Explanation Given: The mass of the crate m is 200 lb . The time t is 2.5 s . The graph for force and time is shown in Figure (1). From Figure (1) write the expression for the force F = 250 lb 2.5 t = ( 100 t ) lb Write the expression for the mass m = 200 lb = 200 32.2 = 6.2 Write the expression for the instantaneous acceleration. a = d υ d t d υ = a d t (I). Here, acceleration is a and rate of change of velocity with respect to time ( t ) is d υ d t . Draw the free body diagram of the crate without equation of motion as shown in Figure (2). Write the formula to calculate vertical force components for Figure (2). ∑ F y = m a y = m ( 0 ) Here, summation of total force acting on vertical direction is ∑ F y . Refer Figure (2). Resolve the forces along y − axis . ∑ F y = F − 200 (II). Draw the free body diagram of the crate with equation of motion as shown in Figure (3). Write the formula to calculate vertical force components for Figure (3). ∑ F y = m a y (III). Refer Figure (3). Apply equation of motion in y − axis and resolve the forces along y − axis . ∑ F y = F − 200 (IV). Conclusion: Substitute 200 32.2 for m in Equation (III). ∑ F x = m a = 200 32.2 a Substitute 0 for ∑ F y , and 100 t for F Equation (II)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 13.4, Problem 29P

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Chapter 13 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

Ch. 13.4 - The 10-kg block is subjected to the forces shown....Ch. 13.4 - The 10-kg block is subjected to the forces shown....Ch. 13.4 - Determine the initial acceleration of the 10-kg...Ch. 13.4 - Prob. 4PP Ch. 13.4 - Prob. 1FP Ch. 13.4 - If motor M exerts a force of F = (10t2 + 100) N on...Ch. 13.4 - A spring of stiffness k = 500 N/m is mounted...Ch. 13.4 - Prob. 5FP Ch. 13.4 - Block B rests upon a smooth surface. If the...Ch. 13.4 - The 6-lb particle is subjected to the action of...

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The velocity of the crate when time t is 2.5 s .

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