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# hmwk03 - Su(ycs73 HW03 Tsoi(58020 This print-out should...

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Su (ycs73) – HW03 – Tsoi – (58020) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find 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 3 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 3 kg 4 kg μ = 0 . 17 What is the magnitude of the acceleration of the system? Correct answer: 1 . 62244 m / s 2 . Explanation: m 1 m 2 m 3 μ a Let : m 1 = 2 kg , m 2 = 3 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) (3 kg) 2 kg 2 kg + 3 kg + 4 kg × (9 . 8 m / s 2 ) = 1 . 62244 m / s 2 . 002 (part 1 of 2) 10.0 points Consider the 65 N weight held by two cables shown below. The left-hand cable is horizon- tal.

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Su (ycs73) – HW03 – Tsoi – (58020) 2 65 N 47 a) What is the tension in the cable slanted at an angle of 47 ? Correct answer: 88 . 8763 N. Explanation: Observe the free-body diagram below. 60 . 6135 N 60 . 6135 N 88 . 8763 N 65 N 65 N Scale: 10 N 47 Note: The sum of the x - and y -components of T 1 , T 2 , and W g are equal to zero. Given : W g = 65 N and θ = 47 . Basic Concept: Vertically, we have F y,net = F 1 sin θ − W g = 0 Solution: F 1 (sin θ ) = W g F 1 = W g sin θ = 65 N sin 47 = 88 . 8763 N 003 (part 2 of 2) 10.0 points b) What is the tension in the horizontal ca- ble? Correct answer: 60 . 6135 N. Explanation: Basic Concept: Horizontally, F x,net = F 1 cos θ F 2 = 0 Solution: F 2 = F 1 cos θ = W g cos θ sin θ = (65 N) cos 47 sin 47 = 60 . 6135 N 004 10.0 points The initial position of the block is the origin; i.e. , x = 0 at t = 0 . Consider up the track to be the positive x -direction. A block with an initial velocity v 0 slides up and back down a frictionless incline. v 0 θ Which graph best represents a description the position of the block versus time? 1. t x
Su (ycs73) – HW03 – Tsoi – (58020) 3 2.

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