Beer Ch13 Prob Sol 02

The 40-lb block is moving downward with a speed of 3 ft/s at t = 0 whenconstant forces P and 2P are applied through the ropes. Knowing that theblock is moving upward with a speed of 2 ft/s when t = 4 s, determine(a) the magnitude of P, (b) the time at which the speed is zero. Neglectthe effect of friction and the masses ofthe pulleys.

f:(5P - 40)dt = (40/32.2)(2 - (-3))

''1 lfO 16

f~(5P - 40)dt = (40/32.2)(0 - (-3))

PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

The coefficients of friction between the three blocks and the horizontalsurfaces are J1s = 0.25 and J1k = 0.20. The weights of the blocks areWA = We = 20 lb, and WB = 10 lb. The velocities of blocks A and C at timet = 0 are VA = 9 ft/s and Ve = 15 ft/s, both to the right. Determine (a) thevelocity of each block at t = 0.5 s, (b) the tension in the cable.

ctJE$ LtJ..io. ~ (;l.o)t ~ i (}.l(),()t

'J,o t 'J.,6t

0~t"i 0.1(10) tlot

20(2) 3Tt - 4t = -(VA - 9)

32.220

(3) -4Tt - 4t = -(ve -15)32.210(4) 2Tt - 2t = -( vB - 16.5)32.2

1.5T - 0.6211 VA = -3.5901

-2T -0.6211 Vc =-7.3168

T - 0.31056 vB = -4.1242

PROPRIETARY MATERIAL 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, without the prior written permission of the publisher. or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual. you are using it without permission.

54 ftfs--.. ;jOfL's4Ir~-------I---_ /-~T. A player hits a 2-oz tennis ball with a horiz~ntal initia~ velocity of ~4 ftls! . I ...v\..... .3 ft at a height of 4.5 ft. The ball bounces at pomt A and nses to a maximum

height of 3 ft where the velocity is 30 ftls. Knowing that the duration ofthe impact is 0.004 s, determine the impulsive force exerted on the ball atpointA.

m = 2 oz (1Ib/16 oz) ( 1 2) = 0.003882 Ibs2/ft32.2 ftls

.... x: 0.003882(54) - FH (0.004) = 0.003882(30), FH = 23.292 lb

+t y: -0.003882(17.0235) + Fv(0.004) = 0.003882(13.8996), Fv= 30.011lb


6 ftis----"120 ton A 120-ton tugboat is moving at 6 ft/s with a slack towing cable attached

to a 100-ton barge which is at rest. The cable is being unwound from adrum on the tugboat at a constant rate of 5.4 ftls and that rate ismaintained after the cable becomes taut. Neglecting the resistance of thewater, determine (a) the velocity of the tugboat after the cable becomestaut, (b) the impulse exerted on the barge as the cable becomes taut.

mB = (100)(2000) (_1_) = 6211.18 lb . s2/ft32.2

mr = (120) (2000) (_1_) = 7453.42lb . s2/ft32.2

lre~

''''''''' B AR.G- E

0+ F!1t = mB vB = (6211.18) vB

mr (6) - F 11/ = mT vr

7

7453.42( 6) = (6211.18) [vr - 5.4] + 7453.42 vr(13664.6) Vr = 78260.9

VB = 5.7273 ftls -5.4 ft/s = 0.3273 ft/s

F!1t = mB VB = (6211.18) (0.3273) = 2032 1bs

PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, without the prior wrillen permission of the publisher, or used beyond the limited distribution to teachers andeducators permi/led by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Vy = ~2g(4.8) = 17.582 ft/s

A 3-oz ball is projected from a height of 4.8 ft with a horizontal velocityof 6 ft/s and bounces from a 14-oz smooth plate supported by springs.Knowing that the height of the rebound is 1.8 ft, determine (a) thevelocity of the plate immediately after the impact, (b) the energy lost dueto the impact.

mball = (~) (_1_) = 0.00582316 32.2

mplate =G:) (3~.2) = 0.027174

V~ = ~2g(1.8) = 10.7666 ft/s

(+Yp

(0.005823) (17.582) + 0 = -(0.005823) (10.7666) + (0.027174) V~late

V~late = 6.07 ft/s ~

A baseball player catching a ball can soften the impact by pulling hishand back. Assuming that a 5-02 ball reaches his glove at 96 mi/h andthat the player pulls his hand back during the impact at an average speedof 25 ftls over a distance of 8 in., bringing the ball to a stop, determinethe average impulsive force exerted on the player's hand.

