(a) Explanation Given information: The weight ( W A ) of block A is 200 lb. The weight ( W B ) of block B is 350 lb. Calculation: Write the general equation of mass ( m ) of the blocks. m = W g Here, W is the mass of the block and g is the acceleration due to gravity. The acceleration ( a B ) of block B is the half of the acceleration ( a A ) of block A . Sketch the free body diagram and kinetic diagram of the block A shown in Figure (1). Here, N is the normal force on the block A , T is the tension in the rope, and m A is the mass of the block A . Refer Figure (1), Apply Newton’s law of equation along x -axis (Consider positive along x -axis). Σ F x = m A a A T − 200 sin 30 ° = m A a A Substitute W A g for m A . T − 200 sin 30 ° = W A g a A Substitute 200 lb for W A and 32.2 ft/s 2 for g . T − 200 sin 30 ° = 200 32.2 a A T − 100 = 6.2112 a A T − 6.2112 a A = 100 (1) Sketch the free body diagram and kinetic diagram of the block B shown in Figure 1. Refer Figure (2), Apply Newton’s law of equation along y -axis (Consider positive along opposite to y -axis). Σ F y = m B a B 350 − 2 T = m B a B Substitute W B g for m A and 1 2 a A for a B (b) To determine The tension in the cable.

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ISBN: 9781259633133

Author: BEER

Publisher: MCG

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Chapter 12.1, Problem 12.13P

(a)

To determine

The acceleration of each block.

(b)

To determine

The tension in the cable.

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Chapter 12.1, Problem 12.12P

Chapter 12.1, Problem 12.14P

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