL = x × (4a + 9 x) − 2 a 27 0 a 0 2 (4a + 9 x)3/ 2 dx 27 a = (13)3/2 2 1 2 (4a + 9 x)5/2 a − 27 27 a 45 0 = 2 a2 3/ 2 5/2 (13) − [(13) − 32] 27 45 = 0.78566a 2 xL = xEL dL : x (1.43971a ) = 0.78566a 2 or x = 0.546a PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 608 PROBLEM 5.48 Determine by direct integration the centroid of the area shown. SOLUTION y = a sin πx L 1 y, dA = y dx 2 L /2 L /2 πx sin A= y dx = a dx 0 0 L xEL = x, yEL = L /2 L π x A = a − cos L 0 π Setting u = πx L , we have x = Integrating by parts, L /2 xEL dA = L u , dx = L π 0 π xydx = L /2 aL = π xa sin 0 πx L du, π /2 L L L π u a sin u π du = a π xEL dA = L xEL dA = a [−u cos u ]π0 /2 + π yEL dA = 0 2 = dx L /2 0 a2 L 2π 2 1 2 1 y dx = a 2 2 2 π /2 1 0 2 L /2 0 π /2 0 sin 2 2 π /2 0 u sin x du 2 aL cos u du = 2 π πx L dx = a2 L 2π π /2 0 sin 2 u du π /2 (1 − cos 2u )du = 2 aL aL xA = xEL dA: x = 2 π π aL 1 2 yA = yEL dA: y = a L π 8 a2 L 1 u − sin 2u 4π 2 0 1 = a2 L 8 x= y= π 8 L π a PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 609 PROBLEM 5.49* Determine by direct integration the centroid of the area shown. SOLUTION 2 2 r cos θ = aeθ cos θ 3 3 2 2 θ = r sin θ = ae sin θ 3 3 xEL = We have yEL dA = and 1 1 (r )(rdθ ) = a 2 e2θ d θ 2 2 A = dA = Then π 0 π 1 2 2θ 1 1 a e dθ = a 2 e 2θ 2 2 2 0 1 2 2π a (e − 1) 4 = 133.623a 2 = and x EL dA = π 0 2 θ 1 ae cos θ a 2 e2θ dθ 3 2 1 = a3 3 π 0 e3θ cos θ dθ To proceed, use integration by parts, with u = e3θ du = 3e3θ dθ and dv = cos θ dθ Then e 3θ then du = 3e3θ dθ dv = sin θ dθ , then Then e 3θ v = sin θ cos θ dθ = e3θ sin θ − sin θ (3e3θ dθ ) u = e3θ Now let and v = − cos θ cos θ dθ = e3θ sin θ − 3 −e3θ cos θ − ( − cos θ )(3e3θ dθ ) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 610 PROBLEM 5.49* (Continued) so that e3θ cos θ dθ = π 1 e3θ (sin θ + 3cos θ ) xEL dA = a 3 3 10 0 = Also, e3θ (sin θ + 3cos θ ) 10 yEL dA = a3 (−3e3π − 3) = −1239.26a3 30 π 0 2 θ 1 ae sin θ a 2 e2θ dθ 3 2 1 = a3 3 π 0 e3θ sin θ dθ Use integration by parts, as above, with u = e3θ and dv = sin θ dθ Then e so that 3θ and v = − cos θ sin θ dθ = −e3θ cos θ − (− cos θ )(3e3θ dθ ) e3θ sin θ dθ = e3θ (− cos θ + 3sin θ ) 10 π 1 e3θ (− cos θ + 3sin θ ) yEL dA = a3 3 10 0 = Hence, du = 3e3θ dθ a 3 3π (e + 1) = 413.09a 3 30 or x = −9.27a or xA = xEL dA: x (133.623a 2 ) = −1239.26a 3 yA = yEL dA : y (133.623a 2 ) = 413.09a3 y = 3.09a PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 611 PROBLEM 5.50 Determine the centroid of the area shown when a = 2 in. SOLUTION xEL = x We have yEL = 1 dA = ydx = 1 − dx x and A = dA = Then and 1 1 1 y = 1 − 2 2 x x EL dA = yEL dA = a a 1 1 a 1 1 dx 2 a 1 − x 2 = [ x − ln x]1 = ( a − ln a − 1) in a a2 1 x 2 1 − a + in 3 x 1 − dx = − x = x 2 2 1 2 1 1 1 1 1 − 1 − dx = 2 x x 2 a 1 2 1 1 − x + 2 dx x a = 1 1 1 1 x − 2ln x − = a − 2ln a − in 3 2 x 1 2 a xA = xEL dA: x = yA = yEL dA: y = a2 2 −a+ 1 2 in. a − ln a − 1 a − 2ln a − 1a 2(a − ln a − 1) in. Find x and y when a = 2 in. We have x= and y= 1 2 (2) 2 − 2 + 1 2 2 − ln 2 − 1 2 − 2ln 2 − 12 2(2 − ln 2 − 1) or x = 1.629 in. or y = 0.1853 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 612 PROBLEM 5.51 Determine the value of a for which the ratio x / y is 9. SOLUTION xEL = x We have yEL = 1 1 1 y = 1 − 2 2 x 1 dA = y dx = 1 − dx x and A = dA = Then a 1 1 dx = [ x − ln x]1a 1 − x 2 = (a − ln a − 1) in 2 and x EL dA = a 1 a 1 x 2 x 1 − dx = − x x 2 1 a2 1 = − a + in 3 2 2 yEL dA = a 1 1 1 1 1 1 − 1 − dx = 2 x x 2 a 1 2 1 1 − x + 2 dx x a = 1 1 x − 2ln x − 2 x 1 = 1 1 a − 2ln a − in 3 2 a xA = xEL dA: x = yA = yEL dA: y = a2 2 −a+ 1 2 in. a − ln a − 1 a − 2ln a − 1a 2(a − ln a − 1) in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 613 PROBLEM 5.51 (Continued) Find a so that x = 9. y x xA = = y yA We have Then or 1 2 1 2 a2 − a + 1 2 ( a − 2 ln a − a1 ) x y EL dA EL dA =9 a 3 − 11a 2 + a + 18a ln a + 9 = 0 Using trial and error or numerical methods, and ignoring the trivial solution a = 1 in., we find a = 1.901 in. and a = 3.74 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 614 PROBLEM 5.52 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.1 about (a) the x-axis, (b) the y-axis. SOLUTION From the solution of Problem 5.1, we have A = 11 in 2 Σ xA = 11.5 in 3 Σ yA = 39.5 in 3 Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the x-axis: Volume = 2π yarea A = 2π Σ yA = 2π (39.5 in 3 ) or Volume = 248 in 3 Area = 2π yline L = 2π Σ( yline ) L = 2π ( y2 L2 + y3 L3 + y4 L4 + y5 L5 + y6 L6 + y7 L7 + y8 L8 ) = 2π [(1)(2) + (2)(3) + (2.5)(1) + (3)(3) + (5.5)(5) + (8)(1) + (4)(8)] or Area = 547 in 2 (b) Rotation about the y-axis: Volume = 2π xarea A = 2π Σ xA = 2π (11.5 in 3 ) or Volume = 72.3 in 3 Area = 2π xline L = 2π Σ( xline ) L = 2π ( x1 L1 + x2 L2 + x3 L3 + x4 L4 + x5 L5 + x6 L6 + x7 L7 ) = 2π [(0.5)(1) + (1)(2) + (2.5)(3) + (4)(1) + (2.5)(3) + (1)(5) + (0.5)(1)] or Area = 169.6 in 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 615 PROBLEM 5.53 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.2 about (a) the line y = 72 mm, (b) the x-axis. SOLUTION From the solution of Problem 5.2, we have A = 2808 mm 2 x = 36 mm y = 48 mm Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the line y = 72 mm: Volume = 2π (72 − y ) A = 2π (72 − 48)(2808) Volume = 423 × 103 mm3 Area = 2π yline L = 2π Σ( yline ) L = 2π ( y1 L1 + y3 L3 ) where y1 and y3 are measured with respect to line y = 72 mm. Area = 2π (36) ( ) 482 + 722 + (36) ( ) 302 + 722 Area = 37.2 × 103 mm 2 (b) Rotation about the x-axis: Volume = 2π yarea A = 2π (48)(2808) Volume = 847 × 103 mm3 Area = 2π yline L = 2π Σ( yline ) L = 2π ( y1 L1 + y2 L2 + y3 L3 ) = 2π (36) ( ) 482 + 722 + (72)(78) + (36) ( ) 302 + 722 Area = 72.5 × 103 mm 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 616 PROBLEM 5.54 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.8 about (a) the line x = −60 mm, (b) the line y = 120 mm. SOLUTION From the solution of Problem 5.8, we have A = 7200 mm 2 Σ x A = −72 × 103 mm3 Σ y A = 629.83 × 103 mm3 Applying the theorems of Pappus-Guldinus, we have (a) Rotation about line x = −60 mm: Volume = 2π ( x + 60) A = 2π (Σ xA + 60 A) = 2π [−72 × 103 + 60(7200)] Volume = 2.26 × 106 mm3 Area = 2π xline L = 2π Σ( xline ) L = 2π ( x1 L1 + x2 L2 + x3 L3 ) 2(60) π (60) 2(60) π (60) = 2π 60 − + 60 + + (60)(120) π 2 π 2 where x1 , x2 , x3 are measured with respect to line x = −60 mm. (b) Area = 116.3 × 103 mm 2 Rotation about line y = 120 mm: Volume = 2π (120 − y ) A = 2π (120 A −Σ yA) = 2π [120(7200) − 629.83 × 103 ] Volume = 1.471 × 106 mm3 Area = 2π yline L = 2π Σ( yline ) L = 2π ( y1 L1 + y2 L2 + y4 L4 ) where y1 , y2 , y4 are measured with respect to line y = 120 mm. 2(60) π (60) 2(60) π (60) Area = 2π 120 − + + (60)(120) π 2 π 2 Area = 116.3 × 103 mm 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 617 PROBLEM 5.55 Determine the volume of the solid generated by rotating the parabolic area shown about (a) the x-axis, (b) the axis AA′. SOLUTION First, from Figure 5.8a, we have 4 ah 3 2 y= h 5 A= Applying the second theorem of Pappus-Guldinus, we have (a) Rotation about the x-axis: Volume = 2π yA 2 4 = 2π h ah 5 3 (b) or Volume = 16 π ah 2 15 or Volume = 16 2 πa h 3 Rotation about the line AA′: Volume = 2π (2a) A 4 = 2π (2a) ah 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 618 PROBLEM 5.56 Determine the volume and the surface area of the chain link shown, which is made from a 6-mm-diameter bar, if R = 10 mm and L = 30 mm. SOLUTION The area A and circumference C of the cross section of the bar are A= π 4 d 2 and C = π d . Also, the semicircular ends of the link can be obtained by rotating the cross section through a horizontal semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have for the volume V, V = 2(Vside ) + 2(Vend ) = 2( AL) + 2(π RA) = 2( L + π R) A or π V = 2[30 mm + π (10 mm)] (6 mm) 2 4 = 3470 mm3 For the area A, or V = 3470 mm3 A = 2( Aside ) + 2( Aend ) = 2(CL) + 2(π RC ) = 2( L + π R)C or A = 2[30 mm + π (10 mm)][π (6 mm)] = 2320 mm 2 or A = 2320 mm 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 619 PROBLEM 5.57 Verify that the expressions for the volumes of the first four shapes in Figure 5.21 on Page 264 are correct. SOLUTION Following the second theorem of Pappus-Guldinus, in each case, a specific generating area A will be rotated about the x-axis to produce the given shape. Values of y are from Figure 5.8a. (1) Hemisphere: the generating area is a quarter circle. We have (2) 2 V = π a3 3 4a π V = 2π y A = 2π ha 3π 4 2 or V = π a 2 h 3 Paraboloid of revolution: the generating area is a quarter parabola. We have (4) or Semiellipsoid of revolution: the generating area is a quarter ellipse. We have (3) 4a π 2 V = 2π y A = 2π a 3π 4 3 2 V = 2π y A = 2π a ah 8 3 1 or V = π a 2 h 2 Cone: the generating area is a triangle. We have a 1 V = 2π y A = 2π ha 3 2 1 or V = π a 2 h 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 620 PROBLEM 5.58 Determine the volume and weight of the solid brass knob shown, knowing that the specific weight of brass is 0.306 lb/in3. SOLUTION Volume of knob is obtained by rotating area at left about the x-axis. Consider area as made of components shown below. Area, in2 1 π 4 (0.75) 2 = 0.4418 y , in. y A, in 3 0.8183 0.3615 2 (0.5)(0.75) = 0.375 0.25 0.0938 3 (1.25)(0.75) = 0.9375 0.625 0.5859 4 −π (0.75) 2 = −0.4418 4 0.9317 Σ −0.4116 0.6296 V = 2π Σ y A = 2π (0.6296 in 3 ) = 3.9559 in 3 V = 3.96 in 3 W = γ V = (0.306 lb/in 3 )(3.9559 in 3 ) W = 1.211 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 621 PROBLEM 5.59 Determine the total surface area of the solid brass knob shown. SOLUTION Area is obtained by rotating lines shown about the x-axis. 1 2 3 4 π 2 π 2 L, in. y , in. yL, in 2 0.5 0.25 0.1250 (0.75) = 1.1781 0.9775 1.1516 (0.75) = 1.1781 0.7725 0.9101 0.25 0.1250 0.5 Σ 2.3117 A = 2π Σ y L = 2π (2.3117 in 2 ) A = 14.52 in 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 622 PROBLEM 5.60 The aluminum shade for the small high-intensity lamp shown has a uniform thickness of 1 mm. Knowing that the density of aluminum is 2800 kg/m3, determine the mass of the shade. SOLUTION The mass of the lamp shade is given by m = ρV = ρ At where A is the surface area and t is the thickness of the shade. The area can be generated by rotating the line shown about the x-axis. Applying the first theorem of Pappus Guldinus, we have A = 2π yL = 2π Σ yL = 2π ( y1 L1 + y2 L2 + y3 L3 + y4 L4 ) or 13 mm 13 + 16 2 2 A = 2π (13 mm) + mm × (32 mm) + (3 mm) 2 2 16 + 28 2 2 + mm × (8 mm) + (12 mm) 2 28 + 33 mm × (28 mm) 2 + (5 mm) 2 + 2 = 2π (84.5 + 466.03 + 317.29 + 867.51) = 10,903.4 mm 2 Then m = ρ At = (2800 kg/m3 )(10.9034 × 10−3 m 2 )(0.001 m) m = 0.0305 kg or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 623 PROBLEM 5.61 The escutcheon (a decorative plate placed on a pipe where the pipe exits from a wall) shown is cast from brass. Knowing that the density of brass is 8470 kg/m3, determine the mass of the escutcheon. SOLUTION The mass of the escutcheon is given by m = (density)V , where V is the volume. V can be generated by rotating the area A about the x-axis. From the figure: L1 = 752 − 12.52 = 73.9510 m L2 = 37.5 = 76.8864 mm tan 26° a = L2 − L1 = 2.9324 mm 12.5 = 9.5941° 75 26° − 9.5941° = 8.2030° = 0.143168 rad α= 2 φ = sin −1 Area A can be obtained by combining the following four areas: Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have V = 2π yA = 2π Σ yA Seg. A, mm 2 y , mm yA, mm3 1 1 (76.886)(37.5) = 1441.61 2 1 (37.5) = 12.5 3 18,020.1 2 −α (75) 2 = −805.32 2(75)sin α sin (α + φ ) = 15.2303 3α −12,265.3 3 1 − (73.951)(12.5) = −462.19 2 1 (12.5) = 4.1667 3 −1925.81 4 −(2.9354)(12.5) = −36.693 1 (12.5) = 6.25 2 −229.33 Σ 3599.7 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 624 PROBLEM 5.61 (Continued) Then V = 2π Σ yA = 2π (3599.7 mm3 ) = 22, 618 mm3 m = (density)V = (8470 kg/m3 )(22.618 × 10−6 m3 ) = 0.191574 kg or m = 0.1916 kg PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 625 PROBLEM 5.62 A 34 - in.-diameter hole is drilled in a piece of 1-in.-thick steel; the hole is then countersunk as shown. Determine the volume of steel removed during the countersinking process. SOLUTION The required volume can be generated by rotating the area shown about the y-axis. Applying the second theorem of Pappus-Guldinus, we have V = 2π x A 3 1 1 1 1 1 = 2π + in. × × in. × in. 8 3 4 2 4 4 V = 0.0900 in 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 626 PROBLEM 5.63 Knowing that two equal caps have been removed from a 10-in.-diameter wooden sphere, determine the total surface area of the remaining portion. SOLUTION The surface area can be generated by rotating the line shown about the y-axis. Applying the first theorem of Pappus-Guldinus, we have A = 2 π XL = 2 π Σ x L = 2π (2 x1 L1 + x2 L2 ) Now tan α = 4 3 or α = 53.130° Then x2 = 5 in. × sin 53.130° π 53.130° × 180 ° = 4.3136 in. and π L2 = 2 53.130° × (5 in.) 180° = 9.2729 in. 3 A = 2π 2 in. (3 in.) + (4.3136 in.)(9.2729 in.) 2 A = 308 in 2 or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 627 PROBLEM 5.64 Determine the capacity, in liters, of the punch bowl shown if R = 250 mm. SOLUTION The volume can be generated by rotating the triangle and circular sector shown about the y-axis. Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have V = 2π xA = 2π Σ xA = 2π ( x1 A1 + x2 A2 ) 1 1 1 1 3 2 R sin 30° π 2 = 2π × R × R × R + R 2 3 × π6 6 3 2 2 2 R3 R3 = 2π + 16 3 2 3 3 3 π R3 8 3 3 = π (0.25 m)3 8 = 0.031883 m3 = Since 103 l = 1 m3 V = 0.031883 m3 × 103 l 1 m3 V = 31.9 l PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 628 PROBLEM 5.65* The shade for a wall-mounted light is formed from a thin sheet of translucent plastic. Determine the surface area of the outside of the shade, knowing that it has the parabolic cross section shown. SOLUTION First note that the required surface area A can be generated by rotating the parabolic cross section through π radians about the y-axis. Applying the first theorem of Pappus-Guldinus, we have A = π xL Now at x = 100 mm, 250 = k (100) and 2 y = 250 mm or k = 0.025 mm −1 xEL = x 2 dy dL = 1 + dx dx where dy = 2kx dx Then dL = 1 + 4k 2 x 2 dx We have xL = xEL dL = 100 0 x ( 1 + 4k x dx ) 2 2 100 1 1 (1 + 4k 2 x 2 )3/2 xL = 2 3 4k 0 1 1 = [1 + 4(0.025)2 (100) 2 ]3/2 − (1)3/2 12 (0.025) 2 { } = 17,543.3 mm 2 Finally, A = π (17,543.3 mm 2 ) or A = 55.1 × 103 mm 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 629 PROBLEM 5.66 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports. SOLUTION (a) 1 RI = (1100 N/m) (6 m) = 2200 N 3 RII = (900 N/m) (6 m) = 5400 N R = RI + RII = 2200 + 5400 = 7600 N XR = Σ xR : X (7600) = (2200)(1.5) + (5400)(3) X = 2.5658 m (b) R = 7.60 kN , X = 2.57 m Σ M A = 0: B (6 m) − (7600 N)(2.5658 m) = 0 B = 3250.0 N B = 3.25 kN Σ Fy = 0: A + 3250.0 N − 7600 N = 0 A = 4350.0 N A = 4.35 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 630 PROBLEM 5.67 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports. SOLUTION 1 (150lb/ft)(9 ft) = 675 lb 2 1 RII = (120 lb/ft)(9ft) = 540lb 2 R = RI + RII = 675 + 540 = 1215 lb RI = XR = Σ x R: X (1215) = (3)(675) + (6)(540) X = 4.3333 ft R = 1215 lb (a) (b) Reactions: X = 4.33 ft Σ M A = 0: B (9 ft) − (1215 lb) (4.3333 ft) = 0 B = 585.00 lb B = 585 lb Σ Fy = 0: A + 585.00 lb − 1215 lb = 0 A = 630.00 lb A = 630 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 631 PROBLEM 5.68 Determine the reactions at the beam supports for the given loading. SOLUTION First replace the given loading by the loadings shown below. Both loading are equivalent since they are both defined by a linear relation between load and distance and have the same values at the end points. 1 (900 N/m) (1.5 m) = 675 N 2 1 R2 = (400 N/m) (1.5 m) = 300 N 2 R1 = Σ M A = 0: − (675 N)(1.4 m) + (300 N)(0.9 m) + B(2.5 m) = C B = 270 N B = 270 N Σ Fy = 0: A − 675 N + 300 N + 270 N = 0 A = 105.0 N A = 105.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 632 PROBLEM 5.69 Determine the reactions at the beam supports for the given loading. SOLUTION R I = (200 lb/ft)(15 ft) R I = 3000 lb 1 (200 lb/ft)(6 ft) 2 R II = 600 lb R II = ΣM A = 0: − (3000 lb)(1.5 ft) − (600 lb)(9 ft + 2 ft) + B(15 ft) = 0 B = 740 lb B = 740 lb ΣFy = 0: A + 740 lb − 3000 lb − 600 lb = 0 A = 2860 lb A = 2860 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 633 PROBLEM 5.70 Determine the reactions at the beam supports for the given loading. SOLUTION R I = (200 lb/ft)(4 ft) = 800 lb R II = 1 (150 lb/ft)(3 ft) = 225 lb 2 ΣFy = 0: A − 800 lb + 225 lb = 0 A = 575 lb ΣM A = 0: M A − (800 lb)(2 ft) + (225 lb)(5 ft) = 0 M A = 475 lb ⋅ ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 634 PROBLEM 5.71 Determine the reactions at the beam supports for the given loading. SOLUTION 1 (4 kN/m)(6 m) 2 = 12 kN RII = (2 kN/m)(10 m) RI = = 20 kN ΣFy = 0: A − 12 kN − 20 kN = 0 A = 32.0 kN ΣM A = 0: M A − (12 kN)(2 m) − (20 kN)(5 m) = 0 M A = 124.0 kN ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 635 PROBLEM 5.72 Determine the reactions at the beam supports for the given loading. SOLUTION First replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a parabolic relation between load and distance and the values at the end points are the same. We have RI = (6 m)(300 N/m) = 1800 N RII = Then 2 (6 m)(1200 N/m) = 4800 N 3 ΣFx = 0: Ax = 0 ΣFy = 0: Ay + 1800 N − 4800 N = 0 or Ay = 3000 N A = 3.00 kN 15 ΣM A = 0: M A + (3 m)(1800 N) − m (4800 N) = 0 4 or M A = 12.6 kN ⋅ m M A = 12.60 kN ⋅ m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 636 PROBLEM 5.73 Determine the reactions at the beam supports for the given loading. SOLUTION We have Then 1 RI = (12 ft)(200 lb/ft) = 800 lb 3 1 RII = (6 ft)(100 lb/ft) = 200 lb 3 ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 800 lb − 200 lb = 0 or Ay = 1000 lb A = 1000 lb ΣM A = 0: M A − (3 ft)(800 lb) − (16.5 ft)(200 lb) = 0 or M A = 5700 lb ⋅ ft M A = 5700 lb ⋅ ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 637 PROBLEM 5.74 Determine the reactions at the beam supports for the given loading when wO = 400 lb/ft. SOLUTION 1 1 wO (12 ft) = (400 lb/ft)(12 ft) = 2400 lb 2 2 1 RII = (300 lb/ft)(12 ft) = 1800 lb 2 RI = ΣM B = 0: (2400 lb)(1 ft) − (1800 lb)(3 ft) + C (7 ft) = 0 C = 428.57 lb C = 429 lb ΣFy = 0: B + 428.57 lb − 2400 lb − 1800 lb = 0 B = 3771 lb B = 3770 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 638 PROBLEM 5.75 Determine (a) the distributed load wO at the end A of the beam ABC for which the reaction at C is zero, (b) the corresponding reaction at B. SOLUTION For wO , 1 wO (12 ft) = 6 wO 2 1 RII = (300 lb/ft)(12 ft) = 1800 lb 2 RI = (a) For C = 0, ΣM B = 0: (6 wO )(1 ft) − (1800 lb)(3 ft) = 0 (b) Corresponding value of R I : wO = 900 lb/ft RI = 6(900) = 5400 lb ΣFy = 0: B − 5400 lb − 1800 lb = 0 B = 7200 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 639 PROBLEM 5.76 Determine (a) the distance a so that the vertical reactions at supports A and B are equal, (b) the corresponding reactions at the supports. SOLUTION (a) We have Then 1 (a m)(1800 N/m) = 900a N 2 1 RII = [(4 − a ) m](600 N/m) = 300(4 − a) N 2 RI = ΣFy = 0: Ay − 900a − 300(4 − a ) + B y = 0 or Ay + B y = 1200 + 600a Now Ay = B y Ay = B y = 600 + 300a (N) Also, (1) a ΣM B = 0: − (4 m) Ay + 4 − m [(900a) N] 3 1 + (4 − a ) m [300(4 − a) N] = 0 3 or Ay = 400 + 700a − 50a 2 Equating Eqs. (1) and (2), 600 + 300a = 400 + 700a − 50a 2 or a 2 − 8a + 4 = 0 (2) 8 ± (−8) 2 − 4(1)(4) 2 Then a= or a = 0.53590 m Now a≤4m a = 7.4641 m a = 0.536 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 640 PROBLEM 5.76 (Continued) (b) We have From Eq. (1): ΣFx = 0: Ax = 0 Ay = By = 600 + 300(0.53590) = 761 N A = B = 761 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 641 PROBLEM 5.77 Determine (a) the distance a so that the reaction at support B is minimum, (b) the corresponding reactions at the supports. SOLUTION (a) We have Then or Then (b) From Eq. (1): and 1 (a m)(1800 N/m) = 900a N 2 1 RII = [(4 − a )m](600 N/m) = 300(4 − a) N 2 RI = a 8+ a ΣM A = 0: − m (900a N) − m [300(4 − a)N] + (4 m) By = 0 3 3 By = 50a 2 − 100a + 800 dBy da = 100a − 100 = 0 By = 50(1)2 − 100(1) + 800 = 750 N (1) or a = 1.000 m B = 750 N ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 900(1) N − 300(4 − 1) N + 750 N = 0 or Ay = 1050 N A = 1050 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 642 PROBLEM 5.78 A beam is subjected to a linearly distributed downward load and rests on two wide supports BC and DE, which