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Capitulo 3 solucionario estatica beer 9 edicion

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Solucionario de mecanica vectorial para ingenieros 9 edicion capitulo 3

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Capitulo 3 solucionario estatica beer 9 edicion

1. 1. CHAPTER 3
2. 2. 80 mm PROBLEM 3.1 A foot valve for a pneumatic system is hinged at B. Knowing that a - 28°, determine the moment of the 1 6-N force about Point B by resolving the force into horizontal and vertical components. SOLUTION Note that and = a- 20° = 28° -20° = 8° Fx = (1 6 N)cos 8° = 1 5.8443 N Fv =(16N)sin8° = 2.2268N ^£*. ©* — tL> Kl * U^r^-^T^P,, /|t^^^u^d ^r^^^^-, Also x = (0. 1 7 m)cos 20° = 0. 1 59748 m y = (0.17 m)sin 20° = 0.058143 m. >n^y c Noting that the direction of the moment of each force component about B is counterclockwise, MB =xFy +yFx = (0.1 59748 m)(2.2268N) +(0.058143 m)(l 5.8443 N) = 1.277 N-m or MB =1.277N-m^)4 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pail of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 153
3. 3. PROBLEM 3.2 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28°, determine the moment of the 1 6-N force about Point B by resolving the force into components along ABC and in a direction perpendicular to ABC. SOLUTION First resolve the 4-lb force into components P and Q, where g = (16 N)sin 28° = 7.5115 N Then MB = rmQ °^7^ = (0.17m)(7.5115N) = 1.277N-m or MB = 1.277 N-m^^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies* Inc. All rights reserved. No part of (his Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 154
4. 4. 200 in.., 25" -100 mm- 200 mm i '"> mm PROBLEM 3.3 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D. SOLUTION (a) O.2. PC 0>T-**> Fv =(300N)cos25° = 27.1.89 N Fy =(300 N) sin 25° = 126.785 N F = (27 1 .89 N)i + (1 26.785 N) j r = ZM = -(0.1m)i-(0.2m)j MD =rxF MD = HO. 1 m)i - (0.2 m)j] x [(271 .89 N)i + (1.26.785 N)j] = -(12.6785 N • m)k + (54.378 N • m)k = (41.700 N-m)k M =41.7N-nO^ (b) The smallest force Q at B must be perpendicular to DB at 45°^£L MD =Q(DB) 41 .700 N m = (2(0.28284 m) Q = 147.4 N ^L 45° < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 155
5. 5. 200 mn. 25° •-lOOnim-* -200 mm *. 125 il • C H PROBLEM3.4 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the magnitude and sense of the horizontal force applied at C that creates the same moment about D, (c) the smallest force applied at C that creates the same moment about D. SOLUTION (a) See Problem 3.3 for the figure and analysis leading to the determination of Md M =41.7N-m^H Cl^n 0>7.Siy c = at. (b) Since C is horizontal C = Ci r = DC = (0.2 m)i - (0. 1 25 m)j MD =rxCi = C(0.l25m)k 4l.7N-m = (0.l25m)(C) C = 333.60 N (c) The smallest force C must be perpendicular to DC; thus, it forms a with the vertical C = 334N < tan6^ 0.125 m 0.2 m a = 32.0° MD = C(£>C); DC = V( - 2 m) 2 + (°- 1 25 m)' = 0.23585 m 41.70 Nm = C(0.23585m) C = 176.8 N^L 5HX)°< PROPRIETARY MATERIAL. €5 20)0 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 156
6. 6. PROBLEM 3.5 An 8-1b force P is applied to a shift lever. Determine the moment of V about B when a is equal to. 259. SOLUTION First note Px = (8 lb) cos 25° = 7.2505 lb /^ =(8 lb) sin 25° = 3.3809 lb Noting that the direction of the moment of each force component about B is clockwise, have = -(8in.)(3.3809 1b) - (22 in.)(7.2505 lb) = -186.6 lb -in. i*22. •**. or Mj =186.6 lb -in. J) ^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 157
7. 7. PROBLEM 3.6 For the shift lever shown, determine the magnitude and the direction of the smallest force P that has a 21 0-lb • in. clockwise moment about B. 22 in. SOLUTION For P to be minimum it must be perpendicular to the line joining Points A. and B. Thus, a = e , 8 in ^T^s*. K 22 in. / = 19.98° i and MB =dP^n p ZZ i«. Where d = rAIB fl \$ u = ^m.y+(22m.y JB w = 23.409 in. Then _210Ib-in. ' min " 23.409 in. -8.97 lb Pmin =8.97 lb ^19.98° < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 158
8. 8. PROBLEM 3.7 An 1 1 -lb force P is applied to a shift lever. The moment of P about B is clockwise and has a magnitude of 250 lb • in. Determine the value ofa. 22 in. SOLUTION By definition where and also Then or or and MB =rmPsm.6 = a + (9Q°-tf>) _i 8 in. <p - tan" 19.9831' 22 in. rf/fl =V(8in.) 2 +(22in.) 2 = 23.409 in. 250lb-in = (23.409in.)(lllb) xsin(tf + 90° -19.9831°) sin (« + 70.01 69°) = 0.97088 a + 70.0 169° = 76. 1391° a+ 70.0169° = 103.861° a = 6.12° 33.8° < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 159
9. 9. PROBLEM 3.8 It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if a ~ .1 0°, (c) the smallest force P that creates the same moment about B. V—Ar 4 in. n SOLUTION (a) We have MB =raB FN (4 in.)(200 lb) 800 lb -in. or MB =H00b-m.)< A- m. (/;) By definition MB ~rA/B Psin 6» = 10° + (180°~70°) = 120° Then 800 lb • in. = (18 in.) x Psin 1 20° 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 or or A*W™. cl = f A IB 800 lb • in. = (18 in.)Pm ^i„=44.4 1b Pmitl =44.41b^l20 ^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it withoutpermission. 160
10. 10. PROBLEM 3.9 A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and length d is 1 ,90 m, determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at Point C, (b) at Point E. 0.2 m 0.875 m SOLUTION (a) Slope of line Then EC = 0.875 m 1.90 m + 0.2 m 12 *abx - ,~ (Tab) 12 13 960 N (1040 N) 0,1«y and *W=-0040N) = 400 N Then (b) We have MD - TABx (0.875 m)~TABy (0.2 m) = (960 N)(0.875 m) - (400 N)(0.2 m) = 760 N • m MD ~rm(y) + TABx (x) = (960 N)(0) + (400 N)(I .90 m) = 760 N • m or M./J =760N-m > )^ or M7> = 760N-m v )<« PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AM rights reserved. No part of this Manual may be displayed, reproduced or. distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 161
11. 11. PROBLEM 3.10 It is known that a force with a moment of 960 N • m about D is required to straighten the fence post CD. If d- 2,80 m, determine the tension that must be developed in the cable of winch puller AB to create the required moment about Point D. 0.875 m 0.2 j.i» SOLUTION *. » £^*AB 'My o. a? 5^i z-acw OiZ^ Slope of line Then and We have EC = 0.875 m 7 7' 2.80 m + 0.2 m 24 24 My r/ffl)> 25 1 25 T/)B rAH 24 7 960N-m= —7^(0) +—7^(2.80*) 7^= 1224 N or r^=1224N ^ PROPRIETARY MATERIAL; © 2010 The McGraw-Hill Companies, Inc. All. rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 162
12. 12. PROBLEM 3.11 It is known that a force with a moment of 960 N • m about D is required to straighten the fence post CD. If the capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified moment about Point D. fe 0.2 in 0.875 in SOLUTION o.mtn czom The minimum value of d can be found based on the equation relating the moment of the force TAB about D: MD ={TABmK ),(d) where Now MD =960N-m (^flmax )y = TAIHmx sin & = (2400 N)sfo . . 0.875m sin # 960 N • m = 2400 N ^(t/ + 0.20) 2 +(0.875) 2 m 0.875 (d) + 0.20) 2 +(0.875) 2 or ^+ 0.20) 2 + (0.875) 2 = 2. ! 875d or (J + 0.20) 2 + (0.875) 2 = 4.7852rf 2 or 3 .7852</ 2 - 0.40c/ - .8056 = Using the quadratic equation, the minimum values of d are 0.51719 m and -.41151 m. Since only the positive value applies here, d — 0.5 1 7 1 9 m or d ~ 5 1 7 mm ^ PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. JVo port o/rt/s Manual may be displayed, reproduced or distributed in any form or by any means, without (he prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill/or their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 163
