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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.

227

PROBLEM 12.1

The acceleration due to gravity on Mars is 23.75 m/s . Knowing that themass of a silver bar has been officially designated as 20 kg, determine, onMars, its weight in newtons.

SOLUTION

220 kg, 3.75 m/sm g= =

( )( )20 3.75W mg= = 75 NW =

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228

PROBLEM 12.2

The value of the acceleration of gravity at any latitude is given by

( )2 29.7087 1 0.0053 sin m/s ,g = + where the effect of the rotation of theearth as well as the fact that the earth is not spherical have been taken into

account. Knowing that the mass of a gold bar has been officiallydesignated as 2 kg, determine to four significant figures its mass inkilograms and its weight in newtons at a latitude of (a) 0 , (b) 45 ,(c) 60 .

SOLUTION

At all latitudes, 2.000 kgm =

(a) ( )2 20 , 9.7807 1 0.0053 sin 9.7807 m/sg = = + =

( )( )2.000 9.7807W mg= =

19.56 NW =

(b) ( )2 245 , 9.7807 1 0.0053 sin 45 9.8066 m/sg = = + =

( )( )2.000 9.8066W mg= = 19.61 NW =

(c) ( )2 260 , 9.7807 1 0.0053 sin 60 9.8196 m/sg = = + =

( )( )2.000 9.8196W mg= = 19.64 NW =

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229

PROBLEM 12.3

A spring scaleAand a lever scaleBhaving equal lever arms are fastened

to the roof on an elevator, and identical packages are attached to the

scales as shown. Knowing that when the elevator moves downward with

an acceleration of 20.6 m/s the spring scale indicates a load of 3.2 kgdetermine (a) the weight of the packages, (b) the load indicated by the

spring scale and the mass needed to balance the lever scale when the

elevator moves upward with an acceleration of 2.6 m/s .

SOLUTION

Assume 232.2 ft/sg=

Wm

g=

:s

WF ma W F ag

= =

3.2 9.811 or

.61 1

9.81

s

s

a FW F W

ag

g

= = =

33.4NW =

33.43.4 kg

9.81

Wm

g= = =

:s

WF ma F W a

g = =

1

.633.4 1

9.81

s

aF W

g

= +

= +

35.4 NsF =

For the balance systemB,

00: 0

w pM bF bF = =

w pF F=

But, 1w w

aF W

g

= +

and 1

p p

aF W

g

= +

so that andp

w p w

WW W m

g= = 3.3.4 kgm =

9.81 m/s2

PROBLEM 12.3

A spring scale Aand a lever scale Bhaving equal lever arms are fastened

to the roof on an elevator, and identical packages are attached to the

scales as shown. Knowing that when the elevator moves downward with

an acceleration of 20.6 m/s the spring scale indicates a load of 3.2 kg,

determine (a) the weight of the packages, (b) the load indicated by the

spring scale and the mass needed to balance the lever scale when the

elevator moves upward with an acceleration of 2.6 m/s .

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230

PROBLEM 12.4

A Global Positioning System (GPS) satellite is in a circular orbit

20157 km above the surface of the earth and completes one orbit every

12 h. Knowing that the magnitude of the linear momentum of the satellite

is 340194 kg. m/sand the radius of the earth is 6370 kmdetermine (a) themass of the satellite, (b) the weight of the satellite before it was launched

from earth.

SOLUTION

Periodic time: 12 h 43200 s = =

Radius of Earth: =R

Radius of orbit: 6370 20157 26527 kmr = + =

Velocity of satellite:( )( )32 26527 102

43200

rv

= =

It is given that =mv

(a)mv

mv

= =

m =

(b) ( ) ( )= = =W mg

=W

6370 km

3858.2 m/s=

340194 kg.s

340194

3858.288.17 kg

88.17 9.81 865 N

88.17 kg

865 N

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231

PROBLEM 12.5

The 80 Nblock starts from rest and moves upward when constant forcesof 20 Nand 40 Nare applied to supporting ropes. Neglecting the massesof the pulleys and the effect of friction, determine the speed of the block

after it has moved .457 m.

SOLUTION

+ :y y y

F ma a= + + + =

( )( )y

a = =

y

dv dy dv dva v

dt dt dy dy= = =

y yv dv a d =

2

0 0

1

2

v v

y y yv dv a d v a y= =

( )( )( )2 2yv a y== v=

20 20 20 40 80 809.81

9.81 20

802.45 m/s

2

2.45 .457 1.5 m/s

80-N

,

=

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232

PROBLEM 12.6

A motorist traveling at a speed of 108 km/h suddenly applies the brakesand comes to a stop after skidding 75 m. Determine (a) the time requiredfor the car to stop, (b) the coefficient of friction between the tires and the

pavement.

SOLUTION

Data:0

108 km/h 30 m/s, 75 mfv x= = =

(a) Assume constant acceleration. constant= = =dv dv

a vdx dt

0

0

0

fx

vv dv a dx=

2

0

1

2 fv a x =

( )( )( )

2

2030

6 m/s2 2 75f

va

x= = =

0

0

0

ft

vdv a dt =

0 fv a t =

030

6f

vt

a

= =

5.00 sft =

(b) + 0: 0yF N W= =

N W=

:xF ma N ma= =

ma ma a

N W g= = =

( )69.81

= 0.612=

acceleration:

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233

PROBLEM 12.7

A 1400-kg automobile is driven down a 4incline at a speed of 88 km/hwhen the brakes are applied, causing a total braking force of 7500 N to beapplied to the automobile. Determine the distance traveled by the

automobile before it comes to a stop.

SOLUTION

(a) : sinfF ma F W ma= + =

sinsin= + = +

f fF FWa g

m m m

( )2 27500 N

9.81 m/s sin 4 4.6728 m/s1400 kg

= + =

2

4.6728 m/sa= 4

0 88 km/h 24.444 m/s= =v

From kinematics,dv

a vdx

=

0

0

0

fx

va dx v dv=

2

0

1

2fa x v=

( )( )( )

22

0 24.4442 2 4.6728

f vxa= =

63.9 mfx =

+

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234

PROBLEM 12.8

In the braking test of a sports car its velocity is reduced from 112.16 km/htozero in a distance of 51.8 m with slipping impending. Knowing that thecoefficient of kinetic friction is 80 percent of the coefficient of static

friction, determine (a) the coefficient of static friction, (b) the stoppingdistance for the same initial velocity if the car skids. Ignore air resistanceand rolling resistance.

SOLUTION

(a) Coefficient of static friction.

0: 0yF N W = = N W=

0v = =

( )02 2 tv

a s s

( )( )

( ) ( )

222

20

0

0 31.16

2 2t

v va

s s

= = =

For braking without skidding , so that | |s s tN m a = =

:t t s t F ma N ma = =

t ts

ma a

W g = = = s =

(b) Stopping distance with skidding.

Use ( )( )0.80k = = =

:t k tF ma N ma = =

27.49 m/s

kt k

Nag

m

= = =

Since acceleration is constant,

( ) ( )

( ) ( )

22 2

0

0

0 31.16

2 2t

v vs s

a

= =

0 64.8 ms s =

112.16 km/h 31.16 m/s.

2 v20

51.89.37 m/s

9.37

9.81.955

.955 .764

7.49

In the braking test of a sports car, its velocity is reduced from 112.16 km/h to

friction:

0,

m/s

v2v20

For braking without skidding:

skidding:

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235

PROBLEM 12.9

A .09 kgmodel rocket is launched vertically from rest at time t=0 with aconstant thrust of 4 Nfor one second and no thrust for t>1 s. Neglectingair resistance and the decrease in mass of the rocket, determine (a) the

maximum height hreached by the rocket, (b) the time required to reachthis maximum height.

