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SOLUTIONxas afunction of v.

_ f -0.00057\/1. c154 V

a as afunction ofx

-0.00057 _ _ I VI54

-0.00057*= In 1 154=-1754.4In

v 2 = 23716l

154

a v L = 11858) 0.00057K-05*A dx 2 ^ Aa) v =20m/s.

From 1),From 2),

0) v =40m/s.From l),From 2),

a =6.75906e-00057*= 6.75906

x = 29.843a =6.64506

x=122.54a =6.30306

1- I 54J

x =a =6

x =a =6

PROPRIETARY MATERIAL. 2007 Th e McGraw-Hill Companies Inc.A llri htsicserad M op rto f thi s Manual may be dispor distributed in anyform or by any means, without th epr iorwrit ten permission ofAefmHiilfi orusedbeyondt he limited distributieducators permitted byMcGraw-Hil lfor their individua l course preparation. Ifyov oreastomla*usingMs Man ual, you a re using i t wi34

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PRO L M 11 57Slider block Bmoves to the left with a constant velocity of 50 mm/st =0,slider blockAis moving to the rightwitha constantacceleand a velocity of 100 mm/s. Knowing that at t =2 s slider block Cmoved 40 mm to the right, determine (a) the velocity of slider block t =0, (b ) the velocity of portion D of the cable at t = 0, (caccelerations ofA andC.

SOLUTIONLe txbeposition relativeto theanchor, positiveto the right.Constraint of cable: X B + xc *g)+3 xc XA ) = constant

4vc- 2vB- 3vA=0 4ac- 2aB- 3aA=0 0 2

When t =0 , VB = -50 mm/s and V O ) Q = 100 mm/s() (vc)0=^[2vB+3 v,)J=i[ 2) -50)+(3)(100)] (vc)fl = 50mm/s

ConstraintofpointD: ( X D - X A) +(xc- X A ) +(xc- XB)- X B =constant

6

V D +2vc- 2vA - 2vB=0V D )O=2(v,)0+2vB-2 vc\ 2) 100)+(2)(-50)- 2 ) 50)

=

ac=30mm/s2Solving 2) foraA

-30)-(2)(0)]=-40mm/s2

a,= 40mm/s2

PROPRIETARY MATERIAL 2007 The McGraw-Hill Companies,Inc.All rights reserved. Nopartof this Manual may be displayed,ordistributed in any form or by any mea ns, without the prior written perm ission of the publisher, or used beyond the limited distribution to teducators perm itted by McGr aw-Hill for their individual course prepar ation. Ifyou are a student using this Manu al, you are using it without p

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SOLUTION

pE A ,r s l

PRO L M 11 76A n elevator starts from rest and moves upward, accelerating at a rate of4ft/s2 unt i l it reaches a speed of 24 ft/s, which it then maintains. Twoseconds after the elevator begins to m ove, a m an standing 40 ft above theinitial position of the top of the elevator throws a ball upward with aninitialvelocityof 64 ft/s. Determine when theballwillhit the elevator.

Construct th e a-t curvesfor the elevator and the ball.Limit on 4 is 24ft/s. Using A } = 4t

4t2 =24 /2 = 6 s

i~ M V

. A,,

i_

Motion of elevator.For 0 2 s,Moment of A ^ about t = t

vg)0 = 6 4 f t / s- 2 f t / s

9-2f

When ball hits elevator, X = XE

18.1?, 2-128.4?, + 1 5 2 . 4 = 0

or

= 1.507s a nd 5 . 5 9 solving the quadratic equation,The smaller rootis out ofrange, henceSince this is less than 6 s, the solution iswithin range.

= 5.59 s

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PROBLEM11 95The motion of a particle is defined by the position vectorr = j4 cos/ + /sin/)i+yl sin/ - /cos/)j, where / is expressed inseconds. Determinethevaluesof / forwhichtheposition vectorand theacceleration vectorare a)perpendicular, b)parallel.

