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Unformatted text preview: Su (ycs73) HW03 Tsoi (58020) 1 This printout should have 30 questions. Multiplechoice 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 = 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 a Let : m 1 = 2 kg , m 2 = 3 kg , m 3 = 4 kg , and = 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 = ma 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 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 m 2 g acts to the left and F net 2 = m 2 a = T 2 T 1 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 m 2 g m 1 g a = m 3 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 lefthand cable is horizon tal. Su (ycs73) HW03 Tsoi (58020) 2 65 N 4 7 a) What is the tension in the cable slanted at an angle of 47 ? Correct answer: 88 . 8763 N. Explanation: Observe the freebody diagram below. 60 . 6135 N 60 . 6135 N 8 8 . 8 7 6 3 N 65N 65N Scale: 10 N 47 Note: The sum of the x and ycomponents of T 1 , T 2 , and W g are equal to zero. Given : W g = 65 N and = 47 . Basic Concept: Vertically, we have F y,net = F 1 sin W g = 0 Solution: F 1 (sin ) = W g F 1 = W g sin = 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 F 2 = 0 Solution: F 2 = F 1 cos = W g cos sin = (65 N) cos47 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 xdirection. A block with an initial velocity v slides up and back down a frictionless incline....
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 Spring '09
 TSOI
 Friction, Heat

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