Explanation Given information: The weight of the panel ( W ) is 40 lb. The weight of the counterweight A ( W A ) is 25 lb. The friction is negligible for the system. Calculation: Case ( a ): Sketch the free body diagram of the panel as shown in Figure (1). Apply the Newton’s second law of motion along horizontal axis: Σ F x = m a x T − F = W g a (1) Here, m is the mass of the panel, T is the tension in the cord, F is the frictional force on the panel, g is the acceleration due to gravity, and a is the acceleration of the panel. The acceleration of counterweight has two components. Sketch the free body diagram of the counterweight A as shown in Figure (2). Refer Figure (2). Apply the Newton’s second law of motion along vertical axis: Σ F x = m a x − F = − W A g a F = W A g a (2) Apply the Newton’s second law of motion along horizontal axis: Σ F y = m a y T − W A = − W A g a W A − T = W A g a (3) Find the acceleration of the panel immediately after the system is released from rest in case ( a ). Add Equation (1), (2), and (3). T − F + F + W A − T = W g a + W A g a + W A g a W A = ( W + 2 W A ) a g a = W A g W + 2 W A Substitute 25 lb for W A , 40 lb for W , and 32.2 lb/ft 2 for g . a = ( 25 ) ( 32.2 ) 40 + 2 × 25 = 8.94 ft/s 2 ← Thus, the acceleration of the panel immediately after the system is released from rest in case ( a ) is 8.94 ft/s 2 ← _ . Find the tension in the cord immediately after the system is released from rest in case ( a ) using Equation (3); W A − T = W A g a Substitute 25 lb for W A , 40 lb for W , 32.2 lb/ft 2 for g , and 8.94 ft/s 2 for a . 25 − T = 25 32.2 ( 8.94 ) T = 18.06 lb Thus, the tension in the cord immediately after the system is released from rest in case ( a ) is 18.06 lb _ . Case ( b ): Sketch the free body diagram of the panel as shown in Figure (3). Refer Figure (3). Apply the Newton’s second law of motion along horizontal axis: Σ F x = m a x T = W g a (4) Sketch the free body diagram of the counterweight A as shown in Figure (2). Refer Figure (4). Apply the Newton’s second law of motion along vertical axis: Σ F y = m a y T − W A = − W A g a W A − T = W A g a (5) Find the acceleration of the panel immediately after the system is released from rest in case ( b )

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 12.1, Problem 12.29P

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Find the acceleration of the panel and the tension in the cord immediately after the system is released from rest in each case.

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Chapter 12.1, Problem 12.28P

Chapter 12.1, Problem 12.30P

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Vector Mechanics for Engineers: Statics and Dynamics

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