SOLUTION

For the thrust phase, :t

WF ma F W ma a

g = = =

( ) 241 1 34.6 m/stFa gW

= = =

At 1 s,t=

( )( )1 34.6m/sv at= = =

( ) ( )221 1

2y at = =

For the free flight phase, 1 s.t> a g= =

( ) ( )( )1 1 1v v a t t = + = +

At 0, 1 s,v t t= = = =

( ) ( )2 21 1 12 2v v a y y g y y = =

( )( )( )

22 2

1

1

0

2 2

v vy y

g

= = =

(a) maxy h= = + h=

(b) As already determined, t=

9.81.09 9.81

34.6

1

2= 34.6 17.3 m

9.81 m/s2

34.6 9.81

34.6

9.813.527 4.527 s

34.6

9.8161 m

61 17.3 78.3 m

4.527 s

PROBLEM 12.9

A .09-kg model rocket is launched vertically from rest at time t=0 with aconstant thrust of 4 N for one second and no thrust for t>1 s. Neglectingair resistance and the decrease in mass of the rocket, determine (a) the

maximum height hreached by the rocket, (b) the time required to reachthis maximum height.

t=4.53 s

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236

PROBLEM 12.10

A 40-kg package is at rest on an incline when a force Pis applied to it.Determine the magnitude of Pif 4 s is required for the package to travel10 m up the incline. The static and kinetic coefficients of friction between

the package and the incline are 0.30 and 0.25, respectively.

SOLUTION

Kinematics: Uniformly accelerated motion. ( )0 00, 0x v= =

( )( )

( )

2 2

0 0 2 2

2 101 2, or 1.25 m/s

2 4

xx x v t at a

t= + + = = =

0: sin 50 cos20 0yF N P mg = =

sin 50 cos20N P mg= +

: cos50 sin 20xF ma P mg N ma = =

or ( )cos50 sin 20 sin50 cos20P mg P mg ma + =

( )sin 20 cos20

cos50 sin50

ma mg P

+ + =

For motion impending, set 0 and 0.30.sa = = =

( )( ) ( )( )( )40 0 40 9.81 sin 20 0.30cos20 593 Ncos50 0.30 sin 50

P+ +

= =

For motion with 21.25 m/s , use 0.25.ka = = =

( )( ) ( )( )( )40 1.25 40 9.81 sin 20 0.25cos20

cos50 0.25sin 50P

+ + =

612 NP =

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237

PROBLEM 12.11

If an automobiles braking distance from 100 km/h is 60 m on levelpavement, determine the automobiles braking distance from 100 km/hwhen it is (a) going up a 6 incline, (b) going down a 2-percent incline.

SOLUTION

Calculation of braking force/mass ( )/bF m from data for level pavement.

0 100 km/hr 27.778 m/sv = =

( )2 2

00

2 2

v va x x =

( )( )

( )( )

22 2

0

0

2

0 27.778

2 2 60

6.43 m/s

v va

x x

= =

=

br:xF ma F ma = =

2br 6.43 m/sF

am

= =

(a) Going up a 6 incline.( )6 =

br: sinF ma F mg ma = =

br sinF

a gm

=

26.43 9.81sin 6 7.455 m/s= =

( )( )( )

22 2

00

0 27.7782 2 7.455

v vx xa

= =

0 51.7 mx x =

(b) Going down a 2% incline. ( )tan 0.02, 1.145 = =

br: sinF ma F mg ma = =

br sinF

a gm

=

( ) 26.43 9.81sin 1.145 6.234 m/s= =

( )

( )( )

22 2

00

0 27.778

2 2 6.234

v vx x

a

= =

0 61.9 mx x =

v0=100 km/h =27.778 m/s

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238

PROBLEM 12.12

The two blocks shown are originally at rest. Neglecting the masses of thepulleys and the effect of friction in the pulleys and between the blocks

and the incline, determine (a) the acceleration of each block, (b) the

tension in the cable.

SOLUTION

Let the positive directions of Ax and Bx be down the incline.

Constraint of the cable: 3 constantA Bx x+ =

13 0 or3

A B B Aa a a a+ = =

For blockA: : sin 30A A AF ma m g T m a = = (1)

For blockB: : sin 30 3B B B B AF ma m g T m a m a = = = (2)

Eliminating Tand solving for ,Aa

g

( )3 sin 30 33

= +

B

A B A A

mm g m g m a

( ) ( )3 sin 30 30 8 sin 30 0.336733 / 3 30 2.667

A BA

A B

m ma

g m m = = =

+ +

(a) ( )( ) 20.33673 9.81 3.30 m/sAa = = 23.30 m/s

A=a 30

( ) 21

3.30 1.101 m/s3

= = B

a 21.101 m/sB

=a 30

(b) Using equation (1),

( )( )( )sin 30 10 9.81 sin30 0.33673AAa

T m gg

= =

16.02 NT =

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239

PROBLEM 12.13

The two blocks shown are originally at rest. Neglecting the masses of thepulleys and the effect of friction in the pulleys and assuming that thecoefficients of friction between both blocks and the incline are s =0.25

and k = 0.20, determine (a) the acceleration of each block, (b) thetension in the cable.

SOLUTION

Let the positive directions of Ax and Bx be down the incline.

Constraint of the cable: 3 constantA Bx x+ =

3 0A Ba a+ =1

3

B Aa a=

BlockA: 0: cos30 0y A AF N m g = =

: sin30x A A A AF ma m g N T m a = =

Eliminate .AN

( )sin 30 cos30A A Am g T m a =

BlockB: 0: cos30 0y B BF N m g = =

: sin 30 33

B AB B B B

m aF ma m g N T m a = + = =

Eliminate .BN

( )sin 30 cos30 33

B AB

m am g T + =

Eliminate T.

( ) ( )3 sin 30 3 cos30 33

BA B A B A A

mm g m g m g m g m a

+ = +

Check the value of s required for static equilibrium. Set 0Aa = andsolve for .

( )( )

( )( )

3 sin30 75 20tan 30 0.334.

3 cos30 75 20

A B

A B

m m

m m

= = =

+ +

Since 0.25 0.334,s = < sliding occurs.

Calculate Aa

gfor sliding. Use 0.20.k = =

continued

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240

PROBLEM 12.13 CONTINUED

( ) ( )

( ) ( )( )

3 sin 30 3 cos30

3 / 3

30 8 sin 30 0.20 30 8 cos300.13525

30 2.667

A B A BA

A B

m m m ma

g m m

+ =

+

+ = =

+

(a) ( )( ) 20.13525 9.81 1.327 m/s= =Aa 21.327 m/s

A=a 30

( ) 21

1.327 0.442 m/s3

= =

Ba 20.442 m/s

B=a 30

(b) ( )

( )( )( ) ( )( )

sin 30 cos30

10 9.81 sin 30 0.20cos30 10 1.327

=

=

A A AT m g m a

18.79 N=T

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241

PROBLEM 12.14

A light train made up of two cars is traveling at 88 km/hwhen the brakesare applied to both cars. Knowing that carAhas a weight of 24947.56 kgandcarBhas a weight of 19958 kgand that the braking force is 31137.5 Non

each car, determine (a) the distance traveled by the train before it comesto a stop, (b) the force in the coupling between the cars while the train isslowing down.

SOLUTION

Data : 24947.56 kgAm =

19958 kgBm =

0v =

(a) Use both cars together as a free body. Consider horizontal force components only. Both cars have sameacceleration.

