This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: khounvivongsy (sk27799) Homework 7 Weathers (17104) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points In the figure below the lefthand cable has a tension T 1 and makes an angle of 41 with the horizontal. The righthand 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 lefthand 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 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...
View
Full
Document
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.
 Spring '09
 weathers
 Physics

Click to edit the document details