Question

The radius of curvature of a highway exit is r = 49.5 m. The surface of the exit road is horizontal, not banked What is the minimum required value of the coefficient of static friction between the tires of the car and the surface of the road so that the car can safely exit the highway at a constant speed of 42.0 km/h without sliding?

Answer 1

Speed of the car = 42.0 km/h = (42 x 1000) / (60 x 60) = 11.67 m/s

Suppose the minimum required value of the coefficient of static friction between the tires of the car and the surface of the road is $\mu s$ .

Now, in order not to slip when negotiating curve at max speed the static friction force must equal the centripetal force.

So, if m = mass of car, then:
$\mu s$ *(mg) = Fc = mV²/R

=> $\mu s$ = V²/(R*g) = 11.67^2 / (49.5 x 9.8) = 0.28

So, the minimum required value of the coefficient of static friction = 0.28 (Answer).