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Unformatted text preview: alexander (jra2623) – homework 11 – Turner – (92510) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A pulley is massless and frictionless. 2 kg, 3 kg, and 7 kg masses are suspended as in the figure. 2 . 8 m 21 . 9 m ω 3 kg 2 kg 7 kg T 2 T 1 T 3 What is the tension T 1 in the string be tween the two blocks on the lefthand side of the pulley? The acceleration of gravity is 9 . 8 m / s 2 . 1. T 1 = parenleftbigg 19 6 kg parenrightbigg (9 . 8 m / s 2 ) 2. T 1 = parenleftbigg 8 3 kg parenrightbigg (9 . 8 m / s 2 ) 3. T 1 = parenleftbigg 11 3 kg parenrightbigg (9 . 8 m / s 2 ) 4. T 1 = parenleftbigg 10 3 kg parenrightbigg (9 . 8 m / s 2 ) 5. T 1 = parenleftbigg 7 2 kg parenrightbigg (9 . 8 m / s 2 ) 6. T 1 = parenleftbigg 5 2 kg parenrightbigg (9 . 8 m / s 2 ) 7. T 1 = parenleftbigg 13 6 kg parenrightbigg (9 . 8 m / s 2 ) 8. T 1 = parenleftbigg 7 3 kg parenrightbigg (9 . 8 m / s 2 ) correct 9. T 1 = (3 kg) (9 . 8 m / s 2 ) 10. T 1 = parenleftbigg 17 6 kg parenrightbigg (9 . 8 m / s 2 ) Explanation: Let : R = 21 . 9 m , m 1 = 2 kg , m 2 = 3 kg , m 3 = 7 kg , and h = 2 . 8 m . Consider the free body diagrams 2 kg 3 kg 7 kg T 1 T 2 T 3 m 1 g T 1 m 2 g m 3 g a a For each mass in the system vector F net = mvectora . Since the string changes direction 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 acceleration rate be a and the tension over the pulley be T ≡ T 2 = T 3 . For the lower lefthand mass m 1 the accel eration is up and T 1 − m 1 g = m 1 a . (1) For the upper lefthand mass m 2 the acceler ation is up and T − T 1 − m 2 g = m 2 a . (2) For the righthand 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) alexander (jra2623) – homework 11 – Turner – (92510) 2 a = m 3 − m 1 − m 2 m 1 + m 2 + m 3 g (5) = 7 kg − 2 kg − 3 kg 2 kg + 3 kg + 7 kg g = 2 kg 12 kg (9 . 8 m / s 2 ) = 1 6 (9 . 8 m / s 2 ) . The tension in the string between block m 1 and block m 2 (on the lefthand side of the pulley) can be determined from Eq. 1: T 1 = m 1 parenleftbigg 1 6 + 1 parenrightbigg g (6) = (2 kg) parenleftbigg 7 6 parenrightbigg (9 . 8 m / s 2 ) = parenleftbigg 7 3 kg parenrightbigg...
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 Summer '08
 Kaplunovsky
 Force, Friction, Work, kg, Alexander

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