exert uniformly distributed upward loads as shown. Determine the values of wBC and wDE corresponding to equilibrium when wA = 600 N/m. SOLUTION We have 1 (6 m)(600 N/m) = 1800 N 2 1 RII = (6 m)(1200 N/m) = 3600 N 2 RBC = (0.8 m) (wBC N/m) = (0.8wBC ) N RI = RDE = (1.0 m) (wDE N/m) = ( wDE ) N Then ΣM G = 0: − (1 m)(1800 N) − (3 m)(3600 N) + (4 m)( wDE N) = 0 wDE = 3150 N/m or and or ΣFy = 0: (0.8wBC ) N − 1800 N − 3600 N + 3150 N = 0 wBC = 2812.5 N/m wBC = 2810 N/m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 643 PROBLEM 5.79 A beam is subjected to a linearly distributed downward load and rests on two wide supports BC and DE, which exert uniformly distributed upward loads as shown. Determine (a) the value of wA so that wBC = wDE, (b) the corresponding values of wBC and wDE. SOLUTION (a) 1 (6 m)( wA N/m) ⋅ (3 wA ) N 2 1 RII = (6 m)(1200 N/m) = 3600 N 2 RBC = (0.8 m) (wBC N/m) = (0.8wBC ) N RI = We have RDE = (1 m) (wDE N/m) = ( wDE ) N ΣFy = 0: (0.8wBC ) N − (3wA ) N − 3600 N + ( wDE ) N = 0 Then or 0.8wBC + wDE = 3600 + 3wA 5 wBC = wDE wBC = wDE = 2000 + wA 3 Now (1) ΣM G = 0: −(1 m)(3wA N) − (3 m)(3600 N) + (4 m)( wDE N) = 0 Also, wDE = 2700 + or 3 wA 4 (2) Equating Eqs. (1) and (2), 5 3 2000 + wA = 2700 + wA 3 4 or (b) Eq. (1) wA = 8400 N/m 11 wA = 764 N/m 5 8400 wBC = wDE = 2000 + 3 11 or wBC = wDE = 3270 N/m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 644 PROBLEM 5.80 The cross section of a concrete dam is as shown. For a 1-m-wide dam section, determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam. SOLUTION (a) Consider free body made of dam and section BDE of water. (Thickness = 1 m) p = (3 m)(10 kg/m3 )(9.81 m/s 2 ) W1 = (1.5 m)(4 m)(1 m)(2.4 × 103 kg/m3 )(9.81 m/s 2 ) = 144.26 kN 1 W2 = (2 m)(3 m)(1 m)(2.4 × 103 kg/m3 )(9.81 m/s 2 ) = 47.09 kN 3 2 W3 = (2 m)(3 m)(1 m)(103 kg/m3 )(9.81 m/s 2 ) = 39.24 kN 3 1 1 P = Ap = (3 m)(1 m)(3 m)(103 kg/m3 )(9.81 m/s 2 ) = 44.145 kN 2 2 ΣFx = 0: H − 44.145 kN = 0 H = 44.145 kN H = 44.1 kN ΣFy = 0: V − 141.26 − 47.09 − 39.24 = 0 V = 227.6 kN V = 228 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 645 PROBLEM 5.80 (Continued) 1 (1.5 m) = 0.75 m 2 1 x2 = 1.5 m + (2 m) = 2 m 4 5 x3 = 1.5 m + (2 m) = 2.75 m 8 x1 = ΣM A = 0: xV − Σ xW + P (1 m) = 0 x(227.6 kN) − (141.26 kN)(0.75 m) − (47.09 kN)(2 m) − (39.24 kN)(2.75 m) + (44.145 kN)(1 m) = 0 x(227.6 kN) − 105.9 − 94.2 − 107.9 + 44.145 = 0 x(227.6) − 263.9 = 0 x = 1.159 m (to right of A) (b) Resultant of face BC: Consider free body of section BDE of water. − R = 59.1 kN R = 59.1 kN 41.6° 41.6° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 646 PROBLEM 5.81 The cross section of a concrete dam is as shown. For a 1-m-wide dam section, determine (a) the resultant of the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam. SOLUTION (a) Consider free body made of dam and triangular section of water shown. (Thickness = 1 m.) p = (7.2 m)(103 kg/m3 )(9.81m/s 2 ) 2 (4.8 m)(7.2 m)(1 m)(2.4 × 103 kg/m3 )(9.81 m/s 2 ) 3 = 542.5 kN 1 W2 = (2.4 m)(7.2 m)(1 m)(2.4 × 103 kg/m3 )(9.81 m/s 2 ) 2 = 203.4 kN 1 W3 = (2.4 m)(7.2 m)(1 m)(103 kg/m3 )(9.81 m/s 2 ) 2 = 84.8 kN 1 1 P = Ap = (7.2 m)(1 m)(7.2 m)(103 kg/m3 )(9.81 m/s 2 ) 2 2 = 254.3 kN W1 = ΣFx = 0: H − 254.3 kN = 0 H = 254 kN ΣFy = 0: V − 542.5 − 203.4 − 84.8 = 0 V = 830.7 kN V = 831 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 647 PROBLEM 5.81 (Continued) 5 x1 = (4.8 m) = 3 m 8 1 x2 = 4.8 + (2.4) = 5.6 m 3 2 x3 = 4.8 + (2.4) = 6.4 m 3 (b) ΣM A = 0: xV − Σ xW + P(2.4 m) = 0 x(830.7 kN) − (3 m)(542.5 kN) − (5.6 m)(203.4 kN) − (6.4 m)(84.8 kN) + (2.4 m)(254.3 kN) = 0 x(830.7) − 1627.5 − 1139.0 − 542.7 + 610.3 = 0 x(830.7) − 2698.9 = 0 x = 3.25 m (to right of A) (c) Resultant on face BC: Direct computation: P = ρ gh = (103 kg/m3 )(9.81 m/s 2 )(7.2 m) P = 70.63 kN/m 2 BC = (2.4) 2 + (7.2) 2 = 7.589 m θ = 18.43° 1 PA 2 1 = (70.63 kN/m 2 )(7.589 m)(1 m) 2 R= R = 268 kN 18.43° Alternate computation: Use free body of water section BCD. − R = 268 kN R = 268 kN 18.43° 18.43° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 648 PROBLEM 5.82 An automatic valve consists of a 9 × 9-in. square plate that is pivoted about a horizontal axis through A located at a distance h = 3.6 in. above the lower edge. Determine the depth of water d for which the valve will open. SOLUTION Since valve is 9 in. wide, w = 9 p = 9γ h, where all dimensions are in inches. w1 = 9γ (d − 9), w2 = 9γ d 1 1 (9 in.) w1 = (9)(9γ )(d − 9) 2 2 1 1 PII = (9 in.) w2 = (9)(9γ d ) 2 2 PI = Valve opens when B = 0. ΣM A = 0: PI (6 in. − 3.6 in.) − PII (3.6 in. − 3 in.) = 0 1 1 2 (9)(9γ )( d − 9) (2.4) − 2 (9)(9γ d ) (0.6) = 0 (d − 9)(2.4) − d (0.6) = 0 1.8d − 21.6 = 0 d = 12.00 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 649 PROBLEM 5.83 An automatic valve consists of a 9 × 9-in. square plate that is pivoted about a horizontal axis through A. If the valve is to open when the depth of water is d = 18 in., determine the distance h from the bottom of the valve to the pivot A. SOLUTION Since valve is 9 in. wide, w = 9 p = 9γ h, where all dimensions are in inches. w1 = 9γ ( d − 9) w2 = 9γ d For d = 18 in., w1 = 9γ (18 − 9) = 81γ w2 = 9γ (18) = 162γ 1 1 (9)(9γ )(18 − 9) = (729γ ) 2 2 1 PII = (9)(9γ )(18) = 729γ 2 PI = Valve opens when B = 0. ΣM A = 0: P1 (6 − h) − PII ( h − 3) = 0 1 729γ (6 − h) − 729( h − 3) = 0 2 1 3− h −h +3= 0 2 6 − 1.5h = 0 h = 4.00 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 650 PROBLEM 5.84 The 3 × 4-m side AB of a tank is hinged at its bottom A and is held in place by a thin rod BC. The maximum tensile force the rod can withstand without breaking is 200 kN, and the design specifications require the force in the rod not to exceed 20 percent of this value. If the tank is slowly filled with water, determine the maximum allowable depth of water d in the tank. SOLUTION Consider the free-body diagram of the side. We have Now where P= 1 1 Ap = A( ρ gd ) 2 2 ΣM A = 0: hT − d P=0 3 h=3m Then for d max , (3 m)(0.2 × 200 × 103 N) − d max 3 1 3 3 2 2 (4 m × d max ) × (10 kg/m × 9.81 m/s × d max ) = 0 3 120 N ⋅ m − 6.54d max N/m 2 = 0 or d max = 2.64 m or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 651 PROBLEM 5.85 The 3 × 4-m side of an open tank is hinged at its bottom A and is held in place by a thin rod BC. The tank is to be filled with glycerine, whose density is 1263 kg/m3. Determine the force T in the rod and the reactions at the hinge after the tank is filled to a depth of 2.9 m. SOLUTION Consider the free-body diagram of the side. We have Then 1 1 Ap = A( ρ gd ) 2 2 1 = [(2.9 m)(4 m)] [(1263 kg/m3 )(9.81 m/s 2 )(2.9 m)] 2 = 208.40 kN P= ΣFy = 0: Ay = 0 2.9 m (208.4 kN) = 0 ΣM A = 0: (3 m)T − 3 T = 67.151 kN or ΣFx = 0: T = 67.2 kN Ax + 208.40 kN − 67.151 kN = 0 Ax = −141.249 kN or A = 141.2 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 652 PROBLEM 5.86 The friction force between a 6 × 6-ft square sluice gate AB and its guides is equal to 10 percent of the resultant of the pressure forces exerted by the water on the face of the gate. Determine the initial force needed to lift the gate if it weighs 1000 lb. SOLUTION Consider the free-body diagram of the gate. Now 1 1 ApI = [(6 × 6) ft 2 ][(62.4 lb/ft 3 )(9 ft)] 2 2 = 10,108.8 lb PI = 1 1 ApII = [(6 × 6) ft 2 ][(62.4 lb/ft 3 )(15 ft)] 2 2 = 16,848 lb PII = Then F = 0.1P = 0.1( PI + PII ) = 0.1(10,108.8 + 16,848) lb = 2695.7 lb Finally ΣFy = 0: T − 2695.7 lb − 1000 lb = 0 or T = 3.70 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 653 PROBLEM 5.87 A tank is divided into two sections by a 1 × 1-m square gate that is hinged at A. A couple of magnitude 490 N · m is required for the gate to rotate. If one side of the tank is filled with water at the rate of 0.1 m3/min and the other side is filled simultaneously with methyl alcohol (density ρma = 789 kg/m3) at the rate of 0.2 m3/min, determine at what time and in which direction the gate will rotate. SOLUTION Consider the free-body diagram of the gate. First note V = A base d and V = rt. Then dW = d MA = Now P= 0.1 m3 / min × t (min) = 0.25t (m) (0.4 m)(1 m) 0.2 m3 / min × t (min) = t (m) (0.2 m)(1 m) 1 1 Ap = A( ρ gh) so that 2 2 1 PW = [(0.25t ) m × (1 m)][(103 kg/m3 )(9.81 m/s 2 )(0.25t ) m] 2 = 306.56t 2 N 1 PMA = [(t ) m × (1 m)][(789 kg/m3 )(9.81 m/s 2 )(t ) m] 2 = 3870t 2 N Now assume that the gate will rotate clockwise and when d MA ≤ 0.6 m. When rotation of the gate is impending, we require 1 1 ΣM A : M R = 0.6 m − d MA PMA − 0.6 m − dW 3 3 PW Substituting 1 1 490 N ⋅ m = 0.6 − t m × (3870t 2 ) N − 0.6 − × 0.25t m × (306.56t 2 ) N 3 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 654 PROBLEM 5.87 (Continued) Simplifying 1264.45t 3 − 2138.1t 2 + 490 = 0 Solving (positive roots only) t = 0.59451 min and t = 1.52411 min Now check assumption using the smaller root. We have d MA = (t ) m = 0.59451 m < 0.6 m ∴ t = 0.59451min = 35.7 s and the gate rotates clockwise. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 655 PROBLEM 5.88 A prismatically shaped gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. The pin is located at a distance h = 0.10 m below the center of gravity C of the gate. Determine the depth of water d for which the gate will open. SOLUTION First note that when the gate is about to open (clockwise rotation is impending), By 0 and the line of action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then a= d − (0.25 − h) 3 and b= 2 8 d (0.4) − 3 15 3 Now a 8 = b 15 so that d − (0.25 − h) 3 2 (0.4) − 158 d3 3 ( ) = 8 15 Simplifying yields 289 70.6 d + 15h = 45 12 (1) Alternative solution: Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water above the gate. Now 1 1 Ap′ = (d × 1 m)( ρ gd ) 2 2 1 = ρ gd 2 (N) 2 1 8 W ′ = ρ gV = ρ g × d × d × 1 m 2 15 4 = ρ gd 2 (N) 15 P′ = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 656 PROBLEM 5.88 (Continued) Then with By = 0 (as explained above), we have 2 1 8 4 d 1 ΣM A = 0: (0.4) − d ρ gd 2 − − (0.25 − h) ρ gd 2 = 0 3 3 15 15 3 2 Simplifying yields 289 70.6 d + 15h = 45 12 as above. Find d: Substituting into Eq. (1), h = 0.10 m 289 70.6 d + 15(0.10) = 45 12 or d = 0.683 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 657 PROBLEM 5.89 A prismatically shaped gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. Determine the distance h if the gate is to open when d = 0.75 m. SOLUTION First note that when the gate is about to open (clockwise rotation is impending), By 0 and the line of action of the resultant P of the pressure forces passes through the pin at A. In addition, if it is assumed that the gate is homogeneous, then its center of gravity C coincides with the centroid of the triangular area. Then a= d − (0.25 − h) 3 and b= 2 8 d (0.4) − 3 15 3 Now a 8 = b 15 so that d − (0.25 − h) 3 2 (0.4) − 158 d3 3 ( ) = 8 15 Simplifying yields 289 70.6 d + 15h = 45 12 (1) Alternative solution: Consider a free body consisting of a 1-m thick section of the gate and the triangular section BDE of water above the gate. Now 1 1 Ap′ = (d × 1 m)( ρ gd ) 2 2 1 2 = ρ gd (N) 2 1 8 W ′ = ρ gV = ρ g × d × d × 1 m 2 15 4 = ρ gd 2 (N) 15 P′ = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 658 PROBLEM 5.89 (Continued) Then with By = 0 (as explained above), we have 2 1 8 4 d 1 ΣM A = 0: (0.4) − d ρ gd 2 − − (0.25 − h) ρ gd 2 = 0 3 15 15 3 2 3 Simplifying yields 289 70.6 d + 15h = 45 12 as above. Find h: Substituting into Eq. (1), d = 0.75 m 289 70.6 (0.75) + 15h = 45 12 or h = 0.0711 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 659 PROBLEM 5.90 The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 ft, determine the force exerted on the gate by the shear pin. SOLUTION First consider the force of the water on the gate. We have 1 Ap 2 1 = A(γ h) 2 P= Then 1 (1.8 ft) 2 (62.4 lb/ft 3 )(1.7 ft) 2 = 171.850 lb PI = 1 (1.8 ft)2 (62.4 lb/ft 3 ) × (1.7 + 1.8cos 30°) ft 2 = 329.43 lb PII = Now or 1 2 ΣM A = 0: LAB PI + LAB PII − LAB FB = 0 3 3 1 2 (171.850 lb) + (329.43 lb) − FB = 0 3 3 FB = 276.90 lb or FB = 277 lb 30.0° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 660 PROBLEM 5.91 A long trough is supported by a continuous hinge along its lower edge and by a series of horizontal cables attached to its upper edge. Determine the tension in each of the cables, at a time when the trough is completely full of water. SOLUTION Consider free body consisting of 20-in. length of the trough and water. l = 20-in. length of free body π W = γ v = γ r 2l 4 PA = γ r P= 1 1 1 PA rl = (γ r )rl = γ r 2l 2 2 2 1 ΣM A = 0: Tr − Wr − P r = 0 3 π 4r Tr − γ r 2 l 4 3π 1 2 1 − 2 γ r l 3 r = 0 1 1 1 T = γ r 2l + γ r 2l = γ r 2l 3 6 2 Data: γ = 62.4 lb/ft 3 r = Then T= 24 20 ft = 2 ft l = ft 12 12 1 20 (62.4 lb/ft 3 )(2 ft)2 ft 2 12 = 208.00 lb T = 208 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 661 PROBLEM 5.92 A 0.5 × 0.8-m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a frictionless stop at B. Determine the reactions at A and B when cable BCD is slack. SOLUTION First consider the force of the water on the gate. We have P= 1 1 Ap = A( ρ gh) 2 2 1 PI = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.45 m)] 2 = 882.9 N so that 1 PII = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.93 m)] 2 = 1824.66 N Reactions at A and B when T = 0: We have ΣM A = 0: 1 2 (0.8 m)(882.9 N) + (0.8 m)(1824.66 N) − (0.8 m)B = 0 3 3 B = 1510.74 N or or B = 1511 N 53.1° A = 1197 N 53.1° ΣF = 0: A + 1510.74 N − 882.9 N − 1824.66 N = 0 or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 662 PROBLEM 5.93 A 0.5 × 0.8-m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a frictionless stop at B. Determine the minimum tension required in cable BCD to open the gate. SOLUTION First consider the force of the water on the gate. We have so that P= 1 1 Ap = A( ρ gh) 2 2 1 PI = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.45 m)] 2 = 882.9 N 1 PII = [(0.5 m)(0.8 m)] × [(103 kg/m3 )(9.81 m/s 2 )(0.93 m)] 2 = 1824.66 N T to open gate: First note that when the gate begins to open, the reaction at B Then or ΣM A = 0: 0. 1 2 (0.8 m)(882.9 N) + (0.8 m)(1824.66 N) 3 3 8 − (0.45 + 0.27)m × T = 0 17 235.44 + 973.152 − 0.33882 T = 0 or T = 3570 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 663 PROBLEM 5.94 A 4 × 2-ft gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 828 lb/ft. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor. SOLUTION First determine the forces exerted on the gate by the spring and the water when B is at the end of the cylindrical portion of the floor. sin θ = We have 2 θ = 30° 4 Then xSP = (3 ft) tan 30° and FSP = kxSP = 828 lb/ft × 3 ft × tan30° = 1434.14 lb Assume d ≥ 4 ft We have P= 1 1 Ap = A(γ h) 2 2 1 PI = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )( d − 4) ft] 2 = 249.6(d − 4) lb Then 1 PII = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )(d − 4 + 4cos 30°)] 2 = 249.6(d − 0.53590°) lb For d min so that the gate opens, W = 0 Using the above free-body diagrams of the gate, we have 4 8 ΣM A = 0: ft [249.6(d − 4) lb] + ft [249.6( d − 0.53590) lb] 3 3 −(3 ft)(1434.14 lb) = 0 or (332.8d − 1331.2) + (665.6d − 356.70) − 4302.4 = 0 d = 6.00 ft or d ≥ 4 ft assumption correct d = 6.00 ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 664 PROBLEM 5.95 Solve Problem 5.94 if the gate weighs 1000 lb. PROBLEM 5.94 A 4 × 2-ft gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 828 lb/ft. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor. SOLUTION First determine the forces exerted on the gate by the spring and the water when B is at the end of the cylindrical portion of the floor. We have sin θ = 2 θ = 30° 4 Then xSP = (3 ft) tan 30° and FSP = kxSP = 828 lb/ft × 3 ft × tan30° = 1434.14 lb Assume d ≥ 4 ft We have P= 1 1 Ap = A(γ h) 2 2 1 PI = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )( d − 4) ft] 2 = 249.6(d − 4) lb 1 PII = [(4 ft)(2 ft)] × [(62.4 lb/ft 3 )(d − 4 + 4cos 30°)] 2 = 249.6(d − 0.53590°) lb Then For d min so that the gate opens, W = 1000 lb Using the above free-body diagrams of the gate, we have 4 8 ΣM A = 0: ft [249.6( d − 4) lb] + ft [249.6( d − 0.53590) lb] 3 3 − (3 ft)(1434.14 lb) − (1 ft)(1000 lb) = 0 or (332.8d − 1331.2) + (665.6d − 356.70) − 4302.4 − 1000 = 0 d = 7.00 ft or d ≥ 4 ft assumption correct d = 7.00 ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 665 PROBLEM 5.96 A hemisphere and a cone are attached as shown. Determine the location of the centroid of the composite body when (a) h = 1.5a, (b) h = 2a. SOLUTION V y yV Cone I 1 2 πa h 3 h 4 1 π a 2 h2 12 Hemisphere II 2 3 πa 3 − 3a 8 1 − π a4 4 1 V = π a 2 ( h + 2a ) 3 1 Σ yV = π a 2 (h 2 − 3a 2 ) 12 (a) For h = 1.5a, 1 7 V = π a 2 (1.5a + 2a ) = π a 2 3 6 Σ yV = 1 1 π a 2 [(1.5a) 2 − 3a 2 ] = − π a 4 12 16 1 3 7 Y V = Σ yV : Y π a3 = − π a 4 Y = − a 16 56 6 Centroid is 0.0536a below base of cone. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 666 PROBLEM 5.96 (Continued) (b) For h = 2a, 1 4 V = π a 2 (2a + 2a) = π a 3 3 3 Σ yV = 1 1 π a 2 [(2a) 2 − 3a 2 ] = π a 4 12 12 1 4 1 Y V = Σ yV : Y π a 3 = π a 4 Y = a 16 3 12 Centroid is 0.0625a above base of cone. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 667 PROBLEM 5.97 Consider the composite body shown. Determine (a) the value of x when h = L/2, (b) the ratio h/L for which x = L. SOLUTION V x xV Rectangular prism Lab 1 L 2 1 2 L ab 2 Pyramid 1 b a h 3 2 1 h 4 1 1 abh L + h 6 4 1 ΣV = ab L + h 6 1 1 Σ xV = ab 3L2 + h L + h 6 4 Then X ΣV = Σ xV Now so that 1 1 1 X ab L + h = ab 3L2 + hL + h 2 6 6 4 h 1 h2 1 h 1 X 1 + L 3 = + + L 4 L2 6 L 6 or (a) L+ X = ? when h = Substituting (1) 1 L. 2 h 1 = into Eq. (1), L 2 2 1 1 1 1 1 1 X 1 + = L 3 + + 6 2 6 2 4 2 or X= 57 L 104 X = 0.548L PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 668 PROBLEM 5.97 (Continued) (b) h = ? when X = L. L Substituting into Eq. (1), or or h 1 h2 1 h 1 L 1 + = L 3 + + L 4 L2 6 L 6 1+ 1 h 1 1 h 1 h2 = + + 6 L 2 6 L 24 L2 h2 = 12 L2 h =2 3 L PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 669 PROBLEM 5.98 Determine the y coordinate of the centroid of the body shown. SOLUTION First note that the values of Y will be the same for the given body and the body shown below. Then V y Cone 1 2 πa h 3 1 − h 4 2 Cylinder 1 a −π b = − π a 2 b 4 2 1 − b 2 Σ We have Then π 12 a 2 (4 h − 3b) yV − 1 π a2 h2 12 1 2 2 πa b 8 − π 24 a 2 (2 h 2 − 3b 2 ) Y ΣV = Σ yV π π Y a 2 (4h − 3b) = − a 2 (2h 2 − 3b 2 ) 24 12 or Y = − 2h 2 − 3b 2 2(4h − 3b) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 670 PROBLEM 5.99 Determine the z coordinate of the centroid of the body shown. (Hint: Use the result of Sample Problem 5.13.) SOLUTION First note that the body can be formed by removing a half cylinder from a half cone, as shown. V 1 2 πa h 6 Half cone Half cylinder Σ − π a 2 24 π 1 − a3h 6 4 a 2a =− 3π 2 3π 1 3 ab 12 − π b = − a 2b 2 2 8 π zV z − a − a 2 (4h − 3b) 1 3 a (2h − b) 12 From Sample Problem 5.13: We have Then Z ΣV = Σ zV 1 π Z a 2 (4h − 3b) = − a3 (2h − b) 12 24 or Z = − a 4h − 2b π 4h − 3b PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 671 PROBLEM 5.100 For the machine element shown, locate the y coordinate of the center of gravity. SOLUTION For half-cylindrical hole, r = 1.25 in. 4(1.25) 3π = 1.470 in. yIII = 2 − For half-cylindrical plate, r = 2 in. 4(2) zIV = 7 + = 7.85 in. 3π V, in 3 y , in. z , in. y V , in 4 z V , in 4 3.5 –7.875 73.50 I Rectangular plate (7)(4)(0.75) = 21.0 –0.375 II Rectangular plate (4)(2)(1) = 8.0 1.0 2 8.000 16.00 III –(Half cylinder) (1.25) 2 (1) = 2.454 1.470 2 –3.607 –4.908 IV Half cylinder (2) 2 (0.75) = 4.712 –0.375 –1.767 36.99 V –(Cylinder) −π (1.25) 2 (0.75) = −3.682 1.381 –25.77 Σ 27.58 –3.868 95.81 − π 2 π 2 –0.375 –7.85 7 Y ΣV = Σ yV Y (27.58 in 3 ) = −3.868 in 4 Y = −0.1403 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 672 PROBLEM 5.101 For the machine element shown, locate the y coordinate of the center of gravity. SOLUTION First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume. V , mm3 x , mm y , mm x V, mm 4 y V, mm 4 I (100)(18)(90) = 162, 000 50 9 8,100,000 1,458,000 II (16)(60)(50) = 48, 000 92 48 4,416,000 2,304,000 III π (12)2 (10) = 4523.9 105 54 475,010 244,290 IV −π (13) 2 (18) = −9556.7 28 9 –267,590 –86,010 Σ 204,967.2 12,723,420 3,920,280 We have Y ΣV = Σ yV Y (204,967.2 mm3 ) = 3,920, 280 mm 4 or Y = 19.13 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 673 PROBLEM 5.102 For the machine element shown, locate the x coordinate of the center of gravity. SOLUTION First assume that the machine element