Solution Manual for Engineering Mechanics Statics, 13th Edition

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SOLUTION MANUALCONTENTSChapter 1 General PrinciplesChapter 2 Force VectorsChapter 3 Equilibrium of a ParticleChapter 4 Force System ResultantsChapter 5 Equilibrium of a Rigid BodyChapter 6Structural AnalysisChapter 7 Internal ForcesChapter 8 FrictionChapter 9 Center of Gravity and CentroidChapter 10Moments of InertiaChapter 11 Virtual Work12217124842151564378793110711190

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1–1.Round off the following numbers to three significantfigures:(a) 58 342 m, (b) 68.534 s, (c) 2553 N, and(d) 7555 kg.SOLUTIONa)58.3 kmb)68.5 sc)2.55 kNd)7.56MgAns.

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1–2.Wood has a densityofexpressed in SI units?SOLUTION4.70 slug>ft3.Whatisitsdensity(4.70 slug>ft3)b(1 ft3)(14.59 kg)2.42Mg>m3Ans.(0.3048 m)3(1 slug)r=

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======1–3.Represent each of thefollowing combinations of unitsinthecorrect SI form using an appropriate prefix:(a),(b) Mg>mN,and (c) MN>(kg#ms).SOLUTIONa)kN>ms=(103)N(10-6)s(109)NsGN>sAns.b)Mg>mN=(106)g(10-3)N(109)gNGg>NAns.c)MN>(kg#ms)=(106) Nkg#(10-3)s(109)Nkg#sGN>(kg#s)Ans.kN>ms

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=====-====-#=-=-#=*1–4.Represent each ofthe following combinations of unitsinthecorrect SI form using an appropriate prefix:(a)(b)mkm, (c) ks>mg,and (d) km#mN.SOLUTIONa) m>ms¢11m02-3s≤¢1023sm≤km>sAns.b)mkm102-6103m103mmmAns.c)ks>mg1023s109sGs>kgAns.d) kmmN101063kgm106kgN103mNmm#NAns.m>ms,11121211212

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AB==AB=A=B=1–5.Represent each of the following quantities in the correctSI formusing an appropriate prefix:(a) 0.000 431 kg,(b) 35.3(103) N,and (c) 0.005 32 km.SOLUTIONa)0.000 431 kg0.000 43110g0.431gAns.b)35.310N35.3kNAns.c)0.005 32 km0.005 3210m5.32mAns.333

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=1–6.If a car is traveling at, determinekilometers per hour and meters per second.SOLUTIONitsspeedin55 mi>h=1 hba88.5 km>h5280 ft1 miba0.3048 m1 ftba1 km1000 mbAns.88.5 km>h=a88.5 km1000 m1 kmba1 h3600sb=24.6 m>sAns.55 mi>ha55mi1 hba

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=A=-B=AB==1–7.Thepascal(Pa) is actually a very small unit of pressure.Toshowthis, convertto. Atmosphericpressure at sea level is 14.7 lb>in2.How many pascals is this?SOLUTIONUsing Table 1–2,we have1 Pa1Nm2a1 lb4.4482 Nba0.30481 ft2m2b20.9103lb>ft2Ans.1 ATM14.7 lbina4.448 N1lbba144 in1ftba1ft0.3048mb101.310N>m2101 kPaAns.1 Pa=1 N>m2lb>ft222222223

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==*1–8.Thespecificweight(wt. vol.)ofbrassis.Determineitsdensity(mass vol.)inSIunits.Useanappropriate prefix.SOLUTION520 lb>ft3ft3ba1 ft0.3048 mb3a4.448 N1 kg9.81 Nb8.33 Mg>m3Ans.>>520 lb>ft3a520lb1 lbba

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ABCA=BDAB==1–9.A rocket has a mass ofslugs on earth. Specify(a)its mass in SI units and (b) its weight in SI units. If therocket ison the moon,where the acceleration due to gravityis,determine to three significant figures(c)its weight in SI units and (d) its mass in SI units.SOLUTIONUsing Table 1–2 and applying Eq.1–3,we havea)25010slugs25010slugsa14.59 kg1slugsb3.647510kg3.65Gg250(103)gm=5.30 ft>s2336

