prob_016 - out is 2 1.6 N 3.2 N = . If we assume that only...

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16. REASONING AND SOLUTION Initially, the stone executes uniform circular motion in a circle of radius r which is equal to the radius of the tire. At the instant that the stone flies out of the tire, the force of static friction just exceeds its maximum value fF s MAX sN = µ (see Equation 4.7). The force of static friction that acts on the stone from one side of the tread channel is, therefore, MAX s 0.90(1.8 N) = 1.6 N f = and the magnitude of the total frictional force that acts on the stone just before it flies out is 2 1.6 N
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Unformatted text preview: out is 2 1.6 N 3.2 N = . If we assume that only static friction supplies the centripetal force, then, F c = 3.2 N . Solving Equation 5.3 ( F m v r c = 2 / ) for the radius r , we have ( ) 3 2 2 c 6.0 10 kg (13 m/s) 0.31 m 3.2 N mv r F = = = _______________________________________________________________________________________ ______...
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