Explanation Given: The weight of a crate, m = 200 kg . The coefficient of static friction, μ s = 0.5 The coefficient of kinetic friction, μ k = 0.4 The acceleration due to gravity, g = 9.81 m/s 2 Show the free body diagram of the crate as in Figure (1). Write the frictional force of the crate when the crate is on the verge of sliding. F f = μ s N (I) Here, the normal force is N . Consider the equilibrium of the crate by referring Figure (1) for vertical and horizontal axes. ∑ F y = 0 N − m g = 0 (II) ∑ F x = 0 2 ( T ) − μ s N = 0 (III) Here, the acceleration due to gravity is g . Write the frictional force of the crate when it is sliding. F f = μ k N (IV) Apply the principle of impulse and momentum equation to calculate the speed of the crate. m v 1 + 2 ∑ ∫ t 1 t 2 F d t = m v 2 m v 1 + 2 ∫ t 1 t 2 ( T − F f ) d t = m v 2 (V) Here, towing force is T , initial and final velocity of the crate is v 1 and v 2 , impulses acting on the crate from time limits t 1 to t 2 is ∑ ∫ t 1 t 2 F d t , and variable in time is dt . Conclusion: Substitute 0.5 for μ s in Equation (I). F f = 0.5 N (VI) Substitute 200 kg for m and 9.81 m/s 2 for g in Equation (II). N − ( 200 kg ) ( 9.81 m/s 2 ) = 0 N = 1962 kg ⋅ m s 2 × 1 N kg ⋅ m s 2 N = 1962 N Substitute 1962 N for N in Equation (VI). F f = 0

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ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Chapter 15, Problem 9P

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The speed of the crate when t = 4 s .

Solution Summary: The author shows the free body diagram of the crate as in Figure (1). Write the frictional force when it is on the verge of sliding.

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Chapter 15 Solutions