:x x b b A x B xF ma F F m a m a= = +

31137.5b bx

A B

F Fa

m m

+ += = =

+ +

x

dva v

dx=

0

20 0

0 2= = f

x

x x fv

va dx v dv a x

( )( )( )

22

0215.7 m

2 2f

x

vx

a= = =

to the left

(b) Use carA as free body.Fc= coupling force.

:x x c b A xF ma F F m a= =

( )( )= = + =c A x bF m a F

tensioncF =

88 kg/h = 24.4 m/s

31137.5

249 7.56 199581.38 m/s

2

24.4

1.38

215.7 m

24947.56 1.38 31137.5 65.6 kN

65.6 kN

4

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242

PROBLEM 12.15

Solve Prob. 12.14, assuming that the brakes of carBfail to operate.

Problem 12.14:A light train made up of two cars is traveling at 55 mi/h

when the brakes are applied to both cars. Knowing that carAhas a weightof 55,000 lb and carBhas a weight of 44,000 lb and that the braking forceis 7000 lb on each car, determine (a) the distance traveled by the train

before it comes to a stop, (b) the force in the coupling between the carswhile the train is slowing down.

SOLUTION

Data:

(a) Use both cars together as a free body. Consider horizontal force components only. Both cars have sameacceleration.

:x x b b A x B xF ma F F m a m a= = +

bx

A B

Fa

m m=

+

x

dva v

dx=

0

20 0

0 2= = fx

x x fv

v

a dx v dv a x

2

0

2f

x

vx

a=

to the left

(b) Use carB as a free body.Fc= coupling force.

:x x c B xF ma F m a= =

( )( )19958 .69 13.8 kNcF = =

compressioncF =

24947.56 kgAm =

19958 kgBm =

0v = 24.4 m/s

31137.5+= =

+

31137.5

249 7.56 19958.69 m/s

2

4

( )( )( )

2

431.4 m2

= =24.4

.69

431.4 m

13.8 kN

PROBLEM 12.15

Solve Prob. 12.14, assuming that the brakes of car Bfail to operate.

Problem 12.14:A light train made up of two cars is traveling at 88 km/hwhen the brakes are applied to both cars. Knowing that car Ahas a weightof 24947.56 kg and car Bhas a weight of 199.58 kg and that the brakingforce is 31137.5 kg on each car, determine (a) the distance traveled by the

train before it comes to a stop, (b) the force in the coupling between thecars while the train is slowing down.

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243

PROBLEM 12.16

BlockAweighs 356 N, and block Bweighs 71 N. The coefficients of

friction between all surfaces of contact are 0.20s = and 0.15.k =

Knowing that 0,P = determine (a) the acceleration of block B, (b) the

tension in the cord.

SOLUTION

Constraint of cable: ( )2 A B A A Bx x x x x+ = + = constant.

0, orA B B Aa a a a+ = =

Assume that blockAmoves down and blockBmoves up.

BlockB: 0: cos 0y AB BF N W = =

: sin Bx AB B BW

F ma T N W ag

= + + =

Eliminate ABN and .Ba

( )sin cos B AB B Ba a

T W W W g g

+ + = =

BlockA: 0: cos 0y A AB AF N N W = =

( )cos cosA AB A B AN N W W W = + = +

: sin A

x A A A AB A A

W

F m a T W F F ag = + =

( )sin cos sin cosAB B A Ba

W W W W g

+ +

( )cos AB A Aa

W W Wg

+ =

( ) ( ) ( )sin 3 cos AA B A B A Ba

W W W W W W g

+ = +

Check the condition of impending motion.

0.20, 0,s A B sa a = = = = =

( ) ( )sin 0.20 3 cos 0A B s A B sW W W W + =

continued

Block Aweighs 356 N and block B weighs 71 N. The coefficients of

0

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244


( ) ( )( )0.20 3 0.20 128tan 0.40

64

A B

s

A B

W W

W W

+= = =

21.8 25 .s = < = The blocks move.

Calculate Aa

gusing 0.15 and 25 .k = = =

( ) ( )sin 3 cosA B k A BAA B

W W W W a

g W W

+=

+

( )( )sin 25 0.15 cos25427

= =

( )( )A

a = =

(a) Ba = B =a 25

(b) ( )sin cos AB Ba

T W Wg

= + +

( ) ( )( )sin 25 0.15 cos 25= + +

T =

285 569.10093

.10093 9.812

.99 m/s

2.99 m/s

2.99 m/s

71 71 .10093

46.8N

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245

PROBLEM 12.17

BlockA weighs 350N, and block B weighs 71N. The coefficients offriction between all surfaces of contact are 0.20s = and 0.15.k =

Knowing that =P , determine (a) the acceleration of block B,

(b) the tension in the cord.

SOLUTION

Constraint of cable: ( )2 constant.A B A A Bx x x x x+ = + =

0, orA B B Aa a a a+ = =

Assume that blockAmoves down and blockBmoves up.

BlockB: 0: cos 0y AB BF N W = =

x: sin Bx AB B B

WF ma T N W a

g = + + =

Eliminate ABN and .Ba

( )sin cos B AB B Ba a

T W W W g g

+ + = =

BlockA: 0: cos sin 0y A AB AF N N W P = + =

cos sinA AB AN N W P = +

( )cos sinB AW W P = +

: sin cos Ax A A A AB A AW

F m a T W F F P ag

= + + =

( )sin cos sin cosAB B A Ba

W W W W g

+ +

( )cos sin cos AB A Aa

W W P P W g

+ + + =

( ) ( ) ( ) ( )sin 3 cos sin cos AA B A B A Ba

W W W W P W W g

+ + + = +

Check the condition of impending motion.

0.20, 0, 25s A Ba a = = = = =

( ) ( ) ( )sin 3 cos sin cos 0A B s A B s sW W W W P + + + =

continued

44.5 N

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246


( ) ( )3 cos sinsin cos

s A B A B

s

s

W W W W P

+ =

+

( )( )0.20 cos25 sin 250.20 sin 25 cos 25

= = a a so that the boxes separate. Boxes are slipping.

k =

0: cos15 0yF N mg = =

cos15N mg=

: sin15x kF ma N mg ma = =

cos15 sin15kmg mg ma =

( )cos15 sin15 ,ka g = independent of m.

For boxA, 0.30k =

( )9.81 0.30cos15 sin15Aa = or20.304 m/sA =a 15

For boxB, 0.32k =

( )9.81 0.32cos15 sin15Ba = or20.493 m/sB =a 15

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248

PROBLEM 12.19

The system shown is initially at rest. Neglecting the masses of the pulleysand the effect of friction in the pulleys, determine (a) the acceleration ofeach block, (b) the tension in each cable.

SOLUTION

Letybe positive downward position for all blocks.

Constraint of cable attached to massA: 3 constantA By y+ =

3 0A Ba a+ = or 3A Ba a=

Constraint of cable attached to mass C: constantC By y+ =

0C Ba a+ = or C Ba a=

For each block :F ma =

BlockA: , or 3A A A A A A A A A A BW T m a T W m a W m a = = =

BlockC: , orC C C C C C C C C C BW T m a T W m a W m a = = =

BlockB: 3B A C B BW T T m a =

( ) ( )3 3B A A B C C B B BW W m a W m a m a =

or3 265

88795.07665

9

B B A C

B A C

a W W W

g W W W

= = =

+ + + +

(a) Accelerations. ( )( )= = Ba 2

.752 m/sB

=a

( )( )3

.07665 9.81

Aa = = 2

2.256 m/sA

=a

( )Ca = = 2

.752 m/sC

=a

(b) Tensions. ( )88 2.256AT = 67.76 NA

T =

( )CT = 81.25 NCT =

265

265

88

2.752 m/s

.752 2

2.256 m/s

.752 2

.752 m/s

88

9.81

8888

9.81.752

Let ybe the positive downward position for all blocks.

