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# solution24_pdf - arellano(aa39398 Homework 15...

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arellano (aa39398) – Homework 15 – weathers – (17101) 1 This print-out should have 10 questions. Multiple-choice 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 . 2 kg mass of the pulley is concentrated on its rim, which is a distance 22 . 2 cm from the axle. The mass on the right is 1 . 33 kg and on the left is 1 . 99 kg. 1 . 9 m 2 . 2 kg 22 . 2 cm ω 1 . 99 kg 1 . 33 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 . 17174 m / s 2 . Explanation: Let : M = 2 . 2 kg , R = 22 . 2 cm , m 1 = 1 . 33 kg , m 2 = 1 . 99 kg , h = 1 . 9 m , and v = ω R . Consider the free body diagrams 1 . 99 kg 1 . 33 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 m a = ( m 1 - m 2 ) g - ( m 1 + m 2 ) a m a + ( m 1 + m 2 ) a = ( m 1 - m 2 ) g a = ( m 1 - m 2 ) g m + m 1 + m 2 = (1 . 33 kg - 1 . 99 kg) (9 . 8 m / s 2 ) 2 . 2 kg + 1 . 33 kg + 1 . 99 kg = 1 . 17174 m / s 2 . 002 (part 1 of 2) 10.0 points A block of mass 3 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 . 26 m, and a mass of 8 kg. In addition, the

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arellano (aa39398) – Homework 15 – weathers – (17101) 2 blocks are allowed to move on a fixed block- wedge of angle 43 , as shown. The coefficient of kinetic friction is 0 . 17 for both blocks. 3 kg 6 kg 43 0 . 26 m 8 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 . 13791 m / s 2 . Explanation: Let : m 1 = 3 kg , m 2 = 6 kg , M = 8 kg , and R = 0 . 26 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 . 17) (3 kg) (9 . 8 m / s 2 ) = 4 . 998 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 perpendicular to the slanted surface N 2 - m 2 g cos θ = m 2 a perp = 0 N 2 = m 2 g cos θ , so the force of friction is f 2 = μ N 2 = μ m 2 g cos θ = (0 . 17) (6 kg) (9 . 8 m / s 2 ) cos 43 = 7 . 31061 N .
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