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HW 9 solution_pdf

# HW 9 solution_pdf - ghei(mg42382 HW 9 opyrchal(11130 This...

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ghei (mg42382) – HW 9 – opyrchal – (11130) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A block of mass 2 kg and one of mass 8 kg are connected by a massless string over a pulley that is in the shape of a disk having a radius of 0 . 17 m, and a mass of 9 kg. In addition, the blocks are allowed to move on a fixed block- wedge of angle 33 , as shown. The coefficient of kinetic friction is 0 . 15 for both blocks. 2 kg 8 kg 33 0 . 17 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 . 06186 m / s 2 . Explanation: Let : m 1 = 2 kg , m 2 = 8 kg , M = 9 kg , and R = 0 . 17 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 . 15) (2 kg) (9 . 8 m / s 2 ) = 2 . 94 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 . 15) (8 kg) (9 . 8 m / s 2 ) cos 33 = 9 . 86277 N . Applying Newton’s law parallel to the surface, m 2 g sin θ - f 2 - T 2 = m 2 a T 2 = - m 2 a + m 2 g sin θ - f 2 (2) Subtraction eq. (1) from, eq. (2), T 2 - T 1 = m 2 g sin θ - ( m 1 + m 2 ) a - f 1 - f 2 . Applying Newton’s law to the pulley, summationdisplay τ = I α + T 1 R - T 2 R = - M R 2 2 a R - T 1 + T 2 = 1 2 M a m 2 g sin θ - ( m 1 + m 2 ) a - f 1 - f 2 = 1 2 M a a = 2 m 2 g sin θ - 2 f 1 - 2 f 2 2 m 1 + 2 m 2 + M . Since

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ghei (mg42382) – HW 9 – opyrchal – (11130) 2 2 m 2 g sin θ - 2 ( f 1 + f 2 ) = 2(8 kg)(9 . 8 m / s 2 )(sin 33 ) - 2(2 . 94 N + 9 . 86277 N) = 59 . 7939 kg · m / s 2 and 2 m 1 + 2 m 2 + M = 2(2 kg) + 2(8 kg) + 9 kg = 29 kg , then a = 2 m 2 g sin θ - 2 f 1 - 2 f 2 2 m 1 + 2 m 2 + M = 59 . 7939 kg · m / s 2 29 kg = 2 . 06186 m / s 2 . 002 (part 2 of 2) 10.0 points Find the tension in the horizontal part of the string. Correct answer: 7 . 06372 N. Explanation: From eq. (1), T 1 = f 1 + m 1 a = 2 . 94 N + (2 kg) (2 . 06186 m / s 2 ) = 7 . 06372 N . 003 10.0 points Consider a thin 42 m rod pivoted at one end. A uniform density spherical object (whose mass is 3 kg and radius is 3 . 5 m) is attached to the free end of the rod and the moment of inertia of the rod about an end is I rod = 1 3 m L 2 and the moment of iner- tia of the sphere about its center of mass is I sphere = 2 5 m r 2 .
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HW 9 solution_pdf - ghei(mg42382 HW 9 opyrchal(11130 This...

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