Accelerations:

Tensions:

.75 m/s2

2.26 m/s2

.75 m/s2

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249

PROBLEM 12.20

Each of the systems shown is initially at rest. Neglecting axle friction andthe masses of the pulleys, determine for each system (a) the accelerationof block A, (b) the velocity of block Aafter it has moved through 1.5 m,

(c) the time required for blockAto reach a velocity of 3 m/s.

SOLUTION

Letybe positive downward for both blocks.

Constraint of cable: constantA By y+ =

0A Ba a+ = or B Aa a=

For blocksAandB, :F ma =

BlockA: AA A

WW T a

g = or A

A A

WT W a

g=

BlockB: B B

B B A

W W

P W T a ag g+ = =

A B

B A A A

W WP W W a a

g g+ + =

Solving for , A BA A

A B

W W Pa a g

W W

=

+ (1)

( ) ( ) ( )22

0 0 02 with 0A A A A A Av v a y y v = =

( )0

2A A A A

v a y y = (2)

( ) ( )0 0

with 0A A A Av v a t v = =

A

A

vt

a= (3)

continued

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(a) Acceleration of blockA.

System (1): 441N, 223 N, 0A B

W W P= = =

By formula (1), ( ) ( )1A

a

= ( 1A =a System (2): , 0,A BW W P= = =

By formula (1), ( ) ( )2

9.81441

Aa

= (

)

2

24.85m/s

A =a

System (3): 4895 N, 4669.5 N, 0A BW W P= = =

By formula (1), ( ) ( )3

4895 669.5 N9.81

4895 4669.5A

a

=

+

( ) 23

.23m/sA

=a

(b) ( )0

at 1.5 m. Use formula (2).A A Av y y =

System (1): ( ) ( )( )( )1

2 3.22

4.85

.23

1.5Av = ( )1 3.1 m/sAv =

System (2): ( ) ( )( ) ( )2

2 1.5A

v = ( )2

3.8 m/sAv =

System (3): ( ) ( )( ) ( )3

2 1.5A

v = ( )3

.83 m/sAv =

(c) Time at 3 m/s. Use formula (3).Av =

System (1): 13

3.22t = 1 0.932 st =

System (2):2

3

4.85t = 2 .618st =

System (3):3

3

.23t = 3 13.04 st =

441 2239.81 3.22 m/s2

)

441 + 223

441 N 223N

441 223

4

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251

PROBLEM 12.21

The flat-bed trailer carries two 1360.8 kg beams with the upper beamsecured by a cable. The coefficients of static friction between the two

beams and between the lower beam and the bed of the trailer are 0.25 and

0.30, respectively. Knowing that the load does not shift, determine (a) themaximum acceleration of the trailer and the corresponding tension in thecable, (b) the maximum deceleration of the trailer.

SOLUTION

(a) Maximum acceleration. The cable secures the upper beam; only the lower beam can move.

For the upper beam,1

0: 0y

F N W = =

1N W mg= =

For the lower beam,2 1

0: 0y

F N N W = = or 2 2N W=

( )1 2: 0.25 0.30 0.25 0.60xF ma N N W ma = + = + =

( ) ( ) 20.85 8.3 m/sW

am

= = = =a

For the upper beam, 1: 0.25xF ma T N ma = =

( )( ) ( )13349.45

0.25 0.25 8.3 14.6 kN9.81

= + = =

T W ma 14.6 kNT =

(b) Maximum deceleration of trailer.

Case 1: Assume that only the top beam slips. As in Part (a) 1 .N mg=

: 0.25F ma W ma = =20.25 2.45 m/sa g= =

continued

0.85 9.81

13349.45 +

28.3 m/s

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252


Case 2: Assume that both beams slip. As before 2 2 .=N W

( ) ( )( ) ( )2 : 0.30 2 2 = =F m a W m a

20.30 2.9 m/sa g= =

The smaller deceleration value governs. 2.45 m/sa = 2

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253

PROBLEM 12.22

The 10-kg blockBis supported by the 40-kg block Awhich is pulled upan incline by a constant 500 N force. Neglecting friction between the

block and the incline and knowing that blockBdoes not slip on block A,

determine the smallest allowable value of the coefficient of static frictionbetween the blocks.

SOLUTION

Since both blocks move together, they have a common acceleration. UseblocksA andBtogether as a free body.

:F ma =

( )sin 30 sin 30A B A BP m g m g m m a = +

500sin 30 9.81 sin 3050A B

Pa gm m

= = +

25.095 m/s=

Use blockBas a free body.

cos30 : cos30B f BF m a F m a = =

( )( )10 5.095 cos 30 44.124 NfF = =

sin 30 : sin 30B B B BF m a N m g m a = =

( ) ( )sin 30 10 9.81 5.095 sin 30B BN m g a= + = +

123.575 N=

Minimum coefficient of static friction:

min

44.124

123.575

f

B

F

N = = min 0.357 =

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254

PROBLEM 12.23

A package is at rest on a conveyor belt which is initially at rest. The beltis started and moves to the right for 1.5 s with a constant acceleration of

23.2 m/s . The belt then moves with a constant deceleration 2a and comes

to a stop after a total displacement of 4.6 m. Knowing that thecoefficients of friction between the package and the belt are 0.35s =

and 0.25,k = determine (a) the deceleration 2a of the belt, (b) thedisplacement of the package relative to the belt as the belt comes to a

stop.

SOLUTION

(a) Kinematics of the belt. 0ov =

1. Acceleration phase with 21

3.2 m/sa =

( )( )1 1 1 0 3.2 1.5 4.8 m/sov v a t = + = + =

( )( )22

1 1 1 1

1 10 0 3.2 1.5 3.6 m

2 2o ox x v t a t= + + = + + =

2. Deceleration phase. 2 0v = since the belt stops.

( )2 22 1 2 2 12v v a x x =

( )

( )

( )

22 2

2 1

2

2 1

0 4.811.52

2 2 4.6 3.6

v va

x x

= = =

2

211.52 m/s=a

2 1

2 1

2

0 4.80.41667 s

11.52

v vt t

a

= = =

(b) Motion of the package.

1. Acceleration phase. Assume no slip. ( ) 21

3.2 m/spa =

0: 0 oryF N W N W mg = = = =

( )1

:x f pF ma F m a = =

The required friction force is .fF

The available friction force is 0.35 0.35sN W mg = =

( ) ( )( ) 21

0.35 9.81 3.43 m/sf s

p s

F Na g

m m

= < = = =

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255


Since 2 23.2 m/s 3.43 m/s ,< the package does not slip.

( ) ( )11 14.8 m/s and 3.6 m.p pv v x= = =

2. Deceleration phase. Assume no slip. ( ) 22

11.52 m/spa =

( )2

:x f pF ma F m a = =

( ) 22

11.52 m/sf

p

Fa

m= =

2 23.43 m/s 11.52 m/ss s sN mg

gm m

= = =

Try 75 N. By Eq. (1 ), 74.34 N 75 N (acceptable)DE DAT a T= =

sin nma P

W W =

sin .6393

140.3

1

9 81102

2

N

m skg

. /. .=

2 7

61215 2

m s

mm s

. /

.. /=

102 1215

1

6

1

. . .( )( )

39.7

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288

PROBLEM 12.51

A 240 Npilot flies a jet trainer in a half vertical loop of 1097 mradius sothat the speed of the trainer decreases at a constant rate. Knowing that the

pilots apparent weights at points A and C are 760 N and 160 N,

respectively, determine the force exerted on her by the seat of the trainerwhen the trainer is at pointB.