13. 13. 5.3 in. PROBLEM 3.12 12.0 in. 2.33 in. mmM1 >iii. i 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 Then and Now where Then dcB == 7(12.0 in.) 2 = 12.224 lin. + (2.33 in.) 2 cos 9 = 12.0 in. 12.2241 in. sin 9- 2.33 in. 12.2241 in. *cb = FCB cos 9 -FCB \$m9l 1251b 12.2241 in. [(12.0 in.)i- (2.33 in.) j] M.A = VB/A X ^CB XBIA = (15.3in.)i-- (12.0 in. + 2.33 in.) j M = (15.3 in.) i--(14.33 in.) j ru-nimn* 1251b «5»,a> im. Z.& >N. 2.33 ttvi. —(12.01 — 2.331) 12.2241 in. (1393.87 lb in.)k (116.156 lb -ft)k or M„ = 116.2 lb- ft *)< PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 164
14. 14. 20.5 in. h — - 4.38 in i *Ti 7.62 1 t§| 1.7.2 hi. PROBLEM 3.13 The tailgate of a car is supported by the hydraulic lift EC. 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 Then dCB := 7(17.2 in.) 2 = 18.8123 in. + (7.62 in.) 2 cos - 17.2 in. 18.8123 in. sin 6 - 7.62 in. 18.8123 in. Z.O.'S «w. and Now where Then fcd = (Ft:v* cos 0) - (FCB sin 6>)j = S(,7' 2i"-)i + (7' 62i^ r^ = (20.5 in.)i- (4.38 in.)j M, = [(20.5 in.)i - (4.38 in.)j] x t . 1251b (1 7.21 - 7.62j) 18.8 123 in. n.z .«»». 4.?.e. >Ni. XKvZ >KJ. (1538.53 lb -in.)k (128.2 lb -ft)k or M^ =128.2 lb- ft ^H IROlRlhTARl MAlhRlAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. M> ,«,#/ «///,& MW,»«> 6* rfwpfe^reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used bevond the. limited d.stnbuUoiMo teachers and educatorspermitted by McGraw-Hillfor their individual course, preparation. ^ you are using it without permission. 165
15. 15. 120mm 65 mm 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 Mc =rwc xFj, Noting the direction of the moment of each force component about C is clockwise. Where and Mc =xFBy +yFBx x - 1 20 mm - 65 mm = 55 mm y - 72 mm + 90 mm - 1 62 mm 4 j, F 65 lix Fa 7(65) 2 +(72) 2 72 V(65) 2 + (72) 3 -(485N) = 325N .(485 N)- 360 N iWc = (55 mm)(360 N) + (1 62)(325 N) = 72450 N- mm = 72.450 N-m or Mc =72.5 N-m J) < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 166
16. 16. By definition: Now and PROBLEM 3.15 Form the vector products B x C and B' * C, where B = B and use the results obtained to prove the identity sin a cos/? = -sin (a + P) + -sin (a - p). SOLUTION N(>te: B = £(cos/?i + sin/?j) B' = 5(cos/?i-sin/?j) C - C(cos ai + sin aj) |BxC| = flCsin(a-jff) |B'xC| = 5Csin(flf + y?) B xC = Z?(cos /?i + sin y9j) x C(cos tfi + sin orj) = BC(cos /?sin « - sin /?cos «)k B'x C - /?(cos /?i - sin /?j)x C(cos ai -f sin #j) - £C(cos yffsin ar + sin /?cos ar)k Equating the magnitudes of BxC from Equations (I ) and (3) yields: BCs'm(a -p)~ BC(cos ps'm a - sin pcos a) Similarly, equating the magnitudes of B'xC from Equations (2) and (4) yields; BCsm(a + p) = BC(cos ps'm a + sln pcos a) Adding Equations (5) and (6) gives: sin(a - p) + s'm(a + p) - 2cos /?sin a 0,) (2) (3) (4) (5) (6) or sin «cos /? - -sin(ar + /?) + -sin(flf - /?) ^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. AV, /«», „/-,/,& AAW ,««, he displayed reproduced or distributed m any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual you are using it without permission. 167
17. 17. PROBLEM 3.16 A line passes through the Points (20 m, 16 m) and (-1 m, ~4 m). Determine the perpendicular distance rffrom the tine to the origin O of the system of coordinates. SOLUTION dAB = V[20m- (-1 m)] 2 + [1 6 m - (-4 m)f - 29 m Assume that a force F, or magnitude F(N), acts at Point A and is directed ixomA to B. Then,, Where By definition Where Then ¥~FX m 'All *B~ rA d.41! = - —<21i + 20j) 29 V MQ = r A xF = dF r,=-(lm)i-(4m)j F E> (Zom, K-rti^ M =[-(-1 m)i-(4 m)j]x—-[(21 m)i + (20 m)j] 29 m a -(20)k + (84)k] ~F|k N-m 29 -*x Finally 64 29 F = </(F) . 64 rf- — m 29 </ = 2.21m ^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manna! may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 168
18. 18. PROBLEM 3.17 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (tf)P--7i + 3j-3kandQ = 2i + 2j + 5k,(6)P = 6i-5j-2kandQ = -2i + 5j~k. SOLUTION (a) We have where Then (b) We have where v4 = |PxQf P = -7i + 3j-3k Q = 2i + 2j + 5k PxQ k -3 > J -7 3 2 2 5 = [(15 + 6)i + (-6 + 35)j + (-14-6)k] = (21)1 + (29)j(-20)k ^ = V(20) 2 +(29) 2 +(-20) 2 A = PxQ P = 6i~5j-2k Q = ~2i + 5in.j~lk or 4 = 41.0 <4 Then PxQ i fci 6 -5 -2 5 = [(5 + 1 0)i + (4 + 6)j + (30-1 0)k] «(15)i + (10)j + (20)k A=j(i5)-+(oy + (2oy or 4 = 26.9 ^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior writ/en permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are astudent using this Manual, you are using it without permission, 169
19. 19. PROBLEM 3.18 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) i + 2j - 5k and 4i - 7j - 5k, (b) 3i - 3j + 2k and -2i + 6j - 4k. SOLUTION {a) We have where Then and (b) We have where Then and AxB A B AxB |AxB| :li + 2j-5k = 4i - 7 j - 5k i J * 1 +2 -5 4 -7 -5 (-1 ~ 35)i + (20 + 5)j + (-7 - 8)k 15(31 -1J -Ik) |AxB| X-- X- A B AxB 15V(-3) 2 +H)2 +(-l) 2 -I5>/ri 15(-3f-lJ-lk) or X 15VH AxB (-3i-j~k) 4 |AxB| 3i-3j + 2k :-2i + 6j-4k I J k 3-3 2 -2 6 -4 (12-12)i + (-4 + l2)j + (18-6)k (8j + 12k) |AxB| 4^(2) 2 +(37 = 4^13 4(2j + 3k) 4^13 or X, V^ (2j + 3k) < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 170
20. 20. PROBLEM 3.19 Determine the moment about the origin O of the force F = 4i + 5j - 3 k that acts at a Point A. Assume that the positioji vector ofA is {a) r - 2i - 3j + 4k, (£) r - 2i + 2.5j - 1 .5k, (c) r - 2i + 5j + 6k. SOLUTION (a) M, (h) M, ' J k 2-3 4 4 5 -3 (9-20)i + (16 + 6)j + (I0 + 12)k i j k 2 2.5 -1.5 4 5 -:s Ma = -lli + 22j + 22k ^ (-7.5 + 7.5)i + (-6 + 6)j + (10-1 0)k M, « (<0 M, ' J 2 5 4 5 (-1 5 - 30)i + (24 + 6)j + (10- 20)k M, -45i + 30j-10k A Note: The answer to Part b could have been anticipated since the elements ofthe last, two rows of the determinant are proportional. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual coursepreparation, ifyou are a student using this Manual, you are using it without permission. 171
21. 21. PROBLEM 3.20 Determine the moment about the origin of the force F == -2i + 3j + 5k that acts at a Point A. Assume that the position vector ofA is (a) r == i + j + k,(6)r == 2i + 3j- 5k, (c)r == -4i + 6j + 10k. SOLUTION i j k (a) M(} = 1 1 1 -2 3 5 = (5-3)i + (-2-5)j + (3 + 2)k M = 2i-7j + 5k A i j k (h) M - 2 3 -f -2 3 5 = (15 + 15)i + (10-10)j + (6 + 6)k Mo =30i + I2k A 1 J k (c) Ma - -4 6 10 -2 3 5 = (30 - 30)i + (-20 + 20)j + (-.12 + 1 2)k M = < Note: The answer to Part c could have been anticipated since the elements of the last two rows of the determinant are proportional. PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted.by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 172
22. 22. -VI 200 N r PROBLEM 3.21 A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A. SOLUTION We have where Then M, = %.i x *c rCIA = (0.06 m)i + (0.075 m)j *c = ~(200 N)cos 30°j + (200 N)sin 30°k i j & MA -. = 200 0.06 0.075 -cos 30° sin 30° = 200[(0.075sin 30°)i - (0.06sin 30°)j ~ (0.06 cos 30°)k] or M,, = (7.50 N m)i- (6.00 N-m)j-(l 0.39 N-m)k <4 PROPRIETARY MATERIAL <Q 201 The McGraw-Hill Companies, inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 173
23. 23. '/ PROBLEM 3.22 £ ^-^ Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O . .'vjB 4.25 in of the resultant force exerted on the tree by the cables at B. / ,.-"r ^J"" 0.75 in/ -^1 "I .T SOLUTION We have where and M, 1 ff/O r l3/Q X F# (7m)j Tdff + Tec 'yf/f ~ ,V BA' AB l J!C -(0.75m)i-(7m)J + (6m)k (.75) 2 +(7) 2 +(6) 2 m *-BC*BC (555 N) (4.25m)i-(7m)j + (lm)k ^(4.25) 2 +(7) 2 +(l) 2 m -(660 N) Fi? = [-(45.00 N)i - (420.0 N)j -f- (360.0 N)k] +[(340.0 N)i - (560.0 N)j + (80.00 M)k] = (295.0 N)i - (980.0 N)j + (440.0 N)k M, i J k 7 295 980 440 Nm (3080 N • m)i - (2070 N • m)k or M,} = (3080 N-m)i- (2070 N-m)k < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 174