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=A=BB=AB#CA=BDAB=====Ansb)WemgC3.647510kgDA9.81 m>s235.79210kgm>s2=35.8 MNc)Wm=mgm=C250A103BslugsDA5.30 ft>s2BAns.1.32510lba4.448 N1lbb5.894Or10N5.89 MNAns.WWeagmb(35.792MN)a5.30ft>s2b5.89Mmg32.2 ft>s2d) Since the mass is independent of its location,thenmm= me=3.65A106Bkg=3.65 GgAns.6666

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AB==A=B=1–10.Evaluate each of the following to three significant figures andexpress each answer in SI units using an appropriate prefix:(a) (0.631 Mm)>(8.60 kg)2,(b) (35 mm)2(48 kg)3.SOLUTIONa)(0.631 Mm)>(8.60 kg)2a0.63110mb8532 m(8.60)kgkg8.5310m>kg28.53km>kg262223

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CA=-BD=#Ans.b)(35 mm)2(48 kg)335103m2(48 kg)135mkg3Ans.32

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CA-BA=BDA=B#=#1–11.(a) 354mg(45km)>(0.0356kN),(b) (0.004 53 Mg) (201 ms),SOLUTIONa)(354 mg)(45 km)>(0.0356 kN)354103gDC45A103m0.035610N0.44710Ngm0.447kgm>NEvaluate each of the following to three significant figures andexpress each answer in Sl units using an appropriate prefix:and (c) 435 MN>23.2 mm.3B3

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A=-BA-BD=#B-=Ans.b)(0.00453 Mg)(201 ms)C4.53103BA10kgDC201103s0.911kgsAns.c)435 MN>23.2 mm=43523.2A6Nm18.75A109BNm=18.8 GN>mAns.3A1010B3

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=*1–12.Convert each of the following and express the answer usinganappropriate prefix:(a)to(b)tomm>s, and (c) 835 lb#ft to kN#m.SOLUTIONa)ft3≤a1 ft0.3048mb3a4.448 Nm3≤=27.5 kN>m3Ans.b)6 ft>h=a6ft0.3048 m1 ftba1 h3600sb=0.5081102-3m>s=0.508 mm>sAns.c) 835 lb#ft=1835 lb#ft4.448 N1 lb0.3048m1 ft1.13103N#m=1.13kN#mAns.175 lb>ft3kN>m3,6 ft>h175 lb>ft3=¢175 lb1 lbb=¢27.511023N1hba

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===1–13.Convert each of the following to three significant figures:(a)to,(b)to,and(c) 15 ft>h to mm>s.SOLUTIONUsing Table 1–2,we havea)20 lb#ft=(20 lb#ft)a4.448N27.1 N#m0.3048 m1 ftbAns.b)450lb>ft31 ft3ba4.448 N1 lbba1 kN1000 Nba1 ft30.30483m3b70.7 kN>m3Ans.c)15 ft>h=a15 ftAns.20 lb#ftN#m450lb>ft3kN>m31 lbbaa450 lb1 hba304.8 mm1 ftba1 h3600sb=1.27 mm>s

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==1–14.Evaluateeachofthefollowing andexpress with an(c)1230 m2prefix:3.(a)1430kg22, (b)10.002 mg22, andSOLUTIONa)b)10.002 mg22=32110-62g420.185 Mg24mg2Ans.Ans.c) 230m3=0.23 103m3=0.0122 km3Ans.appropriate1430 kg22=0.18511062kg2

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AB-#===AB#===AB#===1–15.Determine the mass of an object that has a weight of (a) 20 mN,(b)150 kN, and (c) 60 MN. Express the answer to threesignificantfigures.SOLUTIONApplying Eq.1–3,we haveW20103kgm>s2a)mg9.81 m>s22.04gAns.W15010kgm>s2b)mg9.81m>s215.3MgAns.W6010kg2c)mg9.81 m>s26.12GgAns.36m>s

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*1–16.What is the weight in newtons of an object that has a massof:(a)10 kg,(b) 0.5 g,and (c) 4.50 Mg? Express the result tothreesignificant figures.Use an appropriate prefix.W =A9.81 m>s2B(10 kg)=98.1Nc)W =A9.81 m>s2B(4.5 Mg)A103kg>MgB=44.1 kNAns.Ans.Ans.SOLUTIONa)b)W =A9.81 m>s2B(0.5 g)(10-3kg>g)=4.90 mN

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1–17.If an object has a mass of 40slugs, determine its massinkilograms.SOLUTION40 slugs (14.59 kg>slug)=584 kgAns.