SOLUTION

At positionA, the vertical component of apparent weight is shown as .AN

:n A n

WF ma N W a

g = =

An

N Wa g

W

= =

2A nva = =

At position C, the vertical component of apparent weight is shown as .CN

:n C n

WF ma N W a

g = + =

C

n

N Wa g

W

+= =

2C nv a= =

Length of arcABC:

( ) 3446.3 mACs = = =

Calculate ,ta using 2 2 2

C A t AC v v a s =

2 2

2

2

.78 m/s

C A

t

AC

v va

s

= =

=

At positionB, ( ) 1723.2 m2 2

ABs

= = =

2 2 2B A t ABv v a s= + =

760 240

2409 81 21 255 2

( )=. . /m s

1097 21255 23316 735 2 2( )( )=. . /m m s

160 240

2409 81 16 35 2

+( )=. . /m s

1097 1635 1793595 2 2( )( )=. . /m s

1097

17935 95 23316 735

2 3446 3

( )( )

. .

.

1097

23316 735 2 78 17232 26004 9 2 2+( ) ( )( )=. . . . /m s

240-N 1097-m

v2

A= ra

n=

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289


Effective forces atB:2

B

n

W vma

g = =

t t

Wma a

g= =

:tF ma =

ort tP W ma P W ma = = + =

:n B nF ma N ma = = =

Force exerted by seat:

2 2BF N P= +

tan = =

620.6N=F t

240

9 81

26004 9

1097580

.

.= N

240

9 8178 19

..( )= N

240 19 221 = N

580 N

= 5 80 221 620 62 .+ = N

221

58020.8

20.8

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290

PROBLEM 12.52

A car is traveling on a banked road at a constant speed v. Determine therange of values of vfor which the car does not skid. Express your answer

in terms of the radius rof the curve, the banking angle , and the angle

of static friction s between the tires and the pavement.

SOLUTION

The road reaction consists of normal componentNand friction

component F. The resultant Rmakes angle s with the normal.

Case 1: maxv v=

( )0: cos 0y sF R mg = + =

( )cos s

mgR

=

+

( ): sinx n s nF ma R ma = + =

( )tann s

ma mg = +

( )2

max tans

vg

r = +

( )max tan sv gr = +

Case 2: minv v=

( )0: cos 0y sF R mg = =

( )coss

mgR

=

( ): sinx n s nF ma R ma = =

( )tann s

ma mg =

( )2

min tans

vg

r =

( )min tan sv gr =

( ) ( )tan tans sgr v gr + t

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291

PROBLEM 12.53

A curve in a speed track has a radius of 200 m and a rated speed of180 km/h. (See Sample Prob. 12.6 for the definition of rated speed.)

Knowing that a racing car starts skidding on the curve when traveling at a

speed of 320 km/h, determine (a) the banking angle , (b) the coefficientof static friction between the tires and the track under the prevailing

conditions, (c) the minimum speed at which the same car could negotiatethat curve.

SOLUTION

Weight. W mg=

Acceleration.2v

a

=

: =x xF ma sin cosF W = ma +

2

cos sinmv

F mg

= (1)

:y y

F ma = cos sinN W ma =

2

sin cosmv

N = mg

+ (2)

(a) Banking angle. Rated speed 180 km/h 50 m/s.v = = 0F = at rated speed.

2

0 cos sinmv

mg

=

2 2(50)tan = 1.2742

(200)(9.81)

v

g

= =

51.875 = 51.9 =

(b) Slipping outward. 320km/h 88.889 m/sv = =

F N=2

2

cos sin

sin cos

F v g

N v g

= =

+

2

2

(88.889) cos51.875 (200)(9.81)sin51.875=

(88.889) sin51.875 (200)(9.81)cos51.875

+ 0.44899= 0.449 = t

continued

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(c) Minimum speed. = F N

2

2cos sinsin cos

v gv g

=+

2 (sin cos )

cos sin

gv

=

+

(200)(9.81)(sin51.875 0.44899cos51.875 )

cos51.875 0.44899sin51.875

=

+

2 21029.87 m /s=

32.09m/sv = 115.5 km/h=v

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PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.No part of this Manual may be displayed, reproduced

or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and

educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

293

PROBLEM 12.54

Tilting trains such as the Acela, which runs from Washington to New

York to Boston, are designed to travel safely at high speeds on curved

sections of track which were built for slower, conventional trains. As it

enters a curve, each car is tilted by hydraulic actuators mounted on itstrucks. The tilting feature of the cars also increases passenger comfort by

eliminating or greatly reducing the side forces

F (parallel to the floor of

the car) to which passengers feel subjected. For a train traveling at

200 km/h on a curved section of track banked at an angle 8= and witha rated speed of 120 km/h, determine (a) the magnitude of the side force

felt by a passenger of weight W in a standard car with no tilt ( 0 = ),(b) the required angle of tilt if the passenger is to feel no side force.(See Sample Problem 12.6 for the definition of rated speed.)

SOLUTION

Rated speed: RvFrom Sample Problem 12.6,

2

2 tan ortan

R

R

vv g

g

= = =

Let thex-axis be parallel to the floor of the car.

( ) ( ): sin cosx x s nF ma F W ma = + + = +

( )2

cosmv

= +

(a) 0. =

( )( )( )

2183.33

cos8 sin832.2 2674

0.247

sF

W

W

=

= 0.247s

F W=

(b) For 0,s

F =

( ) ( )2

cos sin 0v

g

+ + =

( )2

tanv

g

+ = =

21.3

+ =

21.3 8= 13.3 =

=120 33 3 200 55 56km h m s km h m s/ . / , / . /=

33 3

9 81 8804 3

2

m.

. tan .

( )=

= 5556

9 81 804 3

8 8

2.

. .

cos sin ( )

( )( )

W

0.390355556

981 8043

2.

. .

( )

( )( )=

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294

PROBLEM 12.55

Tests carried out with the tilting trains described in Prob. 12.54 revealedthat passengers feel queasy when they see through the car windows thatthe train is rounding a curve at high speed, yet do not feel any side force.

Designers, therefore, prefer to reduce, but not eliminate that force. For thetrain of Prob. 12.54, determine the required angle of tilt if passengersare to feel side forces equal to 12 percent of their weights.

Problem 12.54: Tilting trains such as the Acela, which runs fromWashington to New York to Boston, are designed to travel safely at high

speeds on curved sections of track which were built for slower,conventional trains. As it enters a curve, each car is tilted by hydraulicactuators mounted on its trucks. The tilting feature of the cars alsoincreases passenger comfort by eliminating or greatly reducing the sideforce sF (parallel to the floor of the car) to which passengers feelsubjected. For a train traveling at 125 mi/h on a curved section of track

banked at an angle 8 = and with a rated speed of 75 mi/h, determine(a) the magnitude of the side force felt by a passenger of weight Win a

standard car with no tilt ( 0 = ), (b) the required angle of tilt if thepassenger is to feel no side force. (See Sample Problem 12.6 for thedefinition of rated speed.)

SOLUTION

Rated speed: Rv =

From Sample Problem 12.6,

2

2 tan ortan

R

R

vv g

g

= = =

Let thex-axis be parallel to the floor of the car.