24. 24. fi in PROBLEM 3.23 The 6-m boom ^5 has a fixed end /*. 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 Then We have where Then ^c-VK>) 2 +(2.4) 2 +(-4) : = 7.6 m V=-?4^(-« + 2.4j-4k) 7.6 ^A ~ r/i//l X TBC l B/A (6 m)i M;4 ^(6m.)ix^4r^-(~6» + 2.4j-4k) 7.6 or M//== (7.89 kN-m)j + (4.74 kN-m)k < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 175
25. 25. 3{ji)i.. C- 48itii :'6-iftl PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 57-lb force directed along BA. Determine the moment about C of that force. SOLUTION We have where and rA/c = (48 in.)i - (6 in.)j + (36 in.)k < BA — xMpM -(5in.)i + (90in.)j-(30in.)k V(5) 2 +(90) 2 + (30) 2 in. = -(31b)i + (541b)j-(181b)k (57 lb) Mc lb -in. i J k 48 6 36 3 54 18 -(1 836 lb • in.)i + (756 lb • in.)j + (2574 lb • in.) or Mc = -(1 53 .0 lb • ft)i + (63.0 lb • ft)j + (2 1 5 lb • ft)k A PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the. prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are astudent using this Manual, you are using it without permission. 176
26. 26. /.) «' 0.6 id 0.6 m "X PROBLEM 3.25 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (b) the cable at C. SOLUTION (a) We have M^riy ,xTM where **/•/,(= (2-3 m)j Tw: = *"DE^f.)E - (0.6m)l + (3.3in)J-(3m)k V(0.6) 2 +(3.3) 2 +(3) 2 m = (1 08 N)i + (594 N)j - (540 N)k i J k M,,= 2.3 108 594 -540 Nm = -(1 242 N - m)i - (248.4 N • m)k or M/i =-(1242N-m)i-(248N-m)k < (b) We have ^A=rG//l xTCG where r( ,,=(2.7m)i + (2.3m)j T — 1 T*CO ~~ *"CG'CG = -(.6m)i + (3.3m)j-(3m)k (810N) V(.6) 2 +(3.3) 2 +(3) 2 m = -(108 N)i + (594 N)j - (540 N)k i J k MA = 2.7 2.3 -108 594 -540 N-m = -(1 242 N • m)i + (1458 N • m)j + (1 852 N • m)k or MA = -(1242 N • m)i + (1 458 N • m)j + (1 852 N • m)k A PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 177
28. 28. PROBLEM 3.27 In Problem 3.22, determine the perpendicular distance from Point O to cable AB. PROBLEM 3.22 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B. SOLUTION We have where Now and IM TBA d d = perpendicular distance from O to line AB. M, r B/0 X *BA rBIO ={lm) BA' AB (0.75m)i-(7m)j + (6m)k *BA ~~ '"BA 1 AB M, and or (555 N) '(0.75) 2 +(7) 2 +(6) 2 m -(45.0 N)i - (420 N)j + (360 N)k i j k 7 N-m -45 -420 360 (2520.0 N • m)i + (315.00 N • m)k |M 1 = V(252o-°) 2 + (31 5.00) ; = 2539.6 N-m 2539.6 N-m = (555 N)rf d = 4.5759 m or d = 4.58 m < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 179
29. 29. PROBLEM 3,28 In Problem 3.22, determine the perpendicular distance from Point O to cable BC. PROBLEM 3.22 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at.#. SOLUTION We have where [M, Tscd d — perpendicular distance from O to line BC. M, l B/0 rB/0 X ^liC 7mj P — 1 T'BC ™ *"BC* HC (4.25m)i-(7m)j + (lm)k (660 N) M, and V(4.25) 2 +(7) 2 +(l) 2 m (340 N)i - (560 N)j + (80 N)k I J k 7 340 -560 80 (560 N • m)i - (2380 N • m)k |M ! = 7(560) 2 +(2380) = 2445.0 N-m 2445.0 N-m = (660 N)rf d = 3.7045 m or d~ 3.70 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 180
30. 30. A."(i in. •IS in. Cj-iii. .90 in,: ()(> in. ~~*S B ~~^~£ PROBLEM 3.29 In Problem 3.24, determine the perpendicular distance from Point D to a line drawn through Points A and B. PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 57-lb force directed along BA. Determine the moment about C of that force. SOLUTION We have where and IMJ^rf d — perpendicular distance from D to line AB. M.l} ^rAlii x¥]iA r,/D =-(6in.)j + (36in.)k ^liA ~ ^BA^BA (-(5in.)i + (90in.)j-(30 in.)k) V(5) 2 +(90) 2 + (30) 2 in. -(31b)i + (541b)j-(181b)k (57 lb) M,;) I J k -6 36 lb -in. -3 54 -18 -(1 836.00 lb • in.)i - (1 08.000 lb • in.)j - (1 8.0000 lb • in.)k IM 836.00) 2 +(108.000) 2 + (18.0000)' = 1839.26 lb -in. 1839.26 lb -in =(57 lb)e/ </ = 32.268 in. or </ = 32.3in. < PROPRIETARY MATERIAL. & 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 181
31. 31. (if) in. B^-< PROBLEM 3.30 In Problem 3.24, determine the perpendicular distance from Point C to a line drawn through Points A and B. PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a small roof, exerts at Point A of the roof a 57-3 b force directed along BA. Determine the moment about C of that force. SOLUTION 4 We have where Mc ~FBA d d - perpendicular distance from C to line AB. 3fc.. . M.c ^rA/c x¥BA ^%f M / >l D ^-. rAIC - (48 in.)i - (6 in,)j + (36 in.)k ^^ //S^fylA ~ ^BA^IIA f* _ (~(5 in.)i + (90 in.)j - (30 in.)k) fe* 7(5) 2 +(90) 2 +(30) 2 in. = ™(3 lb)i + (54 lb)j - (1 8 lb)k i J k Mc = 48 -6 36 -3 54 -18 lb -in. and = -(1 836.001b -i |M n.)i - (756.00 lb • in.)j + (2574.0 lb • in.)k 2 c | = V(l 836.00) 2 + (756.00) 2 + (2574.0) -3250.8 lb -in. 3250.8 lb in. = 57 lb d- 57.032 in. or d = 57.0 in. ^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. ;Vt> />flrf o//Afc Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 182
32. 32. 0.6 m PROBLEM 3.31 In Problem 3.25, determine the perpendicular distance from Point A to portion DE of cable DEF. PROBLEM 3.25 The ramp ABCD is supported by cables at comers C and D, The tension in each of the cables is 810 N. Determine the momentabout A of the force exerted by (a) the cable at D, (b) the cable at C. . SOLUTION We have where IM TDEd and d — perpendicular distance from A to line DE, M r„HIA (2.3 m)j 'DE ~ A"DE*-DE (0.6m)i + (3.3m)j-(3m)k (8I0N) M V(0.6) 2 + (3.3) 2 +(3) 2 m (108 N)i + (594 N)j - (540 N)k i J b 2.3 N-m 108 594 540 ~(1 242.00 N • m)i - (248.00 N • m)k |MJ =- N /(1242.00) 2 +(248.00) : = 1266.52 N-m 1266.52 Nm= (8 10 N)d d = 1 .56360 m O'fcn, or d = 1.564 m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it withoutpermission. 183
33. 33. PROBLEM 3.32 In Problem 3.25, determine the perpendicular distance from Point A to a line drawn through Points C and G. PROBLEM 3.25 The ramp ABCD is supported by cables at corners C and D. The tension in each of the cables is 810 N. Determine the moment about A of the force exerted by (a) the cable at D, (h) the cable at C. 0.6 m SOLUTION We have where d - perpendicular distance from A to line CG. rG/A X IceM IVu/A l CG M (810 N) (2.7m)i + (2.3m)j -(0.6.m)i + (3.3m)j-(3 m)k A/(0.6) 2 +(3.3) 2 + (3) 2 m -(108N)i + (594N)j-(540N)k i J k 2.7 2.3 N-m -108 594 -540 -(1242.00 N • m)i + (1458.00 N • m)j +(1 852.00 N m)k and = 2664.3 N-m 2664.3 N-m = (8 ION)d d = 3.2893 m or d = 3.29 m 4 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-HillJbr their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 184
34. 34. PROBLEM 3.33 .In Problem 3.26, determine the perpendicular distance from Point C to portion AD of the line ABAD. PROBLEM 3.26 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force R4 exerted on the davit at A. SOLUTION First compute the moment about C of the force FDA exerted by the line on D: From Problem 3.26: = -(48 lb)i + (62 lb)j + (24 lb)k Mc =rm; xF^ = +(6 ft)i x[-(48 lb)i + (62 lb)j + (24 lb)k] = -(1441b-ft)j + (372 1b-ft)k Mc =V044> 2 +(372) 2 = 398.90 lb • ft Then Mc = VMd Since F,w = 82 1b 398.90 lb -ft 82 1b d = 4.86 ft < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, yon are using it without permission. 185
