Explanation Given: The mass of the crate is m = 20 kg . The velocity of the cable is v = ( 0.05 s 3 / 2 ) m/s . The coefficient of kinetic friction is μ k = 0.2 . The displacement of the crate is s = 10 m . The free body diagram of the blocks as shown in Figure 1. Write the differential relation between the position ( s ) , velocity ( v ) and acceleration ( a ) . a d s = v d v a = v d v d s (I) Here, d v is the change in velocity and d s is the change is displacement. Write the formula for Newton’s second law of motion along x − a x i s . ∑ F x = m a x (II) Here, m is the mass of the block, a x is the acceleration of the block along x − a x i s and ∑ F x is the resultant of all the forces acting on the block along x − a x i s . Write the formula for Newton’s second law of motion along y − a x i s . ∑ F y = m a y (III) Here, m is the mass of the block, a y is the acceleration of the block along y − a x i s and ∑ F y is the resultant of all the forces acting on the block along y − a x i s . Write the formula for friction force F f . F f = μ k N (IV) Here, μ k is the coefficient of kinetic friction and N is the normal reaction. Conclusion: Calculate the acceleration of the crate. Substitute 0.05 s 3 / 2 for v in Equation (I). a = ( 0.05 s 3 / 2 ) d d s ( 0.05 s 3 / 2 ) = 0.05 s 3 / 2 ( 0.05 3 2 s 1 / 2 ) When, s = 10 m . Substitute 10 m for s . a = 0

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Engineering Mechanics: Statics & Dynamics (14th Edition)

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The tension developed in the cable when s = 10 m .

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