SOLUTIONGiven: r =A cost+/sin/)i+/l sin/- /cos/)j

v= =A -sint +sin/+/cos/)i+^4 cos/- cost+/sin/)jdt =yl /cos/)i + j4 /sin/)jd\/ . ./ . \a = = /4 cos/- /sm/)i+ /4 sm/+/cos/jjd t

a) Whenr and a areperpendicular, r a = 0os /+/sin/)i+ sin/- /cos/)jj/lUcos/- /sin/)i+ sin/+ /cos/)jl = 0/42[ cos/+/sin/) cos/- /sin/)+ sin/- /cos/) sin/+/cos/)]= 0

cos2/ - t2sin2/= ls-/2= 0

b) Whenr and a areparallel, r x a = 0y4J cos/+/sin/)i+ sin/ /cos/)j xA\ /sin/)i+ sin/+/cos/)j I = 0

y4 2[ cos/+/sin/) sin/+/cos/)- sin/- /cos/) cos/-/sin/)lk=0sin/cos/+/sin2/+/cos2/+/2sin/cos/J- sin/cos/- /cos2/- /sin2/+/2sin/cos/j= 0

2/ =0 / = 0

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PROBLEM11 114' ' The initial velocity v0 of a hockey puck is 170km/h.Determlargest value lessthan45)oftheanglea for whichthepuc

the net, b)the corresponding time required for the puck to reac

SOLUTION

Horizontalmotion:

Vertical motion:

or t = v0cosaI v0sina)f--g*

=xtana- co s2ajctana--- l+tan2a2v2 V t

Data:

a)

b)

v0= 170 km/h =47.222m/s, j = 4.8m atpointC,y=1.22m atpointC.

9.81) 4.8)

t

x2 4.8tan2a - 94.712a+25.073= 0

tana=0.26547 and 94.45a=14.869 or 89.4x 4.8v0cosa ~ 47.222)cosl4.869

a

=

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PROBLEM11 139Amotorist istravelingon acurved portionofhighwayofradiusa speed of 72km/h.The brakes are suddenly applied, causing thdecreaseat aconstant rateof 1.25m/s2.Determine themagnitutotal accelerationof theautomobile or)immediatelyafter thebrbeen applied, b)4 slater.

SOLUTIONInitial speed.Tangentialacceleration. a) Total acceleration at t =0.

v0= 72 km/h = 20m/sa,=-1.25m/s2

p 350

a=^a f +a\>/(-1.25)2 +(1.14286)2b) Total acceleration at t =4 s .

v= v0+a, t =20 +(-1.25)(4)=15m/s

a= 1.69

35

=a n=^/(-1.25)2+(0.6426)2 a=1.40

PROPRIETARY MATERIAL 2007 Th eMcGraw-Hill Companies, Inc. All rightsreserved. Nopart ofthis Manual may bedisplor distributedin anyform or by anymeans, withouttheprior written permissiono f thepublisher,o rused beyond thelimited distributieducatorspermittedby McGra w-Hill for theirindividualcoursepreparation. Ifyou are astudent using thisManual , you areusingitwi164

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PROBLEM11 164The oscillation of rod OA about O is defined by the relation6 = 4/7r) sinnt , where6 andtareexpressedinradiansand seconds,respectively. CollarBslidesalongthe rod sothatitsdistance fromOisr = 10/(/ + 6), where r and t are expressed in mm and seconds,respectively. When t =1 s, determine a) the velocity of the collar, b) the total acceleration of the collar, c) the acceleration of the collarrelativeto therod.

SOLUTIONDifferentiate the expressions forrand 6with respect to time.

100r mm,

4 .= sm n t rad,r = mm/s,('+ 6)

= 4cos7T7rad/s

2 0 , 2r = r mm/s( +6)

9 =47rsin;r?rad/s2

At t =1 s,

a) Velocity of the collar.

10 10 .. 20 . 2r = m m : r = mm/s, r = mm/s7 49 3436= 0, 0 = -4 rad/s, 0 = 0

vr =r =0.204 mm/s, ve=r =-5.71 mm/sVB= 0.204 mm/s)er - 5.71mm/s)ee

343

Acceleration of thecollar.

ae=r

c) Accelerationof thecollar relativeto therod. 20

_ 1 = _- =1-633mm /s2

B O -as =-(22.8mm/s2)er+(l.633mm/s2)ee

= 0.0583mm/s2)er4

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PROBLEM 12 13Th e two blocks shown are originally at rest. Neglecting the masses of thepulleys and the effect of friction in the pulleys and assuming that thecoefficients offriction between both blocksand theinclineare p .s - 0.25and /J .k = 0.20, determine a) the acceleration of each block, Z > ) thetensionin the cable.

