Page 84 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 6] FRICTION 69

2
2
We begin by finding the ball’s acceleration. From Chapter 3, v = v + 2as. Here v = 0, v 0 = 3 m/s, and
0
s = 50 m, and so
v 2 0 (3 m/s) 2 2
a =− =− =−0.09 m/s
2s (2)(50 m)
The force that corresponds to this acceleration is ma, which is equal and opposite to the frictional force µ r N = µ r mg.
Hence

a −0.09 m/s 2
ma =−µ r mg µ r =− =− = 0.0092
g 9.8 m/s 2

SOLVED PROBLEM 6.5
The coefficient of kinetic friction between a rubber tire and a wet concrete road is 0.5. (a) Find the
minimum time in which a car whose initial velocity is 50 km/h can come to a stop on such a road. (b)
What distance will the car cover in this time?
(a) The maximum available frictional force is µ k N = µ k mg, and so

F f = µ k mg =−ma
2
a =−µ k g =−(0.5)(9.8 m/s ) =−4.9 m/s 2
km 1000 m/km

Here v 0 = 50 = 13.9 m/s
h 3600 s/h
and the final velocity of the car is v = 0. Hence

v = v 0 + at = 0 v 0 =−at

and the time required for the car to come to a stop is

v 0 13.9 m/s
t =− =− = 2.84 s
a −4.9 m/s 2
(b) The distance covered by the car in coming to a stop is

1 2 2 2
1
s = v 0 t + at = (13.9 m/s)(2.84 s) + (−4.9 m/s )(2.84 s) = 19.7m
2 2
2
2
Another way to obtain this result is to use the formula v = v + 2as. Here v = 0, so
0
v 2 0 13.9 m/s 2
s =− =− = 19.7m
2
2a (2)(−4.9 m/s )
SOLVED PROBLEM 6.6
A block at rest on an adjustable inclined plane begins to move when the angle between the plane and the
horizontal reaches a certain value θ, which is known as the angle of repose. How is this angle related to
the coefficient of static friction between the block and the plane?

The weight w of the block can be resolved into a component F w parallel to the plane and another component N
perpendicular to the plane. From Fig. 6-2 the magnitudes of F w and N are