is homogeneous so that its center of gravity will coincide with the centroid of the corresponding volume. V , mm3 x , mm y , mm xV , mm 4 yV, mm 4 I (100)(18)(90) = 162, 000 50 9 8,100,000 1,458,000 II (16)(60)(50) = 48, 000 92 48 4,416,000 2,304,000 III π (12)2 (10) = 4523.9 105 54 475,010 244,290 IV −π (13) 2 (18) = −9556.7 28 9 –267,590 –86,010 Σ 204,967.2 12,723,420 3,920,280 We have X ΣV = Σ xV X (204,967.2 mm3 ) = 12, 723, 420 mm 4 X = 62.1 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 674 PROBLEM 5.103 For the machine element shown, locate the z coordinate of the center of gravity. SOLUTION For half-cylindrical hole, r = 1.25 in. 4(1.25) 3π = 1.470 in. yIII = 2 − For half-cylindrical plate, r = 2 in. 4(2) zIV = 7 + = 7.85 in. 3π V, in 3 Now y , in. z , in. yV , in 4 z V , in 4 3.5 –7.875 73.50 I Rectangular plate (7)(4)(0.75) = 21.0 –0.375 II Rectangular plate (4)(2)(1) = 8.0 1.0 2 8.000 16.00 III –(Half cylinder) (1.25) 2 (1) = 2.454 1.470 2 –3.607 –4.908 IV Half cylinder (2) 2 (0.75) = 4.712 –0.375 –1.767 36.99 V –(Cylinder) −π (1.25) 2 (0.75) = −3.682 1.381 –25.77 Σ 27.58 –3.868 95.81 − π 2 π 2 –0.375 –7.85 7 Z ΣV = zV Z (27.58 in 3 ) = 95.81 in 4 Z = 3.47 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 675 PROBLEM 5.104 For the machine element shown, locate the x coordinate of the center of gravity. SOLUTION Dimensions in mm V, mm3 x , mm z , mm x V , mm 4 z V , mm 4 19 45 649.8 × 103 1539 × 103 46.5 20 292.17 × 103 125.664 × 103 −π (12)2 (10) = −4.5239 × 103 38 20 Rectangular prism (30)(10)(24) = 7.2 × 103 5 78 36 × 103 561.6 × 103 Triangular prism 1 (30)(9)(24) = 3.24 × 103 2 13 78 42.12 × 103 252.72 × 103 Σ 46.399 × 103 848.18 × 103 2388.5 × 103 I Rectangular plate II Half cylinder III –(Cylinder) IV V (10)(90)(38) = 34.2 × 103 π 2 (20)2 (10) = 6.2832 × 103 −171.908 × 103 −90.478 × 103 X ΣV = Σ xV X= Σ xV 848.18 × 103 mm4 = ΣV 46.399 × 103 mm3 X = 18.28 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 676 PROBLEM 5.105 For the machine element shown, locate the z coordinate of the center of gravity. SOLUTION Dimensions in mm V, mm3 x V , mm 4 z V , mm 4 19 45 649.8 × 103 1539 × 103 46.5 20 292.17 × 103 125.664 × 103 −π (12)2 (10) = −4.5239 × 103 38 20 (30)(10)(24) = 7.2 × 103 5 78 36 × 103 561.6 × 103 Triangular prism 1 (30)(9)(24) = 3.24 × 103 2 13 78 42.12 × 103 252.72 × 103 Σ 46.399 × 103 848.18 × 103 2388.5 × 103 I Rectangular plate II Half cylinder III –(Cylinder) IV Rectangular prism V x , mm z , mm (10)(90)(38) = 34.2 × 103 π 2 (20)2 (10) = 6.2832 × 103 −171.908 × 103 −90.478 × 103 Z ΣV = Σ zV Z= Σ zV 2388.5 × 103 mm4 = ΣV 46.399 × 103 mm3 Z = 51.5 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 677 PROBLEM 5.106 Locate the center of gravity of the sheet-metal form shown. SOLUTION Y = 80.0 mm By symmetry, 4(80) = 33.953 mm 3π 2(60) zII = − = −38.197 mm zI = π A, mm 2 I π 2 (80) 2 = 10, 053 II π (60)(160) = 30,159 Σ 40,212 x , mm z , mm xA, mm3 zA, mm3 0 33.953 0 341.33 × 103 60 −38.197 1809.54 × 103 −1151.98 × 103 1809.54 × 103 −810.65 × 103 X Σ A = Σ xA: X (40, 212) = 1809.54 × 103 X = 45.0 mm Z Σ A = Σ zA: Z (40, 212) = −810.65 × 103 Z = −20.2 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 678 PROBLEM 5.107 Locate the center of gravity of the sheet-metal form shown. SOLUTION First assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area. 1 yI = 0.18 + (0.12) = 0.22 m 3 1 zI = (0.2 m) 3 2 × 0.18 0.36 = m xII = yII = π xIV I II IV Σ A, m 2 x, m y, m z, m x A, m3 yA, m3 1 (0.2)(0.12) = 0.012 2 0 0.22 0.2 3 0 0.00264 0.0008 0.36 0.36 π π 0.1 0.00648 0.00648 0.005655 0.26 0 0.1 0.00832 0 0.0032 0.31878 0 0.1 –0.001258 0 –0.000393 0.013542 0.00912 0.009262 π (0.18)(0.2) = 0.018π 2 (0.16)(0.2) = 0.032 III − π 4 × 0.05 = 0.34 − 3π = 0.31878 m π 2 (0.05) 2 = −0.00125π 0.096622 z A, m3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 679 PROBLEM 5.107 (Continued) We have X ΣV = Σ xV : X (0.096622 m 2 ) = 0.013542 m3 or X = 0.1402 m Y ΣV = Σ yV : Y (0.096622 m 2 ) = 0.00912 m3 or Y = 0.0944 m Z ΣV = Σ zV : Z (0.096622 m 2 ) = 0.009262 m3 or Z = 0.0959 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 680 PROBLEM 5.108 A window awning is fabricated from sheet metal of uniform thickness. Locate the center of gravity of the awning. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides with the centroid of the corresponding area. (4)(25) = 14.6103 in. 3π (4)(25) 100 in. zII = zVI = = 3π 3π (2)(25) yIV = 4 + = 19.9155 in. yII = yVI = 4 + π zIV = (2)(25) π AII = AVI = AIV = A, in 2 I (4)(25) = 100 II 490.87 III (4)(34) = 136 IV 1335.18 V (4)(25) = 100 VI Σ 490.87 y , in. 2 π 2 19.9155 in. (25) 2 = 490.87 in 2 (25)(34) = 1335.18 in 2 zA, in 3 12.5 200 1250 100 3π 14.6103 π yA, in 3 π 12.5 2 4 50 z , in. 100 3π 25 50 14.6103 2 π = 7171.8 272 3400 26,591 21,250 200 1250 7171.8 41,607 2652.9 5208.3 5208.3 37,567 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 681 PROBLEM 5.108 (Continued) X = 17.00 in. Now, symmetry implies and Y Σ A = Σ yA: Y (2652.9 in 2 ) = 41,607 in 3 or Y = 15.68 in. Z Σ A = Σ zA: Z (2652.9 in 2 ) = 37,567 in 3 or Z = 14.16 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 682 PROBLEM 5.109 A thin sheet of plastic of uniform thickness is bent to form a desk organizer. Locate the center of gravity of the organizer. SOLUTION First assume that the plastic is homogeneous so that the center of gravity of the organizer will coincide with the centroid of the corresponding area. Now note that symmetry implies Z = 30.0 mm x2 = 6 − 2× 6 x4 = 36 + x8 = 58 − π = 2.1803 mm 2×6 π 2× 6 π x10 = 133 + = 39.820 mm = 54.180 mm 2×6 π = 136.820 mm y2 = y4 = y8 = y10 = 6 − y6 = 75 + 2×5 π 2×6 π = 2.1803 mm = 78.183 mm A2 = A4 = A8 = A10 = π × 6 × 60 = 565.49 mm 2 2 A6 = π × 5 × 60 = 942.48 mm 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 683 PROBLEM 5.109 (Continued) A, mm 2 We have x , mm 1 (74)(60) = 4440 2 565.49 3 (30)(60) = 1800 4 565.49 5 (69)(60) = 4140 42 6 942.48 7 (69)(60) = 4140 8 565.49 9 y , mm 0 xA, mm3 yA, mm3 0 190,920 1233 1233 43 2.1803 2.1803 21 0 37,800 0 39.820 2.1803 22,518 1233 40.5 173,880 167,670 47 78.183 44,297 73,686 52 40.5 215,280 167,670 54.180 2.1803 30,638 1233 (75)(60) = 4500 95.5 0 429,750 0 10 565.49 136.820 2.1803 77,370 1233 Σ 22,224.44 1,032,766 604,878 X Σ A = Σ xA: X (22, 224.44 mm 2 ) = 1,032, 766 mm3 or X = 46.5 mm Y Σ A = Σ yA: Y (22, 224.44 mm 2 ) = 604,878 mm3 or Y = 27.2 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 684 PROBLEM 5.110 A wastebasket, designed to fit in the corner of a room, is 16 in. high and has a base in the shape of a quarter circle of radius 10 in. Locate the center of gravity of the wastebasket, knowing that it is made of sheet metal of uniform thickness. SOLUTION By symmetry, X =Z For III (Cylindrical surface), x= A= For IV (Quarter-circle bottom), x= A= 2r π π 2 = 2(10) rh = π π 2 = 6.3662 in. (10)(16) = 251.33 in 2 4r 4(10) = = 4.2441 in. 3π 3π π A, in 2 4 r2 = π 4 (10) 2 = 78.540 in 2 x , in. x , in. xA, in 3 yA, in 3 I (10)(16) = 160 5 8 800 1280 II (10)(16) = 160 0 8 0 1280 III 251.33 6.3662 8 4.2441 0 IV Σ X Σ A = Σ xA: 78.540 649.87 333.33 2733.3 2010.6 0 4570.6 X (649.87 in 2 ) = 2733.3 in 3 X = 4.2059 in. Y Σ A = Σ yA: 1600.0 X = Z = 4.21 in. Y (649.87 in 2 ) = 4570.6 in 3 Y = 7.0331 in. Y = 7.03 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 685 PROBLEM 5.111 A mounting bracket for electronic components is formed from sheet metal of uniform thickness. Locate the center of gravity of the bracket. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with the centroid of the corresponding area. Then (see diagram) 4(0.625) 3π = 1.98474 in. zV = 2.25 − AV = − π (0.625) 2 2 = −0.61359 in 2 A, in 2 x , in. I (2.5)(6) = 15 1.25 II (1.25)(6) = 7.5 2.5 III (0.75)(6) = 4.5 2.875 IV 5 − (3) = − 3.75 4 1.0 0 3.75 V − 0.61359 1.0 0 1.98474 0.61359 Σ 22.6364 We have y , in. z , in. 0 xA, in 3 yA, in 3 zA, in 3 0 45 3 18.75 –0.625 3 18.75 –4.6875 22.5 –1.25 3 12.9375 –5.625 13.5 3.75 46.0739 0 –14.0625 0 –1.21782 10.3125 65.7197 X ΣA = Σ xA X (22.6364 in 2 ) = 46.0739 in 3 or X = 2.04 in. Y ΣA = Σ yA Y (22.6364 in 2 ) = −10.3125 in 3 or Y = − 0.456 in. Z ΣA = Σ zA Z (22.6364 in 2 ) = 65.7197 in 3 or Z = 2.90 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 686 PROBLEM 5.112 An 8-in.-diameter cylindrical duct and a 4 × 8-in. rectangular duct are to be joined as indicated. Knowing that the ducts were fabricated from the same sheet metal, which is of uniform thickness, locate the center of gravity of the assembly. SOLUTION Assume that the body is homogeneous so that its center of gravity coincides with the centroid of the area. By symmetry, z = 0. 1 2 A, in 2 x , in. y , in. xA, in 3 yA, in 3 π (8)(12) = 96π 0 6 0 576π 10 −128 −160π 4(4) 16 =− 3π 3π 12 − 42.667 96π 6 12 576 1152 6 8 576 768 4(4) 16 = 3π 3π 8 − 42.667 − 64π 6 10 288 480 6 10 288 480 1514.6 4287.4 − π 2 π 3 Then (4) 2 = 8π 4 2 (8)(12) = 96 5 (8)(12) = 96 6 − 2(4) (8)(4) = −16π π (4) 2 = − 8π 7 2 (4)(12) = 48 8 (4)(12) = 48 Σ 539.33 π − = 8 π X = Σ xA 1514.67 = in. 539.33 ΣA or X = 2.81 in. Y = Σ yA 4287.4 = in. 539.33 ΣA or Y = 7.95 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 687 PROBLEM 5.113 An elbow for the duct of a ventilating system is made of sheet metal of uniform thickness. Locate the center of gravity of the elbow. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the centroid of the corresponding area. Also, note that the shape of the duct implies Y = 38.0 mm Note that xI = zI = 400 − xII = 400 − zII = 300 − 2 π 2 π 2 π (400) = 145.352 mm (200) = 272.68 mm (200) = 172.676 mm xIV = zIV = 400 − 4 (400) = 230.23 mm 3π 4 (200) = 315.12 mm 3π 4 zV = 300 − (200) = 215.12 mm 3π xV = 400 − Also note that the corresponding top and bottom areas will contribute equally when determining x and z . Thus, A, mm 2 x , mm z , mm xA, mm3 zA, mm3 (400)(76) = 47,752 145.352 145.352 6,940,850 6,940,850 (200)(76) = 23,876 272.68 172.676 6,510,510 4,122,810 III 100(76) = 7600 200 350 1,520,000 2,660,000 IV π 2 (400)2 = 251,327 4 230.23 230.23 57,863,020 57,863,020 V π −2 (200)2 = −62,832 4 315.12 215.12 –19,799,620 –13,516,420 VI −2(100)(200) = −40,000 300 350 –12,000,000 –14,000,000 Σ 227,723 41,034,760 44,070,260 I II π 2 π 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 688 PROBLEM 5.113 (Continued) We have X Σ A = Σ xA: X (227, 723 mm 2 ) = 41,034, 760 mm3 or X = 180.2 mm Z Σ A = Σ zA: Z (227, 723 mm 2 ) = 44,070, 260 mm3 or Z = 193.5 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 689 PROBLEM 5.114 Locate the center of gravity of the figure shown, knowing that it is made of thin brass rods of uniform diameter. SOLUTION X =0 By symmetry, L, in. y , in. z , in. yL, in 2 zL, in 2 AB 302 + 162 = 34 15 0 510 0 AD 302 + 162 = 34 15 8 510 272 AE 302 + 162 = 34 15 0 510 0 BDE π (16) = 50.265 0 0 512 Σ 152.265 1530 784 2(16) π = 10.186 Y ΣL = Σ y L : Y (152.265 in.) = 1530 in 2 Y = 10.048 in. Y = 10.05 in. Z ΣL = Σ z L : Z (152.265 in.) = 784 in 2 Z = 5.149 in. Z = 5.15 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 690 PROBLEM 5.115 Locate the center of gravity of the figure shown, knowing that it is made of thin brass rods of uniform diameter. SOLUTION Uniform rod: AB 2 = (1 m) 2 + (0.6 m) 2 + (1.5 m) 2 AB = 1.9 m L, m x, m y, m z, m xL, m 2 yL, m 2 ΣL , m AB 1.9 0.5 0.75 0.3 0.95 1.425 0.57 BD 0.6 1.0 0 0.3 0.60 0 0.18 DO 1.0 0.5 0 0 0.50 0 0 OA 1.5 0 0.75 0 0 1.125 0 Σ 5.0 2.05 2.550 0.75 X Σ L = Σ x L : X (5.0 m) = 2.05 m 2 X = 0.410 m Y Σ L = Σ y L : Y (5.0 m) = 2.55 m 2 Y = 0.510 m Z Σ L = Σ z L : Z (5.0 m) = 0.75 m 2 Z = 0.1500 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 691 PROBLEM 5.116 A thin steel wire of uniform cross section is bent into the shape shown. Locate its center of gravity. SOLUTION First assume that the wire is homogeneous so that its center of gravity will coincide with the centroid of the corresponding line. x2 = z2 = 1 2 We have π 2 2 × 2.4 π = 4.8 π m L, m x, m y, m z, m xL, m 2 yL, m 2 zL, m 2 2.6 1.2 0.5 0 3.12 1.3 0 π 5.76 0 5.76 × 2.4 = 1.2π 4.8 π 0 4.8 3 2.4 0 0 1.2 0 0 2.88 4 1.0 0 0.5 0 0 0.5 0 Σ 9.7699 8.88 1.8 8.64 X Σ L = Σ x L : X (9.7699 m) = 8.88 m 2 or X = 0.909 m Y Σ L = Σ y L : Y (9.7699 m) = 1.8 m 2 or Y = 0.1842 m Z Σ L = Σ z L : Z (9.7699 m) = 8.64 m 2 or Z = 0.884 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 692 PROBLEM 5.117 The frame of a greenhouse is constructed from uniform aluminum channels. Locate the center of gravity of the portion of the frame shown. SOLUTION First assume that the channels are homogeneous so that the center of gravity of the frame will coincide with the centroid of the corresponding line. 2×3 x8 = x9 = π y8 = y9 = 5 + L, ft 6 π 2×3 π y , ft ft = 6.9099 ft z , ft xL, ft 2 yL, ft 2 zL, ft 2 1 2 3 0 1 6 0 2 2 3 1.5 0 2 4.5 0 6 3 5 3 2.5 0 15 12.5 0 4 5 3 2.5 2 15 12.5 10 5 8 0 4 2 0 32 16 6 2 3 5 1 6 10 2 7 3 1.5 5 2 4.5 15 6 6.9099 0 9 32.562 0 π 6.9099 2 9 32.562 9.4248 0 8 1 0 16 2 8 9 10 Σ We have x , ft = π 2 π 2 × 3 = 4.7124 × 3 = 4.7124 2 6 π 6 39.4248 69 163.124 53.4248 X Σ L = Σ x L : X (39.4248 ft) = 69 ft 2 or X = 1.750 ft Y Σ L = Σ y L : Y (39.4248 ft) = 163.124 ft 2 or Z Σ L = Σ z L : Z (39.4248 ft) = 53.4248 ft 2 or Z = 1.355 ft Y = 4.14 ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 693 PROBLEM 5.118 Three brass plates are brazed to a steel pipe to form the flagpole base shown. Knowing that the pipe has a wall thickness of 8 mm and that each plate is 6 mm thick, determine the location of the center of gravity of the base. (Densities: brass = 8470 kg/m3; steel = 7860 kg/m3.) SOLUTION Since brass plates are equally spaced, we note that the center of gravity lies on the y-axis. x =z =0 Thus, V= π [(0.064 m) 2 − (0.048 m) 2 ](0.192 m) 4 = 270.22 × 10−6 m3 Steel pipe: m = ρ V = (7860 kg/m3 )(270.22 × 10−6 m3 ) = 2.1239 kg Each brass plate: 1 (0.096 m)(0.192 m)(0.006 m) = 55.296 × 10−6 m3 2 m = ρ V = (8470 kg/m3 )(55.296 × 10−6 m3 ) = 0.46836 kg V= Flagpole base: Σm = 2.1239 kg + 3(0.46836 kg) = 3.5290 kg Σ y m = (0.096 m)(2.1239 kg) + 3[(0.064 m)(0.46836 kg)] = 0.29382 kg ⋅ m Y Σm = Σ y m : Y (3.5290 kg) = 0.29382 kg ⋅ m Y = 0.083259 m Y = 83.3 mm above the base PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 694 PROBLEM 5.119 A brass collar, of length 2.5 in., is mounted on an aluminum rod of length 4 in. Locate the center of gravity of the composite body. (Specific weights: brass = 0.306 lb/in3, aluminum = 0.101 lb/in3) SOLUTION Aluminum rod: W = γV π = (0.101 lb/in 3 ) (1.6 in.)2 (4 in.) 4 = 0.81229 lb Brass collar: W = γV π = (0.306 lb/in.3 ) [(3 in.) 2 − (1.6 in.) 2 ](2.5 in.) 4 = 3.8693 lb yW (lb ⋅ in.) Component W(lb) Rod 0.81229 2 1.62458 Collar 3.8693 1.25 4.8366 Σ 4.6816 y (in.) 6.4612 Y ΣW = Σ y W : Y (4.6816 lb) = 6.4612 lb ⋅ in. Y = 1.38013 in. Y = 1.380 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 695 PROBLEM 5.120 A bronze bushing is mounted inside a steel sleeve. Knowing that the specific weight of bronze is 0.318 lb/in3 and of steel is 0.284 lb/in3, determine the location of the center of gravity of the assembly. SOLUTION X =Z =0 First, note that symmetry implies Now W = ( ρ g )V π yI = 0.20 in. WI = (0.284 lb/in 3 ) [(1.82 − 0.752 ) in 2 ](0.4 in.) = 0.23889 lb 4 π yII = 0.90 in. WII = (0.284 lb/in 3 ) [(1.1252 − 0.752 ) in 2 ](1 in.) = 0.156834 lb 4 π yIII = 0.70 in. WIII = (0.318 lb/in 3 ) [(0.752 − 0.52 ) in 2 ](1.4 in.) = 0.109269 lb 4 We have Y ΣW = Σ yW (0.20 in.)(0.23889 lb) + (0.90 in.)(0.156834 lb) + (0.70 in.)(0.109269 lb) Y = 0.23889 lb + 0.156834 lb + 0.109269 lb Y = 0.526 in. or (above base) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 696 PROBLEM 5.121 A scratch awl has a plastic handle and a steel blade and shank. Knowing that the density of plastic is 1030 kg/m3 and of steel is 7860 kg/m3, locate the center of gravity of the awl. SOLUTION Y = Z = 0 First, note that symmetry implies 5 xI = (12.5 mm) = 7.8125 mm 8 2π 3 WI = (1030 kg/m3 ) (0.0125 m) 3 = 4.2133 × 10−3 kg xII = 52.5 mm π WII = (1030 kg/m3 ) (0.025 m) 2 (0.08 m) 4 −3 = 40.448 × 10 kg xIII = 92.5 mm − 25 mm = 67.5 mm π WIII = −(1030 kg/m3 ) (0.0035 m) 2 (0.05 m) 4 −3 = −0.49549 × 10 kg xIV = 182.5 mm − 70 mm = 112.5 mm π WIV = (7860 kg/m3 ) (0.0035 m) 2 (0.14 m) 2 = 10.5871 × 10−3 kg 4 1 xV = 182.5 mm + (10 mm) = 185 mm 4 π WV = (7860 kg/m3 ) (0.00175 m) 2 (0.01 m) = 0.25207 × 10−3 kg 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 697 PROBLEM 5.121 (Continued) W, kg We have x, mm xW, kg ⋅ mm I 4.123 × 10−3 II 40.948 × 10−3 52.5 2123.5 × 10−3 III −0.49549 × 10−3 67.5 −33.447 × 10−3 IV 10.5871 × 10−3 112.5 1191.05 × 10−3 V 0.25207 × 10−3 185 46.633 × 10−3 Σ 55.005 × 10−3 7.8125 32.916 × 10−3 3360.7 × 10−3 X ΣW = Σ xW : X (55.005 × 10−3 kg) = 3360.7 × 10−3 kg ⋅ mm X = 61.1 mm or (from the end of the handle) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 698 PROBLEM 5.122 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A hemisphere SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, xEL = x The equation of the generating curve is x 2 + y 2 = a 2 so that r 2 = a 2 − x 2 and then dV = π ( a 2 − x 2 ) dx V1 = Component 1: = and 1 xEL dV = a/2 0 a/2 x3 π (a − x )dx = π a 2 x − 3 0 2 2 11 3 πa 24 a/2 0 x π (a 2 − x 2 ) dx a/2 x2 x4 = π a2 − 2 4 0 7 = π a4 64 Now x1V1 = x 1 EL dV : 11 7 x1 π a 3 = π a 4 24 64 or x1 = 21 a 88 Component 2: a x3 π (a − x )dx = π a 2 x − V2 = a /2 3 a/2 a 2 2 a 3 a3 2 a ( 2 ) 2 = π a (a ) − − a − 3 2 3 5 = π a3 24 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 699 PROBLEM 5.122 (Continued) and 2 xEL dV = a x 2 x4 − x π ( a 2 − x 2 )dx = π a 2 a/2 2 4 a/2 a 2 4 a a 2 ( (a ) 4 2 ( 2 ) 2 (a) 2 ) = π a − − − a 2 4 2 4 9 = π a4 64 Now x2V2 = 2 5 9 xEL dV : x2 π a3 = π a 4 24 64 or x2 = 27 a 40 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 700 PROBLEM 5.123 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A semiellipsoid of revolution SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, xEL = x x2 y2 + = 1 so that h2 a 2 The equation of the generating curve is r2 = a2 2 (h − x 2 )dx 2 h dV = π and then V1 = Component 1: = and a2 2 (h − x 2 ) 2 h 1 xEL dV = h/2 0 h/2 a2 a2 x3 π 2 (h 2 − x 2 )dx = π 2 h 2 x − 3 0 h h 11 2 πa h 24 h/2 0 a2 x π 2 (h 2 − x 2 ) dx h a2 =π 2 h = Now Component 2: x1V1 = V2 = h/2 2 x2 x4 − h 4 0 2 7 π a 2 h2 64 x 1 EL dV : h h/2 11 7 x1 π a 2 h = π a 2 h 2 24 64 or x1 = 21 h 88 h π a2 2 a2 x3 (h − x 2 )dx = π 2 h 2 x − 2 3 h/2 h h h 3 a 2 2 ( h)3 2 h ( 2 ) = π 2 h ( h ) − − h − 3 2 3 h 5 = π a2h 24 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 701 PROBLEM 5.123 (Continued) and 2 xEL dV = a2 x π 2 (h 2 − x 2 ) dx h/2 h h a2 =π 2 h h 2 x2 x4 − h 4 h/2 2 h 2 h 4 ( a 2 2 ( h) 2 ( h) 4 2 ( 2 ) 2 ) = π 2 h − − − h 2 4 2 4 h 9 = π a 2 h2 64 Now x2V2 = 5 9 xEL dV : x2 π a 2 h = π a 2 h 2 2 24 64 or x2 = 27 h 40 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 702 PROBLEM 5.124 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Figure 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A paraboloid of revolution SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, xEL = x The equation of the generating curve is x = h − h 2 a2 y so that r 2 = (h − x). 2 h a a2 (h − x)dx h dV = π and then V1 = Component 1: h/2 0 π a2 =π h a2 ( h − x) dx h h/2 x2 hx − 2 0 3 = π a2 h 8 and 1 xEL dV = h/2 0 a2 x π (h − x)dx h h/2 a 2 x 2 x3 1 2 2 =π h − = π a h h 2 3 0 12 Now x1V1 = x 1 EL dV : 3 1 x1 π a 2 h = π a 2 h 2 8 12 or x1 = 2 h 9 Component 2: h a2 a2 x2 V2 = hx π (h − x)dx = π − h/2 h h 2 h/2 h h 2 ( h) 2 h ( 2 ) a2 =π h( h ) − − h − h 2 2 2 1 2 = πa h 8 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 703 PROBLEM 5.124 (Continued) and 2 xEL dV = a2 =π h = Now x2V2 = h a2 a 2 x 2 x3 x π ( h − x)dx = π h − h/2 h 2 3 h/2 h h h 2 ( h )3 2 ( h )3 ( 2 ) ( h) − − 2 h − h 2 3 2 3 1 π a 2 h2 12 2 1 1 xEL dV : x2 π a 2 h = π a 2 h 2 8 12 or x2 = 2 h 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 704 PROBLEM 5.125 Locate the centroid of the volume obtained by rotating the shaded area about the x-axis. SOLUTION y =0 First note that symmetry implies z =0 Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, xEL = x r = kx1/3 Now dV = π k 2 x 2/3 dx so that at x = h, y = a, a = kh1/3 or k= a h1/3 a 2 2/3 x dx h 2/3 dV = π Then V= and h 0 π a 2 2/3 x dx h 2/3 a2 h 2/3 =π h 3 5/3 5 x 0 3 = π a2 h 5 Also x EL dV = h 0 h a 2 2/3 a 2 3 8/3 x π 2/3 x dx = π 2/3 x h 8 0 h 3 = π a 2 h2 8 Now xV = xdV : 3 3 x π a 2 h = π a 2 h2 5 8 5 or x = h 8 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 705 PROBLEM 5.126 Locate the centroid of the volume obtained by rotating the shaded area about the x-axis. SOLUTION y =0 First, note that symmetry implies z =0 Choose as the element of volume a disk of radius r and thickness dx. dV = π r 2 dx, xEL = x Then Now r = 1 − 2 1 dV = π 1 − dx x 2 1 = π 1 − + 2 dx x x 1 so that x V= Then 3 1 π 1 − 3 2 1 1 + dx = π x − 2 ln x − x x2 x 1 1 1 = π 3 − 2ln 3 − − 1 − 2 ln1 − 3 1 = (0.46944π ) m3 and x EL dV = 3 1 3 x2 2 1 x π 1 − + 2 dx = π − 2 x + ln x x x 2 1 32 13 = π − 2(3) + ln 3 − − 2(1) + ln1 2 2 = (1.09861π ) m Now xV = xEL dV : x (0.46944π m3 ) = 1.09861π m 4 or x = 2.34 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 706 PROBLEM 5.127 Locate the centroid of the volume obtained by rotating the shaded area about the line x = h. SOLUTION x =h First, note that symmetry implies z =0 Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dy, yEL = y Now x 2 = Then and Let Then h2 2 h 2 (a − y 2 ) so that r = h − a − y2 . 