3

UT ONLet the positive directions of XA and XB bedownthe incline.Constraintof thecable: XA +3xB =constant

O A +3as = 0 aB =--aABlock^:+\y =0: NA - /w4gcos30=0

/IF,=ma: mAgsm30- nNA -T =mAaAEliminate NA.

mAg(sm3Q- ncos30)-T =mAaABlockB:+\y =0: NB - mBgcos30=0

+/ZF = ma: wfigsin30 + f j :NB -3T =mBaB = ^ -X Eliminate NB.

m ?g sin30+ncos30)-3T = -m0a'B AEliminate T.

3mAg- wsg)sin30-n(3mAg wgg)cos30= ^ ]a ACheckthevalueo f fis required forstatic equilibrium. Se taA =0 andsolve for j. .

_ 75-20) 3mA + /wB )cos30 75 + 20)

Since M s = -25

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PRO LEM12 13CONTINUEDA _

g m B)sin30- /i 3m^ mB)cos30

_ 30- 8)sin30- 0.20) 30 8)cos3030+ 2.667

(a ) aA 0.13525) 9.81)=1.327m/s2 A 1.327m/s2a B -[- Vl.327)=-0.442m/s2 as=0.442

(b) T=m^g sin30- u cos30)-mAaA= I0) 9.8l) sin30-0.20cos30)- 10) 1.327)

=18.7

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PROBLEM 12 32TheweightsofblocksA,B,andCare WA =Wc =20Ib ,and WB = 10Ib.Knowing thatP - 50Ibandneglecting themassesof thepulleys and theeffect of friction, determine a) the acceleration of each block, b) thetension in thecable.

UT ON the positive direction forpo sition coordinates, velocities,andaccelerationsbe to the right. Let theorigin;at the fixedanchor.

aintofcable: 3 xc - XA) + xc - XB) + ~XB) =constant4ac- 2a B - 3aA = 0 1)

ZT HT

3T 3Tck.A: 3T= mAaA or aA = = mA 202T 2TckB: 2T =mBaB or aB = = mB 10

ckC: P- 4T =mcac or ac=tituting 2), 3),and 4)into1),

P-4T P-4T20 20

2)3)4)

203T_2 -3 10 20 =0

20+10+2020T = 0.12121)P= 0.12121) 50)=6.0605Ib

.)From 2),aj= P) 6.0605) 32.2)=293ftjs2

(3),a PXX ).39 0ft/s[50- 4) 6.0605)1 32.2)From 4), ac = -^ -=41.5ft/s2

19 Asdetermined above,

a -29.3ft/s2

aR = 39.0 ft/s2

a =41.5 m/s2^=6.06Ib

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PRO LEM 12 39During a h amm er throwe r s practice swings, the 16-lb head A of thehammer revolves at a constant speed vin a horizontal circle as shown. Ifp= 3 ft and 6 =60, determine a)the tension in wire BC, b)the speedof the hamm er s head.

Na If,, =0: Tsin0 W =0

16sinO sin 60

= man: Tcos =mP

2 _ pTcosd _ pWcosdmsind

or T = 18.48 Ib

tan0 tan 6055.77tv = 7.47 ft/s

ETARY MATER IAL 2007 Th e McGraw-Hill Companies, Inc. All rights reserved. Nopart of this Manual may be displayed, reproducedi in anyform or by any means, without thepriorwritten permission of thep ublisher, or used beyond thelimited distribution toteachers an dpermitted byMcGraw-Hillfor their individual course prep aration. Ifyou are a student using this Manual,you areusing it without permission.

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PRO L M 12 48During ahigh-speed chase, an 1100-kg sports car traveling at aspeed of160 km/h just loses contact w ith the road as it reaches the crestAof ahill, a)Determine the radiusofcurvature p of theverticalprofile of the roadat A. b) Using the value of p found in part a, determine the forceexerted on a 70-kg driver by the seat of his 1400-kg car as the car,traveling at a constant speed of 80 km/h,passes throughA .

