hw13 - hinojosa (jlh3938) homework 13 Turner (58185) This...

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hinojosa (jlh3938) – homework 13 – Turner – (58185) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 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 . 11 What is the magnitude oF the acceleration oF the system? Correct answer: 1 . 5288 m / s 2 . 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 . 11 . Basic Concepts: The acceleration a oF each mass is the same, but the tensions in the two strings will be di±erent. F net = ma n = 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 ²or 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) ²or 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 μm 2 g acts to the leFt and F net 2 = m 2 a = T 2 T 1 μm 2 g . (2) ²or 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 μm 2 g m 1 g a = m 3 μm 2 m 1 m 1 + m 2 + m 3 g = 4 kg (0 . 11) (4 kg) 2 kg 2 kg + 4 kg + 4 kg × (9 . 8 m / s 2 ) = 1 . 5288 m / s 2 . 002 10.0 points The suspended 2 . 1 kg mass on the right is moving up, the 1 . 5 kg mass slides down the ramp, and the suspended 8 . 5 kg mass on the leFt is moving down. There is Friction between the block and the ramp.
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hinojosa (jlh3938) – homework 13 – Turner – (58185) 2 The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 1 . 5 kg μ = 0 2 25 8 . 5 kg 2 . 1 kg What is the tension in the cord connected to the 8 . 5 kg block? Correct answer: 36 . 7481 N. Explanation: Let : m 1 = 2 . 1 kg , m 2 = 1 . 5 kg , m 3 = 8 . 5 kg , and θ = 25 . Basic Concept: F net = ma n = 0 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be diFerent. 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 N a m 2 g ±or 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) ±or 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
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