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solution21 - fullwood(tf5349 Homework 6 Dynamics...

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fullwood (tf5349) – Homework 6 Dynamics – Catala – (49033) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A block oF mass 3 . 26 kg lies on a Frictionless horizontal surFace. The block is connected by a cord passing over a pulley to another block oF mass 7 . 47 kg which hangs in the air, as shown on the Following picture. Assume the cord to be light (massless and weightless) and unstretchable and the pulley to have no Friction and no rotational inertia. The acceleration oF gravity is 9 . 8 m / s 2 . 3 . 26 kg 7 . 47 kg Calculate the acceleration oF the frst block. Correct answer: 6 . 82255 m / s 2 . Explanation: Given : m 1 = 3 . 26 kg , and m 2 = 7 . 47 kg . m 1 m 2 a T N m 1 g a T m 2 g Since the cord is unstretchable, the frst block accelerates to the right at exactly the same rate a as the second (hanging) block ac- celerates downward. Also, the cord’s tension pulls the frst block to the right with exactly the same tension T as it pulls the second block upward. The only horizontal Force acting on the frst block is the cord’s tension T , hence by New- ton’s Second Law m 1 a = F net 1 = T . The second block Feels two vertical Forces: The cord’s tension T (upward) and the block’s own weight W 2 = m 2 g (downward). Conse- quently, m 2 a = F net 2 = m 2 g T . Adding the two equations together, we arrive at ( m 1 + m 2 ) a = m 2 g , and hence a = m 2 m 1 + m 2 g = 7 . 47 kg 3 . 26 kg + 7 . 47 kg (9 . 8 m / s 2 ) = 6 . 82255 m / s 2 . 002 (part 2 oF 2) 10.0 points Calculate the tension in the cord. Correct answer: 22 . 2415 N. Explanation: T = m 1 a = (3 . 26 kg) (6 . 82255 m / s 2 ) = 22 . 2415 N . 003 (part 1 oF 2) 10.0 points A child holds a sled on a Frictionless, snow- covered hill, inclined at an angle oF 27 . 79 N F 27 IF the sled weighs 79 N, fnd the Force ex- erted on the rope by the child. Correct answer: 35 . 8653 N. Explanation:
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fullwood (tf5349) – Homework 6 Dynamics – Catala – (49033) 2 Given : W = 79 N and θ = 27 . Consider the free body diagram for the block mg sin θ N = cos F W Basic Concepts: If we “tilt” our world, and consider the forces parallel to the hill, F net = s F up s F down = 0 then the forces perpendicular to the hill, F net = s F out s F in = 0 Solution: Consider the free body diagram for the sled: The weight of the sled has compo- nents W sin θ acting down the hill and W cos θ acting straight into the hill. The system is in equilibrium, so for forces parallel to the hill, F net = T −W sin θ = 0 = T = W sin θ = (79 N) sin27 = 35 . 8653 N 004 (part 2 of 2) 10.0 points What force is exerted on the sled by the hill? Correct answer: 70
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solution21 - fullwood(tf5349 Homework 6 Dynamics...

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