( ) ( ): sin cosx x s nF ma F W ma = + + = +

( )2

cosmv

= +

Solving for ,sF ( ) ( )2

cos sins

vF W

g

= + +

Now2

0.39035 and 0.12s

vF W

g= = = so that

( ) ( )0.12 0.39035 cos sinW W = + +

Let ( ) ( ) 2

sin . Then, cos 1 .u u = + + = 2 20.12 0.39035 1 or 0.39035 1 0.12u u u u= = +

120 33 3 200 55 56km h m s km h m s/ . / , / . /= =

33 3

9 81 8804 3 m

.

. tan .

( )=

5556

981 8043

2.

. .

( )

( )( )

200 km/hbanked at an angle q=8 and with a rated speed of 120 km/h, determine

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295


Squaring both sides, ( )2 20.15237 1 0.0144 0.24u u u = + +

or 21.15237 0.24 0.13797 0u u+ + =

The positive root of the quadratic equation is 0.2572.u =

Then, 1sin 14.90u + = =

14.90 14.90 8 = = 6.90 =

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296

PROBLEM 12.56

A small 250-g collar Ccan slide on a semicircular rod which is made torotate about the verticalABat a constant rate of 7.5 rad/s. Determine the

three values of for which the collar will not slide on the rod, assuming

no friction between the collar and the rod.

SOLUTION

If the collar is not sliding, it moves at constant speed on a circle of radius sin .r = v =

Normal acceleration.

2 2 2

2

( sin )nv

a r

= = =

:y y

F ma =

cos 0N mg =

cos

mgN

=

:x xF ma =

sinN ma =

2sin

( sin )cos

mg

m r

=

Either sin 0 =

or2

cosg

r

=

0 or 180 =

or2

9.81cos 0.3488

(0.5)(7.5) = =

0 ,180 , and 69.6 =

acceleration:

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297

PROBLEM 12.57

For the collar and rod of Prob. 12.56, and assuming that the coefficients

of friction are s= 0.25 and k= 0.20, indicate whether the collar will

slide on the rod if it is released in the position corresponding to

(a) = 75, (b) = 40. Also, determine the magnitude and direction ofthe friction force exerted on the collar immediately after release.

SOLUTION

If the collar is not sliding, it moves at constant speed on a circle of radius sin .r = v =

From Prob. 12.56 500mm 0.500m, 7.5rad/s,r = = =

250 g 0.250 kg.m= =

Normal acceleration:2 2 2

2( sin )n

va r

= = =

F = :ma

2sin ( sin ) cosF mg m r =

2( cos )sinF m g r =

F = :ma

2cos ( sin ) sinN mg m r =

2 2( cos ) sin )N m g r = +

(a) 75 . = 2(0.25) 9.81 (0.500 cos75 )(7.5) sin75F = 0.61112 N=

2 2(0.25) 9.81cos75 (0.500sin 75 )(7.5)N = + 7.1950 N=

(0.25)(7.1950) 1.7987 N= =sN

Since ,s

F N< the collar does not slide.0.611 NF = 75t

continued

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298


(b) 40 . = 2(0.25) 9.81 (0.500cos40 )(7.5) sin40F = 1.8858 N=

2 2(0.25) 9.81cos40 (0.500sin 40 )(7.5)N = + 4.7839 N=

(0.25)(4.7839) 1.1960 N= =sN

Since ,> sF N the collar slides.

Since the collar is sliding, .kF N=

nF = + :ma

cos sinnN mg ma =2cos ( sin ) sinN mg m r = +

2 2cos ( sin )m g r = +

2 2(0.25) 9.81cos40 (0.500sin 40 )(7.5) = +

4.7839 N=

(0.20)(4.7839) 0.957 N= = =k

F N

0.957 N=F 40t

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299

PROBLEM 12.58

A small blockBfits inside a slot cut in arm OAwhich rotates in a verticalplane at a constant rate. The block remains in contact with the end of theslot closest toAand its speed is 1.28 m/sfor 0 150 . Knowing that

the block begins to slide when 150 , = determine the coefficient ofstatic friction between the block and the slot.

SOLUTION

Draw the free body diagrams of the blockBwhen the arm is at 150 . =

20, 9.81 m/st

v a g= = =

: sin 30 0t tF ma W N = + =

sin30N W=

2 2

: cos30n n

v WvF ma W F m

g = = =

2

cos30Wv

F Wg

=

Form the ratioF

N, and set it equal to s for impending slip.

( )22 cos30 /( )( )cos30 /

sin 30 sin 30s

F v g

N

= = =

s = t

1.28 .3 9.81

.618

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300

PROBLEM 12.59

A 2.7 kgblock is at rest relative to a parabolic dish which rotates at aconstant rate about a vertical axis. Knowing that the coefficient of staticfriction is 0.5 and that r =1.8 m, determine the maximum allowable speed v

of the block.

SOLUTION

Let be the slope angle of the dish.1

tan6

dyr

dr = =

At , tan 0.3 or 16.7r = = =

Draw free body sketches of the sphere.

0: cos sin 0y s

F N N W = =

cos sins

WN

=

2 2

: sin cosn n s

mv WvF ma N N

g

= + = =

( ) 2sin coscos sin

s

s

W N Wv

g

+=

2 sin cos

cos sin

s

s

v g

+= =

v = t

1.8 m

1 8 9 8 1 16 7 1 5 16 7

16 7 0 5 16 716 6 2 2. . sin . . cos .

cos . . sin . . /( )( ) +

= m s

4.07 m/s

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301

PROBLEM 12.60

Four seconds after a polisher is started from rest, small tufts of fleecefrom along the circumference of the 25.4 cm-diameter polishing pad areobserved to fly free of the pad. If the polisher is started so that the fleece

along the circumference undergoes a constant tangential acceleration of23.7 m/s, determine (a) the speed vof a tuft as it leaves the pad, (b) the

magnitude of the force required to free the tuft if the average weight of atuft is

SOLUTION

Uniformly accelerated motion on a circular path. 12.7 cm.2

D = =

14.715 1010 NW =

29.81 m/sg =

61.5 10 kgW

mg

= =

(a) For uniformly accelerated motion,

( )( )0 0 4tv v a t = + = +

v = t

(b) :t t tF ma F = =

2

:n n n n

mvF ma F ma

= = = =

=

Magnitude of force:

2 2 610t n

F F F = + =

F = t

1.5 10 kg.6

6

3.7

14.8 m/s

1 5 10 3 7 5 55 106 6. . .( )( )= N

1 5 10 14 8

127

6. .

.

( )( )

2 587 10 3. N

2587 5 552 2

.( ) +( )

2587 10 6 N

25.4-cm-diameter

path: cm

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302

PROBLEM 12.61

A turntable Ais built into a stage for use in a theatrical production. It isobserved during a rehearsal that a trunk Bstarts to slide on the turntable12 s after the turntable begins to rotate. Knowing that the trunk undergoes

a constant tangential acceleration of , determine the coefficientof static friction between the trunk and the turntable.

SOLUTION

Uniformly accelerated motion on a circular path. 2.4m =

0 tv v a t = +

0= +

: t

t t t t

W a

F ma a F Wg g= = = =

2

:n n nWv

F ma Fg

= = =

2 2.324t nF F F W= + =

This is the friction force available to cause the trunk to slide.

The normal forceNis calculated from equilibrium of forces in thevertical direction.

0: 0yF N W = =

N W=

Since sliding is impending, sF

W = = s = t

0.23 m/s2

. . /23 12 2 76( )( ) = m s

.

.

23

9.81 0 0233=W W

.

