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Unformatted text preview: lim (kl9356) – Homework 14 – Weathers – (17104) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2 . 4 kg mass of the pulley is concentrated on its rim, which is a distance 20 . 8 cm from the axle. The mass on the right is 1 . 06 kg and on the left is 1 . 58 kg. 2 . 5 m 2 . 4 kg 20 . 8 cm ω 1 . 58 kg 1 . 06 kg What is the magnitude of the linear acceler ation of the hanging masses? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 01111 m / s 2 . Explanation: Let : M = 2 . 4 kg , R = 20 . 8 cm , m 1 = 1 . 06 kg , m 2 = 1 . 58 kg , h = 2 . 5 m , and v = ω R . Consider the free body diagrams 1 . 58 kg 1 . 06 kg T 2 T 1 m 2 g m 1 g a a The net acceleration a = r α is in the direc tion of the heavier mass m 2 . For the mass m 1 , F net = m 1 a = m 1 g T 1 T 1 = m 1 g m 1 a and for the mass m 2 , F net = m 2 a = T 2 m 2 g T 2 = m 2 a + m 2 g . The pulley’s mass is concentrated on the rim, so I = M r 2 , and τ net = summationdisplay τ ccw summationdisplay τ cw = I α T 1 r T 2 r = ( m r 2 ) parenleftBig a r parenrightBig = m r a m a = T 1 T 2 ma = m 1 g m 1 a m 2 a m 2 g ma + m 1 a + m 2 a = m 1 g m 2 g a = ( m 1 m 2 ) g m + m 1 + m 2 = (1 . 06 kg 1 . 58 kg) (9 . 8 m / s 2 ) 2 . 4 kg + 1 . 06 kg + 1 . 58 kg = 1 . 01111 m / s 2 . 002 (part 1 of 2) 10.0 points A block of mass 2 kg and one of mass 6 kg are connected by a massless string over a pulley that is in the shape of a disk having a radius of 0 . 38 m, and a mass of 4 kg. In addition, the lim (kl9356) – Homework 14 – Weathers – (17104) 2 blocks are allowed to move on a fixed block wedge of angle 46 ◦ , as shown. The coefficient of kinetic friction is 0 . 21 for both blocks. 2 kg 6 k g 46 ◦ . 38 m 4 kg What is the acceleration of the two blocks? The acceleration of gravity is 9 . 8 m / s 2 . As sume the positive direction is to the right. Correct answer: 2 . 96035 m / s 2 . Explanation: Let : m 1 = 2 kg , m 2 = 6 kg , M = 4 kg , and R = 0 . 38 m . m 1 m 2 θ M T 1 T 2 Applying Newton’s law to m 1 , N 1 m 1 g = m 1 a y = 0 N 1 = m 1 g , where the force of friction on m 1 is f 1 = μ N 1 = μ m 1 g = (0 . 21) (2 kg) (9 . 8 m / s 2 ) = 4 . 116 N and T 1 f 1 = m 1 a T 1 = m 1 a + f 1 . (1) For the mass m 2 , applying Newton’s law...
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 Fall '08
 Weathers
 Magnetism, Force, Friction, Mass, Work, kg

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