# (solucionario) estatica hibbeler 10edicion

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Aunque de una edicion pasada... Sé que les servira mucho!TRANSCRIPT

- 1. Instructor's Solutions ManualENGINEERING MECHANICSSTATICS TENTH EDITIONR. C. HibbelerPearson Education, Inc. Upper Saddle River, New Jersey 07458

2. Executive Editor: Eric Svendsen Associte Editor: Dee Bernhard Executive Managing Editor: Vince O'Brien Managing Editor: David A. George Production Editor: Barbara A. Till Director of Creative Services: Paul Belfanti Manufacturing Manager: Trudy Pisciotti Manufacturing Buyer: Ilene KahnAbout the cover: The forces within the members of this truss bridge must be determined if they are to be properly designed. Cover Image: R.C. Hibbeler. 2004 by Pearson Education, Inc. Pearson Prentice Hall Pearson Education, Inc. Upper Saddle River, NJ 07458All rights reserved. No part of this book may be reproduced in any form or by any means, without permission in writing from the publisher. The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event 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 0-13-141212-4 Pearson Education Ltd., London Pearson Education Australia Pty. Ltd., Sydney Pearson Education Singapore, Pte. Ltd. Pearson Education North Asia Ltd., Hong Kong Pearson Education Canada, Inc., Toronto Pearson Educaci6n de Mexico, S.A. de c.v. Pearson Education-Japan, Tokyo Pearson Education Malaysia, Pte. Ltd. Pearson Education, Inc., Upper Saddle River, New Jersey 3. Contents 1General Principles12Force Vectors53Equilibrium of a Particle774Force System Resultants1295Equilibrium of a Rigid Body2066Structural Analysis2617Internal Forces3918Friction4769Center of Gravity and Centroid55610Moments of Inertia61911Virtual Work680iii 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) 2.77Ma......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' (4.70a1ug/tt') { (0.3048 m)3(1 slug) = .Aas1-----------------------_.__. . . .__------_.----------. . 13. Represent each of the following quantities in the correct SI form using an appropriate prefix: (a) 0.000431 kg. (b) 35.3(10~) N, (c) 0.00532 km.a) 0.00043lkg=0.000431(IOl ) 11=0.431 1Ansb) 3S.3( 10l) N = 35.3 kN......c) 0.00532 Ian ---.~----------=0.OOS32( lal)-----.. -......- ..--.- .....---.--.-.-.......--.... - - - - :...... .. """'Idm = 5.32 m---------~------( m ) (10>' m)Ans(a) mlms = - - ~ -S- - laD/s (10)-' S.1.... in the correct SI form using an appropriate prefix: each of the following combinations of uniiS...-.-AnsRep~~sentS) ((10)9 S) kg(10)' (c) ts/1111 = ( ~ ..(aJ mlms, (b) JLkm, (c) kslmg. and (d) km JLN.AnsOs/kgAns------------------------------------------------.----t SS mi/h .. (~)(S280ft)(~)(~) Ih I IDI I ft 1000 m '" 88.S km/h Ans1.5. If a car is traveling at 55 mi/h, determine its speed in.kilometers per hour and meters per second.88.Skm)(IOOOm)( Ih 88.5 km/h =( - - - - __ ) - 246 mls Ih Ikm 3600s - .1-----------_._----_._------------------_._-= 0.185(10') kg' = O.ISS Mi(a)'(O'(lO2 q)'(c)(c) (230m)3,(430 kg)'(b)1-6. EvaIuateeach of the follOwing and express with an appropriate prefix: (a) (430 kg)2 (b) (0.002 mg)2 and(230 m)' = [0.23(10') m]' = 0.0122 laD'ADI= [2(10"') sf = 4p.,-~----~'~----~---------------------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 have a)lSO( lal).lup =[2S0( 10l)4.4482N) =[ 1.32' (10') lb]( l i i ).IUIS](14.S938=3.6484'( 10') kJkJ)='"' ,.894( 10') N '.89 MNI slulIS=3.6'01AnsW.('. )=(3'.791 MN) (5.30 ftlS2) = S....9 MN ---, I 32.2ftlsW. = W. b)= ml =[3.6484'( 10') kl ]( 9.81 m/s2) d) Since the mass is= 35.791 (10) k8' m/s2=35.8 MNAidOrAns.0independent of its locaIion. thenm.. = m. = 3.6S( 10') k81=3.65 01Ans 5. (a)I 300 =51X) Ibsinn" 500', ,", 'sin> = 0.5796 , >= 35.42'45'+ 9 ""ADS3IXI Ib75','" so t~at the resultant f?rce is directed along the ositive x 8XlS and has a magn~tude of FR = 20 lb. P~ _ Sin; -30:ri;;8(30)2 = (30)2 + (20)2 ;=8=70.5- 2(30)(20)cos8 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 in Fig_ (a). Trillonometry : Using law of sines [Fig. (b). we have-2....