14 - 2.3 kg a m1 g m2 g 1.12 kg The net acceleration a = r...

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roofner (bar784) – Homework 14 – Weathers – (17104) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. 001 10.0 points An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2 . 3 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 . 12 kg and on the left is 1 . 53 kg. 2 . 5 m 2 . 3 kg 20 . 8 cm ω 1 . 53 kg 1 . 12 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: 0 . 811717 m / s 2 . Explanation: Let : M = 2 . 3 kg , R = 20 . 8 cm , m 1 = 1 . 12 kg , m 2 = 1 . 53 kg , h = 2 . 5 m , and v = ω R . Consider the free body diagrams 1 . 53 kg 1 . 12 kg T 2 1 m g a The net acceleration a = r α is in the direc- tion of the heavier mass m 2 . ±or 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 = s τ ccw - s τ cw = I α T 1 r - T 2 r = ( m r 2 ) p a r P = 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 . 12 kg - 1 . 53 kg) (9 . 8 m / s 2 ) 2 . 3 kg + 1 . 12 kg + 1 . 53 kg = 0 . 811717 m / s 2 . 002 (part 1 of 2) 10.0 points A block of mass 3 kg and one of mass 7 kg are connected by a massless string over a pulley that is in the shape of a disk having a radius of 0 . 25 m, and a mass of 9 kg. In addition, the
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roofner (bar784) – Homework 14 – Weathers – (17104) 2 blocks are allowed to move on a Fxed block- wedge of angle 47 , as shown. The coe±cient of kinetic friction is 0 . 19 for both blocks. 3 kg 7 kg 47 0 . 25 m 9 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 . 46177 m / s 2 . Explanation: Let : m 1 = 3 kg , m 2 = 7 kg , M = 9 kg , and R = 0 . 25 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 . 19) (3 kg) (9 . 8 m / s 2 ) = 5 . 586 N and T 1 - f 1 = m 1 a T 1 = m 1 a + f 1 . (1) ²or the mass m 2 , applying Newton’s law
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This document was uploaded on 10/26/2011 for the course PHYS 1710 at North Texas.

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14 - 2.3 kg a m1 g m2 g 1.12 kg The net acceleration a = r...

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