Explanation Given: The mass of the block A is 60 lb . The mass of the block B is 40 lb . The displacement of the block B is 2 ft . The friction between the blocks and incline is μ k = 0.1 . Draw the free body diagram of blocks as shown in Figure (1). Refer Figure (1). The forces N A , N B , R do not create any displacement of the blocks. Hence the work done by these forces is zero. From Figure (1). The total length of the cord is equal to the addition of two times of length of the cord of block A and the length of the cord of block B . 2 s A + s B = l Calculate the change in lengths of the cord. 2 s A − s B = 0 2 s A = s B (I) Similarly, calculate the velocities of the cords. 2 v A − v B = 0 2 v B = v B (II) Refer Figure (1). Consider free body diagram of blocks A and B only. Resolve the forces along y − axis to calculate the force N A . ∑ F y = m a y Here, the acceleration along y − axis is zero. a y = 0 Substitute 0 for a y . ∑ F y = m ( 0 ) ∑ F y = 0 (III) Write the formula for friction force F . F = μ k N (IV) Here, μ k is the friction coefficient and N is the normal reaction force. Write the formula for work done by the friction ( F f ) . U F = F s (V) Here, F f is the friction force and s is the displacement due to friction force ( F f ) . Write the formula for work done by their weights. U W = W s (VI) Here, W is the weight and s is the displacement due to force ( W ) . Write the formula for kinetic energy ( T ) T = 1 2 m v 2 (VII) Here, m is the mass of the object and v is the velocity of the object. Write the formula for Principle of Work and Energy. [ ( T 1 ) A + ( T 1 ) B ] + ∑ U = ( T 2 ) A + ( T 2 ) B (VIII) Here, ( T 1 ) A , ( T 1 ) B is initial kinetic energy, ( T 2 ) A , ( T 2 ) B is final kinetic energy of the block A and B respectively and ∑ U is total work done by all the forces on the block. Conclusion: Calculate the change in length of the cord which is connected to block A . Substitute 2 ft for s B in Equation (I). 2 s A = 2 ft s A = 2 ft 2 s A = 1 ft Refer Figure (1), consider free body diagram of block A only. Calculate the normal reaction force ( N A ) . Resolve the forces along y − axis and substitute in Equation (III). N A − 60 cos 60 ° = 0 N A = 30 lb Calculate the friction force ( F A ) . Substitute 0.1 for μ k and 30 lb for N in Equation (IV). F A = 0.1 ( 30 lb ) = 3 lb Calculate the work done by the friction force ( F A ) . Substitute 3 lb for F and 1 ft for s in Equation (V). U F A = 3 lb ( 1 ft ) = 3 ft ⋅ lb Calculate the work done by the weight of the block A . The weight of the block 60 lb is acting at an angle of 60 ° while resolving along the x − axis is 60 sin 60 ° . Substitute 60 sin 60 ° for W and 1 ft for s in Equation (VI). U W A = 60 sin 60 ° lb ( 1 ft ) = 51.9615 ft .lb Refer Figure (1), consider free body diagram of block B only. Calculate the normal reaction force ( N B ) . Resolve the forces along y − axis and substitute in Equation (III)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14.3, Problem 21P

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Chapter 14.3, Problem 20P

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Chapter 14 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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The velocity of the block A .

Solution Summary: The author explains how to determine the velocity of the block A and the friction between the blocks and incline.

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