. ..

2 76

9 8 1 2 43235

2( )

( )( ) =

WW

.324 .324

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303

PROBLEM 12.62

The parallel-link mechanism ABCD is used to transport a component Ibetween manufacturing processes at stationsE,F, and Gby picking it upat a station when 0 = and depositing it at the next station when

180 . = Knowing that member BC remains horizontal throughout itsmotion and that links AB and CD rotate at a constant rate in a vertical

plane in such a way that 0.7Bv = m/s, determine (a) the minimum valueof the coefficient of static friction between the component and BCif thecomponent is not to slide onBCwhile being transferred, (b) the values of for which sliding is impending.

SOLUTION

For constant speed, 0ta =

2

with 0.7 m/s, 0.2 mBn Bv

a v

= = =

: cosx x nF ma F ma = =

: sin siny y n nF ma N W ma N mg ma = = =

Ratio

2

cos cos cos

sin sin sin

n

n

n B

F ma

g gN mg ma

a v

= = =

With ( )( )

( )2 2

9.81 0.2 cos4.0041, the ratio becomes

4.0041 sin0.7B

g F

Nv

= = =

For no impending slide,cos

4.0041 sins

F

N

=

To find the value of for which the ratio is maximum set the derivativewith respect to equal to zero.

( )2

cos 1 4.0041sin0

4.0041 sin 4.0041 sin

d

d

= =

1sin 0.24974

4.0041 = =

cos14.44614.446 , 0.258

4.0041 0.24974

F

N

= = =

180 14.446 165.554 , 0.258F

N = = =

(a) Minimum value of s for no slip. ( )min 0.258s = t(b) Corresponding values of . 14.5 and 165.5 = t

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304

PROBLEM 12.63

Knowing that the coefficients of friction between the component I andmember BC of the mechanism of Prob. 12.62 are 0.35s = and

0.25,k = determine (a) the maximum allowable speed Bv if the

component is not to slide onBCwhile being transferred, (b) the values of for which sliding is impending.

SOLUTION

For constant speed, 0ta =

2

with 0.7 m/s, 0.2 mBn Bv

a v

= = =

: cosx x nF ma F ma = =

: sin siny y n nF ma N W ma N mg ma = = =

Ratio

2

cos cos cos

sin sin sin

n

n

n B

F ma

g gN mg ma

a v

= = =

Let2

cosso that

sinB

g Fu

N uv

= =

Determine the value of at whichF/Nis maximum.

( )( )

( ) ( )

2

2 2

cos sin sincos 1 sin0

sin sin sin

ud u

d u u u

= = =

The corresponding ratio .F

N

2 1

1 2

1 sintan

cos1

F u u

N u u u

= = = =

(a) For impending sliding to the left: tan 0.35sF

N = = =

( )2

1arctan 0.35 19.29 , sin ,Bv

ug

= = = =

( )( )2 2 29.81 0.2 sin19.29 0.648 m /sBv = =

0.805 m/sBv = t

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305


For impending motion to the right: tan 0.35s

F

N = = =

( )arctan 0.35 160.71 = =

2

1 2 sin ,v

ug

= = ( )( )2 2 29.81 0.2 sin160.71 0.648 m /sBv = =

0.805 m/s= t

(b) For impending sliding to the left, 19.3 = t

For impending sliding to the right, 160.7 = t

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306

PROBLEM 12.64

In the cathode-ray tube shown, electrons emitted by the cathode andattracted by the anode pass through a small hole in the anode and thentravel in a straight line with a speed 0v until they strike the screen at A.

However, if a difference of potential V is established between the twoparallel plates, the electrons will be subjected to a force Fperpendicularto the plates while they travel between the plates and will strike thescreen at pointB, which is at a distance fromA. The magnitude of theforce Fis / ,F eV d= where e is the charge of an electron and dis thedistance between the plates. Neglecting the effects of gravity, derive anexpression for the deflection in terms of V, 0 ,v the charge e and themass mof an electron, and the dimensions d, , andL.

SOLUTION

Consider the motion of one electron. For the horizontal motion, let

0x = at the left edge of the plate and x = at the right edge of the

plate. At the screen,

2x L= +

Horizontal motion: There are no horizontal forces acting on the electron

so that 0.xa =

Let 1 0t = when the electron passes the left edge of the plate, 1t t=

when it passes the right edge, and 2t t= when it impacts on the screen.

For uniform horizontal motion,

0 1 2

0 0 0

, so that and .2

Lx v t t t

v v v= = = +

Vertical motion: The gravity force acting on the electron is neglected

since we are interested in the deflection produced by the electric force.

While the electron is between plates ( )10 ,t t the vertical force on

the electron is / .y

F eV d= After it passes the plates ( )1 2 ,t t t it iszero.

For 10 ,t t : y

y y y

F eVF ma a

m md = = =

( )0

0y y y

eVtv v a t

md= + = +

( )2

20

0

1 0 02 2

y yeVty y v t a t

md= + + = + +

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307


At ( )2

1 1

1 11

, and2

y

eVt eVt t t v y

md md = = =

For1 2

, 0y

t t t a =

( )( )1 11

yy y v t t= +

At 2t t= ( )( )2 1 2 11

yy y v t t= = +

( )2

1 1 1

2 1 2 1

1

2 2

eVt eVt eVt t t t t

md md md

= + =

0 0 0 0

1

2 2

eV L

mdv v v v

= +

or

2

0

eV

mdv

L =

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308

PROBLEM 12.65

In Prob. 12.64, determine the smallest allowable value of the ratio /d interms of e, m, 0v , and V if at x = the minimum permissible distance

between the path of the electrons and the positive plate is 0.075 .d

Problem 12.64:In the cathode-ray tube shown, electrons emitted by thecathode and attracted by the anode pass through a small hole in the anodeand then travel in a straight line with a speed 0v until they strike thescreen atA. However, if a difference of potential Vis established betweenthe two parallel plates, the electrons will be subjected to a force F

perpendicular to the plates while they travel between the plates and willstrike the screen at point B, which is at a distance from A. Themagnitude of the force F is / ,F eV d= where e is the charge of anelectron and dis the distance between the plates. Neglecting the effects ofgravity, derive an expression for the deflection in terms of V, 0 ,v thecharge e and the mass mof an electron, and the dimensions d, , andL.

SOLUTION

Consider the motion of one electron. For the horizontal motion, let 0x =

at the left edge of the plate and x = at the right edge of the plate. At

the screen,

2x L= +

Horizontal motion: There are no horizontal forces acting on the electron

so that 0.xa =

Let 1 0t = when the electron passes the left edge of the plate, 1t t=

when it passes the right edge, and 2t t= when it impacts on the screen.

For uniform horizontal motion,

0 1 2

0 0 0

, so that and .2

Lx v t t t

v v v= = = +

Vertical motion: The gravity force acting on the electron is neglectedsince we are interested in the deflection produced by the electric force.

While the electron is between the plates ( )10 ,t t the vertical force

on the electron is / .yF eV d= After it passes the plates ( )1 2 ,t t t

it is zero.

For 10 ,t t : y

y y y

F eVF ma a

m md = = =

( )0

0y y y

eVtv v a t

md= + = +

( )2

2

00

10 0

2 2y y

eVty y v t a t

md= + + = + +

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309


At2

1 2

0 0

, ,2

eVt t y

v mdv= =

But 0.075 0.4252

dy d d< =

So that2

2

0

0.4252

eVd

mdv =

2

0

1.085d eV

mv>

t

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310

PROBLEM 12.66

A 0.5-kg blockBslides without friction inside a slot cut in arm OAwhichrotates in a vertical plane at a constant rate, =2 rad/s. At the instantwhen 30 , = r=0.6 m and the force exerted on the block by the arm is

zero. Determine, at this instant, (a) the relative velocity of the block withrespect to the arm, (b) the relative acceleration of the block with respectto the arm.

