Explanation Given: The weight W of cylinder plug is 2 lb . The distance d is 0.5 ft . The stiffness k is 14 lb/ft . Constant speed υ is 15 ft/s . The stiffness and distance of the system is shown in Figure (1). Write the expression for acceleration. a n = υ 2 r (I). Here, acceleration is a , the velocity is υ and the radius is r . Write the expression for mass ( m ) m = W g (II). Write the expression for force of the spring. F s = k x (III). Here, stiffness is k and the distance is x . Refer Figure (1) and write the equation of radius. r = ( 3 − x ) (IV). Refer Figure (1) and write the equation of distance. x = ( d − x ) (V). Draw the free body diagram as shown in Figure (2). Refer Figure (2) write the formula for equation of motion for block A in the horizontal direction. ∑ F n = m a n (VI). Here, summation of total force acting on horizontal direction is ∑ F n mass is m and acceleration in the horizontal direction is a n . Refer Figure (2) resolve the forces in the horizontal direction. ∑ F n = F s (VII). Conclusion: Substitute ( 3 − x ) for r and 15 ft/s for υ in Equation (I). a n = ( 15 ) 2 ( 3 − x ) Substitute 14 lb/ft for k and ( d − x ) for x in Equation (III). F s = k x = 14 ( d − x ) (VIII). Substitute 2 lb for W A and 32.2 for g in Equation (II). m = W g m = 2 32.2 Substitute 2 32

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 13.5, Problem 77P

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The force Fs of the spring on the plug when the plug is at rest with respect to the pipe.

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Chapter 13.5, Problem 76P

Chapter 13.5, Problem 78P

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Chapter 13 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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The force F s of the spring on the plug when the plug is at rest with respect to the pipe.

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The force Fs of the spring on the plug when the plug is at rest with respect to the pipe.

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