To determine The smallest allowable coefficient of static friction between the bags. Explanation Given information: The mass ( m A ) of the duffel bag A is 10 kg. The mass ( m B ) of the suitcase B is 20 kg. The velocity ( v ) of the conveyor is 0.5 m/s. The coefficient of friction ( μ B ) between the belt and B is 0.3. Calculation: The bag A does not slide on suitcase B . Therefore, Bag A and suitcase B have the same acceleration. Sketch the free body diagram and kinetic diagram of bag A and suitcase B as shown in Figure (1). Here, F is the friction force, N is the normal force, and a is the acceleration of suitcase B . Apply Newton’s law of equation along y -axis (Consider positive along y -axis). Σ F y = 0 − ( m A + m B ) g cos 20 ° + N = 0 N = ( m A + m B ) g cos 20 ° Substitute 10 kg for m A , 20 kg for m B and 9.81 m/s 2 for g . N = ( 10 + 20 ) ( 9.81 ) cos 20 ° = 276.55 N Find the friction force ( F ) using the equation: F = μ B N Substitute 0.30 for μ B and 276.55 N for N . F = ( 0.30 ) ( 276.55 ) = 82.965 N Find the acceleration ( a ) of suitcase B . Apply Newton’s law of equation along x -axis (Consider positive along opposite to x -axis). Σ F x = m a F − ( m A + m B ) g sin 20 ° = ( m A + m B ) a a = F ( m A + m B ) − g sin 20 ° Substitute 82.965 N for F , 10 kg for m A , 20 kg for m B , and 9.81 m/s 2 for g . a = 82.965 ( 10 + 20 ) − ( 9.81 ) sin 20 ° = 0.58972 m/s 2 Sketch the free body diagram and kinetic diagram of bag A alone as shown in Figure (2)

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Chapter 12.1, Problem 12.21P

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The smallest allowable coefficient of static friction between the bags.

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