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Unformatted text preview: Overton, Mays – Oldquiz 2 – Due: Mar 8 2005, noon – Inst: Turner 1 This printout should have 31 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (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 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 1 . 4 k g μ = . 1 5 34 ◦ 7 . 2 kg 2 . 6 kg What is the acceleration of the three block system? Correct answer: 4 . 55768 m / s 2 . Explanation: Let : m 1 = 2 . 6 kg , m 2 = 1 . 4 kg , m 3 = 7 . 2 kg , and θ = 34 ◦ . Basic Concept: F net = ma 6 = 0 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be different. Let T 1 be the tension in the right string and T 3 the tension in the left string. Consider the free body diagrams for each mass T 3 m 3 g a T 1 m 1 g a T 3 T 1 N μ N a m 2 g For the mass m 1 , T 1 acts up and the weight m 1 g acts down, with the acceleration a di rected upward F net 1 = m 1 a = T 1 m 1 g (1) For the mass on the table, the parallel compo nent of its weight is mg sin θ and the perpen dicular component of its weight is mg cos θ . ( N = mg cos θ from equilibrium). The accel eration a is directed down the table, T 3 and the parallel weight component m 2 g sin θ act down the table, and T 1 and the frictional force μ N = μm 2 g cos θ act up the table F net 2 = m 2 a (2) = T 3 + m 2 g sin θ T 1 μm 2 g cos θ . For the mass m 3 , T 3 acts up and the weight m 3 g acts down, with the acceleration a di rected downward F net 3 = m 3 a = m 3 g T 3 . (3) Adding Eqs. (1), (2), & (3) yields ( m 1 + m 2 + m 3 ) a = m 3 g + m 2 g sin θ μm 2 g cos θ m 1 g . Solving for a , we have a = [ m 2 sin θ μm 2 cos θ + ( m 3 m 1 )] g m 1 + m 2 + m 3 = (1 . 4 kg)(9 . 8 m / s 2 ) sin34 ◦ 2 . 6 kg + 1 . 4 kg + 7 . 2 kg (0 . 15)(1 . 4 kg)(9 . 8 m / s 2 ) cos34 ◦ 2 . 6 kg + 1 . 4 kg + 7 . 2 kg + (7 . 2 kg 2 . 6 kg)(9 . 8 m / s 2 ) 2 . 6 kg + 1 . 4 kg + 7 . 2 kg = 4 . 55768 m / s 2 . Overton, Mays – Oldquiz 2 – Due: Mar 8 2005, noon – Inst: Turner 2 002 (part 2 of 3) 10 points What is the tension in the cord connected to the 2 . 6 kg block? Correct answer: 37 . 33 N. Explanation: Using Eq. 1, we have T 1 = m 1 g + m 1 a = (2 . 6 kg)(9 . 8 m / s 2 + 4 . 55768 m / s 2 ) = 37 . 33 N . 003 (part 3 of 3) 10 points What is the tension in the cord connected to the 7 . 2 kg block? Correct answer: 37 . 7447 N....
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This note was uploaded on 03/20/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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