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514

PROBLEM 13.7

Determine the maximum theoretical speed that may be achieved over a distance of 110 m by a car starting from rest assuming there is no slipping. The coefficient of static friction between the tires and pavement is 0.75, and 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume (a) front-wheel drive, (b) rear-wheel drive.

SOLUTION

Let W be the weight and m the mass. W mg=

(a) Front wheel drive: 0.60 0.60

0.75s

N W mg

μ= ==

Maximum friction force without slipping:

1 2

21 2 2

(0.75)(0.60 ) 0.45

0.45

10,

2

sF N W mg

U Fd mgd

T T mv

μ

→

= = == =

= =

Principle of work and energy: 1 1 2 2T U T→+ =

22

2 2 2 22

10 0.45

2

(2)(0.45 ) (2)(0.45)(9.81 m/s )(110 m) 971.19 m /s

mgd mv

v gd

+ =

= = =

2 31.164 m/sv = 2 112.2 km/hv =

(b) Rear wheel drive: 0.40 0.40

0.75s

N W mg

μ= ==

Maximum friction force without slipping:

1 2

21 2 2

(0.75)(0.40 ) 0.30

0.30

10,

2

sF N W mg

U Fd mgd

T T mv

μ

→

= = == =

= =

Principle of work and energy: 1 1 2 2T U T→+ =

22

2 2 2 22

10 0.30

2

(2)(0.30) (2)(0.30)(9.81 m/s )(110 m) 647.46 m /s

mgd mv

v gd

+ =

= = =

2 25.445 m/sv = 2 91.6 km/hv =

Note: The car is treated as a particle in this problem. The weight distribution is assumed to be the same for static and dynamic conditions. Compare with sample Problem 16.1 where the vehicle is treated as a rigid body.

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526

PROBLEM 13.18

The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of cars A, causing it to slide on the track, but are not applied on the wheels of cars A or B. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the distance required to bring the train to a stop, (b) the force in each coupling.

SOLUTION

(a) Entire train:

1

(0.35)(80 kips) 28 kips

30 mi/h 44 ft/s A AF N

v

μ= = == = ← 2 20 0v T= =

1 1 2 2

22

1 (80 kips 100 kips 80 kips)(44 ft/s) (28 kips) 0

2 32.2 ft/s

T V T

x

−+ =+ + − =

279.1 ftx = 279 ftx =

(b) Force in each coupling:

Car A: Assume ABF to be in tension

1 1 2 2T V T−+ =

22

1 80 kips(44 ft/s) (28 kips )(279.1 ft) 0

2 32.2 ft/sABF− + =

28 kips 8.62 kipsABF+ = +

19.38 kipsABF = − 19.38 kips (compression)ABF =

Car C: 1 1 2 2T V T−+ =

22

1 80 kips(44 ft/s) (279.1 ft) 0

2 32.2 ft/sBCF+ =

8.617 kipsBCF = − 8.62 kips (compression)BCF =

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537

PROBLEM 13.26

A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.

SOLUTION

Call blocks A and B. 2 kg, 3 kgA Bm m= =

(a) Position 1: Block B has just been removed.

Spring force: ( )S A BF m m g k x= − + = −

Spring stretch: 2

1( ) (5 kg)(9.81 m/s )

1.22625 m40 N/m

A Bm m gx

k

+= − = − = −

Let position 2 be a later position while the spring still contacts block A.

Work of the force exerted by the spring: 2

1

2

1

1 2

2 2 21 2

2 2 22 2

( )

1 1 1

2 2 2

1 1(40)( 1.22625) (40) 30.074 20

2 2

x

ex

x

x

U k x dx

k x k x k x

x x

→ = −

= − = −

= − − = −

Work of the gravitational force: 1 2 2 1

2 2

( ) ( )

(2)(9.81)( 1.22625) 19.62 24.059

g AU m g x x

x x

→ = − −

= − + = − −

Total work: 21 2 2 220 19.62 6.015U x x→ = − + +

Kinetic energies: 1

2 2 22 2 2 2

0

1 1(2)

2 2A

T

T m v v v

=

= = =

Principle of work and energy: 1 1 2 2T U T→+ =

2 22 2 20 20 19.62 6.015x x v+ − + =

Speed squared: 2 22 2 220 19.62 6.015v x x= − − + (1)

At maximum speed, 2

2

0dv

dx=

PROPRIETARY MATERIAL. © 2013 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.