35. 35. *-* PROBLEM 3.34 Determine the value of a that minimizes the i 16 ft 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 MrA/c xF^F(d) Where d - perpendicular distance from C to line AB %ABF (24ft)i + (24ft)j-(28)k F = W 7(24) 2 +(24) 2 +(18) 2 ft •F (6)i + (6)j-(7)k AK; = (3 ft)i - (1 ft)j - {a - 1 ft)k i J k 3 -10 10a 6 6-7 Mc .[(10 + 6fl)i + (81-6.ai)j + 78k] 11 Since 12! ^/c xr or r^xF2 | = W (1 + 6a) 2 +(81- 6a) 1 + (78) 2 = d* d / j2 Setting -j-{d )- to find a to minimize c/ 1 [2(6)(l + 6a) + 2(-6)(8 1 - 6a)] = Solving 121 a = 5.92 ft or o - 5.92 ft ^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Afo par/ o//Aw Minna/ »m>' fc displayed reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 186
36. 36. PROBLEM 3.35 Given the vectors P =3i - P S, and Q • S. -J + 2k,Q == 4i + 5j-3k, and S = -2i + 3j - k, compute the scalar products P • Q, SOLUTION P-Q = (31-1j + 2k)-(4i-5J-3k) = (3)(4) + (-l)(~5) + (2)(-3) = 1 or P Q = I A P • S = (3i - lj + 2k) • (-2i + 3j - Ik) = (3)(-2) + H)(3) + (2)H) = -U. or P-S = -U < Q-S-(4i-5j~3k)-(-2i + 3j-l.k) = (4K-2) + (5)(3) + (-3X-l) = 10 or Q-S = 10 < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermittedby McGraw-Hillfor their individual coursepreparation, ifyou are a student using this Manual, you are using it without permission. 187
37. 37. PROBLEM 3.36 Form the scalar products B • C and B' • C, where B = B', and use the results obtained to prove the identity cos a cos P ~ —cos (a + /?) +—cos (a - ft). SOLUTION y By definition where (1) B-C = £Ccos(a~/?) B = /?[(cos/?)i + (sin/?)j] C = C[(cos a) + (sin a)j] (B cos /?)(Ccos a) + (B sin /?)(Csin a) - BCcos(a~ (3) or cos /?cos a+ sin /?sin a = cos(a - j3) By definition B' • C - BCcos (or + /?) where B' = [(cos fi)i ~ (sin /?)j] (B cos p) (C cos or) + (~B sin /?)(C sin or) = BCcos (a + /?) or cos /?cos « - sin ft sin a = cos {a + /?) Adding Equations (1) and (2), 2 cos /?cos a = cos (a-fi) + cos (a + /?) or cosacosfi = ~-cos(a + fi) + ~cos(a~ ft) A (2) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. JVo /?«// o/7to Manual may he displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 188
38. 38. PROBLEM 3,37 Section AB of a pipeline lies in the yz plane and forms an angle of 37° with the z axis. Branch lines CD and EF join AB as shown. Determine the angle formed by pipes AB and CD. SOLUTION First note AB = AB{\$m 37°j - cos 37°k) CD = CD(-cos 40° cos 55°j + sin 40°j- cos 40° sin 55°k) fc.b Now or or AB- CD = (AB)(CD) cos AB(sm 37°j - cos 37°k) • CD(-cos 40° cos 55°i + sin 40°j - cos 40°sin 55°k) ~(AB)(CD) cos cose? = (sin 37°)(sin 40°) + (~cos 37°)(--cos 4()°sin 55°) = 0.88799 or (9 = 27.4° A PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies., Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 189
39. 39. PROBLEM 3.38 Section AB of a pipeline lies in the yz plane and forms an angle of 37° with the z axis. Branch lines CD and EF join AB as shown. Determine the angle formed by pipes AB and EF. SOLUTION First note AB = /f£?(sm37 j-cos37 k) __. = £F(cos 32° cos 45°i + sin. 32°j - cos 32° sin 45°k) 1 £ F l/^(€t *W A Now AB-EF--= (AB)(EF)cosO or AB(sm : 17°j - cos 37°k) - £F(cos 32° cos 45°j + sin 32°j - cos 32° sin 45°k) = (AB)(EF)cos0 or cos = (sin 37°)(s.in 32°) + (-cos 37°)(-cos 32° sin 45°) = 0.79782 or (9 = 37.1° ^ PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No parr of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 1 90
40. 40. PROBLEM 3.39 Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC, SOLUTION First note and By definition or or AB - V(-6.5) 2 + (-8) 2 + (2) 2 = 1 0.5 ft AC^yj(0f+(~S)2 + (6) 2 =]0 ft AB = -(6.5 ft)i - (8 ft)j + (2 ft)k 7c = -(8ft)j + (6ft)k AB-AC = (ABXAC)cos0 (-6,51 - 8j + 2k) • (-8j + 6k) = (1 0.5)0 0) cos - (-6.5)(0) + (-8)(-8) + (2)(6) - 1 05 cos 6 cos = 0.72381 or = 43.6° < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it withoutpermission. 191
41. 41. sit PROBLEM 3.40 Consider the volleyball net shown. Determine the angle formed by guy wires AC and AD. SOLUTION First note and By definition or or ^C = V(0) 2 +(~8) 2 +(6) 2 -10 ft AD = J(4) 2 +(-&) 2 + (l? = 9 ft IC = ~(8ft)j + (6ft)k 75 = (4ft)j-(8ft)j + (lft)k AC AD = (AC)(AD)cos0 (-8j + 6k) • (4i - 8j + k) = (1 0)(9) cos t (0)(4) + (-8)(-8) + (6)(1) = 90cos 6 cos0 = 0.77778 or # = 38.9° < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may he displayed reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it withoutpermission. 192
42. 42. 1.2 in 2.-1 PROBLEM 3.41 Knowing that the tension in cable AC is 1260 N, determine (a) the angle between cable AC and the boom AB, (b) the projection on AB ofthe force exerted by cable AC at Points. -f 2.6 in 2.4 m SOLUTION (a) First note and By definition or or or (b) We have AC~-=^j(~2A) 2 + (0.8) 2 +(1.2) 2 = 2.8m AB = yj(-2 A) 2 +(-].&? +(Q) 2 = 3.0m ^C = -(2.4 m)i + (0.8 m)j + (1 .2 m)k I/? = -(2.4m)i-(K8m)j AC -AB = (AC){AB) cos (-2.41 + 0.8j + 1 .2k) • (-2.4i - 1 .8j) = (2.8)(30) x cos (-24X-2.4) + (0.8X-1 .8) + (1 .2X0) - 8.4cos cos# = 0.51429 = 7^c cos<9 = (1260N)(0.51429) or = 59.O C or C^cU=648N PROPRIETARY MATERIAL. © 20 JO The McGraw-Hill Companies, Inc. All rights reserved. ;Vo />«/•/ o//Afe Muiua/ /w/y to? displayed, reproduced or distributed in any form or by any weans, without (he prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 193
44. 44. PROBLEM 3.43 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Knowing that the distance from O to P is 6 in. and that the tension in the cord is 3 lb, determine (a) the angle between the elastic cord and the rod OA, (b) the projection on OA of the force exerted by cord PC at Point P. SOLUTION First note Then ^= >/(12) 2 +(12) 2 +(-6) 2 =18in. » OA 1 , = i(2i + 2j~k) Now OP = 6 in, => OP = -(04) The coordinates ofPoint P are (4 in., 4 in., -2 in.) PC = (5 in.)i + (1 .1 in..)j + (14 in.)k PC = 7(5) 2 + (1 1) 2 + (14) 2 = V342 in. ~PCXOA ^(PC)cosO so that and (a) We have or or (5i + llj + 14k)--(2i + 2(i-k) = 7342cos^ cos ] 3V342 0.32444 [(5)(2) + (ll)(2) + (14)(~l)j (b) We have = {Tpc'K]>c )-XOA PC ft pc OA — Tpr COS & = (3 1b)(0.32444) or = 71.1° ^ or (Tpc)oa =0-973 lb ^ =^^S~S5£=SS£=f£S195
45. 45. PROBLEM 3.44 Slider P can move along rod OA. An elastic cord PC is attached to the slider and to the vertical member BC. Determine the distance from O to P for which cord PC and rod OA are perpendicular. SOLUTION First note Then (M^/(12) 2 +(12) 2 +(-6)- =18 in. OA 1 XCH ~ —(12i + 12j-6k)OA OA 18 (2i + 2j-k) Let the coordinates of Point P be (x in., j> in., z in.). Then PC = [(9 - x)in.]i + (15- y)in.]j + [(12- z)in.]k Also, and OP = rfo^o, = -^(21 + 2J-k) OP ~ (x in.)i + (>' in.)j + (z in.)k 2 2 dOP The requirement that CM and PC" be perpendicular implies that ^PC-0 or -(2j + 2j-k)-[(9-.v)i + (15-y)j + (12-2)k] = or (2)|9--^>| + (2) 'l5-|rfw | + H) 12 -rf0P or Jnp = 12.00 in. ^W PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Alt rights reserved. ,Vo par/ <?///"* Mihha/ may fee displayed, reproduced or distributed in anvform or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using tins Manual, yon are using it without permission. 196
46. 46. PROBLEM 3.45 Determine the volume of the parallelepiped of Fig. 3.25 when (a) P = 4i - 3j + 2k, Q = -2i - 5j + k, and S = 7i + j - k, (/;) P = 5i - j + 6k, Q = 2i + 3j + k5 and S = -3i - 2j + 4k. SOLUTION Volume of a parallelepiped is found using the mixed triple product. O) VoI = P-(QxS) 4-3 2 -2 -5 1 in. 3 7 1 -1 (20-21.-4 + 70 + 6-4) 67 or Volume =67.0 4 (b) Vol=P-(QxS) 5 -1 6 2 3 I in. 3 -3 -2 4 (60 + 3-24 + 54 + 8+10) 111 or Volunie = 111.0 A PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, inc. AH rights reserved. No part of this Manual may be displayed, reproduced or distributed in any Jorm or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual you are using it without permission. 197