=~ sin 50" ~Fasin 110=774NAns950sin 20 = ;;;;!i'i)OF,=346NAns12 16. 221. The truck is to be towed using two ropes. If the resultant force is to be 950 N, directed along the Qositive x axis, determine the magnitudes of forces FA C!!1d Fn acting on each rope and the angle of e of F n so that the magnitude of F H is a minimum. FA acts at 20 from the x axis as shown.:::-_-+___ xParallelogramu.w : In order to produce a minim..". force Fs , Fshasto act perpendicular to FA .The parallelogram law of addition is shown in Fig. (a).Trigonometry: Fig. (b).F;, = 950sin 20 = 325 NA ns=9S0c0s 200 =893 NAns~~The angle 8 is8 = 900 - 200 =70.009501'1Ans(b)f h2-22. Determine the magnitude and dire t'CIon 0 t e resultant F - F h R - I + F2 + F3 of the three forces by first fi d In 109 t e resultant F' = F + F d h . FR = F' + F3. 1 2 an t en fonrung FI= 30NF'y= ,1(20)230.85 sin73.13---=---=~~1--. --xFR+ (30)2 - 2(20)(30) cos73.1330 sin(70 - 8')'8'=1.470F'= ,,(30.85)2 + (50)2 - 2(30.85)(50) cos 1.47 = 19.18 = 19.2 N19.18sm1.4730.85. 8 = 2.37 ~= ~;13~= 30.85 N x"",, l'IAml.W ~~SOleN_ x _ Am30.~5N- -F" 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'yF'= 1(20)2 + (50)2 - 2(20)(50) cos70 = 47.07 Ne = 23.5320 47.07 ainU = sln700;F3= SON__~~~&-_-.. __ xFR =1(47.07)2 + (30)2 - 2(47.07)(30) cos13.34 = 19.18 = 19.2 N ,2019.18 30 sln13.34 = sin;; 8 = 23.53 - 21.15; = 21.15= 2.37"lAIlS QOol3.S3o('OOo3U7) 13. !&to &Ans*2-24. Resolve the SO-Ib force into components acti along (a) the x and yaxes, and (b) the x and y' axes. ng(b)F.= 50 cos4SO = 35.4 IbAnsF,.(a)= 50 s1n45 = 35.4 lbAns50/1.1~ J'J F~'-F. 50 sin 15 = sin1200 F.= 14.9 JbE sln45Ans50 sin 120~=-F,.' = 40.8 Jb~. ;,('Ans.1)0~ 18. 2-25. The log is being towed by two tmctors A and B. Detennine the magnitude of the two towing forces FA and F B if it is required that the resultant force have a magnitude F R = 0 kN and be directed along the x axis. Set {I = 15.BParallelogram lAw: The parallelogram law of addition is shown in Fig. (a). Trigonometry: Using law of sines [Fig. (b)]. we have~=_1_0_ sin 15 sin 135'fA = 3.66 kNF8~ ~ IOkN(a)(b)Ans~=_I_O_ sin 30 sin 135'Fa = 7.07 kNAns2-26. If the resultant F R of the two forces acting on the log is to be directed along the positive x axis and have a magnitude of 10 kN. determine the angle (J of the cable, attached to B such that the force F B in this cable is minimum. What is the magnitude of the force in each cable for this situation?Parallelogram lAw: In order to produce a millimum force FR. FH has to act perpendicular to F.. The parallelog... m law of addition is shown in Fig. (a). Trigonometry: Fig. (b). .Fa= JOsin30 = 5.00 kNf~.= lOcos 30' = 8.66 kN~ 'FEA,..Ans~ lOkN(a)The angle IJ iso = 90' -30 = 60.0Ans(bi 19. 2-27. The beam is to be hoisted using two chains. Determine the magnitudes of forces FA and F n acting on each chain in order to develop a resultant force of 600 N directed along the positive y axis. Set 0 = 450.~-~; sin 10$"lia 4'F.,-439NA_------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 positive y axis, determine the magnitudes of forces FA and Fn acting on each chain and the orientation 0 of FIJ so that the magnitude of FI/ is a minimum. FA acts at 30 from the y axis as shown. ----------- xFor III iaim um 8-6/Y'F,. requireAuF, - 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. If two 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 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.lb~~;;;::----r---xFCOIiDe Jaw:2001b.,F. , .; 3O()Z + 200Z - 2(JOO)(ZOO)cos6O" .. 264.61b Sine Jaw: sin(3O" + 9) sin6O" --::;;-- -264.6 ZOOIJ. 10.9"Au:00.IF F., +F 500. 264.6+ F.;. F Ib'.,-2],230. Three cables pull on the pipe such that they create a resultant force having a magnitude of 900 lb. (f two of the cables are subjected to known forces. as shown in the figure. determine the direction 8 of the third cable so that the magnitude of force F in this cable 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.F" .. /(600)2 +sin, .. ~: 802.64F(400)2 - 2(600)(400) cos 10So