Solution Manual for Engineering Mechanics Dynamics, 13th Edition

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R1–1.An automobile is traveling with aconstant speedalong ahorizontal circular curve that has a radiusIf themagnitude of acceleration isdetermine thespeed at which the automobile is traveling.a=8 ft>s2,r=750 ft.SOLUTIONAns.v=77.5 ft>s8=v2750a=an=8=v2r

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R1–2.SOLUTIONa) The maximum friction force between blocksAandBisThus, both blocks move together.Ans.b) In this caseBlockA:Ans.BlockB:Ans.aB=3.86 ft>s2:+aFx=max;20(0.3)=5032.2aBaA=70.8 ft>s2:+aFx=max;20(0.3)=2032.2aA8 lb6F=50 lbaB=aA=a=2.76 ft>s2:+aFx=max;6=7032.2aFmax=0.4(20)=8 lb76 lbBlockBrests on a smooth surface. If the coefficients offrictionbetweenAandBareanddetermine the acceleration of each block if (a)and (b)F=50 lb.F=6 lb,mk=0.3,ms=0.420 lbAB50 lbF

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R1–3.The small 2-lb collar starting from rest atAslides downalong the smooth rod. During the motion, the collar is actederehwecrofaybnopux, y, zarein feet. Determine the collar’s speed when it strikes the wallatB.F=510i+6yj+2zk6lb,yxz8 ft10 ftF1 ftB4 ftASOLUTIONAns.vB=47.8 ft>s0+2(10-1)+L0410dx+L806ydy+L1102zdz=12a232.2bv2BT1+ ©LFds=T2rAB=rB-rA= -4i+8j-9k

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*R1–4.SOLUTIONSinceSinceAns.a=2(-0.217)2+(0)2+(0)2=0.217 m>s2az=z$ =0au=r#u$ +2ru=10(0)+2(0)(0.1473)=0ar=r-r#u2=0-10(0.1473)2= -0.217u=0vu=ru#=1.473u=1.47310=0.1473r=10r#=0r=0vz= -1.5 sin 10.81°= -0.2814 m>svu=1.5 cos 10.81°=1.473 m>svr=0v=1.5 m>sf=tan-1a122p(10)b=10.81°The automobile travels from a parking deck down along acylindrical spiral ramp at a constant speed of. Iftherampdescendsadistanceof12mforeveryfullrevolution,rad, determine the magnitude of the car’sacceleration as it moves along the ramp,.Hint:Forpart of the solution, note that the tangent to the ramp at anypoint is at an angle offrom the horizontal. Use this to determine the velocitycomponentsand, which in turn are used to determineand.z#u#vzvuf=tan-1(12>32p(10)4)=10.81°r=10 mu=2pv=1.5m>s10 m12 m

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R1–5.SOLUTIONAns.(vR)2=0.840 m>s0+0=0.0015(1400)-2.5(vR)2:+©mv1= ©mv2A rifle has a mass of 2.5 kg. If it is loosely gripped and a1.5-g bullet is fired from it with a horizontal muzzle velocityof 1400 m/s, determine the recoil velocity of the rifle justafter firing.

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R1–6.SOLUTIONAns.v2=21.5 ft>s0+150 sin 30°(30)-(0.3)129.9(30)=12a15032.2bv22T1+ ©U1-2=T2NC=129.9 lb+a©Fy=0;NC-150 cos 30°=0If a 150-lb crate is released from rest atA, determine itsspeed after it slides 30 ft down the plane. The coefficient ofkinetic friction between the crate and plane ismk=0.3.30 ft30°A

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R1–7.Thevanis traveling at 20km hwhenthe coupling ofthetraileratAfails. Ifthe trailerhas amass of250kg andcoasts 45mbefore coming torest,determine the constanthorizontalforceFcreatedbyrollingfrictionwhichcausesthe trailerto stop.>A20 km/hFSOLUTIONAns.:+ ©Fx=max;F=250(0.3429)=85.7 Na= -0.3429 m>s2=0.3429 m>s2:0=5.5562+2(a)(45-0)a;+by2=y20+2ac(s-s0)20 km>h=20(103)3600=5.556 m>s