47. 47. PROBLEM 3.46 Given the vectors P - 4i - 2j + 3k,Q = 2i + 4j - 5k, and S = SJ - j + 2k, determine the value of Sx for which the three vectors are coplanar. SOLUTION If P, Q, and S are coplanar, then P must be perpendicular to (Qx S). P-(QxS) = () (or, the volume of a parallelepiped defined by P, Q, and S is zero). 4-2 3 Then 2 4 -5-0 Sx -1 2 or 32 + lO.S: -6-20 + 8-125;. =0 S =7 < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or bv any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 198
48. 48. 0.1 1 in... PROBLEM 3.47 The 0.61x].00-m lid ABCD of a storage 'bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 66 N, determine the moment about each of the coordinate axes of the force exerted by the cord at D. SOLUTION First note Then and Now where Then z = Vi°-61) 2 -(0.Jl) 2 0.60 m 0.11m dm = V(0.3) 2 +(0.6) 2 +(-0.6r = 0.9 m 1 l)E 66 N 0.9 (0.3i + 0.6j-0.6k) = 22[(lN)i + (2N)j-(2N)k] MA = rD/A xTDf: r/)//( = (0.ll m)j + (0.60m)k i J k M/( = 220 0.11 0.60 1 2 -2 = 22[(-0.22 ~ 1 .20)i + 0.60j - 0. 1 1 k ] = - (3 1 .24 N • m)i + (1 3.20 N • m)j - (2.42 N • m)k -3 1 .2 N • m, Mv = 1 3.20 N • m, A/2 = -2.42 N - m A PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. JVb /*»•/ o/ 1 //™ Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are. a student using this Manual, you are using it without permission. 199
49. 49. PROBLEM 3.48 The 0.61xl.00-m lid ABCD of a storage bin is hinged along side AB and is held open by looping cord DEC over a frictionless hook at E. If the tension in the cord is 66 N, determine the moment about each of the coordinate axes of the force exerted by the cord at C. SOLUTION First note Then and Now where Then x6y-(o.uy 0.60 m = l.lm l CE 66 N 1.1 (-0.7i + 0.6j-0.6k) = 6[-(7N)i + (6N)j-(6N)k] MA ^vm xE r£M = (0.3m)I + (0.71m)J i J k MA -6 0.3 0.71 -7 6 -6 = 6[-4.26i + 1 .8j + (1 .8 + 4.97)k] = - (25.56 N • m)i + (1 0.80 N • m)j + (40.62 N • m)k Mx - -25.6 N • m, My = 1 0.80 N • m, Afz = 40.6 N • m -4 PROPRIETARY MATERIAL. © 201.0 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 200
50. 50. PROBLEM 3,49 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the moments about the y and the z axes of the force exerted at B by portion AB of the rope are, respectively, 120 N • m and -460 N • m, determine the distance a. SOLUTION First note Now where Then Thus ^= (2.2m)i-(3.2 m)j-(am)k ™*D ~ VAID X '«/j 'AID (2.2m}i + (1.6m)j T T,BA I! A dISA M, T,BA dBA (2.2i-3.2j-ak)(N) i J k 2.2 1.6 2.2 -3.2 -a T,BA dt . {- 1 .6a + 22a + [(2.2)(~3 .2) - (1 .6)(2.2)]k} Mv =2.2-^-a BA Then forming the ratio M, it M. -I0.56- 7*4 d (N • m) (N • m) HA 120 N-m 2-2 7Z"(N-m)dB,t -460 N-m -10.56-^- (N-m) or a = 1 .252 m ^j PROPRI&IARl MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of ,his Manual may be displayed reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual you are using it without permission. 201
51. 51. PROBLEM 3.50 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the man applies a 195-N force to end A of the rope and that the moment of that force about they axis is 132 N • m, determine the distance a. SOLUTION First note and Now where Then Substituting for My and dtHA dBA - J(22f+(-3.2) 2 + (-af = Vl5.08 + « 2 m 195 N Ts ,=^^-! -(2.2i-3.2j-ok) My ~ -(rA/D xTw) Vf/0 M, (2.2m)i + (1.6m)j 195 195 d 1 2.2 1.6 2.2 -3.2 -a (2.2a) (N • m) HA 132 N-m 195 JJm+a* (2.2a) or 0.30769^1 5.08 + a* - a Squaring both sides of the equation 0.094675(15.08 + a 2 ) = a 2 or a = 1.256 m A PROPRIETARY MATERIAL. €3 2010 The McGraw-Hill Companies, Inc. Alt rights reserved. No pari of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 202
53. 53. PROBLEM 3.52 For the davit of Problem 3.51, determine the largest allowable distance x when the tension in line ABAD is 60 lb. SOLUTION From the solution of Problem 3.51, TAD is now l AI) AD AD 60 lb yjX 2 +(-lJS) 2 +(-lf (xi-7.75j-3k) Then M = k • (rA/c xT,(0 ) becomes 279 279. 60 V* 2 +(-7.75) 2 +(-3)' 60 1 7.75 3 jc -7.75 -3 six 2 +69.0625 (D(7.75)(x) 279>/x 2 + 69^0625 = 465.x 0.6Vjc 2 + 69.0625 =x Squaring both sides: 0.36x 2 + 24.8625 = r x 1 =38.848 x = 6.23 ft 4 PROPRIETARY MATERIAL. CO 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 204
54. 54. PROBLEM 3,53 To loosen a frozen valve, a force F of magnitude 70 lb is applied to the handle of the valve. Knowing that 6 = 25°, Mx - -6.1 lb • ft, and Mz ~ -43 lb • ft, determine (j) and d. SOLUTION We have where Pol- and From Equation (3) From Equation (1) rivi ( VAIO ' F F ML: M. My M. ^ = cos -(4in.)i + (llin.)j-(rf)k F(cos #eos fi - sin \$ + cos #sin <J)k) 70 1b, 9 = 25° - (70 lb)[(0.9063 1 cos 0)i - 0.42262j + (0.9063 1 sin 0)k] i j k (701b) -4 11 -J in. -0.90631 cos -0.42262 0.90631 sin (70 lb)[(9.9694sin - 0.42262^)1 + (-0.9063 Wcos + 3.6252sin <p) j + (1.69048 -9.9694 cos 0)k] in. (70 lb)(9.9694sin - 0.42262d)m. = -(61 lb • ft)(I2 in./ft) (1) (70 lb)(-0.9063 Irfcos (j) + 3.6252 sin <f>) in. (2) (70 lb)(l .69048 - 9.9694cos ^ in. - -43 lb • ft(l 2 in./ft) (3) U 634.33 697.86 24.636' d 1022.90 29.583 = 34.577 in. or (p = 24.6° < or c/ = 34.6in. ^ PROPRIETARY MATERIAL. €5 2010 The McGraw-Hill Companies, inc. All righis reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are astudent using this Manual, you are using it without permission. 205
55. 55. i I in. PROBLEM 3.54 When a force F is applied to the handle of the valve shown, its moments about the x and z axes are, respectively, Mx ~ -11 lb • ft and M.z - -81 lb • ft. For d~21 in., determine the moment My of F about the y axis. SOLUTION We have Where and ZM : rw xF = M( r^=-(4in.)i + (lim.)j-(27in.)k F = F(cos 0cos <p - sin. \$ + cos sin flk) i J k -4 1 .1 -27 cos 0cos (p -sin 6 cos #sin F[(11 cos sin 0-27 sin 0)i + (-27 cos Bcos <p+ 4 cos #sin 0)j + (4sin - 1 1 cos 6* cos ^)k](lb • in.) Ma =F lb in. Mx = F(l. 1 cos (9 sin 0-27 sin 0)(lb • in.) MJ? = F(-21 cos 0cos0 + 4cos 0sin 0) (lb • in.) Mz = F(4 sin <9 - 1 .1 cos cos (j)) (lb • in.) Now, Equation (1) cos #sin <f>- 11 M F ^ + 27sin# cos 0cos 0- — 4sin — — - n{ F and Equation (3) Substituting Equations (4) and (5) into Equation (2), 0) (2) (3) (4) (5) M, =/N-27 4sin0- F + 4 1 ( M 111 F *- + 27sin0 or M. 11 (27MZ +4.MV ) PROPRIETARY MATERIAL. & 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 206
56. 56. PROBLEM 3.54 (Continued) Noting that the ratios yp and ± are the ratios of lengths, have Mv =—(-81 lb -ft) +—(-77 lb -ft) - 11 11 = 226.82 lb • ft or Mv -= -227 lb - ft A PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved No part of ihis Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are. using it without permission. 207
57. 57. 0.35 in 0.75 m 0.75 in * PROBLEM 3.55 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 ofthe cable. SOLUTION MAD ~ "'AD ' VBIA X *Bll) Where X „,D ---(4i-3k) 'a/a (0.5 m)i and Then dm - vv0.375) 2 +(0.75) 2 +(-0.75) 2 = 1.125 m T.SH 450 N —(0.375i + 0.751 - 0,75k) 1.1.25 J (1 50 N)i + (300 N)j - (300 N)k Finally Hu> = 4 -3 | 0.5 ! 150 300 -300; [(-3X0.5X300)] or MAD 90.0 N-m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Aif rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 208
58. 58. 0.35 m 0.75 m PROBLEM 3.56 In Problem 3.55, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable. SOLUTION Where and Then Finally mad=^ad-(*biaXTbg) M/) =-(4i-3k) r»MA (0.5 m)j BG - V(-°-5) + (0.925) 2 + (-0.4) 2 = 1.125 ra 450 N TliC! =- -(-0.5i + 0.925j-0.4k) 1.125 J ' = -(200 N)i + (370 N)j - (1 60 N)k -3 u »'i 4 0.5 -200 370 -160 [(-3X0.5X370)] MAD -Ul.ON-m PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of (his Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 209
