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qz2sol_6942s11 - to the right and positive y direction...

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TA: Tomoyuki Nakayama Monday, February 7, 2011 PHY 2048: Physic 1, Discussion Section 6942 Quiz 2 (Homework Set #4) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ In the figure below right, blocks A and B have weights of 50.0 N and 30.0 N, respectively. a) Determine the minimum weight of block C to keep A from sliding if μ s between A and the table is 0.400. Taking downward as positive, we apply Newton’s 2 nd law to block B 0 = W B – T T = W B For blocks A and C, we take our positive x direction
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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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