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*R1–8.SOLUTIONAns.Ans.ymax=3(10)2-18(10)+15=135 ft>symaxoccurs whent=10 samax=6(10)-18=42 ft>s2amaxoccurs att=10 s,a=dydt=6t-18y=dsdt=3t2-18t+15s=t3-9t2+15tThe position of a particle along a straight line is given bywheretis in seconds. Determine itsmaximum acceleration and maximum velocity during thetime interval 0…t…10 s.s=1t3-9t2+15t2ft,

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R1–9.The spool, which has a mass of 4 kg, slides along the rotatingrod. At the instant shown, the angular rate of rotation of therodisandthisrotationisincreasingatAt this same instant, the spool has a velocityof 3 m/s and an acceleration ofboth measuredrelative to the rod and directed away from the centerOwhenDetermine the radial frictional force andthe normal force, both exerted by the rod on the spool atthis instant.r=0.5 m.1 m>s2,u$ =2 rad>s2.u#=6 rad>sSOLUTIONAns.Ans.N=2(148)2+(39.24)2=153 NFr= -68 NNz=39.24 N©Fz=maz;Nz-4(9.81)=0©Fu=mau;Nu=4(37)=148 N©Fr=mar;Fr=4(-17)= -68 Nau=ru$ +2r#u#=0.5(2)+2(3)(6)=37ar=r$ -ru#2=1-0.5(6)2= -17r$ =1 m>s2u$ =2 rad>sr#=3 m>su#=6 rad>sr=0.5 mθ••= 2 rad/s2θ•= 6 rad/svs= 3 m/sas= 1 m/s2r= 0.5 mO

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R1–10.Packages having a mass of 6 kg slide down a smooth chuteand land horizontally with a speed of 3 m/s on the surfaceof a conveyor belt. If the coefficient of kinetic frictionbetween the belt and a package isdetermine thetime needed to bring the package to rest on the belt if thebelt is moving in the same direction as the package with aspeedv=1 m/s.mk=0.2,SOLUTIONAns.t=1.02 s)3(6-0.2(58.86)(t)=6(1)(:+)m(v1)x+ ©LFxdt=m(v2)xNp=58.86 N0+Np(t)-58.86(t)=0(+c)m(v1)y+ ©LFydt=m(v2)y1 m/s3 m/s

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R1–11.SOLUTIONThe crate starts moving whenFrom the graph sinceThe time needed for the crate to start moving isHence,theimpulseduetoFisequaltotheareaunderthecurvefromAns.v2=31.7 m>s384.436=20v2(0412t2)252.943+200(10-5)-692.292=20v20+L52.9432005tdt+L105200dt-(0.5)196.2(10-2.943)=20v2:+m(vx)1+ ©LFdt=m(vx)22.943 s…t…10 st=5200(117.72)=2.943 sF=2005t,0…t…5 sF=Fr=0.6(196.2)=117.72 NA 20-kg block is originally at rest on a horizontal surfacefor which the coefficient of static friction isandthe coefficient of kinetic friction isIf a horizontalforceFis applied such that it varies with time as shown,determinethespeedoftheblockin10s.Hint:Firstdetermine the time needed to overcome friction and startthe block moving.mk=0.5.ms=0.6FF(N)t(s)200510x

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*R1–12.SOLUTIONAns.x=0.654 ft=7.85 in.0+12 (20)(12)(x2)=12a632.2b(6)2+8(6)T1+V1=T2+V2The 6-lb ball is fired from a tube by a spring having astiffnessDetermine how far the spring mustbe compressed to fire the ball from the compressed positionto a height of 8 ft, at which point it has a velocity of 6 ft/s.k=20 lb>in.8 ftk= 20 lb/in.v= 6 ft/s

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R1–13.A train car, having a mass of 25 Mg, travels up a 10° inclinewith a constant speed of 80 km/h. Determine the powerrequired to overcome the force of gravity.SOLUTIONAns.P=946 kWP=F#v=25(103)(9.81)(22.22)(sin 10°)v=80 km>h=22.22 m>s