SOLUTIONa) v =160 km/h = 44.44m/s

Wheels do nottouchthe road.= man: mg = -m v Ip

g 9.81p = 201m

b) v = 80 km/h = 22.22m /sm = 70 kg for passenger

IF,, = ma

mo, N mg = mv

N =m g

= (70) 9.81 - 22.22201.4N = 5 1 5 N J - 4

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PRO L M 12 61AturntableAisbuiltintoastagefor use in a theatrical production. It observed duringarehearsal that atrunkB startstoslideon the turntab12saftertheturntable beginstorotate. Knowing thatthe trunk undergoa constant tangential accelerationof 0.75 ft/s2, determine the coefficieofstaticfrictionbetween thetrunkand the turntable.

SOLUTION

FtT k

rvn aB

Uniformlyaccelerated motionon acircular path, p = 8 f tv = v0 a,t

=0 + 0.75) 12)=9 ft/s

F . =ma.=a : F .= W - = W =0.0233WI t A g g 32.2

Fn

F =

BVSP 32.2) 8) =0.3144 fF

+ F2 =0.315WThisis thefriction force availabletocausethetrunktoslide.The normalforceNis calculatedfromequilibrium of forces in thevertical direction.

ZF,, = 0: N - W =0 N = WSince sliding is impending, F = = 0.315W =0.315

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PRO L M 13 12The 7-kg block A is released in the position shown with a velocity of1.5 m /s up the incline. Knowing that the velocity of the block is 3 m /safter it hasmoved0.6 m up the incline, determine thework doneby thefrictionforce exertedon theblock. Neglectthemassesof thepulleys.

ION

30

t>

=7 9.81) =68.67 N

Given:BlockA isreleasedat theposition showna t avelocityof1. 5 m /s up .After mo ving 0.6 m the velocity is 3m/s.

Find:work donebyfriction forceon theblock, V fJFrom the Law ofCosines

d1= 1.2)2+ 0.6)2- 2 1.2) 0.6) cos15)d2= 0.4091m2d = 0.63958m

Uc =140N - 1.2- 0.63958) m =39.229 N mU A =-68.67 N s in l5 ) 0 .6 m ) =-10.664N - m

JA~Ufric,ion=T2 Tl=>A~ *}]39.229- 10.664-Ufriction= 1 7kg)[ 3)2- 1.5)2]m

~ ufrictwr =-39.229+ 10.664 +23.625=-4.94 J

MtOPRI T RY MATERIAL. 2007 T he McGraw-Hill Companies, Inc. A ll rights reserved.N o part of this Manual may be displayed reproduc edordistributed in a ny form or by any means without th epriorwrittenpermission of th epublisher,o r used beyond the limited distribution to teachers an deducatorspermittedb y McGraw-Hill for their individual course preparation. If you are a student using this Manual you are using it withoutpermission.

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PRO L M 13 44A bag is gently pushed off the top of a wall atA and swings in a verticalplane at the end of a rope of length / Determine the angle 9for which therope will break, know ing that it can withstand a m aximum tension equal totwicetheweightof thebag.

l s work-energy:position 1 is atA position2 is atB.(1)

Where r,=0; /,_,.Substitute

0 +mgl sin = mv\2B =2 g / s i n 0

Fo rJ=2 fFus eNewtons2ndlaw.T-2.W

= mv

f maSubstitute (2)into(3)

(2 )

3

m io /sing2 = 3sin0

o r s i n 0=-=0=41.81c3=41.8^

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PRO L M 13 50A pow er specificationformula is to be derived for electric motors7 drive conveyor belts moving solid material at different rates to difj heights anddistances. Denoting the efficiency of the motors by

neglecting the power needed to drive the belt itself, derive a fora ) in the SI system of units, for the power P in kW, in terms of the iflow rate m inkg/h, the height b , and the horizontal distance / inand b) inU.S. customary units, for thepower in hp, interm sof the iflow ratem intons/h,and theheightbandhorizontal distance/ infeet

