= 0.141## Good Job!

What must be the coefficient of friction between the tires and the level roadway to allow a car to make a curve of radius r = 350 m at a speed of 80 km/h?

For a level curve, the force of friction is the

onlyhorizontal force on a car and provides the centripetal force. This can be seen from the free-body diagram:The net force must be horizontal -- pointing toward the center of the circle -- and only the friction force is available to provide it. The normal force and the weight simply cancel each other (n = w = M g).

Calculations are easier if the speed is expressed in m/s so we shall convert that first.

km 1000 m h v = 80 -- [------] [------] h km 3600 s v = 22.2 m/s Now we can calculate the necessary centripetal force,

F _{c}= M v^{2}/ r(22.2 m/s) ^{2 }Fc = M ------------ 350 mF _{c}= 1.41 M (m/s^{2})n

_{}= M g = (M) (9.8 m/s^{2})For this flat curve, the centripetal force is supplied by the friction force, F

_{f},F _{f }= nF

_{f }= M 9.8 m/s^{2}F

_{f }= F_{net}= F_{c}M 9.8 m/s

^{2}= 1.41 M (m/s^{2})= 0.141

Of course, any coefficient of friction

greaterthan 0.141 will keep the car from slipping; this is the minimum value for . Notice that the force of friction is perpendicular to the velocity. But it still is in the direction necessary to oppose the motion that would occur without friction. The car will slide away from the center (to the right in this diagram) if it encounters a patch of low-friction surface (like ice or oil); therefore the friction force is directed toward the center.Notice, too, that we are dealing with

static friction.The piece of the tire in contact with the road is momentarily at rest with respect to the road! The force of static friction is somewhat greater than the force provided by kinetic friction, if the tires start to skid.

(c) 2001, Doug Davis; all rights reserved.