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R1–14.The rocketsled hasa massof 4 Mgand travelsfrom restalongthesmooth horizontal tracksuch that it maintainsaconstant power output of. Neglect the lossof fuelmassand air resistance, and determine how far it musttravel to reach aspeed of.v=60 m>s450 kWSOLUTIONAns.s=4 (103)(60)33(450)(103)=640 ms=mv33PPs=mv33PLs0ds=mLv0v2dvLPds=mLv2dvP=Fv=mav2dvdsbF=ma=mavdvdsb©Fx=max;+:vT

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■R1–15.SOLUTIONatUsing the Runge–Kutta method:t=0s0=0n=3 (8e-1+t)1>2Aprojectile,initiallyattheorigin,movesverticallydownward along a straight-line path through a fluid mediumsuch that its velocity is defined aswheretis in seconds. Plot the positionsof the projectileduring the first 2 s. Use the Runge-Kutta method to evaluateswith incremental values ofh=0.25 s.v=318e-t+t21>2m>s,ts002.010.253.830.505.490.757.031.008.481.259.871.5011.21.7512.52.00

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*R1–16.The chain has a mass of 3 kg/m. If the coefficient of kineticfrictionbetweenthechainandtheplaneisdetermine the velocity at which the endAwill pass pointBwhen the chain is released from rest.mk=0.2,SOLUTIONAns.Also,Ans.v=4.38 m>s0+2(3)(9.81)(2 sin 40°)-0.2(45.09)(2)=12 (2)(3)(v2)T1+ ©U1-2=T2v2=4.38 m>sv22=0+2(4.80)(2)+ bv22=v21+2asa=4.80 m>s2+ b©Fx=max;2(3)(9.81) sin 40°-0.2(45.09)=2(3)aNC=45.09 N+ a©Fy=may;-2(3)(9.81) cos 40°+NC=02 mAB40°

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R1–17.The motorMpulls in its attached rope with an accelerationDetermine the towing force exerted byMonthe rope in order to move the 50-kg crate up the inclinedplane. The coefficient of kinetic friction between the crateand the plane isNeglect the mass of the pulleysand rope.mk=0.3.ap=6 m/s2.SOLUTION(1)Kinematics,Taking two time derivatives, yieldsThus,Substituting into Eq. (1) and solving,Ans.T=158 NaC=63=23aC=ap2sC+(sC-sp)=lQ+ ©Fx=max;3T-0.3(424.79)-50(9.81) sin 30°=50aCNC=424.79a+ ©Fy=may;NC-50(9.81) cos 30°=0aP= 6 m/s2PMθ= 30°

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R1–18.The drinking fountain is designed such that the nozzle islocated from the edge of the basin as shown. Determine themaximum and minimum speed at which water can beejected from the nozzle so that it does not splash over thesides of the basin atBandC.SOLUTION(1)(2)Substituting Eq. (1) into (2) yields:At pointB,.Ans.At pointC,.Ans.vA=A4.905 sin 40°(0.35)2sin240°(0.35 cos 40°+0.05 sin 40°)=1.76 m>sR=0.35 mvA=A4.905 sin 40°(0.1)2sin240°(0.1 cos 40°+0.05 sin 40°)=0.838 m>sR=0.1 mvA=A4.905 sin 40°R2sin240°(Rcos 40°+0.05 sin 40°)-0.05=vAcos 40°¢RvAsin 40°≤+12(-9.81)¢RvAsin 40°≤2-0.05=0+vAcos 40°t+12(-9.81)t2(+c)sy=AsyB0+vyt+12act2R=vAsin 40°tt=RvAsin 40°a:+bsx=vxtvA50 mmABC40°250 mm100 mm

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R1–19.The 100-kg crate is subjected to the action of two forces,andas shown. If it is originally atrest, determine the distance it slides in order to attain aspeed of 6 m/s. The coefficient of kinetic friction betweenthe crate and the surface ismk=0.2.F2=1.5 kN,F1=800 NSOLUTIONAns.s=0.933 ms(1928.7)=18000+800cos 30°(s)-0.2(867.97)(s)+1500 cos 20°(s)=12 (100)(6)2T1+aU1-2=T2NC=867.97 N+ caFy=0;NC-800 sin 30°-100(9.81)+1500 sin 20°=0F2F130°20°

