<ul><li>1.CHAPTER 3 </li></ul><p>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&gt; 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. &gt;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. 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. 200 in.., 25' -100 mm- 200 mm i '&gt; 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&gt;T-**&gt; 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 &lt; PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or 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. 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&gt;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 &lt; tan6^ 0.125 m 0.2 m a = 32.0 MD = C(&gt;C); DC = V( - 2 m) 2 + (- 1 25 m)' = 0.23585 m 41.70 Nm = C(0.23585m) C = 176.8 N^L 5HX)&lt; 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. 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. 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 &lt; PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or 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. 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&gt;) _i 8 in. </p><p> )^ or M7&gt; = 760N-m v )&lt; 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. 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)&gt; 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. 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 &amp; = (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. 5.3 in. PROBLEM 3.12 12.0 in. 2.33 in. mmM1 &gt;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&gt; im. Z.&amp; &gt;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 *)&lt; PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or 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. 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&gt;)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. &gt;Ni. XKvZ &gt;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&gt; ,#/ ///,&amp; MW,&gt; 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. 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) &lt; PROPRIETARY MATERIAL. 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or 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. 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(&gt;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, /, /-,/,&amp; 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. 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! = - (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. 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 &gt; 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 /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...</p>