Explanation Given information: The length ( L ) of the beam is 15 ft. The weight ( W ) of the beam is 500 lb. The deceleration ( a A ) of the cable A is 20 ft/s 2 . The deceleration ( a B ) of the cable B is 2 ft/s 2 . Calculation: Consider the acceleration due to gravity ( g ) as 32.2 ft / s 2 . Show the free body diagram as in Figure (1). Calculate the mass of the beam ( m ) : m = W g Substitute 500 lb for W , and 32.2 ft / s 2 for g . m = 500 32.2 lb ⋅ s 2 / ft Calculate the mass moment of inertia of the beam ( I ¯ ) using the relation: I ¯ = 1 12 m L 2 Substitute 500 32.2 lb ⋅ s 2 / ft for m and 15 ft for L . I ¯ = 1 12 × 500 32.2 × ( 15 ) 2 = 291.15 s lb ⋅ s 2 ⋅ ft Calculate the angular acceleration ( α ) using the relation. a B = a A + L α Substitute − 20 ft/s 2 for a B , − 2 ft/s 2 for a B , and 15 ft for L . − 2 = − 20 + ( 15 ) α α = ( − 2 + 20 ) 15 = 1.2 rad/s 2 Calculate the acceleration ( a ¯ ) using the relation: a ¯ = 1 2 ( a A + a B ) Substitute − 20 ft/s 2 for a B and − 2 ft/s 2 for a B . a ¯ = 1 2 ( ( − 20 ) + ( − 2 ) ) = − 11 ft/s ≈ 11 ft/s ( ↑ ) Show the free body diagram and kinetic diagram of the beam as in Figure (2). 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 beam, and α is the angular acceleration of the crate

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 16.1, Problem 16.60P

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The tension in each cable A (TA) and B (TB).

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The tension in each cable A ( T A ) and B ( T B ) .

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The tension in each cable A (TA) and B (TB).

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