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*R1–20.SOLUTIONAns.s=20.0 fts=0+12(10)+12 (-2)(10)2(:+)s=s0+v0t+12act2If a particle has an initial velocityto the right,and a constant acceleration ofto the left, determinethe particle’s displacement in 10 s. Originallys0=0.2 ft>s2v0=12 ft>s

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R1–21.SOLUTIONAns.h=0.390 mh=0+8.3328(0.048112)-12 (9.81)(0.048112)2(+ c)s=s0+v0t+12act2t=0.048112 s0.75=0+15.5885t(:+)s=s0+v0t(vy)2=8.3328 m>s(+ c)e=0.8=(vy)210.4160(vy)1=10.4160 m>s(vy)1=18 sin 30°+9.81(0.14434)(+ T)v=v0+act(vx)1=(vx)2=18cos 30°=15.5885 m>st=0.14434 s2.25=0+18 cos 30°t(:+)s=s0+v0tThe ping-pong ball has a mass of 2 g. If it is struck with thevelocity shown, determine how highhit rises above the endof the smooth table after the rebound. Takee=0.8.h2.25 m18 m/s300.75 m

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R1–22.SOLUTIONTime to accelerate to 60 m/s,Time to decelerate to a stop,Time to travel at 60 m/s,Total timeAns.t=10+7.5+6.25=23.8 st=6.25 s573=0+60t(:+)s=s0+v0t900-300-225=375 ms=225 ms=0+60(7.5)-12(8)(7.52)(:+)s=s0+v0t+12act2t=7.5 s0=60-8t(:+)v=v0+acts=300 ms=0+0+12(6)(102)(:+)s=s0+v0t+12act2t=10 s06=0+6t(:+)v=v0+actA sports car can accelerate atand decelerate atIf the maximum speed it can attain is 60 m/s,determine the shortest time it takes to travel 900 m startingfrom rest and then stopping when s=900 m.8 m/s2.6 m/s2

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R1–23.A 2-kg particle rests on a smooth horizontal plane and is acteduponbyforcesandIfandwhendetermine the equationwhich describes the path.y=f1x2t=0,vy=2 m/svx=6 m/s,y=0,x=0,Fy=3 N.Fx=0SOLUTION(1)(2)(3)(4)Eliminatingtfrom Eq. (3) and (4) yields:Ans.y=0.0208x2+0.333x(Parabola)x=6tLx0dx=Lt06dtvx=dxdt=6Lvx6dvx=Lt00dtax=dvxdt=0y=0.75t2+2tLy0dy=Lt0(1.5t+2)dtvy=dydt=1.5t+2Lvy2dvy=1.5Lt0dtay=dvydt=1.5:+©Fx=max;0=2axax=0+ c©Fy=may;3=2ayay=1.5 m>s2

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*R1–24.SOLUTIONAns.Ns=240 lb+ c ©Fn=man;Ns-120=12032.2c(31.08)230dr=3[1(dydx)2]3>2d2ydx23=(1+0)3>2130=30 ftd2ydx2=130dydx2x=0=130x2x=0=0y=160x2-15vB=31.08 ft>s0+(120)(15)=12a12032.2bv2B+0TA+VA=TB+VBA skier starts from rest atA(30 ft, 0) and descends thesmooth slope, which may be approximated by a parabola. Ifshe has a weight of 120 lb, determine the normal force sheexerts on the ground at the instant she arrives at pointB.30 ft15 ftyBAxy=x2– 151__60

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R1–25.SOLUTIONBlockA:(1)BlockB:(2)(3)BlockC:(4)Solving Eqs. (1)–(4) fora,Ans.v=2.68 ft>sv2=0+2(1.79)(2-0)(+ T)v2=v20+2ac(s-s0)a=1.79 ft>s2+ c ©Fy=may;T2-6=632.2a+ c ©Fy=may;NB-20=0;+ ©Fx=max;-T2+T1-0.1NB=2032.2a+ T ©Fy=may;-T1+10=1032.2aThe 20-lb blockBrests on the surface of a table for whichthe coefficient of kinetic friction isDetermine thespeed of the 10-lb blockAafter it has moved downward 2 ftfrom rest. Neglect the mass of the pulleys and cords.mk=0.1.A10 lbC6 lbB20 lb

