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Unformatted text preview: kaplan (hmk378) – 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 55 ◦ with the horizontal. The righthand cable has a tension T 3 and makes an angle of 41 ◦ 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 76 N and is horizontal. The acceleration of gravity is 9 . 8 m / s 2 . M 1 M 2 76 N T 1 T 3 4 1 ◦ 5 5 ◦ Determine the mass M 2 . Correct answer: 6 . 74141 kg. Explanation: Given : W 1 = M 1 g , W 2 = M 2 g , θ 1 = 55 ◦ , θ 3 = 41 ◦ , and T 2 = 76 N . T 3 T 1 θ 3 θ 1 T 3 cos θ 3 T 1 cos θ 1 W 2 W 1 Note: T 1 cos θ 1 = T 2 = T 3 cos θ 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 θ 3 acts up, so F net = W 2 − T 3 sin θ 3 = 0 = ⇒ T 3 sin θ 3 = W 2 . (1) Horizontally, T 2 acts to the left and T 3 cos θ 3 acts to the right, so F net = T 2 − T 3 cos θ 3 = 0 = ⇒ T 3 cos θ 3 = T 2 . (2) Dividing Eq. 1 by Eq. 2, we have tan θ 3 = W 2 T 2 . W 2 = T 2 tan θ 3 = (76 N) tan41 ◦ = 100 . 701 N M 2 = W 2 g = 66 . 0658 N 9 . 8 m / s 2 = 6 . 74141 kg and by symmetry , we have T 1 cos θ 1 − T 3 cos θ 3 = 0 , so W 1 = T 2 tan θ 1 = (76 N) tan55 ◦ = 132 . 502 N M 1 = W 1 g = 108 . 539 N 9 . 8 m / s 2 = 11 . 0754 kg . 002 (part 1 of 2) 10.0 points A pulley is massless and frictionless. 2 kg, 2 kg, and 5 kg masses are suspended as in the figure. kaplan (hmk378) – Homework 7 – Weathers – (17104) 2 2 . 7 m 23 . 9 cm ω 2 kg 2 kg 5 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: 21 . 7778 N. Explanation: Let : R = 23 . 9 cm , m 1 = 2 kg , m 2 = 2 kg , m 3 = 5 kg , and h = 2 . 7 m . Consider the free body diagrams 2 kg 2 kg 5 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...
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This document was uploaded on 10/26/2011 for the course PHYS 1710 at North Texas.
 Fall '08
 Weathers
 mechanics, Work

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