A biology student rides her bike around a corner of radius 25 meter at a steady

Question

A biology student rides her bike around a corner of radius 25 meter at a steady speed of 8.2 m/sec. The combined mass of the student and the bike is 87 kg. The coefficient of static friction between the bike and the road is ?s = 0.37.

a) If she is not skidding, what is the magnitude of the force of friction on her bike from the road?

b) What is the minimum value the coefficient of static friction can have before the bike tire will skid?

c) What is the magnitude of the total force between the bike tire and the road?

Explanation / Answer

Part A)

When travelling in a circle, the force needed is the centripetal force. In this case, friction is the centripetal force. The formula for centripetal force is...

Fc = mv2/r

Fc = (87)(8.2)2/25

Fc = 234 N, which is the frictional force

Part B)

Since the friction equation is Ff = Fn and Fn = mg, we have

mg = Ff

()(87)(9.8) = 234

= .274 for the minimum

Part C)

The total force is the normal force (mg) + the centripetal force (mv2/r). Those two forces act at 90o from each other, so to find the total, we will use the Pythagoream Theorem

F = [(234)2 + [(87)(9.8)]2]

F = 884 N

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