Explanation Given information: The weight of the crate ( W ) is 400 lb . The deceleration of the cable A ( a A ) y is 3 ft / s 2 . The deceleration of the cable B ( a B ) y is 1 ft / s 2 . The height of the crate ( w ) is 6.6 ft . The width of the crate ( h ) is 3.6 ft . Calculation: Consider the acceleration due to gravity ( g ) as 32.2 ft / s 2 . Calculate the mass of the crate ( m ) : m = W g Substitute 400 lb for W , and 32.2 ft / s 2 for g . m = 400 32.2 lb ⋅ s 2 / ft Calculate the mass moment of inertia of the crate ( I ¯ ) : I ¯ = 1 12 m ( w 2 + h 2 ) Substitute 400 32.2 lb ⋅ s 2 / ft for m , 6.6 ft for w , and 3.6 ft for h . I ¯ = 1 12 × 400 32.2 × ( 6.6 2 + 3.6 2 ) = 58.5093 lb ⋅ s 2 ⋅ ft Show the free body diagram of the crate as in Figure 1. Here, T A is the tension of the cable A , T B is the tension of the cable B , a is the acceleration of the crate, and α is the angular acceleration of the crate. Refer to Figure 1. Calculate the vertical forces by applying the equation of equilibrium: ∑ F y = m a T A + T B − W = m a Substitute 400 lb for W and 400 32.2 lb ⋅ s 2 / ft for m . T A + T B − 400 = 400 a 32.2 T A + T B = 400 a 32.2 + 400 T A + T B = 400 × ( a 32.2 + 1 ) (1) Calculate the moment about point G by applying the equation of equilibrium: ∑ M G = I G α ( T A − T B ) × 3.3 = I ¯ α Substitute 58.5093 lb ⋅ s 2 ⋅ ft for I ¯ . ( T A − T B ) × 3.3 = 58.5093 × α T A − T B = 58.5093 α 3.3 T A − T B = 17.73 α (2) Show the kinematic diagram of the crate as in Figure 2. Here, ( a A ) y is the deceleration of the cable A at y direction, ( a A ) x is the deceleration of the cable A at x direction, ( a B ) y is the deceleration of the cable B at y direction, and ( a B ) x is the deceleration of the cable B at x direction

12th Edition

ISBN: 9781264095032

Author: BEER

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Find the tension in each cable A (TA) and B (TB)

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Find the tension in each cable A ( T A ) and B ( T B )

Solution Summary: The author calculates the tension in each cable A (T_A) and B

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