Unformatted text preview: block B. Newton’s 2 nd law yields A & C x : 0 = T – f A & C y : 0 = N – W A – W C B: 0 = W B – T ⇒ W B = f If the minimum weight of block C is on A, the blocks A and C are on the verge of sliding. Thus we have f = f s, max = μ N = μ (W A + W C ) W B = μ (W A + W C ) W C = W B / μ W A = 73.3 N b) Block C suddenly is lifted off A . What is the acceleration of block A if μ k between A and the table is 0.150? In our coordinate system, a A = a B = a. Newton’s 2 nd law yields A x : m A a = T – f A y : 0 = N – W A B: m B a = W B – T (m A + m B )a = W B μ k (W A ) a= (W B μ k W A )g/( W B + W A ) = 3.04 m/s 2...
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 Summer '08
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 Work, Kilogram, Polar coordinate system, wA, Tomoyuki Nakayama, WA – WC

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