Explanation Given: The weight of the collar is 2 lb . The velocity is 15 ft/s . The unstretched length is x B = 1 ft Write the equation for weight of the boy. m = W g (I). The stretched length of the spring is shown in Figure (1). From Figure (1) calculate the unstretched length of the spring x B . x B = 1 ft Use Pythagoras theorem and calculate the stretched length or change in length of the spring x A from Figure (1). x A = ( B C ) 2 + ( A B ) 2 (II). From Figure (1) calculate the cosine angle θ . cos θ = Adjacent Hypotenuse = s 1 + s 2 Write the expression for the force of the spring. F s = k ( x A − x B ) (III). Here, Stiffness is k and stretch of the spring is x Write the expression for acceleration. a = υ d υ d s (IV). Here acceleration is a , velocity is υ and rate of change of velocity with respect to distance is d υ d s . Draw the free body diagram as shown in Figure (2). Refer Figure (2) write the equation of motion in x direction. ∑ F x = m a (V). Here, summation of total force acting on horizontal direction is ∑ F x . Refer Figure (2) resolve the forces along x − axis . ∑ F x = − F s cos θ (VI). Conclusion: Substitute 2 lb for W and 32.2 ft/s 2 for g in Equation (I). m = W g m = 2 32.2 Substitute s for B C and 1 ft for A B in Equation (II). x A = ( B C ) 2 + ( A B ) 2 x A = s 2 + 1 ft Substitute 4 lb/ft 2 for F , s 2 + 1 for x A and 1 ft for x B in Equation (III). F s = k ( x A − x B ) F s = 4 ( s 2 + 1 − 1 ) lb Substitute 2 32

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 13.4, Problem 36P

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The velocity of the collar when s=1ft .

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Engineering Mechanics: Statics & Dynamics (14th Edition)

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The velocity of the collar when s = 1 ft .

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