SOLUTION

a) Material islifted to aheight bat arate,Thus,

m k g / h ) g m / s2 )=

At 3600s/h) ^36001 0 0 0 N - m / s = l k w

Thus, including motor efficiency, r\ )=

P k w ) =0.278x10

b) At/ [fF(tons/h)(20001b/ton)][6(ft)]At ~ 3600s/hW b=ft-lb/s; lhp = 550f t - lb/s1.8

With hn IfMhyO p Ti l550 f Mb/s J | _ r ? Ji .oioxicr3n

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Sf t / s- 2 001B

PRO L M 13.65A spring isused to stop a200-lb package which is moving downa 20incline. The spring has a constant k = 125Ib/in.and is held by cables sothat it is initially compressed 6 in. Knowing that the velocity of thepackage is 8 f t / s when it is 25 ft from th e spring an d neglecting friction,determine th e maximum additional deformationof the spring in bringingthe package to rest.

UTIONervation o f energyion 1) is at the top of theincline; position 2) iswhent hespringhasmaximum deformation

fc = 15001b/ f t7; + F, = T2 +F2

At l) 1 2\8)2= 198.76 f t - l b2^32.2f

At 2)

F, = F + F e] = mgzl =kx\m atpoint 2

= 2 0 0 2 5- x)sin20+- l500) 0.5)2x = Deformationof thespringFj = 1710.1+68.404*+1 87.5

T 2 =0; F2 =/ 2+Ve2 = kx\i l500) 0.5+x}2Substituting into 1) 198.78+1710 .1+68.404*+187.5= 750 0.5+ x 2Solve 750x2+681 .596* - 1908.9= 0

x=-2.11 or +1.2044 ftx =1.204 f t-

= 14.45in.

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300 mm

PROBLEM 13 69A thin circular rod is supported in avertical plane by abracket at A.Attached to thebracket andloosely wound around the rod is aspringofconstant k =40 N/m and undeformed length equalto the arc ofcircleAB. A200-g collar C isunattached to thespring and canslide withoutfriction alongtherod.Knowing thatthecollarisreleased fromrest whenQ =30, determine a) the velocity of the collar as it passes throughpoint B, b) the force exerted by the rod on the collar as is passesthroughB.

I D

=0.3 7o

a) vr = 0, Tc =0 WML ~2 V

TB=1(0.2kg)v2rB = o . i v 2 vc

arc BC AL5C = /?/

AL B C = 0.3 m) 30) 18 0CAL B C =0.15708m

(K c)g =-/t(ALsc)2= -(40N/m)(0.15708m)2 =0.493481

(Kc) = 7Z ( l-cos0)=(0.2kg)(9.81m/s2)(0.3m)(l-cos30)(Fc) =0.078857J

Kc = Vc)e + VC) =0.49348J +0.078857J =0.57234JVB= VB}e+ VB}g =0 +0 = 0

Tc + Vc = TB +VB; 0 +0.57234= 0. lv|v2 =5.7234m2/s 2 B vs =2.39m/s

1 -W= B_R R=1.962N+(0.2 (5.7234m2/s2)(0.3 m)=1.962N +3.8156N =5.7776N F R =5.

JET RY MATERIAL. 2007The McGraw-Hill Companies, Inc.All rightsreserved. Nopart of this Manual may bedisplayed reproducedr^Mlributedin anyform or by anymeans withoutthepriorwrittenpermissionof thepublisher orused beyondthelimited distribution toteachers and

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- , j PROBLEM 13.75A ePenduhim shown is releasedt y before the cord touches the fixed

| / ~^\ afor w hich thependulum bob w i:i_L_i.

\

\ /from rest atAand swings through 90pegB.Determine the smallest value of1describe a circle about t he peg.

SO UT ONUse conservation of energy from the po int o f release (^4) and the top of the circle.

7j + F, = T2 V 2 (1) (datum at lowest point)where

=0 ; F, = m giAt 2 = W v2; V2 = mgz =

1a

0+mgl =-mv2+2mg l - a]ubstituting into (1)Weneed another equation - use Newton s 2 ndlaw at the top. (Tension, T0 =0 attop)

Substituting into (2) =gp=g t-a)mgl=-mg l-a)+2mg l - a ] = l-a -4a

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PROBLEM13.134The system shown isreleased fromrest. Determine the time ittakes forthe velocity ofA to reach 0.6m/s. Neglect frictionand themassof thepulleys.

