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  • 1. Instructors Solutions ManualENGINEERING MECHANICSSTATICSTENTH EDITIONR. C. HibbelerPearson Education, Inc. Upper Saddle River, New Jersey 07458

2. Executive Editor: Eric SvendsenAssocite Editor: Dee BernhardExecutive Managing Editor: Vince OBrienManaging Editor: David A. GeorgeProduction Editor: Barbara A. TillDirector of Creative Services: Paul BelfantiManufacturing Manager: Trudy PisciottiManufacturing Buyer: Ilene KahnAbout the cover: The forces within the members of this truss bridge must bedetermined if they are to be properly designed. Cover Image: R.C. Hibbeler. 2004 by Pearson Education, Inc.Pearson Prentice HallPearson Education, Inc.Upper Saddle River, NJ 07458All rights reserved. No part of this book may be reproduced in any form or by anymeans, without permission in writing from the publisher.The author and publisher of this book have used their best efforts in preparing thisbook. These efforts include the development, research, and testing of the theories andprograms to determine their effectiveness. The author and publisher make no warran-ty of any kind, expressed or implied, with regard to these programs or the documen-tation contained in this book. The author and publisher shall not be liable in anyevent for incidental or consequential damages in connection with, or arising out of,the furnishing, performance, or use of these programs.Pearson Prentice Hall is a trademark of Pearson Education, Inc.Printed in the United States of America10 9 8 7 6 5 4 3 2 1ISBN 0-13-141212-4Pearson Education Ltd., LondonPearson Education Australia Pty. Ltd., SydneyPearson Education Singapore, Pte. Ltd.Pearson Education North Asia Ltd., Hong KongPearson Education Canada, Inc., TorontoPearson Educaci6n de Mexico, S.A. de c.v.Pearson Education-Japan, TokyoPearson Education Malaysia, Pte. Ltd.Pearson Education, Inc., Upper Saddle River, New Jersey 3. Contents1General Principles 12Force Vectors53Equilibrium of a Particle 774Force System Resultants1295Equilibrium of a Rigid Body2066Structural Analysis2617Internal Forces3918Friction 4769Center of Gravity and Centroid 55610 Moments of Inertia 61911 Virtual Work 680iii 4. 1.1. Round off the following numbers to three significant figure~: (a) 4.65735 m, (b) 55.578 S, (c) 4555 N, (d) 2768 kg. a) 4.66 mb) 5.6 sc) 4.56 kNd) 2.77 Ma......I----------------..-.------..---.--------------~--------II 1.2. Wood has a density of 470 sIugjft3. . density expressed in SI units? . What IS Its(1it)(14.5938 kg) } 242 Mslm Aas(4.70a1ug/tt) { (0.3048 m)3(1 slug) = .1-----------------------_.__. . . .__------_.----------. .13. Represent each of the following quantities in thecorrect SI form using an appropriate prefix: (a) 0.000431 kg. a) 0.00043lkg=0.000431(IOl ) 11=0.431 1 Ans(b) 35.3(10~) N, (c) 0.00532 km.b) 3S.3( 10l) N = 35.3 kN ......c) 0.00532 Ian =0.OOS32( lal)m = 5.32 m "Id---.~----------""..--.... - - - - :...... ..-----.. -......- ..--.- .....---.--.-.-..... ---------~------ ...-.- ( m ) (10> m) (a) mlms = - - ~ -S- - laD/s(10)- S Ans .1.... in the correct SI form using an appropriate prefix: uniiS Rep~~senteach of the following combinations of Ans(aJ mlms, (b) JLkm, (c) kslmg. and (d) km JLN. (10) (c) ts/1111 = ( ~ ..S) ((10)9 S) - kg Os/kg Ans Ans------------------------------------------------.