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Unformatted text preview: banuelos (ojb93) – Homework 07 – florin – (58140) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A person weighing 0 . 6 kN rides in an elevator that has a downward acceleration of 2 . 3 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . What is the magnitude of the force of the elevator floor on the person? Correct answer: 0 . 459184 kN. Explanation: Basic Concepts: summationdisplay vector F = mvectora vector W = mvectorg Solution: Since W = mg , F net = ma = W − f f = W − ma = W parenleftbigg 1 − a g parenrightbigg = (0 . 6 kN) parenleftbigg 1 − 2 . 3 m / s 2 9 . 8 m / s 2 parenrightbigg = 0 . 459184 kN . 002 (part 1 of 2) 10.0 points A pulley is massless and frictionless. 2 kg, 2 kg, and 6 kg masses are suspended as in the figure. 2 . 1 m 24 . 3 cm ω 2 kg 2 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 lefthand side of the pulley? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 23 . 52 N. Explanation: Let : R = 24 . 3 cm , m 1 = 2 kg , m 2 = 2 kg , m 3 = 6 kg , and h = 2 . 1 m . Consider the free body diagrams 2 kg 2 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 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) banuelos (ojb93) – Homework 07 – florin – (58140) 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) a = m 3 − m 1 − m 2 m 1 + m 2 + m 3 g (5) = 6 kg − 2 kg − 2 kg 2 kg + 2 kg + 6 kg (9 . 8 m / s 2 ) = 1 . 96 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 ( a + g ) (6) = (2 kg) (1 . 96 m / s 2 + 9 . 8 m / s 2 ) = 23 . 52 N . 003 (part 2 of 2) 10.0 points What is the magnitude of the acceleration of the lower lefthand block? Correct answer: 1 . 96 m / s 2 . Explanation: The acceleration is the same for every mass, since the string is inextensible. keywords: 004 10.0 points A 7 kg object hangs at one end of a rope that is attached to a support on a railroad boxcar....
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This note was uploaded on 04/11/2010 for the course PHYSICS 58100 taught by Professor Florin during the Spring '10 term at University of Texas.
 Spring '10
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