538

PROBLEM 13.26 (Continued)

Differentiating Eq. (1), and setting equal to zero,

22 2

2

2 40 19.62 0

19.620.4905 m

40

dvv x

dx

x

= − = − =

= − = −

Substituting into Eq. (1), 2 2 2 22 (20)( 0.4905) (19.62)( 0.4905) 6.015 10.827 m /sv = − − − − + =

Maximum speed: 2 3.29 m/sv =

(b) Position 3: Block A reaches maximum height. Assume that the block has separated from the spring. Spring force is zero at separation.

Work of the force exerted by the spring:

1

02 2

1 3 11 1

( ) (40)(1.22625) 30.074 J2 2e

xU kxdx kx→ = − = = =

Work of the gravitational force:

1 3( ) (2)(9.81) 19.62 g AU m gh h h→ = − = − = −

Total work: 1 3 30.074 19.62 U h→ = −

At maximum height, 3 30, 0v T= =

Principle of work and energy: 1 1 3 3T U T→+ =

0 30.074 19.62 0h+ − =

Maximum height: 1.533 mh =

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585

PROBLEM 13.58

A 3-lb collar is attached to a spring and slides without friction along a circular rod in a horizontal plane. The spring has an undeformed length of 7 in. and a constant 1.5 lb/in.k = Knowing that the collar is in equilibrium at A and is given a slight push to get it moving, determine the velocity of the collar (a) as it passes through B, (b) as it passes through C.

SOLUTION

0 7 in., 20 in.DAL L= =

2 2(8 6) 6 15.23 in.DBL = + + =

8 in.DCL =

20 7 13 in.DALΔ = − =

15.23 7 8.23 in.DBLΔ = − =

8 7 1 in.DCLΔ = − =

(a) 2 21 10, ( ) (1.5)(13) 126.75 lb in.

2 2A A DAT V k L= = Δ = = ⋅

10.5625 lb ft= ⋅

2 21 1.5

2B B BT mv vg

= =

21(1.5)(8.23) 50.8 lb in. 4.233 lb ft

2BV = = ⋅ = ⋅

21.5

: 0 10.5625 4.23332.2

BA A B B

vT V T V+ = + + = + 11.66 ft/sBv =

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586

PROBLEM 13.58 (Continued)

(b) 21.50, 10.5625 lb ft,

32.2A A C CT V T v= = ⋅ =

21(1.5)(1) 0.75 lb in. 0.0625 lb ft

2CV = = ⋅ = ⋅

21.5: 0 10.5625 0.0625

32.2A A C C cT V T V v+ = + + = + 15.01 ft/sCv =

PROPRIETARY MATERIAL. © 2013 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.

599

PROBLEM 13.65

A 1-kg collar can slide along the rod shown. It is attached to an elastic cord anchored at F, which has an undeformed length of 250 mm and a spring constant of 75 N/m. Knowing that the collar is released from rest at A and neglecting friction, determine the speed of the collar (a) at B, (b) at E.

SOLUTION

2 2 2

2 2

2 2

(0.5) (0.4) (0.3)

0.70711 m

(0.4) (0.3)

0.5 m

(0.5) (0.3)

0.58309 m

AF

AF

BF

BF

FE

FE

e g

L

L

L

L

L

L

V V V

= + +=

= +=

= +== +

(a) Speed at B: 0, 0A Av T= =

Point A:

20

1( ) ( ) 0.70711 0.25

2A e AF AF AFV k L L L L= Δ Δ = − = −

0.45711 mAFLΔ =

2

2

1( ) (75 N/m)(0.45711 m)

2( ) 7.8355 N m

( ) ( )(0.4)

(1.0 kg)(9.81 m/s )(0.4 m)

3.9240 N m

( ) ( )

7.8355 3.9240

11.7595 N m

A e

A e

A g

A A e A g

V

V

V mg

V V V

=

= ⋅=

== ⋅= +

= += ⋅

PROPRIETARY MATERIAL. © 2013 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.

