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# hw10 - Overton Mays Homework 10 Due 4:00 am Inst Turner...

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Overton, Mays – Homework 10 – Due: Feb 14 2005, 4:00 am – Inst: Turner 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A(n) 864 N crate is being pushed across a level floor by a force of 396 N at an angle of 25 above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.20. The acceleration of gravity is 9 . 81 m / s 2 . 864 N μ k = 0 . 20 396 N 25 What is the acceleration of the box? Correct answer: 1 . 73295 m / s 2 . Explanation: Let : W = m g = 864 N , F applied = 396 N , α = 25 , μ k = 0 . 20 , and g = 9 . 81 m / s 2 m F applied α m g N μ k N Basic Concepts: F applied,x = F applied cos α F applied,y = F applied sin α F y,net = F n + F applied,y - W = 0 F k = μ k F n F x,net = m a x = F applied,x - F k Solution: F applied,x = (396 N) cos 25 = 358 . 898 N F applied,y = (396 N) sin 25 = 167 . 357 N The normal force is F n = W + F applied,y = 864 N + 167 . 357 N = 1031 . 36 N . The mass of the crate is m = W g = 864 N 9 . 81 m / s 2 = 88 . 0734 kg . From the horizontal motion, m a x = F applied,x - μ k F n a x = F applied,x - μ k F n m = 358 . 898 N - (0 . 2) (1031 . 36 N) 88 . 0734 kg = 1 . 73295 m / s 2 across the floor. 002 (part 1 of 1) 10 points A 3.70 kg block is pushed along the ceiling with a constant applied force of 90.0 N that acts at an angle of 62.0 with the horizontal. The block accelerates to the right at 6.80 m/s 2 . The acceleration of gravity is 9 . 81 m / s 2 . 3 . 7 kg 90 N 62 μ 6 . 8 m / s 2 What is the coefficient of kinetic friction between the block and the ceiling? Correct answer: 0 . 395949 . Explanation:

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Overton, Mays – Homework 10 – Due: Feb 14 2005, 4:00 am – Inst: Turner 2 Basic Concepts: F applied,x = F applied cos θ F applied,y = F applied sin θ F y,net = F applied,y - mg - F n = 0 F x,net = ma x = F applied,x - F k F k = μ k F n m F θ μ a Given: m = 3 . 70 kg θ = 62 . 0 F applied = 90 N a x = 6 . 80 m g = 9 . 81 m / s 2 Solution: F applied,x = (90 N) cos 62 = 42 . 2524 N F applied,y = (90 N) sin 62 = 79 . 4653 N The normal force is F n = F applied,y - mg = 79 . 4653 N - (3 . 7 kg)(9 . 81 m / s 2 ) = 43 . 1683 N From the horizontal motion, ma x = F applied,x - μ k F n μ k = F applied,x - ma x F n = 42 . 2524 N - (3 . 7 kg)(6 . 8 m / s 2 ) 43 . 1683 N = 0 . 395949 003 (part 1 of 3) 10 points The suspended 2 . 6 kg mass on the right is moving up, the 1 . 4 kg mass slides down the ramp, and the suspended 7 . 2 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 15 .
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