SOLUTION

30 , 2 rad/s, 0 = = =

0.6 m,r W mg = =

BlockB: Only force is weight

cos30 , sin30rF W F W= =

(a) ( )2 :F ma m r r = = +

sin302 sin 30

F mgr r r g r

m m

= = =

( ) ( )( )( ) ( )

9.81 sin 30 0.6 0sin301.226 m/s

2 2 2

g rr

+ += = =

/ rod 1.226 m/sB =v 60

(b) ( )2 :r rF ma m r r= =

( )( ) ( )

2 2

2 2

cos30cos30

0.6 2 9.81 cos30 10.90 m/s

rF mg

r r r r g m m

= + = + = +

= + =

2

/ rod10.90 m/s

B =a 60 t

weight.

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311

PROBLEM 12.67

A 0.5-kg blockBslides without friction inside a slot cut in arm OAwhichrotates in a vertical plane. The motion of the rod is defined by the relation

=10 2rad/s , constant. At the instant when = 45 , r=0.8 m and the

velocity of the block is zero. Determine, at this instant, (a) the forceexerted on the block by the arm, (b) the relative acceleration of the block

with respect to the arm.

SOLUTION

245 , 0.8 m, 10 rad/sr = = =

0, 0,r

v r v r W mg = = = = =

(a) ( ): cos45 2F ma N W m r r = = +

( )cos45 2N mg m r r = + +

( )( )(0.5)(9.81)cos 45 0.5 0.8 10 0 = + + 7.468= 7.47 NN= 45 t

(b) ( )2: sin 45r rF ma mg m r r = =

2 2sin 45 sin 45mg

r r g r m

= + = +

( ) 29.81 sin 45 0 6.937 m/s= + =

2

/ rod 6.94 m/sB =a 45 t

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312

PROBLEM 12.68

The motion of a 1.8 kgblock B in a horizontal plane is defined by the

relations 2 3

3r t t= and2

2t= , where ris expressed in meter, t in

seconds, and in radians. Determine the radial and transversecomponents of the force exerted on the block when (a) t= 0, (b) t= 1 s.

SOLUTION

Use radial and transverse components of acceleration.

2 3 23 2 radr t t t = =

26 3 4 rad/sr t t t = =

2

6 6 4 rad/sr t = =

2 2 3 26 6 (3 )(16 )r

a r r t t t t = =

5 4 216 48 6 6t t t= +

2 3 22 (3 )(4) (2)(6 3 )(4 )a r r t t t t t = + = +

2 3 260 28t t=

Mass:17.658

9.81= = =

Wm

g

(a) 0.t = 6r

a =

0a =

Apply Newtons second law.

( )(6)r rF ma= = rF = t

( )(0)F ma == F = t

(b) 1t = s. 232r

a =

232a =


( )( 32)r rF ma= = rF = t

( )(32)F ma = = F = t

m

m/s

m/s

m/s

m/s

1.8 kg

m/s2

1.8

1.8

m/s

m/s

1.8

1.8 57.6 N

57.6 N

10.8 N

0

m,

m/s,

m/s,

PROBLEM 12.68

The motion of a 1.8-kg block B in a horizontal plane is defined by the

relations2 3

3r t t= and2

2t= , where r is expressed in meters, t in

seconds, and in radians. Determine the radial and transversecomponents of the force exerted on the block when (a) t= 0, (b) t= 1 s.

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313

PROBLEM 12.69

The motion of a .45 kgblock B in a horizontal plane is defined by therelations ( )6 1 cos2r t= + and 2 t = , where ris expressed int in seconds, and in radians. Determine the radial and transverse

components of the force exerted on the block when (a) 0,t =(b) 0.75 s.t =

SOLUTION

Use radial and transverse components of acceleration.

6(1 cos2 ) 2 radr t t = + =

12 sin 2 2 rad/sr t = =

2

24 cos2 0r t = =

2 2 224 cos2 (6 6cos2 )(2 )r

a r r t t = = +

2 224 48 cos 2 t =

2 0 (2)( 12 sin 2 )(2 )a r r t = + = +

248 sin 2 t =

Mass:4.4

.45 kg9.81

Wm

g= = =

(a) 0.t = 2710.61r

a =

0a =


( )(710.61)r rF ma= = rF = t

0F ma

= = 0F

= t

(b) 0.75 s.t = 2236.87r

a =

2473.74a

=


( )( 236.87)r rF ma= = rF = t

( )(473.74)F ma = = F = t

meter,

m

m/s

m/s2

2m/s

m/s

.45

m/s

m/s

.45

.45

319.7 N

106.6 N

213.2 N

PROBLEM 12.69

The motion of a .45-kg block B in a horizontal plane is defined by therelations ( )6 1 cos2r t= + and 2 t = , where ris expressed int in seconds, and in radians. Determine the radial and transverse

components of the force exerted on the block when (a) 0,t =(b) 0.75 s.t =

meters,

m,

m/s,

m/s2,

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314

PROBLEM 12.70

The 2.7 kgcollarBslides on the frictionless arm .AA The arm is attached

to drumDand rotates about Oin a horizontal plane at the rate 0.8 ,t=

where

and tare expressed in rad/s and seconds, respectively. As thearm-drum assembly rotates, a mechanism within the drum releases cord

so that the collar moves outward from Owith a constant speed of .457 m/s.

Knowing that at 0,t= 0,r= determine the time at which the tension in

the cord is equal to the magnitude of the horizontal force exerted onBby

arm .AA

SOLUTION

Kinematics: , 0dr

r rdt

= = =

0 0 or .457 m

r t

dr r dt r = =

20.8 rad/s, 0.8 rad/st = =

( )( )22 30 0.8

ra r r t = = =

( )( ) ( )( )( ) 22 0.8 2 0.8 1.0968 m/sa r r t = + = + =

Kinetics: Sketch the free body diagrams for the collar.

:r r rF ma T ma = =

:F ma Q ma = =

Set T Q= to obtain the required time.

orr rma ma a a = =

Using the calculated expressions

3 2 21.0968, 3.75 s.292

t t= = =

1.936 st = t

.457 m/s

t

.457 .292 m/st 2

.457 .457

.292 0968

2.7-kg

expressions,

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PROPRIETARY MATERIAL. 2007 The McGraw-Hill Companies, Inc. All rights reserved.No part of this Manual may be displayed, reproduced

or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and

educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

315

PROBLEM 12.71

The horizontal rod OA rotates about a vertical shaft according to the

relation = 10t, where and t are expressed in rad/s and seconds,

respectively. A .23 kg collar Bis held by a cord with a breaking strength

of 1.8 kg. Neglecting friction, determine, immediately after the cord breaks,(a) the relative acceleration of the collar with respect to the rod, (b) the

magnitude of the horizontal force exerted on the collar by the rod.

SOLUTION

210 rad/s, 10 rad/st = =

2.23 kg.m = =

Before cable breaks: and 0.r

F T r= =

( )2:r rF ma T m r r= =

2 2 ormr T

mr mr T mr

= + = =

13 rad/s =

Immediately after the cable breaks: 0, 0r

F r= =

(a) Acceleration ofBrelative to the rod.

( )2 20 orm r r r r = = =

/ rodB =a radially outward

(b) Transverse component of the forc

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