NKinematicsLengthofcableisconstant.

L = 2XA + XB

=2vA +VB =0dt

(v A =0.6m/s

CollarAm A =15kg

( iM+(2T)(/,_2)- V,_2=(vJ20+[27- (15X 9.8l)]/,_2 = 15) 0.6)

T - 73.575)^_2= 4.5

CollarBm B= 10kg

(vB)2= 2(vA)2=\.2 m/s+( *BvA\T(t,_2} WB(t}_2)(mBvB\+[(10x9.81 - r](f,_2)=1

AddEquation(1) andEquation(2) eliminating7)(98.1-73.575)? _2 =4.5+ 12

165

(1)

(2)

1-2 24.52 =0.673s = 0.673s

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PROBLEM 13 149Tw o swimmersA andB, ofmass 75 kg and 50 kg, respectively,diveofftheendofa200-kgboat. Each swimmerhas arelative horizontal velocityof 3 m/swhenleaving the boat. If theboatisinitiallyatrest,determine itsfinal velocity, assuming that a) the two swimmers dive simultaneously,b) swimmerAdives first, c)swimmerBdives first.

UTIONmA=75kg, mB=50kg, mc=200 kg Boat)

a) Swimm ers dive simultaneouslyQQ

c

0=mcvc+ mA+mB)v2Relative velocityofswimmers with respectto theboatis 3m/s

v2- vc= 3m/s >v2=vc+3Substitute into 1)

Q = mcvc+ mA+mB) vc+3)Solve

_ l mA+mB)_ -3 75+ 50)

1)

mc 75 + 50 +200)vr=1.154m/s

Adives first and thenBA SQO _Q O

jc-dirQ = mc+mB}vC2-mAv2 2)

continued

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PROBLEM 13 165Tw oidentical billiard ballscanmovefreely on ahorizontal table. BallAhas avelocity v0asshownandhits ballB,whichis atrest at apointCdefined by Q = 45. Knowing that the coefficient of restitution betweenthe twoballs is e = 0.8 andassumingno friction, determinethevelocityof each ballafterimpact.

Ba At-dirBa Bt-dir

Ball,4+Bn-dir

Coefficient ofrestitution

Solve(1) and (2)

With numbers

= jttvAt vAt=v0

= (1)

2)

e=0.8; 0= 45v Al=v0sin45= 0.707v0

1-0.8VAn V cos45 =0.0707v0

cos45=0.6364vcontinued

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k = 34 Ib/ft

PROBLEM 13 183A0.6-lbcollarA isreleased fromrest, slides downa frictionlessstrikes a 1.8-lb collar B which is at rest and supported by a spconstant 34 Ib/ft. Knowing that the velocity of collar Aimmediately after impact, determine a) the coefficient of rebetween the two collars, b) the energy lost in the impact,maximum distance collarBmoves downthe rodafterimpact.

SOLUTION

After impact

Velocity ofAjust b efore impact, v0= ^2(32.2ft/s2)(3.6ft)sin30

v0 =^2(32.2X3.6X0.5) =10.7666ft/sConservationofmomentum

+/ mv =mv -mv (Ml =lll BB g J g )Bg'scancel

Restitution

From 1) vFrom 2)fl)b) Energy loss

=(M=(M\.7666ft/s)=3.5889ft/s[ i s ) l^i.s/ =(Vvo)> =~

e

AEnergy =mxg(3.6)sin30 -- 0 .6 )_ I _LL\ .5889f t; 2^32.2/=1.08-0.36=0.72ft-lb

Loss=0.720/c) Static deflection =x0,Bmoves down ? f i

Conservation ofenergy D to Position D-springdeflected, x0

kx 0 =wflgsin30

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PROBLEM 13 183 CONTINUED

T = m a I =0

m Bgd Bsin3QV =

fcco+ /wga ffis in30 +mBv\k d\2dBxQ + X Q ] + 0 +

kdl =B mBVB> 32.2 3.S889)2dB =0 .1455ft

dn =1.746 n .

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