hw10 - Muraj Hamza Homework 10 Due 4:00 am Inst Florin This...

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Muraj, Hamza – Homework 10 – Due: Feb 13 2006, 4:00 am – Inst: Florin 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 0 points The system is in equilibrium and the pulleys are frictionless and massless. The acceleration of gravity is 9 . 8 m / s 2 . 8 . 03032 kg 6 . 76542 kg 6 . 00038 kg 1 2 3 T Find the tension T . Correct answer: 388 . 587 N. Explanation: Let : m 1 = 8 . 03032 kg , m 2 = 6 . 76542 kg , and m 3 = 6 . 00038 kg . m 1 m 2 m 3 1 2 3 T T 4 T 3 T 3 T 3 T 2 T 1 T 1 The mass m 1 defines the tension T 1 : T 1 = m 1 g . At pulley 3, T 1 acts down on either side of the pulley and T 2 acts up, so T 2 = 2 T 1 = 2 m 1 g . At the mass m 2 , T 3 acts up, and m 2 g and T 2 act down, so T 3 = m 2 g + T 2 = m 2 g + 2 m 1 g . At pulley 2, T 3 acts up on either side of the pulley and T 4 acts down, so T 4 = 2 T 3 = 2 m 2 g + 4 m 1 g . At the mass m 3 , T 4 = T + m 3 g T = T 4 - m 3 g = (2 m 2 + 4 m 1 - m 3 ) g = [2 (6 . 76542 kg) + 4 (8 . 03032 kg) - (6 . 00038 kg)] (9 . 8 m / s = 388 . 587 N . 002 (part 1 of 2) 10 points The pulley is massless and frictionless. A massless inextensible string is attached to these masses: 1 kg, 1 kg, and 9 kg. The acceleration of gravity is 9 . 8 m / s 2 . 4 . 1 m 20 . 3 cm ω 1 kg 1 kg 9 kg T 2 T 1 T 3 What is the tension T 1 in the string be- tween the block with mass 1 kg and the block with mass 1 kg (on the left-hand side of the pulley)? Correct answer: 16 . 0364 N. Explanation:
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Muraj, Hamza – Homework 10 – Due: Feb 13 2006, 4:00 am – Inst: Florin 2 Let : R = 20 . 3 cm , m 1 = 1 kg , m 2 = 1 kg , m 3 = 9 kg , h = 4 . 1 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 kg 1 kg 9 kg T 1 T 2 T 3 m 1 g T 1 m 2 g m 3 g a a Basic Concept : For each mass in the system ~ F net = m~a . Solution : Since the string changes direc- tion around the pulley, the forces due to the tensions T 2 and T 3 are in the same direction (up). The acceleration of the system will be down to the right ( m 3 > m 1 + m 2 ), and each mass in the system accelerates at the same rate (the string does not stretch). Let this ac- celeration rate be a and the tension over the pulley be T T 2 = T 3 . In free-body diagram for the lower left-hand mass m 1 the acceleration is up and T 1 - m 1 g = m 1 a . (1) In free-body diagram for the upper left-hand mass m 2 the acceleration is up and T - T 1 - m 2 g = m 2 a . (2) In free-body diagram for the right-hand mass m 3 the acceleration is down and - T + m 3 g = m 3 a . (3) Adding Eqs. (1), (2), and (3), we have ( m 3 - m 1 - m 2 ) g = ( m 1 + m 2 + m 3 ) a . (4)
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