2 a a dV = π V= ( h2 a − a2 − y 2 a2 a 0 π ) dy 2 ( h2 a − a2 − y2 a2 ) dy 2 y = a sin θ dy = a cos θ dθ h2 a2 h2 =π 2 a = π ah 2 V =π π /2 0 π /2 0 π /2 0 ( a − a 2 − a 2 sin 2 θ ) a cosθ dθ 2 a 2 − 2a (a cos θ ) + (a 2 − a 2 sin 2 θ ) a cos θ dθ (2cos θ − 2 cos 2 θ − sin 2 θ cos θ ) dθ π /2 θ sin 2θ 1 3 = π ah 2sin θ − 2 + − sin θ 4 3 2 0 2 π 1 = π ah 2 2 − 2 2 − 2 3 = 0.095870π ah 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 707 PROBLEM 5.127 (Continued) and yEL dV = 0 ( h2 y π 2 a − a 2 − y 2 a ( 2a y − 2ay ) dy 2 ) =π h2 a2 =π h2 a2 1 4 2 2 2 2 2 3/2 a y + 3 a (a − y ) − 4 y 0 =π h2 a2 2 2 1 4 2 2 3/2 a (a ) − a − a( a ) 4 3 = Now a a 0 2 a 2 − y 2 − y 3 dy a 1 π a 2 h2 12 y (0.095870π ah2 ) = yV = yEL dV : 1 π a2 h2 12 or y = 0.869a PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 708 PROBLEM 5.128* Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the x-axis. SOLUTION y =0 First, note that symmetry implies z =0 Choose as the element of volume a disk of radius r and thickness dx. dV = π r 2 dx, xEL = x Then r = b sin Now πx 2a dV = π b 2 sin 2 so that V= Then 2a a πx 2a dx π b 2 sin 2 πx 2a dx 2a x sin π x = π b − π2 a 2 a 2 a 2 = π b 2 ( 22a ) − ( a2 ) 1 = π ab 2 2 and x EL dV = 2a a πx x π b 2 sin 2 dx 2a Use integration by parts with u=x dV = sin 2 du = dx V= x − 2 πx 2a sin πax 2π a PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 709 PROBLEM 5.128* (Continued) Then 2a 2a x sin π x x sin πax xEL dV = π b x − 2π − − 2π a dx a 2 2 a a a 2a 2 π x 2a a 1 2 a 2 = π b 2a − a − x + 2 cos a a 2π 2 4 2 2 3 1 a2 1 a 2 = π b 2 a 2 − (2a) 2 + 2 − (a) 2 + 2 4 2π 2π 2 4 3 1 = π a 2b 2 − 2 4 π = 0.64868π a 2b 2 Now 1 xV = xEL dV : x π ab 2 = 0.64868π a 2b 2 2 or x = 1.297a PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 710 PROBLEM 5.129* Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the y-axis. (Hint: Use a thin cylindrical shell of radius r and thickness dr as the element of volume.) SOLUTION x =0 First note that symmetry implies z =0 Choose as the element of volume a cylindrical shell of radius r and thickness dr. Then Now so that Then dV = (2π r )( y )(dr ), y = b sin 2a a 1 y 2 πr 2a dV = 2π br sin V= yEL = πr 2a dr 2π br sin πr 2a dr Use integration by parts with u = rd dv = sin dr 2a 2a πr v = − cos 2a π du = dr Then πr 2a 2a π r V = 2π b ( r ) − cos − π 2a a 2a π r π cos 2a dr 2a a 2a 2a 4a 2 π r = 2π b − [ (2a)(−1) ] + 2 sin 2a a π π 4a 2 4 a 2 − 2 V = 2π b π π 1 = 8a 2 b 1 − π = 5.4535a 2b PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 711 PROBLEM 5.129* (Continued) Also y EL dV π r πr 1 2π br sin b sin dr 2 a 2a 2a 2a πr r sin 2 dr = π b2 a 2a = 2a Use integration by parts with u=r dv = sin 2 du = dr Then v= r − 2 πr dr 2a sin πar 2π a 2a 2a r sin π r r sin πar yEL dV = π b ( r ) − 2π − − 2π a dr a 2 2 a a a 2a 2 π r a2 2a a r 2 = π b (2a) − ( a) − + 2 cos a a 2 2 4 2π 2 3 (2a) 2 (a ) 2 a2 a 2 = π b2 a 2 − + 2− + 2 4 2π 2π 2 4 3 1 = π a 2b 2 − 2 4 π = 2.0379a 2 b 2 Now yV = yEL dV : y (5.4535a 2b) = 2.0379a 2b 2 or y = 0.374b PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 712 PROBLEM 5.130* Show that for a regular pyramid of height h and n sides (n = 3, 4, ) the centroid of the volume of the pyramid is located at a distance h/4 above the base. SOLUTION Choose as the element of a horizontal slice of thickness dy. For any number N of sides, the area of the base of the pyramid is given by Abase = kb 2 where k = k ( N ); see note below. Using similar triangles, we have s h− y = b h s= or b (h − y ) h dV = Aslice dy = ks 2 dy = k Then V= and h k 0 b2 (h − y ) 2 dy h2 b2 b2 2 ( h y ) dy k − = h2 h2 h 1 3 − 3 (h − y ) 0 1 = kb 2 h 3 yEL = y Also, so that yEL dV = h 0 =k Now Note: b2 b2 y k 2 ( h − y ) 2 dy = k 2 h h b2 h2 h 0 ( h 2 y − 2hy 2 + y 3 ) dy h 1 2 2 1 2 2 2 3 1 4 2 h y − 3 hy + 4 y = 12 kb h 0 1 1 yV = yEL dV : y kb 2 h = kb 2 h 2 3 12 1 Abase = N × b × 2 N = b2 4 tan πN b 2 tan πN or y = 1 h Q.E.D. 4 = k ( N )b 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 713 PROBLEM 5.131 Determine by direct integration the location of the centroid of one-half of a thin, uniform hemispherical shell of radius R. SOLUTION x = 0 First note that symmetry implies The element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the xy plane. Now dA = (π r )( Rdθ ) 2r yEL = − π r = R sin θ where dA = π R 2 sin θ dθ 2R yEL = − sin θ so that π A= Then π /2 0 π R 2 sin θ dθ = π R 2 [− cos θ ]π0 /2 = π R2 and y EL dA = π /2 0 2R sin θ (π R 2 sin θ dθ ) − π π /2 θ sin 2θ = −2 R3 − 4 0 2 =− Now Symmetry implies π 2 R3 yA = yEL dA: y (π R 2 ) = − π 2 R3 1 or y = − R 2 1 z = − R 2 z=y PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 714 PROBLEM 5.132 The sides and the base of a punch bowl are of uniform thickness t. If t < < R and R = 250 mm, determine the location of the center of gravity of (a) the bowl, (b) the punch. SOLUTION (a) Bowl: x =0 First note that symmetry implies z =0 for the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the center of gravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl, an element of area is obtained by rotating the arc ds about the y-axis. Then dAwall = (2π R sin θ )( R dθ ) ( yEL ) wall = − R cos θ and Awall = Then π /2 π /6 2π R 2 sin θ dθ π /2 = 2π R 2 [ − cos θ ]π /6 = π 3R2 π = π ywall Awall = ( yEL ) wall dA and /2 /6 (− R cos θ )(2π R 2 sin θ dθ ) = π R3 [cos 2 θ ]ππ /2 /6 3 = − π R3 4 π By observation, Abase = Now y ΣA = Σ yA or or 4 R2 , ybase = − 3 R 2 π 3 π 3 y π 3R 2 + R 2 = − π R3 + R 2 − R 4 4 4 2 y = − 0.48763R R = 250 mm y = −121.9 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 715 PROBLEM 5.132 (Continued) (b) Punch: x =0 First note that symmetry implies z =0 and that because the punch is homogeneous, its center of gravity will coincide with the centroid of the corresponding volume. Choose as the element of volume a disk of radius x and thickness dy. Then dV = π x 2 dy, Now yEL = y x2 + y 2 = R2 dV = π ( R 2 − y 2 )dy so that V= Then 0 − 3/2 R π ( R 2 − y 2 ) dy 0 1 = π R 2 y − y3 3 − 3/2 R 3 3 1 3 3 2 R − − R = π 3R3 = −π R − 2 3 2 8 and y EL dV = 0 − 3/2 R ( ) ( y ) π R 2 − y 2 dy 0 1 1 = π R2 y 2 − y4 2 4 − 3/2 R 4 1 1 3 3 15 2 R − − R = − π R4 = −π R − 2 2 4 2 64 2 Now 15 3 yV = yEL dV : y π 3 R3 = − π R 4 64 8 y =− or 5 8 3 R R = 250 mm y = − 90.2 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 716 PROBLEM 5.133 Locate the centroid of the section shown, which was cut from a thin circular pipe by two oblique planes. SOLUTION x =0 First note that symmetry implies Assume that the pipe has a uniform wall thickness t and choose as the element of volume a vertical strip of width adθ and height ( y2 − y1 ). Then dV = ( y2 − y1 )ta dθ , Now y1 = = and h 3 2a z+ h 6 yEL = 1 ( y1 + y2 ) z EL = z 2 y2 = − h ( z + a) 6a = 2h 3 2a z+ 2 h 3 h ( − z + 2a ) 3a z = a cos θ h h (−a cos θ + 2a) − (a cos θ + a ) 3a 6a h = (1 − cos θ ) 2 Then ( y2 − y1 ) = and ( y1 + y2 ) = h h (a cos θ + a) + (− a cos θ + 2a) 6a 3a h = (5 − cos θ ) 6 aht h dV = (1 − cos θ )dθ yEL = (5 − cos θ ), z EL = a cos θ 2 12 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 717 PROBLEM 5.133 (Continued) and V =2 π =2 π aht (1 − cos θ )dθ = aht[θ − sin θ ]π0 0 2 = π aht Then y EL dV h aht (5 − cos θ ) (1 − cos θ ) dθ 12 2 0 = ah 2 t 12 = ah 2 t θ sin 2θ 5θ − 6sin θ + + 12 2 4 0 = 11 π ah2 t 24 π 0 (5 − 6 cos θ + cos 2 θ ) dθ π z EL dV =2 π 0 aht a cos θ (1 − cos θ ) dθ 2 π θ sin 2θ = a ht sin θ − − 2 4 0 1 = − π a 2 ht 2 2 11 π ah 2t 24 Now yV = yEL dV : y (π aht ) = and 1 zV = zEL dV : z (π aht ) = − π a 2 ht 2 or y = 11 h 24 1 or z = − a 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 718 PROBLEM 5.134* Locate the centroid of the section shown, which was cut from an elliptical cylinder by an oblique plane. SOLUTION x=0 First note that symmetry implies Choose as the element of volume a vertical slice of width zx, thickness dz, and height y. Then dV = 2 xy dz , x= and y=− Then V= Then h/2 h h z+ = (b − z ) b 2 2b a 2 h b − z 2 (b − z ) dz 2 −b b 2b b z = b sin θ V= 1 , z EL = z 24 a 2 b − z2 b Now Let yEL = ah b2 π /2 π = abh /2 dz = b cos θ dθ (b cos θ )[b(1 − sin θ )] b cos θ dθ π /2 π − /2 (cos 2 θ − sin θ cos 2 θ ) dθ π /2 θ sin 2θ 1 = abh + + cos3 θ 4 3 2 −π /2 1 V = π abh 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 719 PROBLEM 5.134* (Continued) and y EL dV Now so that Then Also, = 1 ah 2 4 b3 b −b z = b sin θ Let Then 1 h a 2 h b − z 2 (b − z ) dz × (b − z ) 2 −b 2 2b b 2b b = dz = b cos θ dθ 1 ah 2 4 b3 1 = abh 2 4 sin 2 θ cos 2 θ = y EL dV z EL dV π /2 π yEL dV = sin 2 θ = (b − z ) 2 b 2 − z 2 dz − /2 [b(1 − sin θ )]2 (b cos θ ) × (b cos θ dθ ) π /2 π − /2 (cos 2 θ − 2sin θ cos 2 θ + sin 2 θ cos 2 θ ) dθ 1 1 (1 − cos 2θ ) cos 2 θ = (1 + cos 2θ ) 2 2 1 (1 − cos 2 2θ ) 4 π /2 = 1 abh 2 4 = θ sin 2θ 1 abh 2 + 4 4 2 = 5 π abh2 32 π − /2 cos 2 θ − 2sin θ cos 2 θ + 1 (1 − cos 2 2θ ) dθ 4 1 1 θ sin 4θ 1 3 + 3 cos θ + 4 θ − 4 2 + 8 π /2 −π /2 a 2 h z 2 a − z 2 (b − z ) dz −b b 2b ah b z (b − z ) b 2 − z 2 dz = 2 b −b = b z = b sin θ Let Then z EL dV = ah b2 π /2 π − /2 = ab 2 h Using sin 2 θ cos 2 θ = z EL dV dz = b cos θ dθ (b sin θ )[b(1 − sin θ )](b cos θ ) × (b cos θ dθ ) π /2 π − /2 (sin θ cos 2 θ − sin 2 θ cos 2 θ ) dθ 1 (1 − cos 2 2θ ) from above, 4 = ab 2 h π /2 π − /2 1 sin θ cos 2 θ − (1 − cos 2 2θ ) dθ 4 1 1 1 θ sin 4θ = ab 2 h − cos3 θ − θ + + 4 4 2 8 3 π /2 1 = − π ab 2 h 8 −π /2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 720 PROBLEM 5.134* (Continued) Now 1 5 yV = yEL dV : y π abh = π abh 2 2 32 or y = and 1 1 z V = z EL dV : z π abh = − π ab 2 h 8 2 1 or z = − b 4 5 h 16 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 721 PROBLEM 5.135 After grading a lot, a builder places four stakes to designate the corners of the slab for a house. To provide a firm, level base for the slab, the builder places a minimum of 3 in. of gravel beneath the slab. Determine the volume of gravel needed and the x coordinate of the centroid of the volume of the gravel. (Hint: The bottom surface of the gravel is an oblique plane, which can be represented by the equation y = a + bx + cz.) SOLUTION The centroid can be found by integration. The equation for the bottom of the gravel is y = a + bx + cz , where the constants a, b, and c can be determined as follows: For x = 0 and z = 0, y = − 3 in., and therefore, − For x = 30 ft and z = 0, y = − 5 in., and therefore, − For x = 0 and z = 50 ft, 3 1 ft = a, or a = − ft 12 4 5 1 1 ft = − ft + b(30 ft), or b = − 12 4 180 y = − 6 in., and therefore, − 6 1 1 ft = − ft + c(50 ft), or c = − 12 4 200 Therefore, y=− Now x= 1 1 1 ft − x− z 4 180 200 x EL dV V A volume element can be chosen as dV = | y| dx dz 1 1 1 1+ x+ z dx dz 4 45 50 or dV = and xEL = x PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 722 PROBLEM 5.135 (Continued) Then x EL dV = 50 30 0 0 x 1 1 1+ x+ z dx dz 4 45 50 30 50 x 2 z 2 1 3 x + x dz + 100 0 2 135 1 = 4 = 1 4 = 1 9 650 z + z 2 4 2 0 0 50 0 (650 + 9 z ) dz 50 = 10937.5 ft 4 The volume is V dV = 50 30 0 0 = 1 4 = 1 4 1 1 1 1+ x+ z dx dz 4 45 50 30 50 0 1 2 z x + 90 x + 50 x dz 0 50 0 3 40 + z dz 5 50 1 3 = 40 z + z 2 4 10 0 = 687.50 ft 3 Then x = x EL dV V = 10937.5ft 4 = 15.9091 ft 687.5 ft 3 V = 688 ft 3 Therefore, x = 15.91 ft PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 723 PROBLEM 5.136 Determine by direct integration the location of the centroid of the volume between the xz plane and the portion shown of the surface y = 16h(ax − x2)(bz − z2)/a2b2. SOLUTION First note that symmetry implies x= a 2 z= b 2 Choose as the element of volume a filament of base dx × dz and height y. Then dV = y dx dz , or Then dV = yEL = 1 y 2 16h (ax − x 2 )(bz − z 2 ) dx dz 2 2 a b b a 0 0 V= V= 16h a 2b2 16h (ax − x 2 )(bz − z 2 ) dx dz a 2b2 a 1 a (bz − z 2 ) x 2 − x3 dz 0 3 0 z b b 16h a 1 1 b = 2 2 ( a 2 ) − ( a )3 z 2 − z 3 3 3 0 a b 2 2 8ah b 2 1 3 (b) − (b) 3 3b 2 2 4 = abh 9 = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 724 PROBLEM 5.136 (Continued) and y EL dV b a 0 0 = = 128h 2 a 4b 4 1 16h 16h ( ax − x 2 )(bz − z 2 ) 2 2 (ax − x 2 )(bz − z 2 ) dx dz 2 2 2 a b a b 128h 2 = 2 4 a b Now b a 0 0 (a 2 x 2 − 2ax3 + x 4 )(b 2 z 2 − 2bz 3 + z 4 )dx dz a a2 a 1 (b z − 2bz + z ) x3 − x 4 + x5 dz 0 2 5 0 3 b 2 2 3 4 b = 128h 2 a 4b4 a2 a 4 1 5 b2 3 b 4 1 5 3 − ( a ) (a) + (a) z − z + z 2 5 z 5 0 3 3 = 64ah 2 15b 4 b3 3 b 4 1 5 32 abh 2 (b) − (b) + (b) = 3 2 5 225 4 32 yV = yEL dV : y abh = abh 2 9 225 8 or y = 25 h PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 725 PROBLEM 5.137 Locate the centroid of the plane area shown. SOLUTION A, in 2 1 Then π 2 (38)2 = 2268.2 2 −20 × 16 = 320 Σ 1948.23 x , in. y , in. x A, in 3 y A, in 3 0 16.1277 0 36,581 3200 −2560 3200 34,021 −10 8 X= Σ xA 3200 = Σ A 1948.23 X = 1.643 in. Y = Σ y A 34, 021 = Σ A 1948.23 Y = 17.46 in. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 726 PROBLEM 5.138 Locate the centroid of the plane area shown. SOLUTION Then A, mm 2 x , mm y , mm x A, mm3 y A, mm3 1 2 (75)(120) = 6000 3 28.125 48 168,750 288,000 2 1 − (75)(60) = −2250 2 25 20 –56,250 –45,000 Σ 3750 112,500 243,000 X ΣA = ΣxA X (3750 mm 2 ) = 112,500 mm3 and or X = 30.0 mm Y ΣA = Σ yA Y (3750 mm 2 ) = 243, 000 mm3 or Y = 64.8 mm PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 727 PROBLEM 5.139 The frame for a sign is fabricated from thin, flat steel bar stock of mass per unit length 4.73 kg/m. The frame is supported by a pin at C and by a cable AB. Determine (a) the tension in the cable, (b) the reaction at C. SOLUTION First note that because the frame is fabricated from uniform bar stock, its center of gravity will coincide with the centroid of the corresponding line. L, m x, m xL, m 2 1 1.35 0.675 0.91125 2 0.6 0.3 0.18 3 0.75 0 0 4 0.75 0.2 0.15 1.07746 1.26936 5 Σ Then π 2 (0.75) = 1.17810 4.62810 2.5106 X ΣL = Σx L X (4.62810) = 2.5106 or X = 0.54247 m The free-body diagram of the frame is then where W = (m′Σ L) g = 4.73 kg/m × 4.62810 m × 9.81 m/s 2 = 214.75 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 728 PROBLEM 5.139 (Continued) Equilibrium then requires 3 ΣM C = 0: (1.55 m) TBA − (0.54247 m)(214.75 N) = 0 5 (a) TBA = 125.264 N or or TBA = 125.3 N 3 ΣFx = 0: C x − (125.264 N) = 0 5 (b) C x = 75.158 N or ΣFy = 0: C y + 4 (125.264 N) − (214.75 N) = 0 5 C y = 114.539 N or C = 137.0 N Then 56.7° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 729 PROBLEM 5.140 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION For the element (EL) shown, at x = a, y = h, h = ka 3 or k = x= Then xEL yEL a 1/3 y dy h1/3 1 1 a 1/3 = x= y 2 2 h1/3 =y A = dA = Then Hence a 1/3 y h1/3 dA = x dy = Now and h a3 xEL dA = yEL dA = h 0 h 0 h 0 ( ) a 1/3 3 a y dy = y 4/3 1/3 1/3 4h h h = 0 3 ah 4 h 1 a 1/3 a 1/3 1 a 3 5/3 3 y 1/3 y dy = y = a2h 1/3 2/3 2h h 2 h 5 0 10 h 3 a 3 a y 1/3 y1/3 dy = 1/3 y 7/3 = ah 2 7 7 h h 0 3 3 x ah = a 2 h 4 10 x= 2 a 5 3 3 y ah = ah2 4 7 y= 4 h 7 xA = xEL dA: yA = yEL dA: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 730 PROBLEM 5.141 Determine by direct integration the centroid of the area shown. SOLUTION xEL = x We have a x x2 1 y = 1 − + 2 L L 2 2 x x2 dA = y dx = a 1 − + 2 dx L L yEL = A = dA = Then 2L 0 2L x x2 x2 x3 a 1 − + 2 dx = a x − + 2 2 L 3L 0 L L 8 = aL 3 and x Hence, EL dA 2L 2L x 2 x3 x x2 x4 x a 1 − + 2 dx = a − + 2 L L 2 3L 4 L 0 = = 10 2 aL 3 yEL dA = 0 2L 0 a x x2 x x2 1 − + 2 a 1 − + 2 dx L L L L 2 EL x x2 x3 x 4 1 − 2 + 3 2 − 2 3 + 4 dx L L L L = a2 2 = a2 2 x 2 x3 x4 x5 + 2 − 3 + 4 x − L L 2L 5L 0 = 11 2 a L 5 0 2L 8 10 xA = xEL dA: x aL = aL2 3 3 1 11 yA = yEL dA: y a = a 2 8 5 x= y= 5 L 4 33 a 40 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 731 PROBLEM 5.142 Three different drive belt profiles are to be studied. If at any given time each belt makes contact with one-half of the circumference of its pulley, determine the contact area between the belt and the pulley for each design. SOLUTION SOLUTION Applying the first theorem of Pappus-Guldinus, the contact area AC of a belt is given by AC = π yL = π Σ yL where the individual lengths are the lengths of the belt cross section that are in contact with the pulley. AC = π [2( y1 L1 ) + y2 L2 ] (a) 0.125 0.125 in. = π 2 3 − + [(3 − 0.125) in.](0.625 in.) in. 2 cos 20° AC = 8.10 in 2 or AC = π [2( y1 L1 )] (b) 0.375 0.375 in. = 2π 3 − 0.08 − in. 2 cos 20° AC = 6.85 in 2 or AC = π [2( y1 L1 )] (c) 2(0.25) = π 3 − in. [π (0.25 in.)] π AC = 7.01 in 2 or PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 732 PROBLEM 5.143 Determine the reactions at the beam supports for the given loading. SOLUTION We have Then 1 (3ft)(480 lb/ft) = 720 lb 2 1 R II = (6 ft)(600 lb/ft) = 1800 lb 2 R III = (2 ft)(600 lb/ft) = 1200 lb R I= ΣFx = 0: Bx = 0 ΣM B = 0: (2 ft)(720 lb) − (4 ft)(1800 lb) + (6 ft)C y − (7 ft)(1200 lb) = 0 or C y = 2360 lb C = 2360 lb ΣFy = 0: −720 lb + B y − 1800 lb + 2360 lb − 1200 lb = 0 or B y = 1360 lb B = 1360 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 733 PROBLEM 5.144 The beam AB supports two concentrated loads and rests on soil that exerts a linearly distributed upward load as shown. Determine the values of ω A and ω B corresponding to equilibrium. SOLUTION 1 RI = ω A (1.8 m) = 0.9ω A 2 1 RII = ωB (1.8 m) = 0.9ωB 2 ΣM D = 0: (24 kN)(1.2 − a) − (30 kN)(0.3 m) − (0.9ω A )(0.6 m) = 0 (1) 24(1.2 − 0.6) − (30)(0.3) − 0.54ωa = 0 For a = 0.6 m, 14.4 − 9 − 0.54ω A = 0 ΣFy = 0: − 24 kN − 30 kN + 0.9(10 kN/m) + 0.9ωB = 0 ω A = 10.00 kN/m ωB = 50.0 kN/m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 734 PROBLEM 5.145 The base of a dam for a lake is designed to resist up to 120 percent of the horizontal force of the water. After construction, it is found that silt (that is equivalent to a liquid of density ρ s = 1.76 × 103 kg/m3 ) is settling on the lake bottom at the rate of 12 mm/year. Considering a 1-m-wide section of dam, determine the number of years until the dam becomes unsafe. SOLUTION First determine force on dam without the silt, 1 1 Apw = A( ρ gh) 2 2 1 = [(6.6 m)(1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 m)] 2 = 213.66 kN = 1.2 Pw = (1.5)(213.66 kN) = 256.39 kN Pw = Pallow Next determine the force P′ on the dam face after a depth d of silt has settled. We have 1 Pw′ = [(6.6 − d ) m × (1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 − d ) m] 2 = 4.905(6.6 − d ) 2 kN ( Ps ) I = [d (1 m)][(103 kg/m3 )(9.81 m/s 2 )(6.6 − d ) m] = 9.81(6.6d − d 2 ) kN 1 ( Ps )II = [ d (1 m)][(1.76 × 103 kg/m3 )(9.81 m/s 2 )(d ) m] 2 = 8.6328d 2 kN P′ = Pw′ + ( Ps ) I + ( Ps ) II = [4.905(43.560 − 13.2000d + d 2 ) + 9.81(6.6d − d 2 ) + 8.6328d 2 ] kN = [3.7278d 2 + 213.66] kN Now it’s required that P′ = Pallow to determine the maximum value of d. (3.7278d 2 + 213.66) kN = 256.39 kN or Finally, d = 3.3856 m 3.3856 m = 12 × 10−3 m ×N year or N = 282 years PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 735 PROBLEM 5.146 Determine the location of the centroid of the composite body shown when (a) h = 2b, (b) h = 2.5b. SOLUTION V x xV Cylinder I π a 2b 1 b 2 1 2 2 πa b 2 Cone II 1 2 πa h 3 b+ 1 h 4 1 2 1 π a hb + h 3 4 1 V = π a2 b + h 3 1 1 1 Σ xV = π a 2 b 2 + hb + h 2 2 3 12 (a) For h = 2b, 1 5 V = π a 2 b + (2b) = π a 2 b 3 3 1 1 1 Σ xV = π a 2 b 2 + (2b)b + (2b) 2 3 12 2 1 2 1 3 = π a 2b 2 + + = π a 2 b2 2 3 3 2 5 3 XV = Σ xV : X π a 2b = π a 2 b 2 3 2 X= 9 b 10 Centroid is 1 b 10 to left of base of cone. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 736 PROBLEM 5.146 (Continued) (b) For h = 2.5b, 1 V = π a 2 b + (2.5b) = 1.8333π a 2 b 3 1 1 1 Σ xV = π a 2 b 2 + (2.5b)b + (2.5b) 2 3 12 2 = π a 2b 2 [0.5 + 0.8333 + 0.52083] = 1.85416π a 2b 2 XV = Σ xV : X (1.8333π a 2 b) = 1.85416π a 2b 2 X = 1.01136b Centroid is 0.01136b to right of base of cone. Note: Centroid is at base of cone for h = 6b = 2.449b. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 737 PROBLEM 5.147 Locate the center of gravity of the sheet-metal form shown. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area. 1 yI = − (1.2) = −0.4 m 3 1 zI = (3.6) = 1.2 m 3 4(1.8) 2.4 xIII = − =− m π 3π A, m 2 x, m y, m z, m xA, m3 I 1 (3.6)(1.2) = 2.16 2 1.5 −0.4 1.2 II (3.6)(1.7) = 6.12 0.75 0.4 0.8 III Σ We have π 2 (1.8)2 = 5.0894 − 2.4 π 13.3694 yA, m3 zA, m3 3.24 −0.864 2.592 1.8 4.59 2.448 11.016 1.8 −3.888 4.0715 3.942 5.6555 9.1609 22.769 X ΣV = Σ xV : X (13.3694 m 2 ) = 3.942 m3 or X = 0.295 m Y ΣV = Σ yV : Y (13.3694 m 2 ) = 5.6555 m3 or Y = 0.423 m Z ΣV = Σ zV : Z (13.3694 m 2 ) = 22.769 m3 or Z = 1.703 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 738 PROBLEM 5.148 Locate the centroid of the volume obtained by rotating the shaded area about the x-axis. SOLUTION First note that symmetry implies y =0 and z =0 We have y = k ( X − h) 2 At x = 0, y = a, a = k ( − h) 2 or k= a h2 Choose as the element of volume a disk of radius r and thickness dx. Then dV = π r 2 dx, X EL = x r= Now a ( x − h) 2 2 h so that dV = π Then V= a2 ( x − h) 4 dx h4 h 0 π a2 π a2 4 x h dx ( − ) = [( x − h)5 ]0h 5 h4 h4 1 = π a2 h 5 and a2 x π 4 ( x − h) 4 dx 0 h 2 a h 5 ( x − 4hx 4 + 6h 2 x3 − 4h3 x 2 + h4 x)dx =π 4 h 0 xEL dV = h h =π = Now a2 1 6 4 5 3 2 4 4 3 3 1 4 2 x − hx + h x − h x + h x 5 2 3 2 h4 6 0 1 π a 2 h2 30 π π 2 2 xV = xEL dV : x a 2 h = a h 5 30 or x= 1 h 6 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 739 CHAPTER 6 PROBLEM 6.1 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Entire truss: ΣFy = 0: By = 0 By = 0 ΣM C = 0: − Bx (3.2 m) − (48 kN)(7.2 m) = 0 Bx = −108 kN B x = 108 kN ΣFx = 0: C − 108 kN + 48 kN = 0 C = 60 kN C = 60 kN Free body: Joint B: FAB FBC 108 kN = = 5 4 3 FAB = 180.0 kN T FBC = 144.0 kN T Free body: Joint C: FAC FBC 60 kN = = 13 12 5 FAC = 156.0 kN C FBC = 144 kN (checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 743 PROBLEM 6.2 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: ΣM A = 0: C = 1260 lb ΣFx = 0: A x = 0 ΣFy = 0: A y = 960 lb Joint B: FAB FBC 300 lb = = 12 13 5 FAB = 720 lb T FBC = 780 lb C Joint A: ΣFy = 0: − 960 lb − 4 FAC = 0 5 FAC = 1200 lb FAC = 1200 lb C 3 ΣFx = 0: 720 lb − (1200 lb) = 0 (checks) 5 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 744 PROBLEM 6.3 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION AB = 32 + 1.252 = 3.25 m BC = 32 + 42 = 5 m Reactions: ΣM A = 0: (84 kN)(3 m) − C (5.25 m) = 0 C = 48 kN ΣFx = 0: Ax − C = 0 A x = 48 kN ΣFy = 0: Ay = 84 kN = 0 A y = 84 kN Joint A: ΣFx = 0: 48 kN − 12 FAB = 0 13 FAB = +52 kN ΣFy = 0: 84 kN − FAB = 52.0 kN T 5 (52 kN) − FAC = 0 13 FAC = +64.0 kN FAC = 64.0 kN T Joint C: FBC 48 kN = 5 3 FBC = 80.0 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 745 PROBLEM 6.4 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: C = D = 600 lb From the symmetry of the truss and loading, we find Free body: Joint B: FAB 5 = FBC 300 lb = 2 1 FAB = 671 lb T FBC = 600 lb C Free body: Joint C: ΣFy = 0: 3 FAC + 600 lb = 0 5 FAC = −1000 lb ΣFx = 0: 4 ( −1000 lb) + 600 lb + FCD = 0 5 FAC = 1000 lb C FCD = 200 lb T From symmetry: FAD = FAC = 1000 lb C , FAE = FAB = 671 lb T , FDE = FBC = 600 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 746 PROBLEM 6.5 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: ΣM D = 0: Fy (24) − (4 + 2.4)(12) − (1)(24) = 0 Fy = 4.2 kips ΣFx = 0: Fx = 0 ΣFy = 0: D − (1 + 4 + 1 + 2.4) + 4.2 = 0 D = 4.2 kips Joint A: ΣFx = 0: FAB = 0 FAB = 0 ΣFy = 0 : −1 − FAD = 0 FAD = −1 kip Joint D: ΣFy = 0: − 1 + 4.2 + ΣFx = 0: 8 FBD = 0 17 FBD = −6.8 kips 15 (−6.8) + FDE = 0 17 FDE = +6 kips FAD = 1.000 kip C FBD = 6.80 kips C FDE = 6.00 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 747 PROBLEM 6.5 (Continued) Joint E: ΣFy = 0 : FBE − 2.4 = 0 FBE = +2.4 kips FBE = 2.40 kips T Truss and loading symmetrical about cL. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 748 PROBLEM 6.6 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION AD = 52 + 122 = 13 ft BCD = 122 + 162 = 20 ft Reactions: ΣFx = 0: Dx = 0 ΣM E = 0: D y (21 ft) − (693 lb)(5 ft) = 0 ΣFy = 0: 165 lb − 693 lb + E = 0 Joint D: D y = 165 lb E = 528 lb ΣFx = 0: 5 4 FAD + FDC = 0 13 5 (1) ΣFy = 0: 12 3 FAD + FDC + 165 lb = 0 13 5 (2) Solving Eqs. (1) and (2) simultaneously, FAD = −260 lb FAD = 260 lb C FDC = +125 lb FDC = 125 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 749 PROBLEM 6.6 (Continued) Joint E: ΣFx = 0: 5 4 FBE + FCE = 0 13 5 (3) ΣFy = 0: 12 3 FBE + FCE + 528 lb = 0 13 5 (4) Solving Eqs. (3) and (4) simultaneously, FBE = −832 lb FBE = 832 lb C FCE = +400 lb FCE = 400 lb T Joint C: Force polygon is a parallelogram (see Fig. 6.11, p. 209). FAC = 400 lb T FBC = 125.0 lb T Joint A: ΣFx = 0: 5 4 (260 lb) + (400 lb) + FAB = 0 13 5 FAB = −420 lb ΣFy = 0: FAB = 420 lb C 12 3 (260 lb) − (400 lb) = 0 13 5 0 = 0 (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 750 PROBLEM 6.7 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Entire truss: ΣFx = 0: Cx + 2(5 kN) = 0 Cx = −10 kN C x = 10 kN ΣM C = 0: D(2 m) − (5 kN)(8 m) − (5 kN)(4 m) = 0 D = +30 kN D = 30 kN ΣFy = 0: C y + 30 kN = 0 C y = −30 kN C y = 30 kN Free body: Joint A: FAB FAD 5 kN = = 4 1 17 FAB = 20.0 kN T FAD = 20.6 kN C Free body: Joint B: ΣFx = 0: 5 kN + 1 5 FBD = 0 FBD = −5 5 kN ΣFy = 0: 20 kN − FBC − 2 5 FBD = 11.18 kN C (−5 5 kN) = 0 FBC = +30 kN FBC = 30.0 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 751 PROBLEM 6.7 (Continued) Free body: Joint C: ΣFx = 0: FCD − 10 kN = 0 FCD = +10 kN FCD = 10.00 kN T ΣFy = 0: 30 kN − 30 kN = 0 (checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 752 PROBLEM 6.8 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: ΣM C = 0: A x = 16 kN ΣFy = 0: A y = 9 kN ΣFx = 0: C = 16 kN Joint E: FBE FDE 3 kN = = 5 4 3 FBE = 5.00 kN T FDE = 4.00 kN C Joint B: ΣFx = 0: 4 (5 kN) − FAB = 0 5 FAB = +4 kN FAB = 4.00 kN T 3 ΣFy = 0: − 6 kN − (5 kN) − FBD = 0 5 FBD = −9 kN FBD = 9.00 kN C Joint D: ΣFy = 0: − 9 kN + 3 FAD = 0 5 FAD = +15 kN FAD = 15.00 kN T 4 ΣFx = 0: − 4 kN − (15 kN) − FCD = 0 5 FCD = −16 kN FCD = 16.00 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 753 PROBLEM 6.9 Determine the force in each member of the Gambrel roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: H x = 0 Because of the symmetry of the truss and loading, A = Hy = 1 total load 2 A = H y = 1200 lb Free body: Joint A: FAB FAC 900 lb = = 5 4 3 FAB = 1500 lb C FAC = 1200 lb T Free body: Joint C: BC is a zero-force member. FBC = 0 FCE = 1200 lb T Free body: Joint B: ΣFx = 0: 24 4 4 FBD + FBE + (1500 lb) = 0 25 5 5 24 FBD + 20 FBE = −30, 000 lb or ΣFy = 0: (1) 7 3 3 FBD − FBE + (1500) − 600 = 0 25 5 5 7 FBD − 15FBE = −7,500 lb or (2) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 754 PROBLEM 6.9 (Continued) Multiply Eq. (1) by 3, Eq. (2) by 4, and add: 100 FBD = −120, 000 lb FBD = 1200 lb C Multiply Eq. (1) by 7, Eq. (2) by –24, and add: 500 FBE = −30,000 lb FBE = 60.0 lb C Free body: Joint D: 24 24 FDF = 0 (1200 lb) + 25 25 ΣFx = 0: FDF = −1200 lb FDF = 1200 lb C 7 7 (1200 lb) − (−1200 lb) − 600 lb − FDE = 0 25 25 ΣFy = 0: FDE = 72.0 lb FDE = 72.0 lb T Because of the symmetry of the truss and loading, we deduce that FEF = FBE FEF = 60.0 lb C FEG = FCE FEG = 1200 lb T FFG = FBC FFG = 0 FFH = FAB FFH = 1500 lb C FGH = FAC FGH = 1200 lb T Note: Compare results with those of Problem 6.11. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 755 PROBLEM 6.10 Determine the force in each member of the Howe roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: H x = 0 Because of the symmetry of the truss and loading, A = Hy = 1 total load 2 A = H y = 1200 lb Free body: Joint A: FAB FAC 900 lb = = 5 4 3 FAB = 1500 lb C FAC = 1200 lb T Free body: Joint C: BC is a zero-force member. FBC = 0 FCE = 1200 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 756 PROBLEM 6.10 (Continued) Free body: Joint B: ΣFx = 0: or FBD + FBE = −1500 lb ΣFy = 0: or 4 4 4 FBD + FBC + (1500 lb) = 0 5 5 5 (1) 3 3 3 FBD − FBE + (1500 lb) − 600 lb = 0 5 5 5 FBD − FBE = −500 lb (2) Add Eqs. (1) and (2): 2 FBD = −2000 lb FBD = 1000 lb C Subtract Eq. (2) from Eq. (1): 2 FBE = −1000 lb FBE = 500 lb C Free Body: Joint D: 4 4 (1000 lb) + FDF = 0 5 5 ΣFx = 0: FDF = −1000 lb FDF = 1000 lb C 3 3 (1000 lb) − ( −1000 lb) − 600 lb − FDE = 0 5 5 ΣFy = 0: FDE = +600 lb FDE = 600 lb T Because of the symmetry of the truss and loading, we deduce that FEF = FBE FEF = 500 lb C FEG = FCE FEG = 1200 lb T FFG = FBC FFG = 0 FFH = FAB FFH = 1500 lb C FGH = FAC FGH = 1200 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 757 PROBLEM 6.11 Determine the force in each member of the Pratt roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: Ax = 0 Due to symmetry of truss and load, Ay = H = 1 total load = 21 kN 2 Free body: Joint A: FAB FAC 15.3 kN = = 37 35 12 FAB = 47.175 kN FAC = 44.625 kN FAB = 47.2 kN C FAC = 44.6 kN T Free body: Joint B: From force polygon: FBD = 47.175 kN, FBC = 10.5 kN FBC = 10.50 kN C FBD = 47.2 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 758 PROBLEM 6.11 (Continued) Free body: Joint C: ΣFy = 0: 3 FCD − 10.5 = 0 5 ΣFx = 0: FCE + FCD = 17.50 kN T 4 (17.50) − 44.625 = 0 5 FCE = 30.625 kN Free body: Joint E: DE is a zero-force member. FCE = 30.6 kN T FDE = 0 Truss and loading symmetrical about cL . PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 759 PROBLEM 6.12 Determine the force in each member of the Fink roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: A x = 0 Because of the symmetry of the truss and loading, Ay = G = 1 total load 2 A y = G = 6.00 kN Free body: Joint A: F FAB 4.50 kN = AC = 2.462 2.25 1 FAB = 11.08 kN C FAC = 10.125 kN FAC = 10.13 kN T Free body: Joint B: ΣFx = 0: 3 2.25 2.25 FBC + FBD + (11.08 kN) = 0 5 2.462 2.462 (1) F 4 11.08 kN FBC + BD + − 3 kN = 0 5 2.462 2.462 (2) ΣFy = 0: − Multiply Eq. (2) by –2.25 and add to Eq. (1): 12 FBC + 6.75 kN = 0 5 FBC = −2.8125 FBC = 2.81 kN C Multiply Eq. (1) by 4, Eq. (2) by 3, and add: 12 12 FBD + (11.08 kN) − 9 kN = 0 2.462 2.462 FBD = −9.2335 kN FBD = 9.23 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 760 PROBLEM 6.12 (Continued) Free body: Joint C: ΣFy = 0: 4 4 FCD − (2.8125 kN) = 0 5 5 FCD = 2.8125 kN, FCD = 2.81 kN T 3 3 ΣFx = 0: FCE − 10.125 kN + (2.8125 kN) + (2.8125 kN) = 0 5 5 FCE = +6.7500 kN FCE = 6.75 kN T Because of the symmetry of the truss and loading, we deduce that FDE = FCD FCD = 2.81 kN T FDF = FBD FDF = 9.23 kN C FEF = FBC FEF = 2.81 kN C FEG = FAC FEG = 10.13 kN T FFG = FAB FFG = 11.08 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 761 PROBLEM 6.13 Using the method of joints, determine the force in each member of the double-pitch roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣM A = 0: H (18 m) − (2 kN)(4 m) − (2 kN)(8 m) − (1.75 kN)(12 m) − (1.5 kN)(15 m) − (0.75 kN)(18 m) = 0 H = 4.50 kN ΣFx = 0: Ax = 0 ΣFy = 0: Ay + H − 9 = 0 Ay = 9 − 4.50 A y = 4.50 kN Free body: Joint A: FAB FAC 3.50 kN = 2 1 5 FAB = 7.8262 kN C = FAB = 7.83 kN C FAC = 7.00 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 762 PROBLEM 6.13 (Continued) Free body: Joint B: ΣFx = 0: 2 5 FBD + 2 5 (7.8262 kN) + 1 2 FBC = 0 FBD + 0.79057 FBC = −7.8262 kN or ΣFy = 0: 1 5 FBD + 1 5 (7.8262 kN) − 1 2 (1) FBC − 2 kN = 0 FBD − 1.58114 FBC = −3.3541 or (2) Multiply Eq. (1) by 2 and add Eq. (2): 3FBD = −19.0065 FBD = −6.3355 kN FBD = 6.34 kN C Subtract Eq. (2) from Eq. (1): 2.37111FBC = −4.4721 FBC = 1.886 kN C FBC = −1.8861 kN Free body: Joint C: 2 ΣFy = 0: 5 FCD − 1 2 (1.8861 kN) = 0 FCD = +1.4911 kN ΣFx = 0: FCE − 7.00 kN + FCE 1 FCD = 1.491 kN T (1.8861 kN) + 2 = +5.000 kN 1 5 (1.4911 kN) = 0 FCE = 5.00 kN T Free body: Joint D: ΣFx = 0: 2 5 FDF + 1 2 FDE + 2 5 (6.3355 kN) − 1 5 (1.4911 kN) = 0 FDF + 0.79057 FDE = −5.5900 kN or ΣFy = 0: or 1 5 FDF − 1 2 FDE + 1 5 (6.3355 kN) − FDF − 0.79057 FDE = −1.1188 kN Add Eqs. (1) and (2): 2 5 (1.4911 kN) − 2 kN = 0 (2) 2 FDF = −6.7088 kN FDF = −3.3544 kN Subtract Eq. (2) from Eq. (1): (1) FDF = 3.35 kN C 1.58114 FDE = −4.4712 kN FDE = −2.8278 kN FDE = 2.83 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 763 PROBLEM 6.13 (Continued) Free body: Joint F: ΣFx = 0: 1 FFG + 2 2 (3.3544 kN) = 0 5 FFG = −4.243 kN ΣFy = 0: −FEF − 1.75 kN + 1 5 FFG = 4.24 kN C (3.3544 kN) − 1 2 FEF = 2.750 kN (−4.243 kN) = 0 FEF = 2.75 kN T Free body: Joint G: ΣFx = 0: 1 FGH − 2 ΣFy = 0: − 1 2 1 2 (4.243 kN) = 0 FGH − 1 2 FEG − 1 2 (1) (4.243 kN) − 1.5 kN = 0 FGH + FEG = −6.364 kN or (2) 2 FGH = −10.607 FGH = −5.303 Subtract Eq. (1) from Eq. (2): 2 FEG + FGH − FEG = −4.243 kN or Add Eqs. (1) and (2): 1 FGH = 5.30 kN C 2 FEG = −2.121 kN FEG = −1.0605 kN FEG = 1.061 kN C Free body: Joint H: FEH 3.75 kN = 1 1 We can also write FGH 2 = FEH = 3.75 kN T 3.75 kN 1 FGH = 5.30 kN C (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 764 PROBLEM 6.14 The truss shown is one of several supporting an advertising panel. Determine the force in each member of the truss for a wind load equivalent to the two forces shown. State whether each member is in tension or compression. SOLUTION Free body: Entire truss: ΣM F = 0: (800 N)(7.5 m) + (800 N)(3.75 m) − A(2 m) = 0 A = +2250 N A = 2250 N ΣFy = 0: 2250 N + Fy = 0 Fy = −2250 N Fy = 2250 N ΣFx = 0: −800 N − 800 N + Fx = 0 Fx = +1600 N Fx = 1600 N Joint D: 800 N FDE FBD = = 8 15 17 FBD = 1700 N C FDE = 1500 N T Joint A: 2250 N FAB FAC = = 15 17 8 FAB = 2250 N C FAC = 1200 N T Joint F: ΣFx = 0: 1600 N − FCF = 0 FCF = +1600 N FCF = 1600 N T ΣFy = 0: FEF − 2250 N = 0 FEF = +2250 N FEF = 2250 N T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 765 PROBLEM 6.14 (Continued) Joint C: ΣFx = 0: 8 FCE − 1200 N + 1600 N = 0 17 FCE = −850 N ΣFy = 0: FBC + 15 FCE = 0 17 FBC = − 15 15 FCE = − ( −850 N) 17 17 FBC = +750 N Joint E: FCE = 850 N C ΣFx = 0: − FBE − 800 N + FBC = 750 N T 8 (850 N) = 0 17 FBE = −400 N ΣFy = 0: 1500 N − 2250 N + FBE = 400 N C 15 (850 N) = 0 17 0 = 0 (checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 766 PROBLEM 6.15 Determine the force in each of the members located to the left of line FGH for the studio roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: A x = 0 Because of symmetry of loading, Ay = L = 1 total load 2 A y = L = 1200 lb Zero-Force Members: Examining joints C and H, we conclude that BC, EH, and GH are zero-force members. Thus, FBC = FEH = 0 Also, FCE = FAC (1) Free body: Joint A: FAB FAC 1000 lb = 2 1 5 FAB = 2236 lb C = FAB = 2240 lb C FAC = 2000 lb T FCE = 2000 lb T From Eq. (1): Free body: Joint B: ΣFx = 0: 2 5 FBD + 2 5 FBE + 2 5 (2236 lb) = 0 FBD + FBE = −2236 lb or ΣFy = 0: 1 5 FBD − 1 5 FBE + 1 FBD − FBE = −1342 lb or (2) 5 (2236 lb) − 400 lb = 0 (3) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 767 PROBLEM 6.15 (Continued) 2 FBD = −3578 lb Add Eqs. (2) and (3): FBD = 1789 lb C Subtract Eq. (3) from Eq. (1): 2 FBE = − 894 lb FBE = 447 lb C Free body: Joint E: ΣFx = 0: 2 5 FEG + 2 5 (447 lb) − 2000 lb = 0 FEG = 1789 lb T ΣFy = 0: FDE + 1 5 (1789 lb) − 1 5 FDE = − 600 lb (447 lb) = 0 FDE = 600 lb C Free body: Joint D: ΣFx = 0: 2 5 FDF + 2 5 FDG + 2 5 (1789 lb) = 0 FDF + FDG = −1789 lb or ΣFy = 0: 1 FDF − 1 FDG + (4) 1 5 5 5 + 600 lb − 400 lb = 0 FDF − FDG = −2236 lb or Add Eqs. (4) and (5): (1789 lb) (5) 2 FDF = − 4025 lb FDF = 2010 lb C Subtract Eq. (5) from Eq. (4): 2 FDG = 447 lb FDG = 224 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 768 PROBLEM 6.16 Determine the force in member FG and in each of the members located to the right of FG for the studio roof truss shown. State whether each member is in tension or compression. SOLUTION Reaction at L: Because of the symmetry of the loading, L= 1 total load, 2 L = 1200 lb (See F.B. diagram to the left for more details.) Free body: Joint L: 9 = 26.57° 18 3 β = tan −1 = 9.46° 18 FJL FKL 1000 lb = = sin 63.43° sin 99.46° sin17.11° α = tan −1 FKL = 3352.7 lb C FJL = 3040 lb T FKL = 3350 lb C Free body: Joint K: ΣFx = 0: − or 2 5 FIK − 2 5 2 FJK − 5 (3352.7 lb) = 0 FIK + FJK = − 3352.7 lb ΣFy = 0: or 1 5 FIK − 1 5 FJK + (1) 1 5 (3352.7) − 400 = 0 FIK − FJK = − 2458.3 lb (2) 2 FIK = − 5811.0 Add Eqs. (1) and (2): FIK = − 2905.5 lb FIK = 2910 lb C Subtract Eq. (2) from Eq. (1): 2 FJK = − 894.4 FJK = − 447.2 lb FJK = 447 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 769 PROBLEM 6.16 (Continued) Free body: Joint J: 2 ΣFx = 0: − 13 3 ΣFy = 0: FIJ + 13 6 FIJ − 37 1 37 6 FGJ + FGJ − 37 1 37 2 (3040 lb) − (3040 lb) − 5 1 5 (447.2) = 0 (447.2) = 0 (3) (4) Multiply Eq. (4) by 6 and add to Eq. (3): 16 13 FIJ − 8 (447.2) = 0 5 FIJ = 360.54 lb FIJ = 361 lb T Multiply Eq. (3) by 3, Eq. (4) by 2, and add: − 16 37 ( FGJ − 3040) − 8 (447.2) = 0 5 FGJ = 2431.7 lb FGJ = 2430 lb T Free body: Joint I: 2 ΣFx = 0: − 5 2 FFI − 5 2 FGI − 5 2 (2905.5) + 13 (360.54) = 0 FFI + FGI = − 2681.9 lb or ΣFy = 0: 1 5 FFI − 1 5 FGI + 1 5 (5) (2905.5) − 3 13 (360.54) − 400 = 0 FFI − FGI = −1340.3 lb or (6) 2 FFI = −4022.2 Add Eqs. (5) and (6): Subtract Eq. (6) from Eq. (5): FFI = −2011.1 lb FFI = 2010 lb C 2 FGI = −1341.6 lb FGI = 671 lb C Free body: Joint F: From ΣFx = 0: FDF = FFI = 2011.1 lb C 1 2011.1 lb = 0 ΣFy = 0: FFG − 400 lb + 2 5 FFG = + 1400 lb FFG = 1400 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 770 PROBLEM 6.17 Determine the force in each of the members located to the left of FG for the scissors roof truss shown. State whether each member is in tension or compression. SOLUTION Free Body: Truss: ΣFx = 0: A x = 0 ΣM L = 0: (1 kN)(12 m) + (2 kN)(10 m) + (2 kN)(8 m) + (1 kN)(6 m) − Ay (12 m) = 0 A y = 4.50 kN FBC = 0 We note that BC is a zero-force member: Also, FCE = FAC Free body: Joint A: ΣFx = 0: 1 ΣFy = 0: 1 (1) 2 2 FAB + 2 FAB + 1 5 5 FAC = 0 (2) FAC + 3.50 kN = 0 (3) Multiply Eq. (3) by –2 and add Eq. (2): − 1 2 FAB − 7 kN = 0 FAB = 9.90 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 771 PROBLEM 6.17 (Continued) Subtract Eq. (3) from Eq. (2): 1 5 FAC − 3.50 kN = 0 FAC = 7.826 kN FCE = FAC = 7.826 kN From Eq. (1): FAC = 7.83 kN T FCE = 7.83 kN T Free body: Joint B: ΣFy = 0: 1 2 FBD + 1 2 (9.90 kN) − 2 kN = 0 FBD = −7.071 kN 1 ΣFx = 0: FBE + 2 (9.90 − 7.071) kN = 0 FBE = −2.000 kN Free body: Joint E: ΣFx = 0: 2 5 FBD = 7.07 kN C FBE = 2.00 kN C ( FEG − 7.826 kN) + 2.00 kN = 0 FEG = 5.590 kN ΣFy = 0: FDE − 1 5 FEG = 5.59 kN T (7.826 − 5.590) kN = 0 FDE = 1.000 kN FDE = 1.000 kN T Free body: Joint D: ΣFx = 0: 2 5 ( FDF + FDG ) + 1 2 (7.071 kN) FDF + FDG = −5.590 kN or ΣFy = 0: or 1 5 ( FDF − FDG ) + 1 2 (4) (7.071 kN) = 2 kN − 1 kN = 0 FDE − FDG = −4.472 (5) Add Eqs. (4) and (5): 2 FDF = −10.062 kN FDF = 5.03 kN C Subtract Eq. (5) from Eq. (4): 2 FDG = −1.1180 kN FDG = 0.559 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 772 PROBLEM 6.18 Determine the force in member FG and in each of the members located to the right of FG for the scissors roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣM A = 0: L(12 m) − (2 kN)(2 m) − (2 kN)(4 m) − (1 kN)(6 m) = 0 L = 1.500 kN Angles: tan α = 1 tan β = 1 2 α = 45° β = 26.57° Zero-force members: Examining successively joints K, J, and I, we note that the following members to the right of FG are zeroforce members: JK, IJ, and HI. FHI = FIJ = FJK = 0 Thus, We also note that and FGI = FIK = FKL (1) FHJ = FJL (2) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 773 PROBLEM 6.18 (Continued) Free body: Joint L: FJL F 1.500 kN = KL = sin116.57° sin 45° sin18.43° FJL = 4.2436 kN FJL = 4.24 kN C FKL = 3.35 kN T From Eq. (1): FGI = FIK = FKL FGI = FIK = 3.35 kN T From Eq. (2): FHJ = FJL = 4.2436 kN FHJ = 4.24 kN C FGH FFH 4.2436 = = sin108.43° sin18.43° sin 53.14° FFH = 5.03 kN C Free body: Joint H: FGH = 1.677 kN T Free body: Joint F: ΣFx = 0: − FDF cos 26.57° − (5.03 kN) cos 26.57° = 0 FDF = −5.03 kN ΣFy = 0: − FFG − 1 kN + (5.03 kN) sin 26.57° − ( −5.03 kN)sin 26.57° = 0 FFG = 3.500 kN FFG = 3.50 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 774 PROBLEM 6.19 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression. SOLUTION ΣFx = 0: Ax = 0 Free body: Truss: Due to symmetry of truss and loading, Ay = G = Free body: Joint A: 1 total load = 6 kips 2 FAB FAC 6 kips = = 5 3 4 FAB = 7.50 kips C FAC = 4.50 kips T Free body: Joint B: FBC FBD 7.5 kips = = 5 6 5 FBC = 7.50 kips T FBD = 9.00 kips C Free body: Joint C: ΣFy = 0: 4 4 (7.5) + FCD − 6 = 0 5 5 FCD = 0 3 ΣFx = 0: FCE − 4.5 − (7.5) = 0 5 FCE = +9 kips FCE = 9.00 kips T Truss and loading is symmetrical about cL. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 775 PROBLEM 6.20 Solve Problem 6.19 assuming that the load applied at E has been removed. PROBLEM 6.19 Determine the force in each member of the Warren bridge truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: Ax = 0 ΣM G = 0: 6(36) − Ay (54) = 0 A y = 4 kips ΣFy = 0: 4 − 6 + G = 0 G = 2 kips Free body: Joint A: FAB FAC 4 kips = = 5 3 4 FAB = 5.00 kips C FAC = 3.00 kips T Free body Joint B: FBC FBD 5 kips = = 5 6 5 FBC = 5.00 kips T FBD = 6.00 kips C Free body Joint C: ΣM y = 0: 4 4 (5) + FCD − 6 = 0 5 5 3 3 ΣFx = 0: FCE + (2.5) − (5) − 3 = 0 5 5 FCD = 2.50 kips T FCE = 4.50 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 776 PROBLEM 6.20 (Continued) Free body: Joint D: 4 4 ΣFy = 0: − (2.5) − FDE = 0 5 5 FDE = −2.5 kips FDE = 2.50 kips C 3 3 ΣFx = 0: FDF + 6 − (2.5) − (2.5) = 0 5 5 Free body: Joint F: FDF = −3 kips FDF = 3.00 kips C FEF FFG 3 kips = = 5 5 6 FEF = 2.50 kips T FFG = 2.50 kips C Free body: Joint G: FEG 2 kips = 3 4 Also, FFG 2 kips = 5 4 FEG = 1.500 kips T FFG = 2.50 kips C (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 777 PROBLEM 6.21 Determine the force in each member of the Pratt bridge truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFz = 0: A x = 0 ΣM A = 0 : H (36 ft) − (4 kips)(9 ft) − (4 kips)(18 ft) − (4 kips)(27 ft) = 0 H = 6 kips ΣFy = 0: Ay + 6 kips − 12 kips = 0 A y = 6 kips Free body: Joint A: FAB FAC 6 kips = = 5 3 4 FAB = 7.50 kips C FAC = 4.50 kips T Free body: Joint C: ΣFx = 0: FCE = 4.50 kips T ΣFy = 0: FBC = 4.00 kips T Free body: Joint B: ΣFy = 0: − 4 4 FBE + (7.50 kips) − 4.00 kips = 0 5 5 FBE = 2.50 kips T ΣFx = 0: 8 3 (7.50 kips) + (2.50 kips) + FBD = 0 5 5 FBD = −6.00 kips FBD = 6.00 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 778 PROBLEM 6.21 (Continued) Free body: Joint D: We note that DE is a zero-force member: FDE = 0 FDF = 6.00 kips C Also, From symmetry: FFE = FBE FEF = 2.50 kips T FEG = FCE FEG = 4.50 kips T FFG = FBC FFG = 4.00 kips T FFH = FAB FFH = 7.50 kips C FGH = FAC FGH = 4.50 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 779 PROBLEM 6.22 Solve Problem 6.21 assuming that the load applied at G has been removed. PROBLEM 6.21 Determine the force in each member of the Pratt bridge truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: A x = 0 ΣM A = 0: H (36 ft) − (4 kips)(9 ft) − (4 kips)(18 ft) = 0 H = 3.00 kips ΣFy = 0: A y + 5.00 kips We note that DE and FG are zero-force members. FDE = 0, Therefore, FFG = 0 Also, FBD = FDF (1) and FEG = FGH (2) Free body: Joint A: FAB FAC 5 kips = = 5 3 4 FAB = 6.25 kips C FAC = 3.75 kips T Free body: Joint C: ΣFx = 0: FCE = 3.75 kips T ΣFy = 0: FBC = 4.00 kips T Free body: Joint B: ΣFx = 0: 4 4 (6.25 kips) − 4.00 kips − FBE = 0 5 5 FBE = 1.250 kips T 3 3 ΣFx = 0: FBD + (6.25 kips) + (1.250 kips) = 0 5 5 FBD = −4.50 kips FBD = 4.50 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 780 PROBLEM 6.22 (Continued) Free body: Joint F: We recall that FFG = 0, and from Eq. (1) that FDF = FBD FDF = 4.50 kips C FEF FFH 4.50 kips = = 5 5 6 FEF = 3.75 kips T FFH = 3.75 kips C Free body: Joint H: FGH 3.00 kips = 3 4 FGH = 2.25 kips T Also, FFH 3.00 kips = 5 4 FFH = 3.75 kips C (checks) From Eq. (2): FEG = FGH FEG = 2.25 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 781 PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression. SOLUTION Free body: Joint A: F FAB 1.2 kN = AC = 2.29 2.29 1.2 FAB = 2.29 kN T FAC = 2.29 kN C Free body: Joint F: FDF F 1.2 kN = EF = 2.29 2.29 2.1 FDF = 2.29 kN T FEF = 2.29 kN C Free body: Joint D: FBD FDE 2.29 kN = = 2.21 0.6 2.29 FBD = 2.21 kN T FDE = 0.600 kN C Free body: Joint B: ΣFx = 0: 4 2.21 (2.29 kN) = 0 FBE + 2.21 kN − 5 2.29 FBE = 0 3 0.6 (2.29 kN) = 0 ΣFy = 0: − FBC − (0) − 5 2.29 FBC = − 0.600 kN FBC = 0.600 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 782 PROBLEM 6.23 (Continued) Free body: Joint C: ΣFx = 0: FCE + 2.21 (2.29 kN) = 0 2.29 FCE = − 2.21 kN ΣFy = 0: − FCH − 0.600 kN − FCH = −1.200 kN FCE = 2.21 kN C 0.6 (2.29 kN) = 0 2.29 FCH = 1.200 kN C Free body: Joint E: ΣFx = 0: 2.21 kN − 2.21 4 (2.29 kN) − FEH = 0 2.29 5 FEH = 0 ΣFy = 0: − FEJ − 0.600 kN − FEJ = −1.200 kN 0.6 (2.29 kN) − 0 = 0 2.29 FEJ = 1.200 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 783 PROBLEM 6.24 For the tower and loading of Problem 6.23 and knowing that FCH = FEJ = 1.2 kN C and FEH = 0, determine the force in member HJ and in each of the members located between HJ and NO. State whether each member is in tension or compression. PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression. SOLUTION Free body: Joint G: FGH F 1.2 kN = GI = 3.03 3.03 1.2 FGH = 3.03 kN T FGI = 3.03 kN C Free body: Joint L: FJL F 1.2 kN = KL = 3.03 3.03 1.2 FJL = 3.03 kN T FKL = 3.03 kN C Free body: Joint J: ΣFx = 0: − FHJ + 2.97 (3.03 kN) = 0 3.03 FHJ = 2.97 kN T 0.6 (3.03 kN) = 0 3.03 = −1.800 kN FJK = 1.800 kN C Fy = 0: − FJK − 1.2 kN − FJK Free body: Joint H: ΣFx = 0: 4 2.97 (3.03 kN) = 0 FHK + 2.97 kN − 5 3.03 FHK = 0 0.6 3 (3.03) kN − (0) = 0 3.03 5 = −1.800 kN FHI = 1.800 kN C ΣFy = 0: − FHI − 1.2 kN − FHI PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 784 PROBLEM 6.24 (Continued) Free body: Joint I: ΣFx = 0: FIK + 2.97 (3.03 kN) = 0 3.03 FIK = −2.97 kN ΣFy = 0: − FIN − 1.800 kN − FIN = −2.40 kN FIK = 2.97 kN C 0.6 (3.03 kN) = 0 3.03 FIN = 2.40 kN C Free body: Joint K: ΣFx = 0: − 4 2.97 (3.03 kN) = 0 FKN + 2.97 kN − 5 3.03 FKN = 0 ΣFy = 0: − FKO − 0.6 3 (3.03 kN) − 1.800 kN − (0) = 0 3.03 5 FKO = −2.40 kN FKO = 2.40 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 785 PROBLEM 6.25 Solve Problem 6.23 assuming that the cables hanging from the right side of the tower have fallen to the ground. PROBLEM 6.23 The portion of truss shown represents the upper part of a power transmission line tower. For the given loading, determine the force in each of the members located above HJ. State whether each member is in tension or compression. SOLUTION Zero-Force Members: Considering joint F, we note that DF and EF are zero-force members: FDF = FEF = 0 Considering next joint D, we note that BD and DE are zero-force members: FBD = FDE = 0 Free body: Joint A: F FAB 1.2 kN = AC = 2.29 2.29 1.2 FAB = 2.29 kN T FAC = 2.29 kN C Free body: Joint B: ΣFx = 0: 4 2.21 (2.29 kN) = 0 FBE − 5 2.29 FBE = 2.7625 kN ΣFy = 0: − FBC − FBE = 2.76 kN T 0.6 3 (2.29 kN) − (2.7625 kN) = 0 2.29 5 FBC = −2.2575 kN FBC = 2.26 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 786 PROBLEM 6.25 (Continued) Free body: Joint C: ΣFx = 0: FCE + 2.21 (2.29 kN) = 0 2.29 FCE = 2.21 kN C ΣFy = 0: − FCH − 2.2575 kN − FCH = −2.8575 kN 0.6 (2.29 kN) = 0 2.29 FCH = 2.86 kN C Free body: Joint E: ΣFx = 0: − 4 4 FEH − (2.7625 kN) + 2.21 kN = 0 5 5 FEH = 0 3 3 ΣFy = 0: − FEJ + (2.7625 kN) − (0) = 0 5 5 FEJ = +1.6575 kN FEJ = 1.658 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 787 PROBLEM 6.26 Determine the force in each of the members connecting joints A through F of the vaulted roof truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣFx = 0: A x = 0 ΣM K = 0: (1.2 kN)6a + (2.4 kN)5a + (2.4 kN)4a + (1.2 kN)3a − Ay (6a) = 0 A y = 5.40 kN Free body: Joint A: FAB FAC 4.20 kN = 2 1 5 FAB = 9.3915 kN = FAB = 9.39 kN C FAC = 8.40 kN T Free body: Joint B: ΣFx = 0: 2 ΣFy = 0: 1 5 5 FBD + 1 FBD − 1 2 2 FBC + 2 FBC + 1 5 5 (9.3915) = 0 (1) (9.3915) − 2.4 = 0 (2) Add Eqs. (1) and (2): 3 5 FBD + 3 5 (9.3915 kN) − 2.4 kN = 0 FBD = − 7.6026 kN FBD = 7.60 kN C Multiply Eq. (2) by –2 and add Eq. (1): 3 2 FB + 4.8 kN = 0 FBC = −2.2627 kN FBC = 2.26 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 788 PROBLEM 6.26 (Continued) Free body: Joint C: ΣFx = 0: 1 ΣFy = 0: 2 5 5 4 FCD + 17 1 FCD + 17 FCE + 1 FCE − 1 2 2 (2.2627) − 8.40 = 0 (3) (2.2627) = 0 (4) Multiply Eq. (4) by −4 and add Eq. (1): − 7 5 FCD + 5 (2.2627) − 8.40 = 0 2 FCD = − 0.1278 kN FCD = 0.128 kN C Multiply Eq. (1) by 2 and subtract Eq. (2): 7 17 FCE + 3 2 (2.2627) − 2(8.40) = 0 FCE = 7.068 kN FCE = 7.07 kN T Free body: Joint D: ΣFx = 0: 2 1 + ΣFy = 0: FDF + 5 5 1 5 + (0.1278) = 0 FDF − 2 5 1 2 FDE + (7.6026) 1.524 5 (5) 1.15 1 FDE + (7.6026) 1.524 5 (0.1278) − 2.4 = 0 (6) Multiply Eq. (5) by 1.15 and add Eq. (6): 3.30 5 FDF + 3.30 5 (7.6026) + 3.15 5 (0.1278) − 2.4 = 0 FDF = − 6.098 kN FDF = 6.10 kN C Multiply Eq. (6) by –2 and add Eq. (5): 3.30 3 FDE − (0.1278) + 4.8 = 0 1.524 5 FDE = − 2.138 kN FDE = 2.14 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 789 PROBLEM 6.26 (Continued) Free body: Joint E: ΣFx = 0: 0.6 4 1 FEF + ( FEH − FCE ) + (2.138) = 0 2.04 1.524 17 (7) ΣFy = 0: 1.95 1 1.15 FEF + ( FEH − FCE ) − (2.138) = 0 2.04 1.524 17 (8) Multiply Eq. (8) by 4 and subtract Eq. (7): 7.2 FEF − 7.856 kN = 0 2.04 FEF = 2.23 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 790 PROBLEM 6.27 Determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Free body: Truss: ΣM F = 0: G (20 ft) − (15 kips)(16 ft) − (40 kips)(15 ft) = 0 G = 42 kips ΣFx = 0: Fx + 15 kips = 0 Fx = 15 kips ΣFy = 0: Fy − 40 kips + 42 kips = 0 Fy = 2 kips Free body: Joint F: ΣFx = 0: 1 5 FDF − 15 kips = 0 FDF = 33.54 kips ΣFy = 0: FBF − 2 kips + 2 5 FDF = 33.5 kips T (33.54 kips) = 0 FBF = − 28.00 kips FBF = 28.0 kips C Free body: Joint B: ΣFx = 0: 5 ΣFy = 0: 2 29 29 FAB + 5 FAB − 6 61 61 FBD + 15 kips = 0 (1) FBD + 28 kips = 0 (2) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 791 PROBLEM 6.27 (Continued) Multiply Eq. (1) by 6, Eq. (2) by 5, and add: 40 29 FAB + 230 kips = 0 FAB = −30.96 kips FAB = 31.0 kips C Multiply Eq. (1) by 2, Eq. (2) by –5, and add: 40 61 FBD − 110 kips = 0 FBD = 21.48 kips FBD = 21.5 kips T Free body: Joint D: 2 ΣFy = 0: 5 2 FAD − 5 (33.54) + 6 61 (21.48) = 0 FAD = 15.09 kips T 1 ΣFx = 0: FDE + 5 (15.09 − 33.54) − 5 (21.48) = 0 61 FDE = 22.0 kips T Free body: Joint A: 5 ΣFx = 0: 29 ΣFy = 0: − 1 FAC + 2 29 5 FAC − FAE + 2 5 5 FAE + 29 (30.36) − 2 29 1 5 (15.09) = 0 2 (30.96) − 5 (15.09) = 0 (3) (4) Multiply Eq. (3) by 2 and add Eq. (4): 8 29 FAC + 12 29 (30.96) − 4 5 FAC = −28.27 kips, (15.09) = 0 FAC = 28.3 kips C Multiply Eq. (3) by 2, Eq. (4) by 5, and add: − 8 5 FAE + 20 29 (30.96) − 12 5 (15.09) = 0 FAE = 9.50 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 792 PROBLEM 6.27 (Continued) Free body: Joint C: From force triangle: FCE 61 = FCG 28.27 kips = 8 29 FCE = 41.0 kips T FCG = 42.0 kips C Free body: Joint G: ΣFx = 0: ΣFy = 0: 42 kips − 42 kips = 0 FEG = 0 (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 793 PROBLEM 6.28 Determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Reactions: ΣFx = 0: Ex = 0 ΣM F = 0: E y = 45 kips ΣFy = 0: F = 60 kips Joint D: FCD FDH 15 kips = = 12 13 5 FCD = 36.0 kips T FDH = 39.0 kips C Joint H: ΣF = 0: FCH = 0 ΣF = 0: FGH = 39.0 kips C ΣF = 0: FCG = 0 ΣF = 0: FBC = 36.0 kips T ΣF = 0: FBG = 0 ΣF = 0: FFG = 39.0 kips C ΣF = 0: FBF = 0 Joint C: Joint G: Joint B: ΣF = 0: FAB = 36.0 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 794 PROBLEM 6.28 (Continued) Joint A: AE = 122 + 152 = 19.21 ft tan 38.7° = 36 kips FAF FAF = 45.0 kips C sin 38.7° = 36 kips FAE FAE = 57.6 kips T Joint E: ΣFx = 0: + (57.6 kips) sin 38.7° + FEF = 0 FEF = −36.0 kips FEF = 36.0 kips C ΣFy = 0: (57.6 kips) cos 38.7° − 45 kips = 0 (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 795 PROBLEM 6.29 Determine whether the trusses of Problems 6.31a, 6.32a, and 6.33a are simple trusses. SOLUTION Truss of Problem 6.31a: Starting with triangle HDI and adding two members at a time, we obtain successively joints A, E, J, and B, but cannot go further. Thus, this truss is not a simple truss. Truss of Problem 6.32a: Starting with triangle ABC and adding two members at a time, we obtain joints D, E, G, F, and H, but cannot go further. Thus, this truss is not a simple truss. Truss of Problem 6.33a: Starting with triangle ABD and adding two members at a time, we obtain successively joints H, G, F, E, I, C, and J, thus completing the truss. Therefore, this is a simple truss. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 796 PROBLEM 6.30 Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b are simple trusses. SOLUTION Truss of Problem 6.31b: Starting with triangle CGM and adding two members at a time, we obtain successively joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus completing the truss. Therefore, this truss is a simple truss. Truss of Problem 6.32b: Starting with triangle ABC and adding two members at a time, we obtain successively joints E, D, F, G, and H, but cannot go further. Thus, this truss is not a simple truss. Truss of Problem 6.33b: Starting with triangle GFH and adding two members at a time, we obtain successively joints D, E, C, A, and B, thus completing the truss. Therefore, this is a simple truss. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 797 PROBLEM 6.31 For the given loading, determine the zero-force members in each of the two trusses shown. SOLUTION Truss (a): FB: Joint B: FBJ = 0 FB: Joint D: FDI = 0 FB: Joint E: FEI = 0 FB: Joint I : FAI = 0 (a) FB: Joint F : FFK = 0 FB: Joint G: FGK = 0 FB: Joint K : FCK = 0 AI , BJ , CK , DI , EI , FK , GK The zero-force members, therefore, are Truss (b): FB: Joint K : FFK = 0 FB: Joint O : FIO = 0 (b) The zero-force members, therefore, are FK and IO All other members are either in tension or compression. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 798 PROBLEM 6.32 For the given loading, determine the zero-force members in each of the two trusses shown. SOLUTION Truss (a): FB: Joint B: FBC = 0 FB: Joint C: FCD = 0 FB: Joint J : FIJ = 0 FB: Joint I : FIL = 0 FB: Joint N : FMN = 0 (a) FB: Joint M : FLM = 0 BC , CD, IJ , IL, LM , MN The zero-force members, therefore, are Truss (b): FB: Joint C: FBC = 0 FB: Joint B: FBE = 0 FB: Joint G: FFG = 0 FB: Joint F : FEF = 0 FB: Joint E: FDE = 0 FB: Joint I : FIJ = 0 (b) FB: Joint M : FMN = 0 FB: Joint N : FKN = 0 BC , BE , DE , EF , FG , IJ , KN , MN The zero-force members, therefore, are PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 799 PROBLEM 6.33 For the given loading, determine the zero-force members in each of the two trusses shown. SOLUTION Truss (a): Note: Reaction at F is vertical ( Fx = 0). Joint G: ΣF = 0, FDG = 0 Joint D: ΣF = 0, FDB = 0 Joint F : ΣF = 0, FFG = 0 Joint G: ΣF = 0, FGH = 0 Joint J : ΣF = 0, FIJ = 0 Joint I : ΣF = 0, FHI = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 800 PROBLEM 6.33 (Continued) Truss (b): Joint A: ΣF = 0, FAC = 0 Joint C: ΣF = 0, FCE = 0 Joint E: ΣF = 0, FEF = 0 Joint F : ΣF = 0, FFG = 0 Joint G: ΣF = 0, FGH = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 801 PROBLEM 6.34 Determine the zero-force members in the truss of (a) Problem 6.26, (b) Problem 6.28. SOLUTION (a) Truss of Problem 6.26: FB : Joint I : FIJ = 0 FB : Joint J : FGJ = 0 FB : Joint G: FGH = 0 GH , GJ , IJ The zero-force members, therefore, are (b) Truss of Problem 6.28: FB : Joint B: FBF = 0 FB : Joint B: FBG = 0 FB : Joint C: FCG = 0 FB : Joint C: FCH = 0 BF , BG, CG, CH The zero-force members, therefore, are PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 802 PROBLEM 6.35* The truss shown consists of six members and is supported by a short link at A, two short links at B, and a ball and socket at D. Determine the force in each of the members for the given loading. SOLUTION Free body: Truss: From symmetry: Dx = Bx and Dy = B y ΣM z = 0: − A(10 ft) − (400 lb)(24 ft) = 0 A = −960 lb ΣFx = 0: Bx + Dx + A = 0 2 Bx − 960 lb = 0, Bx = 480 lb ΣFy = 0: B y + Dy − 400 lb = 0 2 By = 400 lb By = +200 lb B = (480 lb)i + (200 lb) j Thus, Free body: C: FCA = FAC FCB = FBC FCD = FCD CA FAC = ( −24i + 10 j) CA 26 CB FBC = (−24i + 7k ) CB 25 CD FCD = (−24i − 7k ) CD 25 ΣF = 0: FCA + FCB + FCD − (400 lb) j = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 803 PROBLEM 6.35* (Continued) Substituting for FCA , FCB , FCD , and equating to zero the coefficients of i, j, k : 24 24 FAC − ( FBC + FCD ) = 0 26 25 i: − j: 10 FAC − 400 lb = 0 26 k: 7 ( FBC − FCD ) = 0 FCD = FBC 25 (1) FAC = 1040 lb T Substitute for FAC and FCD in Eq. (1): − 24 24 (10.40 lb) − (2 FBC ) = 0 FBC = −500 lb 26 25 FBC = FCD = 500 lb C Free body: B: FBC FBA FBD CB = (500 lb) = −(480 lb)i + (140 lb)k CB BA FAB = FAB = (10 j − 7k ) BA 12.21 = − FBD k ΣF = 0: FBA + FBD + FBC + (480 lb)i + (200 lb) j = 0 Substituting for FBA , FBD , FBC and equating to zero the coefficients of j and k: j: 10 FAB + 200 lb = 0 FAB = −244.2 lb 12.21 k: − 7 FAB − FBD + 140 lb = 0 12.21 FBD = − From symmetry: 7 (−244.2 lb) + 140 lb = +280 lb 12.21 FAD = FAB FAB = 244 lb C FBD = 280 lb T FAD = 244 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 804 PROBLEM 6.36* The truss shown consists of six members and is supported by a ball and socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P = (−2184 N)j and Q = 0. SOLUTION Free body: Truss: From symmetry: Dx = Bx and Dy = By ΣFx = 0: 2 Bx = 0 Bx = Dx = 0 ΣFz = 0: Bz = 0 ΣM c z = 0: − 2 By (2.8 m) + (2184 N)(2 m) = 0 By = 780 N B = (780 N) j Thus, Free body: A: AB FAB FAB = FAB = ( −0.8i − 4.8 j + 2.1k ) AB 5.30 AC FAC = FAC = FAC (2i − 4.8 j) AC 5.20 AD FAD = FAD = FAD (+0.8i − 4.8 j − 2.1k ) AD 5.30 ΣF = 0: FAB + FAC + FAD − (2184 N) j = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 805 PROBLEM 6.36* (Continued) Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k : i: − 0.8 2 ( FAB + FAD ) + FAC = 0 5.30 5.20 (1) j: − 4.8 4.8 ( FAB + FAD ) − FAC − 2184 N = 0 5.30 5.20 (2) 2.1 ( FAB − FAD ) = 0 5.30 k: FAD = FAB Multiply Eq. (1) by –6 and add Eq. (2): 16.8 − FAC − 2184 N = 0, FAC = −676 N 5.20 FAC = 676 N C Substitute for FAC and FAD in Eq. (1): 0.8 2 − 2 FAB + (−676 N) = 0, FAB = −861.25 N 5.30 5.20 Free body: B: FAB = FAD = 861 N C AB = −(130 N)i − (780 N) j + (341.25 N)k FAB = (861.25 N) AB 2.8i − 2.1k FBC = FBC = FBC (0.8i − 0.6k ) 3.5 FBD = − FBD k ΣF = 0: FAB + FBC + FBD + (780 N) j = 0 Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k , i: k: −130 N + 0.8FBC = 0 FBC = +162.5 N 341.25 N − 0.6 FBC − FBD = 0 FBD = 341.25 − 0.6(162.5) = +243.75 N From symmetry: FBC = 162.5 N T FCD = FBC FBD = 244 N T FCD = 162.5 N T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 806 PROBLEM 6.37* The truss shown consists of six members and is supported by a ball and socket at B, a short link at C, and two short links at D. Determine the force in each of the members for P = 0 and Q = (2968 N)i. SOLUTION Free body: Truss: From symmetry: Dx = Bx and Dy = By ΣFx = 0: 2 Bx + 2968 N = 0 Bx = Dx = −1484 N ΣM cz ′ = 0: − 2 By (2.8 m) − (2968 N)(4.8 m) = 0 By = −2544 N B = −(1484 N)i − (2544 N) j Thus, Free body: A: FAB = FAB AB AB FAB (−0.8i − 4.8 j + 2.1k ) 5.30 AC FAC = FAC = (2i − 4.8 j) AC 5.20 AD = FAD AD FAD = (−0.8i − 4.8 j − 2.1k ) 5.30 = FAC FAD ΣF = 0: FAB + FAC + FAD + (2968 N)i = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 807 PROBLEM 6.37* (Continued) Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k , 0.8 2 FAC + 2968 N = 0 ( FAB + FAD ) + 5.30 5.20 (1) 4.8 4.8 FAC = 0 ( FAB + FAD ) − 5.30 5.20 (2) i: − j: − k: 2.1 ( FAB − FAD ) = 0 5.30 FAD = FAB Multiply Eq. (1) by –6 and add Eq. (2): 16.8 − FAC − 6(2968 N) = 0, FAC = −5512 N 5.20 FAC = 5510 N C Substitute for FAC and FAD in Eq. (2): 4.8 4.8 − 2 FAB − (−5512 N) = 0, FAB = +2809 N 5.30 5.20 FAB = FAD = 2810 N T Free body: B: FAB FBC FBD BA = (2809 N) = (424 N)i + (2544 N) j − (1113 N)k BA 2.8 i − 2.1k = FBC = FBC (0.8i − 0.6k ) 3.5 = − FBD k ΣF = 0: FAB + FBC + FBD − (1484 N)i − (2544 N) j = 0 Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k , i: +24 N + 0.8FBC − 1484 N = 0, FBC = +1325 N k: −1113 N − 0.6 FBC − FBD = 0 From symmetry: FBC = 1325 N T FBD = −1113 N − 0.6(1325 N) = −1908 N, FBD = 1908 N C FCD = FBC FCD = 1325 N T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 808 PROBLEM 6.38* The truss shown consists of nine members and is supported by a ball and socket at A, two short links at B, and a short link at C. Determine the force in each of the members for the given loading. SOLUTION Free body: Truss: From symmetry: Az = Bz = 0 ΣFx = 0: Ax = 0 ΣM BC = 0: Ay (6 ft) + (1600 lb)(7.5 ft) = 0 Ay = −2000 lb A = −(2000 lb)j By = C From symmetry: ΣFy = 0: 2 By − 2000 lb − 1600 lb = 0 By = 1800 lb B = (1800 lb) j ΣF = 0: