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(tcs639) schneiter HW07 Ross (89152) This print-out should have 9 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 Consider the 693 N weight held by two cables shown below. The left-hand cable had tension T2 and makes an angle of 2 with the ceiling. The right-hand cable had tension 540 N and makes an angle of 42 with the ceiling. The right-hand cable makes an angle of 42 with the ceiling and has a tension of 540 N . 1 Basic Concept: Vertically and Horizontally, we have x x x Fnet = F1 - F2 = 0 = F1 cos 1 - F2 cos 2 = 0 (1) y y y Fnet = F1 + F2 - Wg = 0 = F1 sin 1 + F2 sin 2 - Wg = 0 (2) Solution: Using Eqs. 1 and 2, we have x F2 = F1 cos 1 = (540 N) cos 42 = 401.298 N , and (1) 2 T2 42 54 0 N y F2 = F3 - F1 sin 1 = 693 N - (540 N) sin 42 = 693 N - 361.331 N = 331.669 N , so (2) 693 N a) What is the tension T2 in the left-hand cable slanted at an angle of 2 with respect to the wall? Correct answer: 520.62 N. Explanation: Observe the free-body diagram below. F2 = = y x (F2 )2 + (F2 )2 (401.298 N)2 + (331.669 N)2 = 520.62 N . 002 (part 2 of 2) 10.0 points b) What is the angle 2 which the left-hand cable makes with respect to the ceiling? Correct answer: 39.5734 . Explanation: Using Eq. 2, we have y F1 2 = arctan x F1 361.331 N = arctan 401.298 N = 39.5734 . 2 F 2 F1 1 Wg Note: The sum of the x- and y-components of F1 , F2 , and Wg are equal to zero. Given : Wg F1 1 2 = 693 N , = 540 N , = 42 , and = 90 - . 003 (part 1 of 2) 10.0 points A block of mass 7.13 kg lies on a frictionless horizontal surface. The block is connected by a cord passing over a pulley to another block of mass 6.26 kg which hangs in the air, as shown on the following picture. Assume the cord to be light (massless and weightless) schneiter (tcs639) HW07 Ross (89152) and unstretchable and the pulley to have no friction and no rotational inertia. The acceleration of gravity is 9.8 m/s2 . 7.13 kg and hence a= m2 g m1 + m2 6.26 kg = (9.8 m/s2 ) 7.13 kg + 6.26 kg = 4.58163 m/s2 . 2 6.26 kg Calculate the acceleration of the first block. Correct answer: 4.58163 m/s2 . Explanation: Given : m1 = 7.13 kg , m2 = 6.26 kg . a m1 N m1 g T and 004 (part 2 of 2) 10.0 points Calculate the tension in the cord. Correct answer: 32.667 N. Explanation: T = m1 a = (7.13 kg) (4.58163 m/s2 ) = 32.667 N . 005 10.0 points A block is pushed up a frictionless incline by an applied horizontal force as shown. The acceleration of gravity is 9.8 m/s2 . F = 33 N T m2 m2 g a 3. 9 kg 22 Since the cord is unstretchable, the first block accelerates to the right at exactly the same rate a as the second (hanging) block accelerates downward. Also, the cord's tension pulls the first block to the right with exactly the same tension T as it pulls the second block upward. The only horizontal force acting on the first block the is cord's tension T , hence by Newton's Second Law net m1 a = F1 =T. What is the magnitude of the resulting acceleration of the block? Correct answer: 4.17426 m/s2 . Explanation: Basic Concepts: F = ma Solution: Consider the forces acting on the block and their components along the incline. The horizontal pushing force F has up-theincline component +F cos while the weight force W = mg has up-the-incline component -mg sin . Consequently, M a = Fnet = F cos - M g sin , F cos a= - g sin M (33 N) cos 22 = 3.9 kg - (9.8 m/s2 ) sin 22 = 4.17426 m/s2 . The second block feels two vertical forces: The cord's tension T (upward) and the block's own weight W2 = m2 g (downward). Consequently, net m2 a = F2 = m2 g - T . Adding the two equations together, we arrive at (m1 + m2 ) a = m2 g , schneiter (tcs639) HW07 Ross (89152) so 006 10.0 points The horizontal surface on which the objects slide is frictionless. T2 T1 F 4 kg 8 kg 12 kg T1 m1 + m2 20 N = 12 kg = 1.66667 m/s2 . 3 a= For the rightmost body, F acts to the right and T1 to the left, so If the tension in string T1 = 20 N, determine F . Correct answer: 40 N. Explanation: m3 a = F - T1 F = T1 + m3 a T1 (m1 + m2 ) m1 + m2 + m3 = T2 m1 + m2 4 kg + 8 kg + 12 kg = (20 N) 4 kg + 8 kg 24 kg = (20 N) 12 kg = 40 N . = T1 + m3 007 (part 1 of 3) 10.0 points Three masses are connected by light strings as shown in the figure. Let : m1 m2 m3 m1 + m2 + m3 T2 = 4 kg , = 8 kg , = 12 kg , = 24 kg . T1 and m1 m2 m3 F Basic Concept: Newton's second law. Fnet = m a for each body. T2 m1 T2 m2 T1 T1 m3 F m2 m1 m3 The string connecting the m1 and the m2 passes over a light frictionless pulley. Given m1 = 6.05 kg, m2 = 7.2 kg, m3 = 9.33 kg, and g = 9.8 m/s2 . The acceleration of gravity is 9.8 m/s2 . Find the downward acceleration of m2 mass. Correct answer: 4.54845 m/s2 . Explanation: Consider the free body diagrams: Solution: For the leftmost body, T2 acts to the right, so m1 a = T2 . For the middle body, T1 acts to the right and T2 to the left, so m2 a = T1 - T2 . Adding these equations, (m1 + m2 ) a = T1 , schneiter (tcs639) HW07 Ross (89152) Explanation: From equation (3), T2 = m3 (g - a) T1 a T2 a m3 m2 m1 T1 a = (9.33 kg) 9.8 m/s2 - 4.54845 m/s2 = 48.997 N . 4 Applying Newton's second law to each of these masses we get m1 a = T1 - m1 g m2 a = T2 + m2 g - T1 m3 a = m3 g - T2 Adding these equations yields (m1 + m2 + m3 ) a = (-m1 + m2 + m3 ) g , so a= -m1 + m2 + m3 g m1 + m2 + m3 -6.05 kg + 7.2 kg + 9.33 kg = 6.05 kg + 7.2 kg + 9.33 kg (9.8 m/s2 ) = 4.54845 m/s2 . 008 (part 2 of 3) 10.0 points Find the tension in the string connecting the m1 and the m2 masses. Correct answer: 86.8081 N. Explanation: From equation (1), T1 = m1 (a + g) = (6.05 kg) 4.54845 m/s2 + 9.8 m/s2 = 86.8081 N . 009 (part 3 of 3) 10.0 points Find the tension in the string connecting the m2 and the m3 masses. Correct answer: 48.997 N. (1) (2) (3) ... View Full Document

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