a 2400-lb (=10.7-kn)car traveling at 30 mi/h (=13.4m/s)attempts to round an unbanked curve with a radius of 200ft (=61.0m).(a) what force of friction is required to keep the car on its path?(b) what minimum coefficient of static friction between the tires and road is required

Answer 1

a)centripetal acceleration is v^2/r. In this case, v is 30 mi/hr and r = 200 ft. The inertial force would be:
F = ma = mv^2/r (1)
This would be the friction force required to keep the car on its path. b) The static friction is:
F = uN (2)
where u is the static friction coefficient and N is the normal force (in this case the weight of the car). Setting (1) equal to (2)
mv^2/r = uN (3a)
m = N (3b)
combine (3a) and (3b) and solve for u.