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R1–26.SOLUTIONBlockA:BlockB:Kinematics:Solving,Ans.vA=26.8 ft>sTvA=6+10.40(2)(+ T)vA=(vA)0+aAtaB= -20.81 ft>s2aA=10.40 ft>s2T=3.38 lb2aA= -aB2sA+sB=l;+ ©Fx=max;-T+0.2(4)=432.2aB+ T ©Fy=may;10-2T=1032.2aAAt a given instant the 10-lb blockAis moving downwardwith a speed of 6 ft/s. Determine its speed 2 s later. BlockBhas a weight of 4 lb, and the coefficient of kinetic frictionbetween it and the horizontal plane isNeglect themass of the pulleys and cord.mk=0.2.AB

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R1–27.Two smooth billiard ballsAandBhave an equal mass ofIfAstrikesBwith a velocity ofas shown, determine their final velocities just after collision.BallBis originally at rest and the coefficient of restitutionise=0.75.1vA21=2 m/sm=200 g.SOLUTION(1)(2)Solving Eqs. (1) and (2)ForA:ForB:HenceAns.Ans.Ans.(uA)2=tan-1a0.19151.285b=8.47°e(vA)2=2(-0.1915)2+(1.285)2=1.30 m>s(vB)2=(vB)x2=1.34 m>s;(vB)y2=0(+ c)mB(vB)y1=mB(vB)y2(vA)y2=1.285 m>s(+ T)mA(vA)y1=mA(vA)y2(vB)x2= -1.3405 m>s(vA)x2= -0.1915 m>s57.0=(vA)x2-(vB)x21.532(:+)e=(vrel)2(vrel)1-2(1.532)+0=0.2(vA)x2+0.2(vB)x2(:+)mA(vA)x1+mB(vB)x1=mA(vA)x2+mB(vB)x2(vA)y1= -2 sin 40°= -1.285 m>s(vA)x1= -2 cos 40°= -1.532 m>syx40°BA(vA)1

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*R1–28.SOLUTIONAns.Ans.U1-2=14.4 ft#kipU1-2=(744.4 cos 15°)(20)=14 380.7 ft#lbNC=1307.3 lbT=744.4 lb=744 lb:+ ©Fx=0;Tcos 15°-0.55NC=0+ c ©Fy=0;NC-1500+Tsin 15°=0A crate has a weight of 1500 lb. If it is pulled along thegroundataconstantspeedforadistanceof20 ft, and the towing cable makes an angle of 15° with thehorizontal, determine the tension in the cable and the workdone by the towing force. The coefficient of kinetic frictionbetween the crate and the ground ismk=0.55.

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R1–29.The collar of negligible size has a mass of 0.25 kg and isattached to a spring having an unstretched length of 100 mm.If the collar is released from rest atAand travels along thesmooth guide, determine its speed just before it strikesB.SOLUTIONAns.vB=10.4 m>s0+(0.25)(9.81)(0.6)+12 (150)(0.6-0.1)2=12 (0.25)(vB)2+12(150)(0.4-0.1)2TA+VA=TB+VB400 mm200 mmBAk150 N/m

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R1–30.Determine the tension developed in the two cords and theacceleration of each block. Neglect the mass of the pulleysand cords.Hint:Since the system consists oftwocords, relatethe motion of blockAtoC, and of blockBtoC. Then, byelimination, relate the motion ofAtoB.SOLUTIONBlockA:(1)BlockB:(2)PulleyC:(3)Kinematics:Taking the two time derivatives:Also,So thatSince(4)Solving Eqs.(1)–(4),Ans.Ans.Ans.Ans.TB=45.3 NTA=90.6 NaB=1.51 m>s2aA=0.755 m>s2aB=2aAaC¿ = -aC2aC¿ =aBsC¿ +(sC¿ -sB)=l¿aA= -aCsA+sC=l+ c ©Fy=0;TA-2TB=0+ c ©Fy=may;TB-4(9.81)=4aB+ T ©Fy=may;10(9.81)-TA=10aACB4 kgA10 kg3 m

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