solutions 15

# solutions 15 - Patel Kinal Homework 15 Due 7:00 pm Inst D...

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Patel, Kinal – Homework 15 – Due: Oct 30 2007, 7:00 pm – Inst: D Weathers 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 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 21 . 5 cm From the axle. The mass on the right is 1 . 04 kg and on the leFt is 1 . 5 kg. 2 . 9 m 2 . 3 kg 21 . 5 cm ω 1 . 5 kg 1 . 04 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 . 931405 m / s 2 . Explanation: Let : m = 2 . 3 kg , R = 21 . 5 cm , m 1 = 1 . 04 kg , m 2 = 1 . 5 kg , h = 2 . 9 m , v = ω R, I = 1 2 M R 2 , and K disk = 1 2 I ω 2 = 1 4 M v 2 . Consider the Free body diagrams 1 . 5 kg 1 . 04 kg T 2 1 m g a The net acceleration a = 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 ±or 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 = mr 2 ±or the pulley, τ net = X τ ccw - X τ cw = I α T 1 r - T 2 r = ( mr 2 ) a r · = mr a. ma = 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 . 04 kg - 1 . 5 kg) 9 . 8 m / s 2 2 . 3 kg + 1 . 04 kg + 1 . 5 kg = 0 . 931405 m / s 2 . keywords: 002 (part 1 oF 2) 10 points A block oF mass 2 kg and one oF mass 5 kg are connected by a massless string over a pulley that is in the shape oF a disk having a radius oF 0 . 28 m, and a mass oF 5 kg. In addition,

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Patel, Kinal – Homework 15 – Due: Oct 30 2007, 7:00 pm – Inst: D Weathers 2 the blocks are allowed to move on a fxed block-wedge oF angle θ = 42 , as shown. The coe±cient oF kinetic Friction is 0 . 24 For both blocks. 2 kg 5 kg 42 0 . 28 m 5 kg Determine the acceleration oF the two blocks. The acceleration oF gravity is 9 . 8 m / s 2 . Assume: The positive direction is to the right. Correct answer: 2 . 03621 m / s 2 . Explanation: Let : m 1 = 2 kg , m 2 = 5 kg , M = 5 kg , and R = 0 . 28 m . Newton’s second low is X ~ F = m~a ²or m 1 N 1 - m 1 g = m 1 a y = 0 N 1 = m 1 g and the Force oF Friction is then f 1 = μN 1 = (0 . 24)(2 kg)(9 . 8 m / s 2 ) = 4 . 704 N . The x - component is - f 1 + T 1 = m 1 a. (1) T 1 = m 1 a - f 1 This is our frst equation. ²or the mass m 2 the component oF the New- ton’s equation perpendicular to the surFace is N 2 - m 2 g cos θ = m 2 a perp = 0 , or N 2 = m 2 g cos θ , and the Force oF Friction is f 2 = μN 2 = (0 . 24)(5 kg)(9 . 8 m / s 2 )(cos 42 ) = 8 . 73938 N The component oF the equation parallel to the surFace is - f 2 - T 2 + m 2 g sin θ = m 2 a. (2) Now we have three equations For the three unknowns a , T 1 and T 2 .
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solutions 15 - Patel Kinal Homework 15 Due 7:00 pm Inst D...

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