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A force of F = 25 lb is applied to the 10-lb ring as shown. If the coefficients of static and kinetic friction are us=0.3 and
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Given data, f=2516 malolb K = 0.25 Ms=013 The mass center of of the ring moment of inertia gravity G is about itle IG = m 2Now, here ring rolls without slipping ag=&r=2(075) - By solving ean Of 2152 (0.75) + 5:6252 = 1.511213 17:52 = 1.511213 a=0.2

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