59. 59. 0.7 m .O.G.Ti] PROBLEM 3.57 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B. SOLUTION First note Then Also Then Now where Then 4E = V(°-9) + (~0-6) + (0-2) 2 =1.1 m «=-yp(0.9i-0.6J + 0.2k) = 5[(9N)i-(6N)j + (2N)k] DB - y[( 2f + (-0.35) 2 + (0) 2 X 1.25m DB OB DB 1.25 1 (1.2I-0.35J) 25 (241 - 7j) MDB ~ '"DB"*DB 'VAID X T Ui) TO/} =-(0.1m)j + (0.2m)k ! M»-^(5) 24 -7 -0.1 0.2 9 -6 2 H1.8- 12.6 + 28.8) or MDfl =2.28N-m ^ PROPRIETARY MATERIAL. £> 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of /his Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 210
60. 60. PROBLEM 3.58 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B. SOLUTION First note dCF -- 0.2) 2 = 1.1 m= V(0.6) 2 +(-0.9) 2 + (- Then :f '- 33 N-_ (0.6i-0.9j + 0.2k) Also DB : = 3[(6N)i~(9N)j-(2N)k] = V0-2) 2 +(-0.35) 2 +(0) 2 ~ 1 .25 m Then ^DB ~~ _ DB ' DB =—(I.2i- 0.35j) 1.25 J = _L(24i-7j) 25 V J; Now Mm -A'OB '(rC/D X *cp) where yan :-(0.2m)j-(0.4m)k 24 -7 Then Mm = -(3) 25 0.2 -0.4 6 -9 -2 = —(-9.6 + 16.8-86.4) 25 or Mm ^~9.50N-m < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any farm or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 211
61. 61. l> PROBLEM 3.59 A regular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA. >- SOLUTION We have where From triangle OBC Since or Then and MoA=*OA<*aoX*) {OA)x 2 f i (OA)z =(OA)x tan W n/3, 2>/3vv-1 / (OA) 2 =(OA)l +(OA)l +{OAz f 2 [ a 2 + {OA)l + ( Ya a , 12 . a 4,0 2 h 2>/3 J. 2. 1 , x- = l ,+ Vl J+ ivr k p = = (a sin30°)i-(a coS 30°)k (/)) = P (| _ ^« 2 rc/0 =oi M(A! 2 V3 I 1 2V3 Kf) ] -73" 2 ' T v - 3 y 0X->/3) = aP Moa = aP_ 72" PROPRIETARY MATERIAL © 20 JO The McGraw-Hill Companies, Inc. All rights reserved. JVo /wirt o///h,s Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 212
62. 62. PROBLEM 3.60 A regular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other, (b) Use this property and the result obtained in Problem 3.59 to determine the perpendicular distance between edges OA and BC. SOLUTION (a) For edge OA to be perpendicular to edge BC OA-BC = Q where From triangle OBC (OA)x (OA)z =(OA)x tm30° = - 04 = I-|I + (CM)J + ( i > V3 J 2^3 k a 2Sj and BC = (asm 30°) i - (a cos 30°) k Then or so that a . 2 2 i+(.oa) v i+ 2 c a ^+ (O^)v (0)-~ = 4 } 4 OA-BC^0 (i-V3k)~ = <9/i is perpendicular to #C. (6) Have A^fW = Pt/, with P acting along BC and d the perpendicular distance from OA to 5C. From the results of Problem 3.57 ft? M0A Pa 4i 4i ~Pd or d = —T^ A 72 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. A? o /«//•/ o/'/Aw Mmmi/ u«iy be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill/or their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 213
63. 63. y PROBLEM 3.61 y^ji 45 in. A sign erected on uneven ground is 4 V guyed by cables EF and EG. If the 11 _^#g& force exerted by cable £7** at E is 46 lb, si /,: .^-^.,-f ?;"" .--.. determine the moment of that force 96 ii r. 8»v jm^M^^^W% about the line join ing Poi nts A and D. ;|*: = i| r . 1 fPI Ml '/ L#j >' M^: ^^i^Ibr>>V - 47 in . l:-;iy';!t'^ " ':>-^ : ': : : '" ' ' <*><. - .--•'- • . •• -'• -/••- ,"- ~-J " "^v "^ .--^ &%8 ift^o^ i 1 ^Xl7in. > • v-ii:—;T*V ->''' T. • -'*'"- I'-i i"»? »*" ^' fT /^ "'' i "x*'' ' SOLUTION First note that EC ~ -V(48) 2 + (36) 2 - 60 in. and that J§ = -g = |. The coordinates of Point E are then(fx48,96, J> 36) or (36 in., 96 in,, 27 in,). Then ^,=>/H5) 2 + H10) 2 +(30) 2 = 115 in. Then T^==^(-l5I-110J + 30k) Also = 2[-(3 lb)i - (22 lb)j + (6 lb)k] AD = 7C48) 2 + (-12) 2 + (36) 2 = 3.2726 in. Then _ = —!—(48i-12j + 36k) 12^26 =ir (4M+3k) Now MAD =-Kiiy(-rEI,4 XTEF) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 214
64. 64. PROBLEM 3.61 (Continued) where Then M rm = (36 in.)i + (96 in.)j + (27 in.)k 1 2 (2) 4 -I 3 36 96 27 -3 -22 6 (2304 + 81-2376 + 864 + 216 + 2376) or M^, =1359 lb -in. <« PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, inc. All rights reserved. JVo /*»/ o//A& M»ma/ may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 215
65. 65. >J PROBLEM 3.62 % in. -1 > in. A sign erected on uneven ground is guyed by cables EF and EG. If the ^0^ force exerted by cable EG at E is 54 lb, ,, ,,^'y^^ determine the moment of that force Is. ^M^M^M:-}-^:-^ about the line joining Points A and D. "&ftK?'?^^T?^i*K; : V^lS ]ff ;1 - : i' p": >^' * *- "' " ^^^x in. XJ7in- ? |V; 36iri7 ^ |12 1 ®^, SOLUTION First note that BC = - 60 in. and that ~~ ~— ~. The coordinates of Point E are BC 60 4=V(48> 2 + (36) 2 = then (Jx 48, 96, f <36) or (36 in , 96 in., 27 in.). Then 4*;=Vai) 2 +(-88) 2 +(-44) 2 = 99 in. Then T*;= ^(lli-88j -44k) Also = 6[(llb)i-(8 1b)j-(4 1.b)k] ^/> - s/(48) 2 + (~1 2) 2 + (36) 2 = 12>/26 in. Then AD AD = —!=(48i~I2j + 36k) 12^26 J = ~=-(4i-j + 3k) V26 Now MAD ^kAD <rm xTEG ) PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course, preparation. Ifyou are a student using this Manual, you are using it without permission. 21fi
66. 66. where Then PROBLEM 3.62 (Continued) rBA = (36 in.)i + (96 in.)j + (27 in.)k (6) (-1536 -27 »864 -.288 -144 + 864) MAD 26 _6_ V26 4 -I 3 36 96 27 I -8 -4 or M,n =~23501b-in. < PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. Ati rights reserved, No part of this Manna! may be displayed, reproduced or distributed hi any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 217
67. 67. PROBLEM 3.63 Two forces F| and F, in space have the same magnitude F, Prove that the moment of F, about the line of action of F2 is equal to the moment of F2 about the line of action of F, , SOLUTION First note that ¥x ~fAy and F2 ~F2 a 2 Let Mj = moment of F2 about the line of action of M, and M2 = moment of F, about the line of action of M, Now, by definition Since Using Equation (3.39) so that itfi=4-(rBM xF2 ) = A-(i :wa x A2 )f2 Fi=F2 =F and xm Mi=^r(»*^xi2 )F M2 =X2 -{-*BIA xAx )F h i?BIA X X2 ) = ^2 HfiM X ^1 ) M2 = A} -(rm xZ2 )F -r»BM M[2 =M2i A PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Afo /««/ «/"//»'i- M»h/«/ »m>' to? displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 218
70. 70. PROBLEM 3.66 In Problem 3.57, determine the perpendicular distance between cable AE and the line joining Points D and B. PROBLEM 3,57 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B, SOLUTION From the solution to Problem 3.57 T„? =55N T^=5[(9N)i-(6N)j + (2N)k] |MD/i | = 2.28N-m X/«-'—(24i-7j) 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 (*AE )parallel ~ ^AE * ^ Dli 5(9i-6j + 2k)-—(24i~7j) [(9)(24) + (-6)(-7)] Also so that 5 = 51. 6N *AE ~ ( *Ae)parallel + ( *AEJperpendicular (''./: ),,, neodfcul* - J(5S) 2 +(51.6) 2 - 1 9.0379 N' perpendicuiar Since A, ra and (T (/i )pcrpendicular are perpendicular, it follows that MDB ^"(T^Operpendieiilar or 2.28 N-m = </(!. 9.0379 N) d~ 0.1 19761 J = 0.1198 m ^ PROPRIETARY MATERIAL. C) 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. Ill
71. 71. PROBLEM 3.67 In Problem 3.58, determine the perpendicular distance between cable CF and the line joining Points D and B. PROBLEM 3.58 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position, shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B. :0.3iJi SOLUTION From the solution to Problem 3.58 Tcr = 33N Tc/ ,=3[(6N)i-(9N)j-(2N)k] jA*D«| = 9.50N-m Xm = 25 (241 ~ 7,) Based on the discussion of Section 3.11, it follows that only the perpendicular component of TCp will contribute to the moment of Tcr about line DB, Now ( i-CF )parallcl ~" *CF " * o« = 3(61-9j-2k)~ (241 - 7j) = ^[(6)(24) + (-9)(-7)] - 24.84 N AlSO Tc y;- — (TcF )jwral le! + ( *-C/' )perpendicular s° that (Ta.Wendicuiar = V(33) 2 - (24.84) 2 = 2.1 .725 N Since XDB and (TCF )pcrpendicillar are perpendicular, it follows that MDB I "V'CY'Vperpcndicperpendicular or 9,50N-m = </x21.725N or d~ 0.437 m A PROPRIETARY MATERIAL. © 20 10 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. Ill
