(a) Explanation Given Information: The weight ( W ) of the block A is 3 lb. The spring constant ( k ) is 2 lb/in . The upward velocity ( x ˙ ) of is 90 in ./s . The acceleration due to gravity value ( g ) is 32.2 m/s 2 . Calculation: Calculate the mass ( m ) using the relation: m = W g Substitute 3 lb for W and 9.81 m/s 2 for g . m = 3 32.2 = 0.09317 lb ⋅ s 2 / ft Calculate the natural circular frequency ( ω n ) using the relation: ω n = k m Substitute 2 lb/in . for k and 0.09317 lb ⋅ s 2 / ft for m . ω n = 2 lb/in . × ( 12 ft 1 in . ) 0.09317 = 16.05 rad/s Write the expression for displacement ( x ) using simple harmonic motion: x = x m sin ( ω n t + ϕ ) (1) Here, x m is amplitude, t is time and ϕ is phase angle. When time ( t ) is 0 the position ( x ) of the block is 0. Calculate the angle ( ϕ ) : Substitute 0 for t and 0 for x 0 = x m sin ( ω n ( 0 ) + ϕ ) ϕ = 0 Differentiate the equation (1), x ˙ = x m ω n cos ( ω n t + ϕ ) (2) Substitute 0 for ϕ and 0 for t in equation (2). x ˙ = x m ω n cos ( ω n ( 0 ) + ( 0 ) ) = x m ω n ( 1 ) = x m ω n (3) Substitute 0 for ϕ in equation (2). x ˙ = x m ω n cos ( ω n t + 0 ) = x m ω n cos ω n t (4) Calculate the amplitude ( x m ) : Substitute 90 in ./s for x ˙ and 16.05 rad/s for ω n in equation (3). 90 in

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259977251

Author: BEER

Publisher: MCG

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Chapter 19.1, Problem 19.12P

(a)

To determine

The position (x), velocity (v) and acceleration (a) of the block 0.90 sec after it has been struck by the hammer.

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Chapter 19.1, Problem 19.11P

Chapter 19.1, Problem 19.13P

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The position ( x ) , velocity ( v ) and acceleration ( a ) of the block 0.90 sec after it has been struck by the hammer.

Solution Summary: The author calculates the mass (m) using the relation: m=Wg Substitute 3 lb for W and 9.81

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The position (x), velocity (v) and acceleration (a) of the block 0.90 sec after it has been struck by the hammer.

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Chapter 19 Solutions