FAB + FAC + FAD − (2000 lb) j = 0 Free body: A: FAB i+k + FAC 2 i−k 2 + FAD (0.6 i + 0.8 j) − (2000 lb) j = 0 Factoring i, j, k and equating their coefficient to zero, 1 2 FAB + 1 FAC + 0.6 FAD = 0 (1) 0.8FAD − 2000 lb = 0 FAD = 2500 lb T 2 1 2 FAB − 1 2 FAC = 0 FAC = FAB PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 809 PROBLEM 6.38* (Continued) Substitute for FAD and FAC into Eq. (1): 2 2 Free body: B: FAB + 0.6(2500 lb) = 0, FAB = −1060.7 lb, FBA FBC FAB = FAC = 1061 lb C BA i+k = FAB = + (1060.7 lb) = (750 lb)(i + k ) BA 2 = − FBC k FBD = FBD (0.8 j − 0.6k ) BE FBE (7.5i + 8 j − 6k ) FBE = FBE = BE 12.5 ΣF = 0: FBA + FBC + FBD + FBE + (1800 lb) j = 0 Substituting for FBA , FBC , FBD , and FBE and equating to zero the coefficients of i, j, k , i: 7.5 750 lb + FBE = 0, FBE = −1250 lb 12.5 FBE = 1250 lb C j: 8 0.8 FBD + (−1250 lb) + 1800 lb = 0 12.5 FBD = 1250 lb C k: 750 lb − FBC − 0.6( −1250 lb) − 6 ( −1250 lb) = 0 12.5 FBC = 2100 lb T FBD = FCD = 1250 lb C From symmetry: Free body: D: ΣF = 0: FDA + FDB + FDC + FDE i = 0 We now substitute for FDA , FDB , FDC and equate to zero the coefficient of i. Only FDA contains i and its coefficient is −0.6 FAD = −0.6(2500 lb) = −1500 lb i: −1500 lb + FDE = 0 FDE = 1500 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 810 PROBLEM 6.39* The truss shown consists of nine members and is supported by a ball and socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0. SOLUTION Free body: Truss: ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dk ) + (0.6i − 0.75k ) × (−1200 j) = 0 −1.8 Ck + 1.8 D y k − 1.8 Dz j + 3D y i − 720k − 900i = 0 Equate to zero the coefficients of i, j, k : i: 3Dy − 900 = 0, D y = 300 N j: Dz = 0, D = (300 N) j k: 1.8C + 1.8(300) − 720 = 0 C = (100 N) j ΣF = 0: B + 300 j + 100 j − 1200 j = 0 B = (800 N) j Free body: B: ΣF = 0: FBA + FBC + FBE + (800 N) j = 0, with BA FAB (0.6i + 3j − 0.75k ) FBA = FAB = BA 3.15 FBC = FBC i FBE = − FBE k Substitute and equate to zero the coefficient of j, i, k : j: 3 FAB + 800 N = 0, FAB = −840 N, 3.315 i: 0.6 (−840 N) + FBC = 0 3.15 FBC = 160.0 N T k: 0.75 − (−840 N) − FBE = 0 3.15 FBE = 200 N T FAB = 840 N C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 811 PROBLEM 6.39* (Continued) Free body: C: ΣF = 0: FCA + FCB + FCD + FCE + (100 N) j = 0, with F CA FCA = FAC = AC (−1.2i + 3j − 0.75 k ) CA 3.317 FCB = −(160 N) i FCD = − FCD k FCE = FCE F CE = CE (−1.8i − 3k ) CE 3, 499 Substitute and equate to zero the coefficient of j, i, k : 3 FAC + 100 N = 0, FAC = −110.57 N 3.317 j: − i: k: − FAC = 110.6 N C 1.2 1.8 FCE = 0, FCE = −233.3 (−110.57) − 160 − 3.317 3.499 FCE = 233 N C 0.75 3 (−110.57) − FCD − (−233.3) = 0 3.317 3.499 FCD = 225 N T Free body: D: ΣF = 0: FDA + FDC + FDE + (300 N) j = 0, with F DA FDA = FAD = AD (−1.2i + 3j + 2.25k ) DA 3.937 FDC = FCD k = (225 N)k FDE = − FDE i Substitute and equate to zero the coefficient of j, i, k : j: 3 FAD + 300 N = 0, 3.937 i: 1.2 − ( −393.7 N) − FDE = 0 3.937 k: 2.25 (−393.7 N) + 225 N = 0 (Checks) 3.937 FAD = −393.7 N, FAD = 394 N C FDE = 120.0 N T Free body: E: Member AE is the only member at E which does not lie in the xz plane. Therefore, it is a zero-force member. FAE = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 812 PROBLEM 6.40* Solve Problem 6.39 for P = 0 and Q = (−900 N)k. PROBLEM 6.39* The truss shown consists of nine members and is supported by a ball and socket at B, a short link at C, and two short links at D. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) Determine the force in each member for P = (−1200 N)j and Q = 0. SOLUTION Free body: Truss: ΣM B = 0: 1.8i × Cj + (1.8i − 3k ) × ( D y j + Dz k ) + (0.6i + 3j − 0.75k ) × (−900N)k = 0 1.8 Ck + 1.8D y k − 1.8Dz j + 3D y i + 540 j − 2700i = 0 Equate to zero the coefficient of i, j, k : 3Dy − 2700 = 0 Dy = 900 N −1.8Dz + 540 = 0 Dz = 300 N 1.8C + 1.8 Dy = 0 Thus, C = − D y = −900 N C = −(900 N) j D = (900 N) j + (300 N)k ΣF = 0: B − 900 j + 900 j + 300k − 900k = 0 B = (600 N)k Free body: B: Since B is aligned with member BE, FAB = FBC = 0, FBE = 600 N T Free body: C: ΣF = 0: FCA + FCD + FCE − (900 N) j = 0, with PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 813 PROBLEM 6.40* (Continued) FCA FCD F CA = FAC = AC ( −1.2i + 3j − 0.75k ) CA 3.317 F CE = − FCD k FCE = FCE = CE (−1.8i − 3k ) CE 3.499 Substitute and equate to zero the coefficient of j, i, k: j: 3 FAC − 900 N = 0, 3.317 FAC = 995.1 N FAC = 995 N T FCE = −699.8 N FCE = 700 N C i: − 1.2 1.8 FCE = 0, (995.1) − 3.317 3.499 k: − 0.75 3 (995.1) − FCD − ( −699.8) = 0 3.317 3.499 FCD = 375 N T Free body: D: ΣF = 0: FDA + FDE + (375 N)k +(900 N)j + (300 N) k = 0 F DA FDA = FAD = AD (−1.2i + 3j + 2.25k ) DA 3.937 with FDE = − FDE i and Substitute and equate to zero the coefficient j, i, k: j: 3 FAD + 900 N = 0, FAD = −1181.1 N 3.937 i: 1.2 − (−1181.1 N) − FDE = 0 3.937 k: FAD = 1181 N C FDE = 360 N T 2.25 (−1181.1 N + 375 N + 300 N = 0) 3.937 (Checks) Free body: E: Member AE is the only member at E which does not lie in the xz plane. Therefore, it is a zero-force member. FAE = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 814 PROBLEM 6.41* The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at E. SOLUTION (a) Check simple truss. (1) Start with tetrahedron BEFG. (2) Add members BD, ED, GD joining at D. (3) Add members BA, DA, EA joining at A. (4) Add members DH, EH, GH joining at H. (5) Add members BC, DC, GC joining at C. Truss has been completed: It is a simple truss. Free body: Truss: Check constraints and reactions. Six unknown reactions—ok; however, supports at A and B constrain truss to rotate about AB and support at G prevents such a rotation. Thus, Truss is completely constrained and reactions are statically determinate. Determination of reactions: ΣM A = 0: 11i × ( By j + Bz k ) + (11i − 9.6k ) × G j + (10.08 j − 9.6k ) × (275i + 240k ) = 0 11By k − 11By j + 11Gk + 9.6Gi − (10.08)(275)k + (10.08)(240)i − (9.6)(275) j = 0 Equate to zero the coefficient of i, j, k: i : 9.6G + (10.08)(240) = 0 G = −252 lb G = (−252 lb) j j: − 11Bz − (9.6)(275) = 0 Bz = −240 lb k : 11By + 11(−252) − (10.08)(275) = 0, By = 504 lb B = (504 lb) j − (240 lb)k PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 815 PROBLEM 6.41* (Continued) ΣF = 0: A + (504 lb) j − (240 lb)k − (252 lb) j + (275 lb)i + (240 lb)k = 0 A = −(275 lb)i − (252 lb) j Zero-force members. The determination of these members will facilitate our solution. FB: C: Writing ΣFx = 0, ΣFy = 0, ΣFz = 0 yields FBC = FCD = FCG = 0 FB: F: Writing ΣFx = 0, ΣFy = 0, ΣFz = 0 yields FBF = FEF = FFG = 0 FB: A: Since FB: H: Writing Az = 0, writing ΣFz = 0 ΣFy = 0 yields FAD = 0 yields FDH = 0 FB: D: Since FAD = FCD = FDH = 0, we need consider only members DB, DE, and DG. Since FDE is the only force not contained in plane BDG, it must be zero. Simple reasonings show that the other two forces are also zero. FBD = FDE = FDG = 0 The results obtained for the reactions at the supports and for the zero-force members are shown on the figure below. Zero-force members are indicated by a zero (“0”). (b) Force in each of the members joined at E. FDE = FEF = 0 We already found that Free body: A: ΣFy = 0 yields FAE = 252 lb T Free body: H: ΣFz = 0 yields FEH = 240 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 816 PROBLEM 6.41* (Continued) ΣF = 0: FEB + FEG + (240 lb)k − (252 lb) j = 0 Free body: E: F FBE (11i − 10.08 j) + EG (11i − 9.6k ) + 240k − 252 j = 0 14.92 14.6 Equate to zero the coefficient of y and k: 10.08 j: − FBE − 252 = 0 14.92 FBE = 373 lb C 9.6 − FEG + 240 = 0 14.6 FEG = 365 lb T k: PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 817 PROBLEM 6.42* The truss shown consists of 18 members and is supported by a ball and socket at A, two short links at B, and one short link at G. (a) Check that this truss is a simple truss, that it is completely constrained, and that the reactions at its supports are statically determinate. (b) For the given loading, determine the force in each of the six members joined at G. SOLUTION See solution to Problem 6.41 for part (a) and for reactions and zero-force members. (b) Force in each of the members joined at G. We already know that FCG = FDG = FFG = 0 Free body: H: Free body: G: ΣFx = 0 yields FGH = 275 lb C ΣF = 0: FGB + FGE + (275 lb)i − (252 lb) j = 0 FBG F (−10.08 j + 9.6k ) + EG (−11i + 9.6k ) + 275i − 252 j = 0 13.92 14.6 Equate to zero the coefficient of i, j, k: 11 i: − FEG + 275 = 0 14.6 FEG = 365 lb T 10.08 j: − FBG − 252 = 0 13.92 FBG = 348 lb C 9.6 9.6 k: (−348) + (365) = 0 13.92 14.6 (Checks) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 818 PROBLEM 6.43 Determine the force in members CD and DF of the truss shown. SOLUTION Reactions: ΣM J = 0: (12 kN)(4.8 m) + (12 kN)(2.4 m) − B(9.6 m) = 0 B = 9.00 kN ΣFy = 0: 9.00 kN − 12.00 kN − 12.00 kN + J = 0 J = 15.00 kN Member CD: ΣFy = 0: 9.00 kN + FCD = 0 FCD = 9.00 kN C Member DF: ΣM C = 0: FDF (1.8 m) − (9.00 kN)(2.4 m) = 0 FDF = 12.00 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 819 PROBLEM 6.44 Determine the force in members FG and FH of the truss shown. SOLUTION Reactions: ΣM J = 0: (12 kN)(4.8 m) + (12 kN)(2.4 m) − B (9.6 m) = 0 B = 9.00 kN ΣFy = 0: 9.00 kN − 12.00 kN − 12.00 kN + J = 0 J = 15.00 kN Member FG: 3 ΣFy = 0: − FFG − 12.00 kN + 15.00 kN = 0 5 FFG = 5.00 kN T Member FH: ΣM G = 0: (15.00 kN)(2.4 m) − FFH (1.8 m) = 0 FFH = 20.0 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 820 PROBLEM 6.45 A Warren bridge truss is loaded as shown. Determine the force in members CE, DE, and DF. SOLUTION Free body: Truss: ΣFx = 0: k x = 0 ΣM A = 0: k y (62.5 ft) − (6000 lb)(12.5 ft) − (6000 lb)(25 ft) = 0 k = k y = 3600 lb ΣFy = 0: A + 3600 lb − 6000 lb − 6000 lb = 0 A = 8400 lb We pass a section through members CE, DE, and DF and use the free body shown. ΣM D = 0: FCE (15 ft) − (8400 lb)(18.75 ft) + (6000 lb)(6.25 ft) = 0 FCE = +8000 lb ΣFy = 0: 8400 lb − 6000 lb − FCE = 8000 lb T 15 FDE = 0 16.25 FDE = +2600 lb FDE = 2600 lb T ΣM E = 0: 6000 lb (12.5 ft) − (8400 lb)(25 ft) − FDF (15 ft) = 0 FDF = −9000 lb FDF = 9000 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 821 PROBLEM 6.46 A Warren bridge truss is loaded as shown. Determine the force in members EG, FG, and FH. SOLUTION See solution of Problem 6.45 for free-body diagram of truss and determination of reactions: A = 8400 lb k = 3600 lb We pass a section through members EG, FG, and FH, and use the free body shown. ΣM F = 0: (3600 lb)(31.25 ft) − FEG (15 ft) = 0 FEG = +7500 lb ΣFy = 0: FEG = 7500 lb T 15 FFG + 3600 lb = 0 16.25 FFG = −3900 lb FFG = 3900 lb C ΣM G = 0: FFH (15 ft) + (3600 lb)(25 ft) = 0 FFH = −6000 lb FFH = 6000 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 822 PROBLEM 6.47 Determine the force in members DF, EF, and EG of the truss shown. SOLUTION tan β = 3 4 Reactions: A=N=0 Member DF: 3 ΣM E = 0: + (16 kN)(6 m) − FDF (4 m) = 0 5 FDF = + 40 kN Member EF: + ΣF = 0: (16 kN) sin β − FEF cos β = 0 FEF = 16 tan β = 16(0.75) = 12 kN Member EG: FDF = 40.0 kN T ΣM F = 0: (16 kN)(9 m) + FEF = 12.00 kN T 4 FEG (3 m) = 0 5 FEG = −60 kN FEG = 60.0 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 823 PROBLEM 6.48 Determine the force in members GI, GJ, and HI of the truss shown. SOLUTION Reactions: A=N=0 Member GI: + ΣF = 0: (16 kN)sin β + FGI sin β = 0 FGI = −16 kN Member GJ: ΣM I = 0: − (16 kN)(9 m) − 4 FGJ (3 m) = 0 5 FGJ = −60 kN Member HI: ΣM G = 0: − (16 kN)(9 m) + FGI = 16.00 kN C FGJ = 60.0 kN C 3 FHI (4 m) = 0 5 FHI = +60 kN FHI = 60.0 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 824 PROBLEM 6.49 Determine the force in members AD, CD, and CE of the truss shown. SOLUTION Reactions: ΣM k = 0: 36(2.4) − B(13.5) + 20(9) + 20(4.5) = 0 ΣFx = 0: −36 + K x = 0 B = 26.4 kN K x = 36 kN ΣFy = 0: 26.4 − 20 − 20 + K y = 0 K y = 13.6 kN ΣM C = 0: 36(1.2) − 26.4(2.25) − FAD (1.2) = 0 FAD = −13.5 kN 8 ΣM A = 0: FCD (4.5) = 0 17 FAD = 13.5 kN C FCD = 0 15 ΣM D = 0: FCE (2.4) − 26.4(4.5) = 0 17 FCE = +56.1 kN FCE = 56.1 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 825 PROBLEM 6.50 Determine the force in members DG, FG, and FH of the truss shown. SOLUTION See the solution to Problem 6.49 for free-body diagram and analysis to determine the reactions at the supports B and K. B = 26.4 kN ; K x = 36.0 kN ; K y = 13.60 kN ΣM F = 0: 36(1.2) − 26.4(6.75) + 20(2.25) − FDG (1.2) = 0 FDG = −75 kN FDG = 75.0 kN C 8 ΣM D = 0: − 26.4(4.5) + FFG (4.5) = 0 17 FFG = +56.1 kN 15 ΣM G = 0: 20(4.5) − 26.4(9) + FFH 17 (2.4) = 0 FFH = +69.7 kN FFG = 56.1 kN T FFH = 69.7 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 826 PROBLEM 6.51 A stadium roof truss is loaded as shown. Determine the force in members AB, AG, and FG. SOLUTION We pass a section through members AB, AG, and FG, and use the free body shown. ΣM G = 0: 40 41 FAB (6.3 ft) − (1.8 kips)(14 ft) − (0.9 kips)(28 ft) = 0 FAB = 8.20 kips T FAB = +8.20 kips 3 ΣM D = 0: − FAG (28 ft) + (1.8 kips)(28 ft) 5 + (1.8 kips)(14 ft) = 0 FAG = 4.50 kips T FAG = +4.50 kips ΣM A = 0: − FFG (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 FFG = 11.60 kips C FFG = −11.60 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 827 PROBLEM 6.52 A stadium roof truss is loaded as shown. Determine the force in members AE, EF, and FJ. SOLUTION We pass a section through members AE, EF, and FJ, and use the free body shown. 8 ΣM F = 0: FAE (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 82 + 92 FAE = +17.46 kips FAE = 17.46 kips T ΣM A = 0: − FEF (9 ft) − (1.8 kips)(12 ft) − (1.8 kips)(26 ft) − (0.9 kips)(40 ft) = 0 FEF = −11.60 kips FEF = 11.60 kips C ΣM E = 0: − FFJ (8 ft) − (0.9 kips)(8 ft) − (1.8 kips)(20 ft) − (1.8 kips)(34 ft) − (0.9 kips)(48 ft) = 0 FFJ = −18.45 kips FFJ = 18.45 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 828 PROBLEM 6.53 Determine the force in members CD and DF of the truss shown. SOLUTION tan α = 5 α = 22.62° 12 sin α = 5 12 cos α = 13 13 Member CD: ΣM I = 0: FCD (9 m) + (10 kN)(9 m) + (10 kN)(6 m) + (10 kN)(3 m) = 0 FCD = 20.0 kN C FCD = −20 kN Member DF: ΣM C = 0: ( FDF cos α )(3.75 m) + (10 kN)(3 m) + (10 kN)(6 m) + (10 kN)(9 m) = 0 FDF cos α = −48 kN 12 FDF = −48 kN 13 FDF = −52.0 kN FDF = 52.0 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 829 PROBLEM 6.54 Determine the force in members CE and EF of the truss shown. SOLUTION Member CE: ΣM F = 0: FCE (2.5 m) − (10 kN)(3 m) − (10 kN)(6 m) = 0 FCE = 36.0 kN T FCE = +36 kN Member EF: ΣM I = 0: FEF (6 m) + (10 kN)(6 m) + (10 kN)(3 m) = 0 FEF = 15.00 kN C FEF = −15 kN PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 830 PROBLEM 6.55 The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members FG, EG, and EH. SOLUTION Reactions at supports. Because of the symmetry of the loading, Ax = 0, Ay = O = 1 total load 2 A = O = 4.48 kN We pass a section through members FG, EG, and EH, and use the free body shown. Slope FG = Slope FI = Slope EG = 1.75 m 6m 5.50 m 2.4 m ΣM E = 0: (0.6 kN)(7.44 m) + (1.24 kN)(3.84 m) − (4.48 kN)(7.44 m) 6 − FFG (4.80 m) = 0 6.25 FFG = −5.231 kN FFG = 5.23 kN C ΣM G = 0: FEH (5.50 m) + (0.6 kN)(9.84 m) + (1.24 kN)(6.24 m) + (1.04 kN)(2.4 m) −(4.48 kN)(9.84 m) = 0 ΣFy = 0: FEH = 5.08 kN T 5.50 1.75 (−5.231 kN) + 4.48 kN − 0.6 kN − 1.24 kN − 1.04 kN = 0 FEG + 6.001 6.25 FEG = −0.1476 kN FEG = 0.1476 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 831 PROBLEM 6.56 The truss shown was designed to support the roof of a food market. For the given loading, determine the force in members KM, LM, and LN. SOLUTION Because of symmetry of loading, O= 1 load 2 O = 4.48 kN We pass a section through KM, LM, LN, and use free body shown. 3.84 ΣM M = 0: FLN (3.68 m) 4 + (4.48 kN − 0.6 kN)(3.6 m) = 0 FLN = −3.954 kN FLN = 3.95 kN C ΣM L = 0: − FKM (4.80 m) − (1.24 kN)(3.84 m) + (4.48 kN − 0.6 kN)(7.44 m) = 0 FKM = +5.022 kN ΣFy = 0: FKM = 5.02 kN T 4.80 1.12 (−3.954 kN) − 1.24 kN − 0.6 kN + 4.48 kN = 0 FLM + 6.147 4 FLM = −1.963 kN FLM = 1.963 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 832 PROBLEM 6.57 A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members DF, EF, and EG. SOLUTION Free body: Truss: ΣFx = 0: N x = 0 ΣM N = 0: (200 lb)(8a ) + (400 lb)(7a + 6a + 5a )+(350 lb)(4a ) + (300 lb)(3a + 2a + a ) − A(8a ) = 0 A = 1500 lb ΣFy = 0: 1500 lb − 200 lb − 3(400 lb) − 350 lb − 3(300 lb) − 150 lb + N y = 0 N y = 1300 lb N = 1300 lb We pass a section through DF, EF, and EG, and use the free body shown. (We apply FDF at F.) ΣM E = 0: (200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) − (1500 lb)(18 ft) 18 FDF (4.5 ft) = 0 − 182 + 4.52 FDF = −3711 lb FDF = 3710 lb C ΣM A = 0: FEF (18 ft) − (400 lb)(6 ft) − (400 lb)(12 ft) = 0 FEF = 400 lb T FEF = +400 lb ΣM F = 0: FEG (4.5 ft) − (1500 lb)(18 ft)+(200 lb)(18 ft) + (400 lb)(12 ft) + (400 lb)(6 ft) = 0 FEG = +3600 lb FEG = 3600 lb T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 833 PROBLEM 6.58 A Polynesian, or duopitch, roof truss is loaded as shown. Determine the force in members HI, GI, and GJ. SOLUTION See solution of Problem 6.57 for reactions: A = 1500 lb , N = 1300 lb We pass a section through HI, GI, and GJ, and use the free body shown. (We apply FHI at H.) 6 ΣM G = 0: FHI (8.5 ft) + (1300 lb)(24 ft) − (300 lb)(6 ft) 2 2 6 +4 − (300 lb)(12 ft) − (300 lb)(18 ft) − (150 lb)(24 ft) = 0 FHI = −2375.4 lb FHI = 2375 lb C ΣM I = 0: (1300 lb)(18 ft) − (300 lb)(6 ft) − (300 lb)(12 ft) − (150 lb)(18 ft) − FGJ (4.5 ft) = 0 FGJ = +3400 lb ΣFx = 0: − FGJ = 3400 lb T 4 6 FGI − (−2375.4 lb) − 3400 lb = 0 2 5 6 + 42 FGI = −1779.4 lb FGI = 1779 lb C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 834 PROBLEM 6.59 Determine the force in members DE and DF of the truss shown when P = 20 kips. SOLUTION Reactions: C = K = 2.5P 7.5 18 β = 22.62° tan β = Member DE: ΣM A = 0: (2.5 P )(6 ft) − FDE (12 ft) = 0 FDE = +1.25 P For P = 20 kips, FDE = +1.25(20) = +25 kips FDE = 25.0 kips T Member DF: ΣM E = 0: P (12 ft) − (2.5 P )(6 ft) − FDF cos β (5 ft) = 0 12 P − 15 P − FDF cos 22.62°(5 ft) = 0 FDF = −0.65 P For P = 20 kips, FDF = −0.65(20) = −13 kips FDF = 13.00 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 835 PROBLEM 6.60 Determine the force in members EG and EF of the truss shown when P = 20 kips. SOLUTION Reactions: C = K = 2.5P 7.5 tan α = 6 α = 51.34° Member EG: ΣM F = 0: P (18 ft) − 2.5 P (12 ft) − P (6 ft) + FEG (7.5 ft) = 0 FEG = +0.8 P; For P = 20 kips, FEG = 16.00 kips T FEG = 0.8(20) = +16 kips Member EF: ΣM A = 0: 2.5 P (6 ft) − P (12 ft) + FEF sin 51.34°(12 ft) = 0 FEF = −0.320 P; For P = 20 kips, FEF = −0.320(20) = −6.4 kips FEF = 6.40 kips C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 836 PROBLEM 6.61 Determine the force in members EH and GI of the truss shown. (Hint: Use section aa.) SOLUTION Reactions: ΣFx = 0: Ax = 0 ΣM P = 0: 12(45) + 12(30) + 12(15) − Ay (90) = 0 A y = 12 kips ΣFy = 0: 12 − 12 − 12 − 12 + P = 0 P = 24 kips ΣM G = 0: − (12 kips)(30 ft) − FEH (16 ft) = 0 FEH = −22.5 kips ΣFx = 0: FGI − 22.5 kips = 0 FEH = 22.5 kips C FGI = 22.5 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 837 PROBLEM 6.62 Determine the force in members HJ and IL of the truss shown. (Hint: Use section bb.) SOLUTION See the solution to Problem 6.61 for free body diagram and analysis to determine the reactions at supports A and P. A x = 0; A y = 12.00 kips ; P = 24.0 kips ΣM L = 0: FHJ (16 ft) − (12 kips)(15 ft) + (24 kips)(30 ft) = 0 FHJ = −33.75 kips FHJ = 33.8 kips C ΣFx = 0: 33.75 kips − FIL = 0 FIL = +33.75 kips FIL = 33.8 kips T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 838 PROBLEM 6.63 Determine the force in members DG and FI of the truss shown. (Hint: Use section aa.) SOLUTION ΣM F = 0: FDG (4 m) − (5 kN)(3 m) = 0 FDG = +3.75 kN FDG = 3.75 kN T ΣFy = 0: − 3.75 kN − FFI = 0 FFI = −3.75 kN FFI = 3.75 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 839 PROBLEM 6.64 Determine the force in members GJ and IK of the truss shown. (Hint: Use section bb.) SOLUTION ΣM I = 0: FGJ (4 m) − (5 kN)(6 m) − (5 kN)(3 m) = 0 FGJ = +11.25 kN FGJ = 11.25 kN T ΣFy = 0: − 11.25 kN − FIK = 0 FIK = −11.25 kN FIK = 11.25 kN C PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 840 PROBLEM 6.65 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading. SOLUTION Free body: Truss: ΣFx = 0: Fx = 0 ΣM H = 0: 4.8(3a) + 4.8(2a) + 4.8a − 2.4a − Fy (2a) = 0 Fy = +13.20 kips F = 13.20 kips ΣFy = 0: H + 13.20 kips − 3(4.8 kips) − 2(2.4 kips) = 0 H = +6.00 kips H = 6.00 kips Free body: ABF: We assume that counter BG is acting. ΣFy = 0: − 9.6 FBG + 13.20 − 2(4.8) = 0 14.6 FBG = +5.475 FBG = 5.48 kips T Since BG is in tension, our assumption was correct. Free body: DEH: We assume that counter DG is acting. ΣFy = 0: − 9.6 FDG + 6.00 − 2(2.4) = 0 14.6 FDG = 1.825 kips T FDG = +1.825 Since DG is in tension, O.K. