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hw 12 - homework 12 HETTIGER CHRISTOF Due 4:00 am Question...

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homework 12 – HETTIGER, CHRISTOF – Due: Feb 18 2008, 4:00 am 1 Question 1, chap 6, sect 1. part 1 of 1 10 points There is friction between the block and the table. The suspended 2 kg mass on the left is moving up, the 4 kg mass slides to the right on the table, and the suspended mass 4 kg on the right is moving down. The acceleration of gravity is 9 . 8 m / s 2 . 2 kg 4 kg 4 kg μ = 0 . 17 What is the magnitude of the acceleration of the system? Correct answer: 1 . 2936 m / s 2 (tolerance ± 1 %). Explanation: m 1 m 2 m 3 μ a Let : m 1 = 2 kg , m 2 = 4 kg , m 3 = 4 kg , and μ = 0 . 17 . Basic Concepts: The acceleration a of each mass is the same, but the tensions in the two strings will be different. F net = m a negationslash = 0 Solution: Let T 1 be the tension in the left string and T 2 be the tension in the right string. Consider the free body diagrams for each mass T 1 m 1 g a T 2 m 3 g a T 1 T 2 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, so F net 1 = m 1 a = T 1 m 1 g . (1) For the mass on the table, a is directed to the right, T 2 acts to the right, T 1 acts to the left, and the motion is to the right so that the frictional force μ m 2 g acts to the left and F net 2 = m 2 a = T 2 T 1 μ m 2 g . (2) For the mass m 3 , T 2 acts up and the weight m 3 g acts down, with the acceleration a di- rected downward, so F net 3 = m 3 a = m 3 g T 2 . (3) Adding these equations yields ( m 1 + m 2 + m 3 ) a = m 3 g μ m 2 g m 1 g a = m 3 μ m 2 m 1 m 1 + m 2 + m 3 g = 4 kg (0 . 17) (4 kg) 2 kg 2 kg + 4 kg + 4 kg × (9 . 8 m / s 2 ) = 1 . 2936 m / s 2 . Question 2, chap 6, sect 1. part 1 of 3 10 points The suspended 2 . 1 kg mass on the right is moving up, the 2 . 1 kg mass slides down the ramp, and the suspended 7 . 5 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 12 .
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homework 12 – HETTIGER, CHRISTOF – Due: Feb 18 2008, 4:00 am 2 The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 2 . 1 kg μ = 0 . 12 29 7 . 5 kg 2 . 1 kg What is the acceleration of the three block system? Correct answer: 5 . 19123 m / s 2 (tolerance ± 1 %). Explanation: Let : m 1 = 2 . 1 kg , m 2 = 2 . 1 kg , m 3 = 7 . 5 kg , and θ = 29 . Basic Concept: F net = m a negationslash = 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 m g sin θ and the perpen- dicular component of its weight is m g cos θ .
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