Page 83 - Schaum's Outline of Theory and Problems of Applied Physics

68 FRICTION [CHAP. 6

frictional force, not the actual frictional force. Up to µ s N, the actual frictional force F f has the same magnitude
as the applied force but is in the opposite direction.
When the applied force exceeds the starting frictional force µ s N, motion begins and now the coefficient of
kinetic friction µ k governs the frictional force. In this case µ k N gives the actual amount of F f , which no longer
depends on the applied force and is constant over a fairly wide range of relative velocities.

ROLLING FRICTION
Ideally there should be no rolling friction, because there should be no relative motion between the surfaces in
contact. In reality, a wheel or ball is slightly flattened when it rests on a surface, which itself is slightly dented. A
resistive force arises when the wheel or ball rolls, partly because it and the surface must be continually deformed
and partly because there is some relative motion between them owing to the deformation. Coefficients of rolling
friction µ r are nevertheless much smaller than those of kinetic friction: For a rubber tire rolling on a concrete
road µ r is about 0.04, for instance, whereas it is 0.7 for the same tire sliding on the road.

SOLVED PROBLEM 6.1
How much force is needed to keep a 1200-kg car moving at constant velocity on a level concrete road?
Assume that the car is moving too slowly for air resistance to be important, and use µ k = 0.04 for the
coefficient of rolling friction.

The normal force here is the car’s weight mg, because weight is a force that acts vertically downward. Hence
2
F = F f = µ k N = µ k mg = (0.04)(1200 kg)(9.8 m/s ) = 470 N

SOLVED PROBLEM 6.2
A force of 200 N is just sufficient to start a 50-kg steel trunk moving across a wooden floor. Find the
coefficient of static friction.
The normal force is the trunk’s weight mg. Hence
F F 200 N
µ s = = = = 0.41
2
N mg (50 kg)(9.8 m/s )
SOLVED PROBLEM 6.3
A 40-kg wooden crate is being pushed across a wooden floor with a force of 160 N. If µ k = 0.3, find the
acceleration of the crate.

The applied force F A = 160 N is opposed by the frictional force
2
F f = µ k N = µ k mg = (0.3)(40 kg)(9.8 m/s ) = 118 N
The net force on the crate is therefore
F = F A − F f = 160 N − 118 N = 42 N
and its acceleration is
F 42 N 2
a = = = 1.05 m/s
m 40 kg

SOLVED PROBLEM 6.4
A bowling ball with an initial velocity of 3 m/s rolls along a level floor for 50 m before coming to a stop.
What is the coefficient of rolling friction?