72. 72. % ill PROBLEM 3.68 In Problem 3.61, determine the perpendicular distance between cable EF and the line joining Points A and D. PROBLEM 3.61 A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the moment of that force about the line joining Points A. and D. SOLUTION From the solution to Problem 3.61 7V„=46 1bEl- M X T^-2H3 1b)i-(22 1b)j + (6 1b)k] lb in. (4i-j + 3k) 1Di -1359 1b-in. 1 AD Based on the discussion of Section 3.11, it follows that only the perpendicular component of TEF will contribute to the moment of TEF about line AD. Now (TeF /parallel ~ ^EF ' ^, AD 2(-3i-22j + 6k) 2 (4i-j + 3k) Also so that [(~3)(4) + (-22)H) + (6)(3)j '26 = 10.9825 lb i EF - ( lEF )para||c] + (lEF )perpemiicular (^perpendicular = V(46) 2 - (10.9825) 2 = 44.670 lb Since AD and OW ^^are perpendicular, it follows that or MAD - d{TEl, )pcrpeiidicular 1359 lb • in. = rfx 44.670 lb or t/ = 30.4in. < PROPRIETARY MATERIAL. €) 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 223
73. 73. PROBLEM 3.69 In Problem 3.62, determine the perpendicular distance between cable EG and the line joining Points A and D. PROBLEM 3.62 A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EG at E is 54 lb, determine the moment of that force about, the line joining Points A and D. SOLUTION From the solution to Problem 3.62 TB0 = 54 lb TfiC =6[(llb)i-(81b)i-(41b)k] |M/JD | = 23501b-in. ^=4^(41 -j + 3k) V26 Based on the discussion of Section 3.11, it follows that only the perpendicular component of TgG will contribute to the moment of Teg about line AD. N»w (TEG ) para„e, = TEG XAD - 6(i - 8j - 4k) • -fL(4i - j + 3k) V26 = ~™L[(I)(4) + (-8X-D + (-4X3)] - Thus, (TBG )perpendiciliar = T£G = 54 lb Since %A0 and (T£6 pe[pcildicutar are perpendicular, it follows that 1 ™AD ~ "*EG /perpendicular or 2350 lb • in. = dx 54 lb or rf = 43.5 in. 4 PROPRIETARY MATERIAL €> 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfar their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 224
74. 74. PROBLEM 3.70 Two parallel 60-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about Points. SOLUTION (a) We have where (b) We have (c) We have SM8 : -t/,Cv +rf2 C,,=M </, = (0.360 m)sin 55° = 0.29489 m d2 =(0.360 m)sin 55° = 0.20649 m C, =(60 N)cos 20° - 56.382 N C,, =(60 N) sin 20° = 20.521 N M = -(0.29489 m)(56.382 N)k + (0.20649 m)(20.52 1 N)k --=-(12.3893 N-m)k M = Fd(-k) = 60 N[(0.360 m)sin(55° - 20°)](-k) = -(12.3893 N-m)k £MA : £(r t x F) = rm x FB + rCIA x Fc = M i J k M = (0.520 m)(60 N) cos 55° sin 55° -cos 20° -sin 20° I J k cos 55° sin 55° cos 20° sin 20° (1 7.8956 N • m - 30.285 N • m)k -(12.3892 N-m)k or M = 12.39 N-mJ^ or M = 12.39 N-m +(0.800 m)(60N) or M = 12.39 N-m PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pari of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorspermitted by McGraw-Hill for their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 225
75. 75. 1211./ A"1 ft 21. II > 1 1 ' lb 16 in. PROBLEM 3.71 A plate in the shape of a parallelogram is acted upon by two couples. Determine (a) the moment of the couple formed by the two 21 -lb forces, (/?) the perpendicular distance between the 12-lb forces if the resultant of the two couples is zero, (c) the value of a if the resultant couple is 72 lb- in. clockwise and dfc 42 in. SOLUTION Z»V? (a) We have Mx =d^ c where dx ~ 1 6 in. Fj=211b M, =(16in.)(21 lb) -336 lb -in. (/?) We have M,+M.2 =0 1 2. Ho or M, -336 lb -in. H </, = 28.0 in. <or 336 lb • in. -^(1 2 lb) = (c) We have Mlolll =M1 +M2 or -72 lb -in. = 336 lb -in. - (42 in.)(sin a){ 2 lb) sin a~ 0.80952 and a = 54.049° or a = 54.0° ^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No pari of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 226
76. 76. 100 imn KiOi 140 mm 160 mm 2-10 mil V^ SOLUTION P D 1 n P A (a) We have or (A) We have (c) ,32- .tv, o r%^ We have PROBLEM 3,72 A couple M of magnitude 18 N-m is applied to the handle of a screwdriver to tighten a screw into a block of wood. Determine the magnitudes of the two smallest horizontal forces that are equivalent to M if they are applied (a) at corners A. and D, (b) -at corners B and C, (c) anywhere on the block. M = Pd 18N-m = P(.24m) P = 75.0N dBC ^yj(BE) 2 +{ECf = V(-24m) 2 +(.08m) 2 = 0.25298 m M = Pd 18N-m = P(0.25298m) P-71J52N dAC =yl(ADf+(DCf or Pmin =75.0N < or P = 71.2N < ).24m) 2 + (0.32 m)2 0.4 m M = PdAC 18N-m = />(0.4m) P = 45.0N or P = 45.0N A PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permittedby McGraw-Hillfar their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 227
77. 77. .',:.) I!> 6 in. 25 lii : 25 ih L Sin.- PROBLEM 3.73 Four 1 -in.-diameter pegs are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated, (a) Determine the resultant couple acting on the board. (/;) 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 ofthat minimum, tension? SOLUTION 35"& sty* (a) +) M = (35 lb)(7 in.) + (25 lb)(9 in.) -245 lb -in. + 225 lb -in. M = 470 lb- in. ^)< (/;) With only one string, pegs A and D, or B and C should be used. We have 6 tan 8 = 36.9< 90°- = 53.1° Direction of forces: With pegs /I andD: With pegs/? and C: (c) The distance between, the centers ofthe two pegs is 53.1° -4 53.1° < F A s <J&+6 2 =10 in. Therefore, the perpendicular distance d between the forces is We must have c/ = 10in. + 2|-in. = 1 1 in. M = Fd 4701b-in. = F(llin.) F = 42.7 lb < PROPRIETARY MATERIAL. © 20 10 The McGraw-Hill Companies, Inc. Alt rights reserved. No part of this Manual may be displayed, reproduced or distributed in anv form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 228
78. 78. 25 lb PROBLEM 3.74 Four pegs of the same diameter are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. Determine the diameter of the pegs knowing that the resultant couple applied to the board is 485 lb in. counterclockwise. SOLUTION M ^ dADpAD+dBCFBC 485 lb • in. = [(6 + rf)in.](35 lb) + [(8 + rf)in.j(25 lb) rf = 1.250 in. < PROPRIETARY MAIERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of Urn Manual mav be displaved reproduced or distributed in any jorm or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifwit are a student using this Manual you are using if withoutpermission. 229
79. 79. SllH'i PROBLEM 3.75 The shafts of an angle drive are acted upon by the two couples shown. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis. filb.fl SOLUTION Based on where I2M M == M,+M2 M,== -(81b-ft)j M2 == -(61b-ft)k M == -(8tb-ft)j-(61b-ft)k |M| == >/(8) 2 +(6) 2 =10 lb -ft 3l = M "|m:| -(8 lb • ft)j - (6 lb • fl)k or M = 10.00 lb- ft < or M 10ib-ft -0.8j-0.6k jMjl = (101b-ft)(~0.8j-0.6k) cos0v =O #v =90° cos<9v =-0.8 6> ( . =143.130° cos0z --O,6 Z =126.870° or 9X - 90.0° 6 = 143. 1 ° Oz = 26.9° < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All lights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 230
80. 80. 170 mm 160 mm 18 N 150 mm .1.50 mm !8N 3-3 N PROBLEM 3.76 If P = 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION We have where Also, M = M, + M, M, l GIC rt?/c x *i -(0.3 m)i (18N)k M,=-(0.3m)ix(18N)k = (5.4N-m)J M2 =rD/r xF2 'D/F -(.15m)i + (.08m)j (.15m)i + (.08m)j + (.17m)k M (.15) 2 +(.08) 2 +(.1.7) 2 m = 141.421 N-m(.15i + .08j + .l 7k) i J k = 141,421 N-m -.15 .08 -.15 .08 .17 - 141 .42 1(.01 361 + 0.0255j)N • m (34 N) PROPRIETARY MATERIAL © 2010 The McGraw-Hill Companies, Inc. Aft rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are astudent using this Manual, you are using it withoutpermission. 231
