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p201f04-homework09-sol

# p201f04-homework09-sol - Version One Homework 9 Heinz 81204...

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Version One – Homework 9 – Heinz – 81204 – Oct 25, 2004 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Two Blocks 02 10:07, calculus, numeric, > 1 min, normal. 001 (part 1 of 3) 4 points The blocks shown in figure are connected by an inextensible string of negligible mass. The string passes over a frictionless pulley without slipping on the rim of the pulley. The acceleration of gravity is 9 . 8 m / s 2 . The block on the frictionless incline is mov- ing with a constant acceleration up the in- cline. 20 kg T 1 R = 0 . 25 m M 6 . 3 kg μ = 0 T 2 6 m / s 2 30 Determine the tension T 1 in the vertical part of the string. Correct answer: 76 N. Explanation: Let : M = mass of pulley , R = 0 . 25 m a = 6 m / s 2 θ = 30 , m 1 = 20 kg , m 2 = 6 . 3 kg , g = 9 . 8 m / s 2 , μ = 0 , and W = m 1 g = 61 . 74 N . T 2 W k N W W m a W k θ From a free-body diagram (not shown) for the hanging mass m 1 = 20 kg , we see m 1 g - T 1 = m 1 a , and we obtain T 1 = m 1 ( g - a ) = (20 kg) (9 . 8 m / s 2 - 6 m / s 2 ) = 76 N . 002 (part 2 of 3) 3 points Determine the tension T 2 in the string parallel to the incline plane. Correct answer: 68 . 67 N. Explanation: From the free-body diagram (shown above) for the mass m 2 = 6 . 3 kg , we can obtain the other tension T 2 - m 2 g sin θ = m 2 a so that T 2 = m 2 a + m 2 g sin θ or T 2 = m 2 ( a + g sin θ ) = (6 . 3 kg) h (6 m / s 2 ) + (9 . 8 m / s 2 ) sin(30 ) i = 68 . 67 N . 003 (part 3 of 3) 3 points Assume: The pulley has uniform density and is shaped like a narrow cylindrical disk. Find the mass M of the pulley. Correct answer: 2 . 44333 kg. Explanation:

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Version One – Homework 9 – Heinz – 81204 – Oct 25, 2004 2 From the torque equation, applied to the pulley ( T 1 - T 2 ) R = = I a R · we obtain I = [ T 1 - T 2 ] R 2 a = [(76 N) - (68 . 67 N)] (0 . 25 m) 2 (6 m / s 2 ) = 0 . 0763542 kg m 2 . Since I disk = 1 2 M R 2 , we have M = 2 I R 2 = 2 (0 . 0763542 kg m 2 ) (0 . 25 m) 2 = 2 . 44333 kg . Cable Over a Pulley 10:07, trigonometry, numeric, > 1 min, nor- mal. 004 (part 1 of 1) 10 points A cable passes over a pulley. Because the cable grips the pulley and the pulley has non- zero mass, the tension in the cable is not the same on opposite sides of the pulley. The force on one side is 120 N, and the force on the other side is 100 N.
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