----t 1.5. If a car is traveling at 55 mi/h, determine its speedSS mi/h .. (~)(S280ft)(~)(~) in.kilometers per hour and meters per second. Ih I IDI I ft 1000 m " 88.S km/h Ans88.5 km/h =( - - - - __ ) - 246 mls 88.Skm)(IOOOm)( Ih IhIkm3600s - .1-----------_._----_._------------------_._-- 1-6. EvaIuateeach of the follOwing and express with an(a) (430 kg) = 0.185(10) kg = O.ISS MiADI appropriate prefix: (a) (430 kg)2 (b) (0.002 mg)2 and (c) (230m)3,(b) (O(lO2 q) = [2(10") sf = 4p., (c) (230 m) = [0.23(10) m] = 0.0122 laD-~----~~----~---------------------4 17. A rocket has a mass of 250(10~) slugs on earth. Specify (a) its mass in SI units, and (b) its weight in SI units. If the rocket is on the moon, where the acceleration due to gravity is gm = 5.30 ft/s2. determine to three significant figures (c) its weight in SI units, and (d) its mass in SI units. c) W. = mI. = [lSO( 103) SIUIS] ( S.30 ftls2)u .ine Table 1 2 IIId i&lPlyinl Eq.1 - 3, we have4.4482N) =[ 1.32 (10) lb]( l i i )a) lSO( lal) .lup =[2S0( 10l).IUIS](14.S938kJ)I slulIS=" ,.894( 10) N .89 MNAidOr=3.6484( 10) kJ=3.601 AnsW. = W. - (. )=(3.791 MN) (5.30 ftlS2) = S....9 MN I---,32.2ftls .0b) W. = ml =[3.6484( 10) kl ]( 9.81 m/s2) d) Since the mass isindependent of its locaIion. then = 35.791 (10) k8 m/s2 =35.8 MN Ans m.. = m. = 3.6S( 10) k8=3.65 01 Ans1 5. (a) Isinn"51X) Ib300 = 500,,", sin> = 0.5796 , 45+ 9 "FAB > = 35.42 ADS" 75,3IXI Ib "so t~at the resultant f?rce is directed along the ositivex 8XlS and has a magn~tude of FR = 20 lb.P~ _30Sin; -:ri;;8 (30)2 = (30)2 + (20)2 - 2(30)(20)cos8 ;=8=70.5 Ans *2.20. The truck is to be towed using two ropes. Determine the magnitude of forces FA and F B acting on each rope in order to develop a resultant force of 950 N directed along the positive x axis. Set 6 = 50. --_-+ ___ xP",alldollrlUfl Law : The parallelollnun law of addition is shown inFig_ (a).Trillonometry : Using law of sines [Fig. (b). we have-2....=~sin 50"sin 110~ =774NAns Fa950sin 20 = ;;;;!ii)OF,=346NAns 12 16. 221. The truck is to be towed using two ropes. If theresultant force is to be 950 N, directed along the Qositivex axis, determine the magnitudes of forces FA C!!1d Fnacting on each rope and the angle of e of F n so that themagnitude of F H is a minimum. FA acts at 20 from the xaxis as shown.:::-_-+___ xParallelogramu.w : In order to produce a minim..". force Fs , Fshasto act perpendicular to FA .The parallelogram law of addition is shown inFig. (a).Trigonometry: Fig. (b).F;, = 950sin 20 = 325 N A ns~ =9S0c0s 200 =893 N Ans~ The angle 8 is8 = 900 - 200 =70.00 Ans 95011 (b)2-22. Determine the magnitude and dire tf hresultant F - F CIon 0 t efi d h R - I + F2 + F3 of the three forces by first In 109 t e resultant F = F + Fd h.FR = F + F3. 12 an t en fonrung FI= 30N ~F = ,1(20)2+ (30)2 - 2(20)(30) cos73.13 = 30.85 N y 30.8530 x ---=F sin73.13sin(70 - 8) 8 = 1.470"",, FR= ,,(30.85)2 + (50)2 - 2(30.85)(50) cos 1.47 = 19.18 = 19.2 N l Am ---=~~1--. --x19.1830.85. l.W ~~SOleN_ x _I sm1.47 = ~;8 = 2.37 ~Am 30.~5N - -F"13 17. ------.- ------_._------_ .... 2-23. Determine the magnitude and direction of the resultant FR = Fl + F2 + F3 of the three forces by ftrst finding the resultant F = F2 + F3 and then forming FR = F + Fl y F = 1(20)2 + (50)2 - 2(20)(50) cos70 = 47.07 N 20 47.07 ainU = sln700; e = 23.53 F3= SON __~~~&-_-.. __ xFR = 1(47.07)2 + (30)2 - 2(47.07)(30) cos13.34 = 19.18 = 19.2 N AIlS, QOol3.S3o(OOo3U7) 19.1830 13. !