81. 81. PROBLEM 3.76 (Continued) and M = [(5.4 N -m)j] + [14 1.42.1 (.01 36i + . 0255j) N • m] = (1 .92333 N • m)i + (9.0062 N • m)j |m:|-^(m,) 2 +(m;v ) 2 = V(l-92333) 2 +(9.0062) 2 = 9.2093 N • m or M = 9,2 1 N • m < M = (1 .92333 N • m)i + (9.0062 N • m)j ~|Mj~ 9.2093 N-m = 0.20885 + 0.97795 cos 6> v =0.20885 0,= 77.945° or 9^11.9° < cos y =0.97795 0, =12.054° or 0, =12.05°* cos Z =0.0 =90° or 6' =90.0° < PROPRIETARY MATERIAL. €> 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 232
82. 82. PROBLEM 3.77 If P~0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION dOE E, M = M, + M2 ; /'J = 16 lb, F2 = 40 lb M, = rc x F, - (30 in.)* x H)6 lb)jj = -(480 lb • in.)k M2 = rm x F2 ; rm = (1 5 in.)i - (5 in.)j >/(0) 2 + (5) 2 +(10) 2 = 5^ in. = 8>/5[(llb)j-(2ib)k] i j k M2 -8n/5 15 ~5 1 -2 - 8>/5[(10 lb • in.)i + (30 lb • in.)j + (1 5 lb • in.)k] -(480 lb in.)k + 8V5[(1 lb • in.)i + (30 lb • in.)j + (1 5 lb • in.)k] (178.885 lb • in.)i + (536.66 lb • in.)j - (21 1 .67 lb • in.)k ^/(178.885) 2 +(536.66) 2 +(-21 1.67) 2 M M K 603.99 lb in M A/ = 604 lb -in. < cos0v COS COS 0.. M 0.29617 0.88852 -0.35045 0.29617! + 0.88852j - 0.35045k 72.8° 0=27.3° 0,,=110.5° A PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, inc. All rights reserved. No part of this Manual may he displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are astudent using this Manual, you are using it without permission. 233
83. 83. PROBLEM 3.78 If P = 20 lb, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION From the solution to Problem. 3.77 M, = -(480 lb • in.)k M2 = 8>/5 [(10 lb • in.)i + (30 lb • in.)j + (1 5 lb • in.)k] 16 lb force: 40 lb force: F - 20 lb M3 = rc x P = (30in.)ix(20lb)k = (600 lb - in.)j M-M, +M2 +M3 = ~(480)k + 8V5 (1 Oi + 30j + 1 5k) + 600j -(178.885 lb -in)i + (1136.66 lb -in.)j -(211.67 lb-in.)k M = ./{178.885) 2 + (1 1 3.66) 2 +(211 .67) 2 = 1169.96 lb in M = 1170 lb- in. < X„ M M 0. 1 52898i + 0.97 1 54j - 0. 1 8092 1 k cos ex = 0.152898 cos ^ = 0.97 154 cos 0.= -0.1 8092.1 9. =81.2° 0=13.70° 6'=100.4° < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 234
84. 84. 160 nun ,- 18 N PROBLEM 3.79 If P = 20 N, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION We have where M-M, + M2 +M3 M^r^xF, i J k 0.3 18 N-m = (5.4N-m)j M3 =r /fXF2 i J k ,15 .08 141.421 N-m .15 .08 .17 141.421(.0136i + .0255j)N-m (See Solution to Problem 3.76.) M3= r(X4 xF3= 0.3 0.17 N-m 20 = -(3.4N-m)i + (6N-m)k M=[(1.92333-3.4)i + (5.4 + 3.6062)j + (6)k]N.m = -(1 .47667 N • m)i + (9.0062 N • m)j + (6 N in) llVff— 1**2 , nj2 , xj2 M; + ML V + m; V(l .47667) + (9.0062) + (6) 2 10.9221 N-m or M = 10.92N-m A PROPRIETARY MATERIAL. CO 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 235
85. 85. PROBLEM 3.79 (Continued) X M -1.47667 + 9.0062 + 6 jM| 10.9221 = -0. 1 35200i + 0.82459J + 0.54934k cos(?T = -0.135200 6X =97.770 or <9 V -97.8° < cos V = 0.82459 #,,=34.453 or y = 34.5° ^ cos#. = 0.54934 Z =56.678 or <9 2 = 56.7° ^ PROPRIETARY MATERIAL. © 20 JO The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 236
86. 86. z A } 1600 N-m ] 200 N-iii 1 1.20 N«m PROBLEM 3.80 Shafts A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and B lie in the vertical yz plane, while shaft C is directed along the x axis. Replace the couples applied to the shafts with a single equivalent couple, specifying its magnitude and the direction of its axis. SOLUTION Represent the given couples by the following couple vectors: M^ = -1 600sin 20°j + 1 600cos20°k M: Mc -(547.232 N m)j + (1 503.5 1 N • m)k 1200sin 20°j + 1200cos 20°k (41 0.424 N • m)j + (11 27.63 N • m)k -(1120N-m)i The single equivalent couple is M^M,, +M5 +MC = -(1 120 N • m)i - (1 36.808 N • m)j + (263 1.1 N- m)k M = yj( 1 20) 2 + (1 36.808) 2 + (263 1 . I) 2 = 2862.8 N-m -1120 cos 6, COS tfy ^ cos 0„ 2862.8 -136.808 2862.8 2631.1 2862.8 M = 2860N-m 0=113.0° B' =92.7* 23.2° < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 237
87. 87. Aim W 20° A*' PROBLEM 3.81 The tension in the cabJe 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 force- couple system (a) at A, (b) at B. SOLUTION {a) Based on ZF: F,,=r = 560lb mK^*^t^^ or ¥A = 560 lb ^ 20° ^ X^^S TscnSo ZMA : M^^sinSO )^,) = (560lb)sin50°(18ft) ^A (b) or Based on = 7721.7 lb -ft M/( =7720.1b-ftjH XF: 7^ = 7* = 560 lb Ffi = 560 lb ^C 20° < ZMB : MB =(Tsm50°)(dB ) I & or «eA = (560 lb) sin 50° (10 ft) 3*6 ^MB c or = 4289.8 lb ft Mg =42901b-ftjH £&<!2t h c*** F* PROPRIETARY MATSRIAL, © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using if without permission. 238
88. 88. PROBLEM 3.82 A 1 60-lb force P is applied at Point A of a structural member. Replace P with (a) an equivalent force-couple system at C, (b) an equivalent system consisting of a vertical force at B and a second force at D. SOLUTION (a) Based on where (b) Based on 2.F: Pc ^P = 60b XMC : Mc —Pxdcy + Py dCx Px =(160 lb) cos 60° = 80 lb /;=(1601b)sin60° = 138.564 lb rf tt =4ft JCl, = 2.75 ft Mc = (80 lb)(2.75 ft) + (1 38.564 lb)(4 ft) = 220 lb -ft + 554.26 lb- ft = 334.26 lb -ft EFV : PDx =Pcos 60° or P, =160 lb ^T.60°^ or Mc =334 lb • ft )< = (160lb)cos60c = 80 lb TM, (Pcos60°)(clDA ) = PB (dDB ) [(160 lb)cos60°](1.5 ft) = PB (fi ft) PB = 20.0 lb or P„=20.01bH PROPRIETARY MATERIAL. €> 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of (his Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the. publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation, ifyou are a student using this Manual, you are using it without permission. 239
89. 89. PROBLEM 3.82 (Continued) ZFV : Psin60o = PB + PI)}, (160 lb) sin 60° = 20.0 lb + PDy PDy =118.564 lb 2 ^=>/('k) 2 +('y = 7(8°) 2 +0 18- 564) 2 = 143.029 lb # = tan _1 (p p J118.564^ - tan ] — — I 80 ) = 55.991° or PD ,= 143.0 lb ^56.0°^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 240
90. 90. -*-r-HS A PROBLEM 3.83 50 nun F,- i «i 100 mm 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 JvJ: A (a) Based on IF: F/? =F = 80N XM: Mn =Fdr 80 N (.05 m) 4.0000 N • m or FB = 80.0 N --•••- ^ or M«-4,00N-m }^ 5 «* ^ 1 (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, XM: MD =Fcd 4.0000 N-m = Fc (. 04m) Fc =100.000 N or Fc =100.0nH 2Fy : = FD -Fc Fft =l 00.000 N or F„ =100.0 nH PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may he displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 241
91. 91. PROBLEM 3.84 ts iSsif^S& &^-fift^i isKSS ......' A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 1 040 N, replace the force exerted by the cable at B with an equivalent system formed by two A B C parallel forces applied at A and C, 6.7 m 4 m 60°/ ......,lA.^ , SOLUTION Require the equivalent forces acting at A and C be parallel and at an angle of awith the vertical. Then for equivalence, XFX : (1 040 N) sin 30° = FA si n a + FB sin a ZFy : -(1 040 N) cos 30° = -FA cos a - FB cos a Dividing liquation (1) by Equation (2), (1040 N) sin 30° __ {FA +FB )sna -(.1040 N) cos 30° ~ -{FA + FB )cosa Simplifying yields a ~ 30° Based on XMC : [(1040 N)cos30°](4 m) = {FA cos30°)(l 0.7 m) FA = 388.79 N (1) (2) or Based on or F,, = 389 N "^ 60° M XM.A : - [(1 040 N) cos 30°](6.7 m) = (Fc cos 30°)(l 0.7 m) Fc =651.21 M FC =651N ^60°^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ali rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without (he prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 242