hw7 - khounvivongsy (sk27799) Homework 7 Weathers (17104) 1...

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Unformatted text preview: khounvivongsy (sk27799) Homework 7 Weathers (17104) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points In the figure below the left-hand cable has a tension T 1 and makes an angle of 41 with the horizontal. The right-hand cable has a tension T 3 and makes an angle of 46 with the horizontal. A W 1 weight is on the left and a W 2 weight is on the right. The cable connecting the two weights has a tension 49 N and is horizontal. The acceleration of gravity is 9 . 8 m / s 2 . M 1 M 2 49 N T 1 T 3 4 6 4 1 Determine the mass M 2 . Correct answer: 5 . 17765 kg. Explanation: Given : W 1 = M 1 g , W 2 = M 2 g , 1 = 41 , 3 = 46 , and T 2 = 49 N . T 3 T 1 3 1 T 3 cos 3 T 1 cos 1 W 2 W 1 Note: T 1 cos 1 = T 2 = T 3 cos 3 Basic Concepts: summationdisplay vector F = mvectora = 0 vector W = mvectorg Solution: Consider the point of attachment of cable 2 and cable 3. Vertically, W 2 = M 2 g acts down and T 3 sin 3 acts up, so F net = W 2 T 3 sin 3 = 0 = T 3 sin 3 = W 2 . (1) Horizontally, T 2 acts to the left and T 3 cos 3 acts to the right, so F net = T 2 T 3 cos 3 = 0 = T 3 cos 3 = T 2 . (2) Dividing Eq. 1 by Eq. 2, we have tan 3 = W 2 T 2 . W 2 = T 2 tan 3 = (49 N) tan46 = 70 . 5383 N M 2 = W 2 g = 50 . 741 N 9 . 8 m / s 2 = 5 . 17765 kg and by symmetry , we have T 1 cos 1 T 3 cos 3 = 0 , so W 1 = T 2 tan 1 = (49 N) tan41 = 64 . 9256 N M 1 = W 1 g = 42 . 5951 N 9 . 8 m / s 2 = 4 . 34643 kg . 002 (part 1 of 2) 10.0 points A pulley is massless and frictionless. 2 kg, 3 kg, and 6 kg masses are suspended as in the figure. 1 . 9 m 24 . 1 cm 3 kg 2 kg 6 kg T 2 T 1 T 3 khounvivongsy (sk27799) Homework 7 Weathers (17104) 2 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 . Correct answer: 21 . 3818 N. Explanation: Let : R = 24 . 1 cm , m 1 = 2 kg , m 2 = 3 kg , m 3 = 6 kg , and h = 1 . 9 m . Consider the free body diagrams 2 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...
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This note was uploaded on 02/08/2010 for the course PHYS 1710 taught by Professor Weathers during the Spring '09 term at University of North Carolina Wilmington.

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hw7 - khounvivongsy (sk27799) Homework 7 Weathers (17104) 1...

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