600

PROBLEM 13.65 (Continued)

Point B: 2 2

2

1 1(1.0 kg)

2 2

0.5

B B B

B B

T mv v

T v

= =

=

20

1( ) ( ) 0.5 0.25

2B e BF BF BFV k L L L L= Δ Δ = − = −

0.25 mBFLΔ =

2

2

1( ) (75 N/m)(0.25 m) 2.3438 N m

2

( ) ( )(0.4) (1.0 kg)(9.81 m/s )(0.4 m) 3.9240 N m

( ) ( ) 2.3438 3.9240 6.2678 N m

B e

B g

B B e B g

V

V mg

V v V

= = ⋅

= = = ⋅

= + = + = ⋅

2

2

2 2 2

0 11.7595 0.5 6.2678

(5.49169)/(0.5)

10.983 m /s

A A B B

B

B

B

T V T V

v

v

v

+ = +

+ = +

=

= 3.31 m/sBv =

(b) Speed at E:

Point A: 0 11.7595 N mA AT V= = ⋅ (from part (a))

Point E:

2 2 2

20

1 1(1.0 kg) 0.5

2 21

( ) ( ) 0.5831 0.252

E E E E

E e FE FE FE

T mv v v

V k L L L L

= = =

= Δ Δ = − = −

0.3331 mFELΔ =

2

2

2

2 2 2

1( ) (75 N/m)(0.3331 m) 4.1607 N m

2( ) 0 4.1607 N m

0 11.7595 0.5 4.1607

7.5988/0.5

15.1976 m /s

E e

E g E

A A E E E

E

E

V

V V

T V T V v

v

v

= = ⋅

= = ⋅

+ = + + = +

=

= 3.90 m/sEv =

PROPRIETARY MATERIAL. © 2013 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.

612

PROBLEM 13.72

A 1-lb collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 5 in. and a constant 10 lb/ft.k = Knowing that the collar is released from being held at A determine the speed of the collar and the normal force between the collar and the rod as the collar passes through B.

SOLUTION

For the collar, 210.031056 lb s /ft

32.2

Wm

g= = = ⋅

For the spring, 010 lb/ft 5 in.k l= =

At A:

0

7 5 5 17 in.

12 in. 1 ftA

Δ

= + + =− = =

At B: 2 2

0

(7 5) 5 13 in.

21.8 in. ft

3

B

B

= + + =

− = =

Velocity of the collar at B.

Use the principle of conservation of energy.

A A B BT V T V+ = +

Where 210

2A AT mv= =

20

2

2 2 2

20

2

2

2 2 2

1( ) (0)

21

(10)(1) 0 5 ft lb21 1

(0.031056) 0.0155282 21

( )2

1 2 5(10) (1)

2 3 12

1.80556 ft lb

0 5 0.015528 1.80556

205.72 ft /s

A A

B B B B

B B

B

B

V k W

T mv v v

V k Wh

v

v

= − +

= + = ⋅

= = =

= − +

= + −

= ⋅

+ = =

=

14.34 ft/sBv =

PROPRIETARY MATERIAL. © 2013 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.

613

PROBLEM 13.72 (Continued)

Forces at B.

0

2

2( ) (10) 6.6667 lb.

3

5sin

135

5 in. ft12

(0.031056)(205.72)

5/1215.3332 lb

s B

Bn

F k

mvma

α

ρ

ρ

= − = =

=

= =

=

=

=

: siny y s nF ma F W N maαΣ = − + =

sin

515.3332 1 (6.6667)

13

13.769 lb

n sN ma W F

N

α= + −

= + −

= 13.77 lb=N