hw11 - Muraj Hamza Homework 11 Due 4:00 am Inst Florin This...

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Muraj, Hamza – Homework 11 – Due: Feb 15 2006, 4:00 am – Inst: Florin 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. 001 (part 1 of 1) 0 points The surfaces between a 6 kg block and the 23 kg wedge and between the 23 kg wedge and the horizontal plane is smooth (without friction). The acceleration of gravity is 9 . 8 m / s 2 . A block is released on the inclined plane (top side of the wedge). 6 kg μ = 0 29 23 kg μ = 0 F 9 . 8 m / s 2 What is the force F which must be exerted on the 23 kg block in order that the 6 kg block does not move up or down the plane? Correct answer: 157 . 535 N. Explanation: Let : m A = 23 kg , block A m B = 6 kg , block B θ = 29 , and μ s = 0 between A and B . Using Newton’s second law, the accelera- tion of the wedge of mass 23 kg and block of mass 6 kg is a AB = F m A + m B (1) Consider the free body diagram for block B m g sin θ N μ N m g a AB The condition that block B does not move with respect to wedge A implies that its ac- celeration down the wedge with respect to the wedge is zero. Applying Newton’s second law for B in the direction parallel and perpendic- ular to the wedge yields X F : N - m B g cos θ = m B a AB sin θ (1) X F k : m B g sin θ - μ N = m B a AB cos θ (2) Using Eq. 1 to solve for the normal force N exerted on block B by block A , we have N = m B ( g cos θ + a AB sin θ ) . Substituting N from Eq. 1 into Eq. 2, but since μ = 0 , we have m B g sin θ = m B a AB cos θ Solving for a AB , we have a AB = g sin θ cos θ = g tan θ = (9 . 8 m / s 2 ) tan 29 = (9 . 8 m / s 2 ) (0 . 554309) = 5 . 43223 m / s 2 . Therefore the force F is F = ( m A + m B ) a AB = (23 kg + 6 kg) (5 . 43223 m / s 2 ) = 157 . 535 N . 002 (part 1 of 1) 0 points A force F acts to the right on a(n) 6 . 67631 kg block. A(n) 2 . 25761 kg block is stacked on top of the 6 . 67631 kg block, with a coefficient of friction of 0 . 2 between the blocks. The table has a coefficient of friction of 0 . 249. The acceleration of gravity is 9 . 8 m / s 2 . The system is in equilibrium.
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Muraj, Hamza – Homework 11 – Due: Feb 15 2006, 4:00 am – Inst: Florin 2 6 . 67631 kg 2 . 25761 kg μ 2 μ 1 F Find the force F required to accelerate the 6 . 67631 kg block at 3 . 2 m / s 2 . Correct answer: 51 . 6082 N. Explanation: Given : m 1 = 2 . 25761 kg , m 2 = 6 . 67631 kg , μ 1 = 0 . 2 , and μ 2 = 0 . 249 . 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.
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