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 841 PROBLEM 6.66 The diagonal members in the center panels of the truss shown are very slender and can act only in tension; such members are known as counters. Determine the forces in the counters that are acting under the given loading. SOLUTION Free body: Truss: ΣFx = 0: Fx = 0 ΣM G = 0: − Fy a + 4.8(2a) + 4.8a − 2.4a − 2.4(2a) = 0 Fy = 7.20 F = 7.20 kips Free body: ABF: We assume that counter CF is acting. ΣFy = 0: 9.6 FCF + 7.20 − 2(4.8) = 0 14.6 FCF = +3.65 FCF = 3.65 kips T Since CF is in tension, O.K. Free body: DEH: We assume that counter CH is acting. ΣFy = 0: 9.6 FCH − 2(2.4 kips) = 0 14.6 FCH = +7.30 FCH = 7.30 kips T Since CH is in tension, O.K. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 842 PROBLEM 6.67 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter. Counters CJ and HE SOLUTION Free body: Portion ABDFEC of tower. We assume that counter CJ is acting and show the forces exerted by that counter and by members CH and EJ. ΣFx = 0: 4 FCJ − 2(1.2 kN)sin 20° = 0 FCJ = +1.026 kN 5 Since CJ is found to be in tension, our assumption was correct. Thus, the answers are (a) CJ (b) 1.026 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 843 PROBLEM 6.68 The diagonal members in the center panels of the power transmission line tower shown are very slender and can act only in tension; such members are known as counters. For the given loading, determine (a) which of the two counters listed below is acting, (b) the force in that counter. Counters IO and KN SOLUTION Free body: Portion of tower shown. We assume that counter IO is acting and show the forces exerted by that counter and by members IN and KO. ΣFx = 0 : 4 FIO − 4(1.2 kN)sin 20° = 0 FIO = +2.05 kN 5 Since IO is found to be in tension, our assumption was correct. Thus, the answers are (a) IO (b) 2.05 kN T PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 844 PROBLEM 6.69 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a) Number of members: m = 16 Number of joints: n = 10 Reaction components: r=4 m + r = 20, 2 n = 20 (a) m + r = 2n Thus, To determine whether the structure is actually completely constrained and determinate, we must try to find the reactions at the supports. We divide the structure into two simple trusses and draw the free-body diagram of each truss. This is a properly supported simple truss – O.K. This is an improperly supported simple truss. (Reaction at C passes through B. Thus, Eq. ΣM B = 0 cannot be satisfied.) Structure is improperly constrained. Structure (b) m = 16 n = 10 r=4 m + r = 20, 2n = 20 (b) m + r = 2n Thus, PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 845 PROBLEM 6.69 (Continued) We must again try to find the reactions at the supports dividing the structure as shown. Both portions are simply supported simple trusses. Structure is completely constrained and determinate. Structure (c) m = 17 n = 10 r=4 m + r = 21, 2n = 20 (c) m + r > 2n Thus, This is a simple truss with an extra support which causes reactions (and forces in members) to be indeterminate. Structure is completely constrained and indeterminate. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 846 PROBLEM 6.70 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): Nonsimple truss with r = 4, m = 16, n = 10, so m + r = 20 = 2n, but we must examine further. FBD Sections: FBD I: ΣM A = 0 T1 II: ΣFx = 0 T2 I: ΣFx = 0 Ax I: ΣFy = 0 Ay II: ΣM E = 0 Cy II: ΣFy = 0 Ey Since each section is a simple truss with reactions determined, structure is completely constrained and determinate. Structure (b): Nonsimple truss with r = 3, m = 16, n = 10, so m + r = 19 < 2n = 20 Structure is partially constrained. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 847 PROBLEM 6.70 (Continued) Structure (c): Simple truss with r = 3, m = 17, n = 10, m + r = 20 = 2n, but the horizontal reaction forces Ax and E x are collinear and no equilibrium equation will resolve them, so the structure is improperly constrained and indeterminate. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 848 PROBLEM 6.71 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): Nonsimple truss with r = 4, m = 12, n = 8 so r + m = 16 = 2n. Check for determinacy: One can solve joint F for forces in EF, FG and then solve joint E for E y and force in DE. This leaves a simple truss ABCDGH with r = 3, m = 9, n = 6 so r + m = 12 = 2n Structure is completely constrained and determinate. Structure (b): Simple truss (start with ABC and add joints alphabetically to complete truss) with r = 4, m = 13, n = 8 so r + m = 17 > 2n = 16 Constrained but indeterminate PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 849 PROBLEM 6.71 (Continued) Structure (c): Nonsimple truss with r = 3, m = 13, n = 8 so r + m = 16 = 2n. To further examine, follow procedure in part (a) above to get truss at left. Since F1 ≠ 0 (from solution of joint F), ΣM A = aF1 ≠ 0 and there is no equilibrium. Structure is improperly constrained. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 850 PROBLEM 6.72 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a) Number of members: m = 12 Number of joints: n=8 Reaction components: Thus, r =3 m + r = 15, 2 n = 16 m + r < 2n Structure is partially constrained. Structure (b) m = 13, n = 8 r =3 m + r = 16, 2n = 16 Thus, m + r = 2n To verify that the structure is actually completely constrained and determinate, we observe that it is a simple truss (follow lettering to check this) and that it is simply supported by a pin-and-bracket and a roller. Thus, structure is completely constrained and determinate. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 851 PROBLEM 6.72 (Continued) Structure (c) m = 13, n = 8 r=4 m + r = 17, 2n = 16 Thus, m + r > 2n Structure is completely constrained and indeterminate. This result can be verified by observing that the structure is a simple truss (follow lettering to check this), therefore it is rigid, and that its supports involve four unknowns. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 852 PROBLEM 6.73 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): Rigid truss with r = 3, m = 14, n = 8, so r + m = 17 > 2n = 16 so completely constrained but indeterminate Structure (b): Simple truss (start with ABC and add joints alphabetically), with r = 3, m = 13, n = 8, so r + m = 16 = 2n so completely constrained and determinate Structure (c): Simple truss with r = 3, m = 13, n = 8, so r + m = 16 = 2n, but horizontal reactions ( Ax and Dx ) are collinear, so cannot be resolved by any equilibrium equation. Structure is improperly constrained. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 853 PROBLEM 6.74 Classify each of the structures shown as completely, partially, or improperly constrained; if completely constrained, further classify as determinate or indeterminate. (All members can act both in tension and in compression.) SOLUTION Structure (a): m = 12 No. of members No. of joints n = 8 m + r = 16 = 2n No. of reaction components r = 4 unknows = equations FBD of EH: ΣM H = 0 FDE ; ΣFx = 0 FGH ; ΣFy = 0 Hy Then ABCDGF is a simple truss and all forces can be determined. This example is completely constrained and determinate. Structure (b): No. of members m = 12 No. of joints n = 8 m + r = 15 < 2n = 16 No. of reaction components r = 3 unknows < equations partially constrained Note: Quadrilateral DEHG can collapse with joint D moving downward; in (a), the roller at F prevents this action. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 854 PROBLEM 6.74 (Continued) Structure (c): No. of members m = 13 No. of joints n = 8 m + r = 17 > 2n = 16 No. of reaction components r = 4 unknows > equations completely constrained but indeterminate PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 855 PROBLEM 6.75 Determine the force in member BD and the components of the reaction at C. SOLUTION We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD. Free body: ABC: BD = (24) 2 + (10) 2 = 26 in. 10 ΣM C = 0: (160 lb)(30 in.) − FBD (16 in.) = 0 26 FBD = +780 lb ΣM x = 0: C x + FBD = 780 lb T 24 (780 lb) = 0 26 C x = −720 lb ΣFy = 0: C y − 160 lb + C x = 720 lb 10 (780 lb) = 0 26 C y = −140.0 lb C y = 140.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 856 PROBLEM 6.76 Determine the force in member BD and the components of the reaction at C. SOLUTION We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD. Free body: ABC: Attaching FBD at D and resolving it into components, we write ΣM C = 0: 450 FBD (240 mm) = 0 (400 N)(135 mm) + 510 FBD = −255 N ΣFx = 0: C x + FBD = 255 N C 240 (−255 N) = 0 510 C x = +120.0 N ΣFy = 0: C y − 400 N + C x = 120.0 N 450 (−255 N) = 0 510 C y = +625 N C y = 625 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 857 PROBLEM 6.77 Determine the components of all forces acting on member ABCD of the assembly shown. SOLUTION Free body: Entire assembly: ΣM B = 0: D (120 mm) − (480 N)(80 mm) = 0 D = 320 N ΣFx = 0: Bx + 480 N = 0 B x = 480 N ΣFy = 0: By + 320 N = 0 B y = 320 N Free body: Member ABCD: ΣM A = 0: (320 N)(200 mm) − C (160 mm) − (320 N)(80 mm) − (480 N)(40 mm) = 0 C = 120.0 N ΣFx = 0: Ax − 480 N = 0 A x = 480 N ΣFy = 0: Ay − 320 N − 120 N + 320 N = 0 A y = 120.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 858 PROBLEM 6.78 Determine the components of all forces acting on member ABD of the frame shown. SOLUTION Free body: Entire frame: ΣM D = 0: (300 lb) − (12 ft) + (450 lb)(4 ft) − E (6 ft) = 0 E = +900 lb E = 900 lb D x = 900 lb ΣFx = 0: Dx − 900 lb = 0 ΣFy = 0: Dy − 300 lb − 450 lb = 0 D y = 750 lb Free body: Member ABD: We note that BC is a two-force member and that B is directed along BC. ΣM A = 0: (750 lb)(16 ft) − (900 lb)(6 ft) − B (8 ft) = 0 B = +825 lb B = 825 lb ΣFx = 0: Ax + 900 lb = 0 Ax = −900 lb A x = 900 lb ΣFy = 0: Ay + 750 lb − 825 lb = 0 Ay = +75 lb A y = 75.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 859 PROBLEM 6.79 For the frame and loading shown, determine the components of all forces acting on member ABC. SOLUTION Free body: Entire frame: ΣM E = 0: − Ax (4) − (20 kips)(5) = 0 Ax = −25 kips, A x = 25.0 kips ΣFy = 0: Ay − 20 kips = 0 Ay = 20 kips A y = 20.0 kips Free body: Member ABC: Note: BE is a two-force member, thus B is directed along line BE and By = 2 Bx . 5 ΣM C = 0: (25 kips)(4 ft) − (20 kips)(10 ft) + Bx (2 ft) + By (5 ft) = 0 −100 kip ⋅ ft + Bx (2 ft) + 2 Bx (5 ft) = 0 5 Bx = 25 kips By = B x = 25.0 kips 2 2 ( Bx ) = (25) = 10 kips 5 5 B y = 10.00 kips ΣFx = 0: C x − 25 kips − 25 kips = 0 C x = 50 kips C x = 50.0 kips ΣFy = 0: C y + 20 kips − 10 kips = 0 C y = −10 kips C y = 10.00 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 860 PROBLEM 6.80 Solve Problem 6.79 assuming that the 20-kip load is replaced by a clockwise couple of magnitude 100 kip ⋅ ft applied to member EDC at Point D. PROBLEM 6.79 For the frame and loading shown, determine the components of all forces acting on member ABC. SOLUTION Free body: Entire frame: ΣFy = 0: Ay = 0 ΣM E = 0: − Ax (4 ft) − 100 kip ⋅ ft = 0 Ax = −25 kips A x = 25.0 kips A = 25.0 kips B x = 25.0 kips Free body: Member ABC: Note: BE is a two-force member, thus B is directed along line BE and By = 2 Bx . 5 ΣM C = 0: (25 kips)(4 ft) + Bx (2 ft) + By (5 ft) = 0 100 kip ⋅ ft + Bx (2 ft) + 2 Bx (5 ft) = 0 5 Bx = −25 kips By = 2 2 Bx = ( −25) = −10 kips; 5 5 ΣFx = 0: − 25 kips + 25 kips + C x = 0 B y = 10.00 kips Cx = 0 ΣFy = 0: +10 kips + C y = 0 C y = −10 kips C y = 10 kips C = 10.00 kips PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 861 PROBLEM 6.81 Determine the components of all forces acting on member ABCD when θ = 0. SOLUTION Free body: Entire assembly: ΣM B = 0: A(200) − (150 N)(500) = 0 A = +375 N A = 375 N B x = 375 N ΣFx = 0: Bx + 375 N = 0 Bx = −375 N ΣFy = 0: By − 150 N = 0 By = +150 N B y = 150 N Free body: Member ABCD: We note that D is directed along DE, since DE is a two-force member. ΣM C = 0: D (300) − (150 N)(100) + (375 N)(200) = 0 D = −200 N D = 200 N ΣFx = 0: C x + 375 − 375 = 0 Cx = 0 ΣFy = 0: C y + 150 − 200 = 0 C y = +50.0 N C = 50.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 862 PROBLEM 6.82 Determine the components of all forces acting on member ABCD when θ = 90°. SOLUTION Free body: Entire assembly: ΣM B = 0: A(200) − (150 N)(200) = 0 A = +150.0 N A = 150.0 N ΣFx = 0: Bx + 150 − 150 = 0 Bx = 0 ΣFy = 0: By = 0 B = 0 Free body: Member ABCD: We note that D is directed along DE, since DE is a two-force member. ΣM C = 0: D (300) + (150 N)(200) = 0 D = −100.0 N D = 100.0 N ΣFx = 0: C x + 150 N = 0 C x = −150 N Cx = 150.0 N ΣFy = 0: C y − 100 N = 0 C y = +100.0 N C y = 100.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 863 PROBLEM 6.83 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D. SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since position of load on its line of action is immaterial. ΣM E = 0: − (750 N)(80 mm) − Ax (200 mm) = 0 Ax = − 300 N A x = 300 N ΣFx = 0: E x − 300 N = 0 E x = 300 N E x = 300 N ΣFy = 0: Ay + E y − 750 N = 0 (a) (1) Load applied at B. Free body: Member CE: CE is a two-force member. Thus, the reaction at E must be directed along CE. Ey 300 N From Eq. (1): = 75 mm 250 mm E y = 90 N Ay + 90 N − 750 N = 0 Ay = 660 N Thus, reactions are (b) A x = 300 N , A y = 660 N E x = 300 N , E y = 90.0 N Load applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 300 N = 125 mm 250 mm Ay = 150 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 864 PROBLEM 6.83 (Continued) Ay + E y − 750 N = 0 From Eq. (1): 150 N + E y − 750 N = 0 E y = 600 N E y = 600 N Thus, reactions are A x = 300 N , E x = 300 N , A y = 150.0 N E y = 600 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 865 PROBLEM 6.84 Determine the components of the reactions at A and E if a 750-N force directed vertically downward is applied (a) at B, (b) at D. SOLUTION Free-body: Entire frame: The following analysis is valid for both parts (a) and (b) since position of load on its line of action is immaterial. ΣM E = 0: − (750 N)(240 mm) − Ax (400 mm) = 0 Ax = −450 N A x = 450 N ΣFx = 0: E x − 450 N = 0 E x = 450 N E x = 450 N ΣFy = 0: Ay + E y − 750 N = 0 (a) (1) Load applied at B. Free body: Member CE: CE is a two-force member. Thus, the reaction at E must be directed along CE. Ey 450 N From Eq. (1): = 240 mm ; E y = 225 N 480 mm Ay + 225 − 750 = 0; A y = 525 N Thus, reactions are A x = 450 N , , E y = 225 N A x = 450 N , A y = 150.0 N E x = 450 N , E x = 450N (b) A y = 525 N Load applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 450 N From Eq. (1): = 160 mm 480 mm A y = 150.0 N Ay + E y − 750 N = 0 150 N + E y − 750 N = 0 E y = 600 N E y = 600 N Thus, reactions are E y = 600 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 866 PROBLEM 6.85 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D. SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since the point of application of the couple is immaterial. ΣM E = 0: −36 N ⋅ m − Ax (0.2 m) = 0 Ax = − 180 N A x = 180 N ΣFx = 0: − 180 N + E x = 0 E x = 180 N E x = 180 N ΣFy = 0: Ay + E y = 0 (a) (1) Couple applied at B. Free body: Member CE: AC is a two-force member. Thus, the reaction at E must be directed along EC. Ey 180 N From Eq. (1): = 0.075 m 0.25 m E y = 54 N Ay + 54 N = 0 Ay = − 54 N A y = 54.0 N Thus, reactions are (b) A x = 180.0 N , A y = 54.0 N E x = 180.0 N , E y = 54.0 N Couple applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along EC. Ay 180 N = 0.125 m 0.25 m Ay = 90 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 867 PROBLEM 6.85 (Continued) From Eq. (1): Ay + E y = 0 90 N + E y = 0 E y = − 90 N E y = 90 N Thus, reactions are A x = 180.0 N E x = −180.0 N A y = 90.0 N , , E y = 90.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 868 PROBLEM 6.86 Determine the components of the reactions at A and E if the frame is loaded by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D. SOLUTION Free body: Entire frame: The following analysis is valid for both parts (a) and (b) since the point of application of the couple is immaterial. ΣM E = 0: −36 N ⋅ m − Ax (0.4 m) = 0 Ax = − 90 N A x = 90.0 N ΣFx = 0: −90 +E x = 0 E x = 90 N E x = 90.0 N ΣFy = 0: Ay + E y = 0 (a) (1) Couple applied at B. Free body: Member CE: AC is a two-force member. Thus, the reaction at E must be directed along EC. Ey 90 N From Eq. (1): = 0.24 m ; E y = 45.0 N 0.48 m Ay + 45 N = 0 Ay = − 45 N A y = 45.0 N Thus, reactions are (b) A x = 90.0 N , A y = 45.0 N E x = 90.0 N , E y = 45.0 N Couple applied at D. Free body: Member AC: AC is a two-force member. Thus, the reaction at A must be directed along AC. Ay 90 N = 0.16 m ; A y = 30 N 0.48 m PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 869 PROBLEM 6.86 (Continued) From Eq. (1): Ay + E y = 0 30 N + E y = 0 E y = − 30 N E y = 30 N A x = 90.0 N Thus, reactions are E x = − 90.0 N A y = 30.0 N , , E y = 30.0 N PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 870 PROBLEM 6.87 Determine all the forces exerted on member AI if the frame is loaded by a clockwise couple of magnitude 1200 lb · in. applied (a) at Point D, (b) at Point E. SOLUTION Free body: Entire frame: Location of couple is immaterial. ΣM H = 0: I (48 in.) − 1200 lb ⋅ in. = 0 I = +25.0 lb (a) and (b) I = 25.0 lb We note that AB, BC, and FG are two-force members. Free body: Member AI: tan α = (a) 20 5 = 48 12 α = 22.6° Couple applied at D. ΣFy = 0: − ΣM G = 0: 5 A + 25 lb = 0 13 A = +65.0 lb A = 65.0 lb 12 (65 lb)(40 in.) − C (20 in.) = 0 13 C = +120 lb ΣFx = 0: − 12 (65 lb) + 120 lb + G = 0 13 G = −60.0 lb 22.6° C = 120 lb G = 60 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 871 PROBLEM 6.87 (Continued) (b) Couple applied at E. ΣFy = 0: − ΣM G = 0: 5 A + 25 lb = 0 13 A = +65.0 lb A = 65.0 lb 22.6° 12 (65 lb) + (40 in.) − C (20 in.) − 1200 lb ⋅ in. = 0 13 C = +60.0 lb C = 60.0 lb ΣFx = 0: − 12 (65 lb) + 60 lb + G = 0 13 G=0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 872 PROBLEM 6.88 Determine all the forces exerted on member AI if the frame is loaded by a 40-lb force directed horizontally to the right and applied (a) at Point D, (b) at Point E. SOLUTION Free body: Entire frame: Location of 40-lb force on its line of action DE is immaterial. ΣM H = 0: I (48 in.) − (40 lb)(30 in.) = 0 I = +25.0 lb (a) and (b) I = 25.0 lb We note that AB, BC, and FG are two-force members. Free body: Member AI: tan α = (a) 20 5 = 48 12 α = 22.6° Force applied at D. ΣFy = 0: − ΣM G = 0: 5 A + 25 lb = 0 13 A = +65.0 lb 12 (65 lb)(40 in.) − C (20 in.) = 0 13 C = +120.0 lb ΣFx = 0: − 12 (65 lb) + 120 lb + G = 0 13 G = −60.0 lb A = 65.0 lb 22.6° C = 120.0 lb G = 60.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 873 PROBLEM 6.88 (Continued) (b) Force applied at E. ΣFy = 0: − ΣM G = 0: 5 A + 25 lb = 0 13 A = +65.0 lb A = 65.0 lb 22.6° 12 (65 lb)(40 in.) − C (20 in.) − (40 lb)(10 in.) = 0 13 C = +100.0 lb C = 100.0 lb ΣFx = 0: − 12 (65 lb) + 100 lb + 40 lb + G = 0 13 G = −80.0 lb G = 80.0 lb PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 874 PROBLEM 6.89 Determine the components of the reactions at A and B, (a) if the 500-N load is applied as shown, (b) if the 500-N load is moved along its line of action and is applied at Point F. SOLUTION Free body: Entire frame: Analysis is valid for either parts (a) or (b), since position of 100-lb load on its line of action is immaterial. ΣM A = 0: By (10) − (100 lb)(6) = 0 By = + 60 lb ΣFy = 0: Ay + 60 − 100 = 0 Ay = + 40 lb ΣFx = 0: Ax + Bx = 0 (a) (1) Load applied at E. Free body: Member AC: Since AC is a two-force member, the reaction at A must be directed along CA. We have Ax 40 lb = 10 in. 5 in. From Eq. (1): A x = 80.0 lb , A y = 40.0 lb B x = 80.0 lb , B y = 60.0 lb − 80 + Bx = 0 Bx = + 80 lb Thus, PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 875 PROBLEM 6.89 (Continued) (b) Load applied at F. Free body: Member BCD: Since BCD is a two-force member (with forces applied at B and C only), the reaction at B must be directed along CB. We have, therefore, Bx = 0 The reaction at B is From Eq. (1): The reaction at A is B y = 60.0 lb Bx = 0 Ax + 0 = 0 Ax = 0 A y = 40.0 lb Ax = 0 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher,