Question
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
Answers
FredR
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.
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.