Posted by **Anonymous** on Thursday, March 11, 2010 at 10:33am.

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

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**FredR**, Thursday, March 11, 2010 at 7:23pm
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.

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