prob_086 - 2 ) = 2 1.00 10 N b. The maximum acceleration...

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86. REASONING AND SOLUTION a. The static frictional force is responsible for accelerating the top block so that it does not slip against the bottom one. The maximum force that can be supplied by friction is f s MAX = μ s F N = μ s m 1 g Newton's second law requires that f s MAX = m 1 a , so a = μ s g The force necessary to cause BOTH blocks to have this acceleration is F = ( m 1 + m 2 )a = ( m 1 + m 2 ) μ s g F = (5.00 kg + 12.0 kg)(0.600)(9.80 m/s
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Unformatted text preview: 2 ) = 2 1.00 10 N b. The maximum acceleration that the two block combination can have before slipping occurs is 17.0 kg F a = Newton's second law applied to the 5.00 kg block is F s m 1 g = m 1 a = (5.00 kg) 17.0 kg F Hence F = 41.6 N ______________________________________________________________________________________ ______...
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