0=....--96 mi/h

I

1r'l-1f = 0

5m = -/g = 0.3l25/g

16

8

t = ~ = 12 = 0.02667 svAVE 25

--- +0.3125 (140.8) - FAVE (0.02667) = 0g

PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hillfor their individual course preparation./fyou are a student using this Manual, you are using it without permission.

4 m/sA----+-Two steel blocks slide without friction on a horizontal surface;immediately before impact their velocities are as shown. Knowing thate = 0.75, determine (a) their velocities after impact, (b) the energy lossduring impact.

DO

Solve 2 equations and 2 unknowns

V~ = - 2.3 m/s; v~= 2.2 m/s

12121 21 21[ = "2mA vA +"2 mB VB ="2 (0.6)(4) +"2 (0.9)(2) = 6.6 J

T2= !mA (v~)2+!mB (v~)2=! (0.6)(2.3)2 +! (0.9)(2.2)2 =3.765 J222 2


Three steel spheres of equal mass are suspended from the ceiling by cordSof equal length which are spaced at a distance slightly greater than thediameter of the spheres. After being pulled back and released, sphere Ahits sphere B, which then hits sphere C. Denoting by e the coefficient ofrestitution between the spheres and by va the velocity of A just before ithits B, determine (a) the velocities of A and B immediately after the firstcollision, (b) the velocities of Band C immediately after the secondcollision. (c) Assuming now that n spheres are suspended from the ceilingand that the first sphere is pulled back and released as described above,determine the velocity of the last sphere after it is hit for the first time.(d) Use the result of part c to obtain the velocity of the last sphere whenn = 8 and e = 0.9.

(a) First collision (between A and B)

The total momentum is conserved

V~ = Va (1 - e) 2

v~ = vo(l + e) 2

(b) Second collision (Between Band C)The total momentum is conserved.

, ",mVB + mvc = mVB + mvc

Using the result from (a) for v~

Va (1 + e) ",----+ 0 = vB +Vc2

Substituting again for v~ from (a)

(l+e)()_, "va--- e - Vc -vB2


Two identical billiard balls can move freely on a horizontal table. Ball Ahas a velocity Voas shown and hits ball B, which is at rest, at a point Cdefined by (J = 45. Knowing that the coefficient of restitution betweenthe two balls is e = 0.8 and assuming no friction, determine the velocityof each ball after impact.

I

Yr1A~h

BallA t-dir

BallB t-dir

, (1 - e )VAn = Vo -2- cos(J . , (l+e} (JVBn= Vo -2- os

V~t = vosin45= 0.707vo

V~n= voC -28COS450)= 0.0707 VoV~t=o

V~n= voC +28}OS450= 0.6364vo


IIVI/AI= [(0.707vO)2 + (0.0707VO)2]2

= 0.711 Vo

{3 = tan- I(0.0707)= 5.710600.707 .

So e = 45 - 5.7106 = 39.30

-I 11 .&..30vA= .7 VoVa= 0.636vo ~


A 17.5-lb sphere A of radius 4.5 in. moving with a velocity Vo ofmagnitude Vo = 6 ft/s strikes a 1.6-lb sphere B of radius 2 in. which wasat rest. Both spheres are hanging from identical light flexible cords.Knowing that the coefficient of restitution is 0.8, determine the velocityof each sphere immediately after impact.

() = cos-1 ~ = 22.626.5

Total momentum conservedBall A:

/1JB'/6---- .....-- ----

PROPRIETARY MATERIAL. 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

(17.5Ig)( 6 ft/s) = (1.6Ig )v~(~)+(17.5Ig )v~6.5105= 1.6(~14.8(~)+ V~(~)~+ 17.5v~6.5 6.5 6.5 U

v~ = 5.22 ft/s ----- ~v~ = 9.25 ft/s L 22.6..

PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.

(a) Show that when two identical spheres A and B with coefficient ofrestitution e = 1 collide while moving with velocities vA and vB whichare perpendicular to each other they will rebound with velocities vA andV'B which are also perpendicular to each other. (b) To verify thisproperty, solve Sample Prob. 13.15, assuming e = 1, and determine theangle formed by vA and v'B'

n-DirectionTotal momentum is conserved.

