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Unformatted text preview: homework 06 – ASH, BEN – Due: Feb 27 2008, 11:00 pm 1 Question 1, chap 6, sect 1. part 1 of 1 10 points A force F acts to the right on a 6 . 06 kg block. A 1 . 49 kg block is stacked on top of the 6 . 06 kg block and can slide on it with a coefficient of friction of 0 . 19 between the blocks. The table has a coefficient of friction of 0 . 17. The acceleration of gravity is 9 . 8 m / s 2 . The system is in equilibrium. 6 . 06 kg 1 . 49 kg μ 2 μ 1 F Find the force F required to accelerate the 6 . 06 kg block at 3 . 5 m / s 2 . Correct answer: 39 . 2536 N (tolerance ± 1 %). Explanation: Given : m 1 = 1 . 49 kg , m 2 = 6 . 06 kg , μ 1 = 0 . 19 , and μ 2 = 0 . 17 . Consider the free body diagrams below for each mass m 2 m 1 F T μ 2 ( m 1 + m 2 ) g 2 T μ 1 m 1 g μ 1 m 1 g Basic Concepts : The acceleration of the two masses will be different because of the pulley system between them. The tensions in the strings will be different. Solution Let T be the tension in the string attached to m 2 on its left. Its acceleration will be a and directed to the right. Because of the pulley system between m 1 and m 2 , 2 T will be the tension in the string attached to m 1 on its left, and the acceleration of that string, and of m 1 , will be a 2 and will be directed to the left. For the mass m 1 , 2 T acts to the left and the motion to the left defines the frictional force μ 1 m 1 g acting to the right, with the acceleration a 2 directed to the left F net 1 = m 1 parenleftBig a 2 parenrightBig = 2 T − μ 1 m 1 g . Multiplying by 2, m 1 a = 4 T − 2 μ 1 m 1 g . (1) For the mass m 2 , notice first that the normal force will be N = ( m 1 + m 2 ) g , and since the frictional force μ 1 m 1 g acted to the right on m 1 , it acts to the left on m 2 . For the block m 2 , the force F acts to the right, and the tension T and the two frictional forces μ 1 m 1 g and μ 2 ( m 1 + m 2 ) g act to the left, with the acceleration a directed to the right F net 2 = m 2 a = F − T − μ 1 m 1 g − μ 2 ( m 1 + m 2 ) g . Multiplying by 4 4 m 2 a = 4 F − 4 T − 4 μ 1 m 1 g − 4 μ 2 ( m 1 + m 2 ) g . (2) Adding equations (1) and (2) yields ( m 1 + 4 m 2 ) a = 4 F − 6 μ 1 m 1 g − 4 μ 2 ( m 1 + m 2 ) g . Therefore F = ( m 1 + 4 m 2 ) a 4 + 3 μ 1 m 1 g 2 + μ 2 ( m 1 + m 2 ) g . = [1 . 49 kg + 4 (6 . 06 kg)] (3 . 5 m / s 2 ) 4 + 3 (0 . 19) (1 . 49 kg) (9 . 8 m / s 2 ) 2 + (0 . 17) (1 . 49 kg + 6 . 06 kg) (9 . 8 m / s 2 ) . = 39 . 2536 N . homework 06 – ASH, BEN – Due: Feb 27 2008, 11:00 pm 2 Question 2, chap 6, sect 1. part 1 of 1 10 points The suspended m 1 mass on the right is moving up, the m 2 mass slides down the ramp, and the suspended m 3 mass on the left is moving down. There is friction between the block and the ramp, with the coefficient of the kinetic friction μ . The acceleration of gravity is g and the acceleration of the three block system is a . The pulleys are massless and frictionless....
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 Spring '08
 ERSKINE/TSOI
 Force, Friction, Work, Sin, Cos

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