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Unformatted text preview: to the right and positive y direction upward. Newton’s 2 nd law yields x: 0 = T – f f = T = 30 N y: 0 = N – (W A + W C ) N = W A + W C When the weight of block C is minimum, the friction on A is equal to the maximum static friction. Therefore, we have f = μ s N μ s (W A + W C ) = W B W C = W B / μ s – W C = 25.0 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 0.200? Two blocks are connected together, thus in our coordinates systems, a A = a B = a. The kinetic friction on block A is f k = μ k N = μ s W A = 10 N Newton’s 2 nd law applied to block A and B yields m A a = T - f k m B a = W B- T (m A + m B )a = W B- f k a = (W B- f k )/((W A + W B )/g) = 2.45 m/s 2...
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This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.
- Spring '08