&to 20; = 21.15& sln13.34 = sin;; 8 = 23.53 - 21.15= 2.37 "l Ans*2-24. Resolve the SO-Ib force into components actialong (a) the x and yaxes, and (b) the x and y axes. ng 50/1.~F.= 50 cos4SO = 35.4 Ib Ans1~(a)- F~F,. = 50 s1n45 = 35.4 lb Ans JJ F.50(b)sin 15 = sin1200F.= 14.9 Jb Ans E 50~. 1)0 ~=- sln45sin 120;,( .F,. = 40.8 JbAns 18. 2-25. The log is being towed by two tmctors A and B.Detennine the magnitude of the two towing forces FAand F B if it is required that the resultant force have amagnitude F R = 0 kN and be directed along the x axis.Set {I = 15. BParallelogram lAw: The parallelogram law of addition is shown inFig. (a).~Trigonometry: Using law of sines [Fig. (b)]. we have~=_1_0_ F8~IOkNsin 15 sin 135(a)(b)fA = 3.66 kNAns~=_I_O_sin 30 sin 135Fa = 7.07 kNAns2-26. If the resultant F R of the two forces acting onthe log is to be directed along the positive x axis andhave a magnitude of 10 kN. determine the angle (J of thecable, attached to B such that the force F B in this cableis minimum. What is the magnitude of the force in eachcable for this situation?Parallelogram lAw: In order to produce a millimum force FR. FH hasto act perpendicular to F.. The parallelog... m law of addition is shownin Fig. (a).Trigonometry: Fig. (b).~.Fa= JOsin30 = 5.00 kN A,..FE~f~. = lOcos 30 = 8.66 kNAnslOkN(a)(biThe angle IJ is o = 90 -30 = 60.0Ans 19. 2-27. The beam is to be hoisted using two chains.Determine the magnitudes of forces FA and F n acting oneach chain in order to develop a resultant force of 600 Ndirected along the positive y axis. Set 0 = 450.~-~;sin 10$"lia 4 F.,-439N A_ ------Q-----------x *2-28. The beam is to be hoisted using two chains. If the resultant force is to be 600 N, directed along the positivey axis, determine the magnitudes of forces FA and Fnacting on each chain and the orientation 0 of FIJ so thatthe magnitude of FI/ is a minimum. FA acts at 30 fromthe y axis as shown.----------- xFor III iaim um F,. require 8-6/Y Au F, - 600 siD 30" -300 NA_16 20. 229. Three chains act on the bracket such that they create a resultant force having a magnitude of 500 lb. Iftwo of the chains are subjected to known forces, as shown. determine the orientation 8 of the third chain. measured clockwise from the positive x axis, so that the magnitude lb of force F in this chain is a minimum. All forces lie in the x-y plane. What is the magnitude of F? Hint: First find the resultant of the two known forces. Force F acts in this direction.~~;;;::----r---x F COIiDe Jaw: 2001b F. , .; 3O()Z + 200Z - 2(JOO)(ZOO)cos6O" .. 264.61b., Sine Jaw:sin(3O" + 9) sin6O"--::;;-- -264.6ZOO IJ. 10.9"Au ~ :00.F F., +F I . 38. 2-61. Determine the magnitude and coordinatedirection angles of the force F acting on the stake.4-F5 = 40,F = 50NF =(40 cos700i + 40 sin700j +~(50)k )F ={13.7i +37.6j +30.0k}NADsF = 1(13.68)2 + (37.59)2 + (30)2 = 50NAns-1 13.68 a= cos(so-)=74.1Ans37.59 f3= cos-1 (so-) = 41.3ADS-1 30 Y =cos (50) = 53.1 Ans2-.62.. Determine the magnltude andd IrectlOn angles of the resultant force. the coordinate z CtU"t.SUut V.ctor Nottllion: 7SIb F2 = SS{cos 3Ocos 6O"i + cos 30"sin 6O"j -sin 30"k} Ib={23.82i+41.25j-27.5k} IbRenlllUlt Force:F=F,+F