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==1–18.UsingtheSIsystemofunits,showthatEq.1–2isadimensionallyhomogeneousequationwhichgivesFinnewtons.Determinetothreesignificantfiguresthegravitationalforceactingbetweentwospheresthataretouching each other. The mass of each sphere is 200 kg andtheradius is 300 mm.SOLUTIONUsing Eq.1–2,Gm1m2F=N=F=Gm1m2r2akgm3Gm1m2r2kg#kgm2bkg#ms2(Q.E.D.)66.73A10-12Bc200(200)=7.41A10-6BN=7.41mNAns.#s2ba0.62d

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===1–19.Water has a density of.Whatisthedensityexpressed in SI units? Express the answer to threesignificantfigures.SOLUTIONrwa1.94 slug1 ftba14.59 kg1slugba1 ft30.3048mb999.6 kg>m31.00 Mg>m3Ans.1.94 slug>ft3333

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=A=-B#==*1–20.Two particles have a mass of 8 kg and 12 kg,respectively.Iftheyare800mmapart, determinetheforceofgravityactingbetween them. Compare this result with the weightof eachparticle.SOLUTIONFGm1mr22WhereG66.731012m3(kgs2)F =66.73A10-12BB(8(12)-9BN=10.0 nNAns.W18(9.81)=78.5NAns.W212(9.81)=118NAns.>0.8)2R=10.0A10

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======1–21.If a man weighs 155 lb on earth, specify (a) his massinslugs, (b)his mass in kilograms, and (c) his weightinnewtons.Ifthemanisonthemoon,wheretheacceleration due to gravity isdetermine(d) his weight in pounds, and (e) his mass in kilograms.SOLUTION155a)m=32.24.81slugAns.b)m=155c14.59 kg70.2kgAns.c)W=1554.4482689 NAns.d)W=e)m=Also,155c5.30155c14.59kg14.59 kg25.5lb70.2kgAns.Ans.m=25.55.3070.2kgAns.gm=5.30 ft>s2,32.2d1232.2d32.2d

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2–5.SOLUTIONAns.Ans.F1v=160 NF1vsin 30°=300sin 110°F1u=205 NF1usin 40°=300sin 110°Resolve the forceinto components acting along theuandvaxes and determine the magnitudes of the components.F1uv703045F1300 NF2500 N

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2–6.Resolve theforceinto components acting along theuandvaxes and determine themagnitudes ofthe components.F2SOLUTIONAns.Ans.F2v=482NF2vsin65°=500sin70°F2u=376NF2usin45°=500sin70°uv703045F1300NF2500N© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected byCopyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,or transmission in any form or by any means, electronic, mechanical,photocopying, recording, or likewise. For information regarding permission(s), write to:Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

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SOLUTIONParallelogram Law:The parallelogram law of addition isshown in Fig.a.Trigonometry:Usingthe law ofsines(Fig.b), we haveAns.Ans.FAC=366 NFACsin 45°=500sin 75°FAB=448 NFABsin 60°=500sin 75°2–7.Thevertical forceactsdownward aton the two-memberedframe. Determine the magnitudesof the two componentsofdirected alongthe axesofand. Set.F=500 NACABFAFFCBA3045© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected byCopyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,or transmission in any form or by any means, electronic, mechanical,photocopying, recording, or likewise. For information regarding permission(s), write to:Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

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*2–8.Solve Prob. 2-7 with.F=350 lbSOLUTIONParallelogram Law:The parallelogram law of addition isshown in Fig.a.Trigonometry:Usingthe law ofsines(Fig.b), we haveAns.Ans.FAC=256 lbFACsin 45°=350sin 75°FAB=314 lbFABsin 60°=350sin 75°FCBA3045© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected byCopyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,or transmission in any form or by any means, electronic, mechanical,photocopying, recording, or likewise. For information regarding permission(s), write to:Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

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2–9.ResolveF1into components along theuandaxes anddetermine the magnitudes of these components.vSOLUTIONSine law:Ans.Ans.F1usin 45°=250sin 105°F1u=183 NF1vsin 30°=250sin 105°F1v=129 NF1250 NF2150 Nuv3030105© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected byCopyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,or transmission in any form or by any means, electronic, mechanical,photocopying, recording, or likewise. For information regarding permission(s), write to:Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

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