CHAPTER 12CHAPTER 12

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PROBLEM 12.1

The value of g at any latitude f may be obtained from the formula

g = +9 78 1 0 0053. ( . ) sin m/s2 2f

which takes into account the effect of the rotation of the earth, as well as the fact that the earth is not truly spherical. Determine to four significant figures (a) the weight in N, (b) the mass in kg, at the latitudes of 0, 45, and 60, of a silver bar, the mass of which has been officially designated as 5 kg.

SOLutiOn

g = +9 78 1 0 0053. ( . ) sin m/s2 2f

f = 0 : g = 9 78. m/s2

f = 45 : g = 9 8059. m/s2

f = 60 : g = 9 8189. m/s2

(a) Weight: W mg=

f = 0 : W = =( ) .5 48 9kg)(9.78 m/s N2 b

f = 45 : W = =( )( . ) .5 9 8059 49 0kg m/s N2 b

f = 60 : W = =( ) .5 49 1kg)(9.8189 m/s N2 b

(b) Mass: At all latitudes: m = 5.000 kg b

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PROBLEM 12.2

The acceleration due to gravity on the moon is 1 62. . m/s2 Determine (a) the weight in newtons, (b) the mass in kilograms, on the moon, of a gold bar, the mass of which has been officially designated as 2 kg.

SOLutiOn

(a) Weight: W mg= = ( )2 kg)(1.62 m/s2 W = 3 24. N b

(b) Mass: Same as on earth: m = 2 00. kg b

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PROBLEM 12.3

A 200-kg satellite is in a circular orbit 1500 km above the surface of Venus. The acceleration due to the gravitational attraction of Venus at this altitude is 5 52. . m/s2 Determine the magnitude of the linear momentum of the satellite knowing that its orbital speed is 23 4 103. km/h.

SOLutiOn

First note v = =23 4 10 65003. km/h m/s

Now L mv= = 200 kg 6500 m/s

or L = 1 300 106. kg m/s b

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PROBLEM 12.4

A spring scale A and a lever scale B having equal lever arms are fastened to the roof of an elevator, and identical packages are attached to the scales as shown. Knowing that when the elevator moves downward with an acceleration of 1.2 m/s2 the spring scale indicates a load of 70 N, 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 1.2 m/s2.

SOLutiOn

Assume g = 9 81. m/s2

(a) mW

g=

+ SF ma W FW

gas= - =:

Wa

gFs1-

=

or WF

ag

s=-

=-1

70

11 29 81

..

W = 79 8. N b

(b) mW

g= = =

79 756

9 818 1300

2

.

..

N

m/s kg

SF ma F W Wg

as= - =:

F Wa

gs= +

= +

1

79 756 11 2

9 81.

.

. Fs = 89 5. N b

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PROBLEM 12.4 (continued)

For the balance system B,

SM bF bFw p0 0 0= - =:

F Fw p=

But F Wa

gw w= +

1

and F Wa

gp p= +

1

so that W Ww p=

and mW

gwp= = 79 756

9 81

.

. mw = 8 13. kg b

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PROPRIETARY MATERIAL. 2010 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.

PROBLEM 12.5

A hockey player hits a puck so that it comes to rest in 9 s after sliding 30 m on the ice. Determine (a) the initial velocity of the puck, (b) the coefficient of friction between the puck and the ice.

SOLutiOn

(a) Assume uniformly decelerated motion.

Then v v at= +0 At t = 9 s: 0 90= +v a( )

or av

= - 09

Also v v a x2 02 2 0= + -( )

At t = 9 s: 0 2 3002= +v a( )

Substituting for a 0 29

30 002 0= + -

=vv

( )

or v0 6 6667= . m/s or m/sv0 6 67= . b

and a = - = -6 66679

0 74074.

. m/s2

(b) We have

+ SF N Wy = - =0 0: or N W mg= =

Sliding: F N mgk k= =m m

+ SF ma F max = - =: or - =mk mg ma

or mka

g= - = - -0 74074

9 81

.

.

m/s

m/s

2

2 or mk = 0 0755. b

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PROPRIETARY MATERIAL. 2010 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.

PROBLEM 12.6

Determine the maximum theoretical speed that an automobile starting from rest can reach after traveling 400 m. Assume that the coefficient of static friction is 0.80 between the tires and the pavement and that (a) the automobile has front-wheel drive and the front wheels support 62 percent of the automobiles weight, (b) the automobile has rear-wheel drive and the rear wheels support 43 percent of the automobiles weight.

SOLutiOn

(a) For maximum acceleration

F F N W

W

F s F= = =

= =max . ( . )

. .

m 0 8 0 62

0 496 0 496 mg

Now + SF ma F max F= =:

or 0 496. mg = ma

Then a = =0 496 9 81 4 86576. ( . ) . m/s m/s2 2

Since a is constant, we have

v a x2 0 2 0= + -( )

When x v= =400 2 4 86576 4002 m: m/s m)2max ( . )(

or vmax .= 62 391 m/s

or vmax = 225 km/h b

(b) For maximum acceleration

F F N W

W

R s R= = =

= =max . )

. .

m 0 8

0 344 0 344

(0.43

mg

Now + SF ma F max R= =:

or 0 344. mg = ma

Then a = =0 344 3 37464. ) .(9.81 m/s m/s2 2

Since a is constant, we have

v a x2 0 2 0= + -( )

When x v= =400 2 3 37464 4002 m: m/s m)2max ( . )(

or vmax .= 51 959 m/s

or vmax .= 187 1 km/h b

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PROPRIETARY MATERIAL. 2010 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.

PROBLEM 12.7

In anticipation of a long 7 upgrade, a bus driver accelerates at a constant rate of 0.9 m/s2 while still on a level section of the highway. Knowing that the speed of the bus is 100 km/h as it begins to climb the grade and that the driver does not change the setting of his throttle or shift gears, determine the distance traveled by the bus up the grade when its speed has decreased to 80 km/h.

SOLutiOn

First consider when the bus is on the level section of the highway.

alevel2 m/s= 0 9.

We have + SF ma P Wg

ax = =: level

Now consider when the bus is on the upgrade.

We have + SF ma P W Wg

ax = - = : sin 7

Substituting for P W

ga W

W

galevel - = sin 7

or = -

= -

= -

a a glevel2

2

m/s

m/s

sin

( . . sin )

.

7

0 9 9 81 7

0 29554For the uniformly decelerated motion

v v a x2 02 2 0= + -( ) ( )upgrade upgrade

Noting that 100250

9 km/h m/s,= then when v v= =

805

0 8 0 km/h

., we have