(v~t = VA COS 0(V~)n = -VB sinO

continuedPROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

vtan a = -.d...VB

V~ = (VB)~+ (V~)~ = ~V~ COS2 (J + V~COS2 (J

a = f3 = tan-1 VA = tan-1 30 = 36.9vB 40

PROPRIETARY MATERIAL. 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproducedor distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual. you are using it without permission.

A boy releases a ball with an initial horizontal velocity at a height of

S=~~-~~

VBy = 0.6(3.431) - 9.81tB= 1.3724 m/s

B---=============~~=======:l0, I?. rY1(, G,i5 mYc = 0.12 + vBic - 4.905t~ = 0 = 0.12 + 1.3724tc - 4.905t~

tc = 0.3497 sXc = 6.75 = eBvotc = eB(21.444)(0.3497)


A 2.5-lb block B is moving with a velocity V 0 of magnitude Vo = 6 ft/sas it hits the l.5-lb sphere A, which is at rest and hanging from a cordattached at O. Knowing that J.1k = 0.6 between the block and thehorizontal surface and e = 0.8 between the block and the sphere,determine after impact (a) the maximum height h reached by the sphere,

--- Vn (b) the distance x traveled by the block.

(VA - vB)e = v~ - v~: (0 - 6)(0.8) = v~ - v~ ~ v~ - v~ = -4.8Solving (1) and (2) simultaneously


T2 = 0V2 = mAgh = 1.5h

1\ + Vi = T2 + V2: 1.06124 = 1.5hh = 0.70749 ft = 8.9899 in.

T2 = 0

T = ~mBv~2 = ~(2.5 \1.95)2 = 0.147612 2 g)'

UI-2 = -Ffx = -J.1kWBx = -D.6(2.5)x = -1.5x

1\ + UI-2 = T2: 0.14761-1.5x = 0x = 0.0984 ft = 1.1808 in.

PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproducedor distributed in any form or by any means, without the prior written permission of the publisher. or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual. you are using it without permission.

A 0.6-lb collar A is released from rest, slides down a frictionless rod, andstrikes a 1.8-lb collar B which is at rest and supported by a spring ofconstant 34 Ib/ft. Knowing that the coefficient of restitution between thetwo collars is 0.9, determine (a) the maximum distance collar A moves upthe rod after impact, (b) the maximum distance collar B moves down therod after impact.

Vo = 2Jih = J2 (32.2 fils2)( 3.6 ft )sin 30= ~2(32.2)(3.6)(0.5) = 10.7666 fils

(0.6} _ (1 8) (0.6 }- 0- . g vB- - Ag g

g's cancel

Substituting for VB from (2) in (1)

0.6vo= 1.8(0.9vo- VA) - 0.6vA; 2AvA= 1.02vB

(a) A moves up the distance d where,

!mAv~ =mAgdsin300; !(4.5758 fils)2 = (32.2 fils2)d(0.5)2 2

INdA = 0.65025 ft = 7.80 m ~dB

(b) Static deflection = xo, B moves down hOConservation of energy (1) to (2)

Position (1) - spring deflected, xo

loco = mBgsin30

PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced ordistributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.

..!..kx5+ mgdBsin30 + ..!..mBV~= ..!..k(d~ + 2dBxO + x5) + 0 + 02 2 2

34d~ = (31~~2}5.1141)2

PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced ordistributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.

~ht

A 340-g ball B is hanging from an inextensible cord attached to a supportC. A 170-g ball A strikes B with a velocity Vo of magnitude 1.5mls at anangle of 60 with the vertical. Assuming perfectly elastic impact (e = 1)and no friction, determine the height h reached by ball B.

Relative velocity in the n-direction

[-VA - (VBt]e = -v'Bcos30 - V~;

(-1.5 -0)(1) = -o.866v~ -v~Solving Equations (1) and (2) simultaneously

v~ = 0.9446 mis, v~ = 0.6820 mlsConservation of energy ball B

1 (' )2II = -mB VB2

II =!WB(3.0232)22 gcontinued

PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced ordistributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

h = (0.9446)2(2)(9.81)

PROPRIETARY MATERIAL. 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced ordistributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permitted by McGraw-Hili for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Documents

Beer Ch13 Prob Sol 02