(a) Explanation Given Information: The weight of the block A ( W A ) is 10 lb. The weight of the plate B ( W B ) is 40 lb. The spring constant ( k ) is 60 lb/ft . The plate B is slowly moved with amplitude ( x m ) is 2.4 in . . The acceleration due to gravity ( g ) value is 32.2 ft/s 2 . Calculation: Calculate the total weight of the block A and plate B using the relation: W t o t a l = W A + W B Substitute 10 lb for W A and 40 lb for W B . W t o t a l = 10 + 40 = 50 lb Calculate the mass ( m ) of the block using the relation: m = W t o t a l g Substitute 50 lb for W t o t a l and 32.2 ft/s 2 for g . m = 50 32.2 = 1.5528 lb ⋅ s 2 / ft Calculate the natural circular frequency ( ω n ) using the relation: ω n = k m Substitute 60 lb/ft for k and 1 (b) To determine The corresponding smallest allowable value of the coefficient ( μ s ) of static friction.

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ISBN: 9781260663778

Author: BEER

Publisher: MCG

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Chapter 19.1, Problem 19.9P

(a)

To determine

The amplitude (xm) and frequency (fn) of the resulting motion.

(b)

To determine

The corresponding smallest allowable value of the coefficient (μs) of static friction.

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Chapter 19.1, Problem 19.8P

Chapter 19.1, Problem 19.10P

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VECTOR MECH. FOR EGR: STATS & DYNAM (LL

Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - Prob. 19.3P Ch. 19.1 - Prob. 19.4P Ch. 19.1 - Prob. 19.5P Ch. 19.1 - Prob. 19.6P Ch. 19.1 - Prob. 19.7P Ch. 19.1 - A simple pendulum consisting of a bob attached to...Ch. 19.1 - Prob. 19.9P Ch. 19.1 - Prob. 19.10P

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The amplitude ( x m ) and frequency ( f n ) of the resulting motion.

Solution Summary: The author calculates the mass of the block A and plate B using the relation: m=W_totalg.

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The amplitude (xm) and frequency (fn) of the resulting motion.

The corresponding smallest allowable value of the coefficient (μs) of static friction.

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Chapter 19 Solutions