hw 11 sol

# hw 11 sol - Patel(pnp223 – homework 11 – Turner...

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Unformatted text preview: Patel (pnp223) – homework 11 – Turner – (92160) 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 (part 1 of 2) 10.0 points A pulley is massless and frictionless. 3 kg, 3 kg, and 6 kg masses are suspended as in the figure. 4 . 2 m 22 . 5 m ω 3 kg 3 kg 6 kg T 2 T 1 T 3 What is the tension T 1 in the string be- tween the two blocks on the left-hand side of the pulley? The acceleration of gravity is 9 . 8 m / s 2 . 1. T 1 = parenleftbigg 19 4 kg parenrightbigg (9 . 8 m / s 2 ) 2. T 1 = (4 kg) (9 . 8 m / s 2 ) 3. T 1 = parenleftbigg 21 4 kg parenrightbigg (9 . 8 m / s 2 ) 4. T 1 = parenleftbigg 17 4 kg parenrightbigg (9 . 8 m / s 2 ) 5. T 1 = parenleftbigg 9 2 kg parenrightbigg (9 . 8 m / s 2 ) 6. T 1 = parenleftbigg 7 2 kg parenrightbigg (9 . 8 m / s 2 ) 7. T 1 = parenleftbigg 13 4 kg parenrightbigg (9 . 8 m / s 2 ) 8. T 1 = parenleftbigg 15 4 kg parenrightbigg (9 . 8 m / s 2 ) 9. T 1 = (5 kg) (9 . 8 m / s 2 ) 10. T 1 = (3 kg) (9 . 8 m / s 2 ) correct Explanation: Let : R = 22 . 5 m , m 1 = 3 kg , m 2 = 3 kg , m 3 = 6 kg , and h = 4 . 2 m . Consider the free body diagrams 3 kg 3 kg 6 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 left-hand mass m 1 the accel- eration is up and T 1- m 1 g = m 1 a. (1) For the upper left-hand mass m 2 the acceler- ation is up and T- T 1- m 2 g = m 2 a. (2) 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) Patel (pnp223) – homework 11 – Turner – (92160) 2 a = m 3- m 1- m 2 m 1 + m 2 + m 3 g (5) = 6 kg- 3 kg- 3 kg 3 kg + 3 kg + 6 kg g = 0 kg 12 kg (9 . 8 m / s 2 ) = 0 (9 . 8 m / s 2 ) ....
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hw 11 sol - Patel(pnp223 – homework 11 – Turner...

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