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Question

# An arrangement of masses and pulleys is shown in the figure. Strings connecting masses A and B with the pulleys are horizontal and all pulleys and strings are light. Friction coefficient between the surface and the block B is 0.2 and between blocks A and B is 0.7. The system is released from rest. Use (g=10 m/s2). Find the magnitude of frictional force (in N) between block A and B.

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Solution

## First, we assume that the system is moving with common acceleration a and that there is no slipping between block A and B. FBD of the system: Here, if block C moves with acceleration a, then block A,B & D will also move with the same acceleration a. (from observation) T= tension in the left string T′= tension in right string f= friction force acting between A & B block f′= friction force between surface and block B. From FBD of 6 kg (block C): 6g−T=6a T=6(g−a) ... (1) From FBD of 1 kg (block D): T′−g=1×a, T′=g+a ... (2) Assume block A & B as a system, making FBD. (we can neglect the internal friction force between A and B. T−T′−f′=(6+3)a ⇒T−T′−0.2(6+3)g=9a ⇒T−T′=1.8g+9a ...... (3) From eqn. (1) and (2), we get 6g−6a−(g+a)=T−T′ i.e 6g−6a−g−a=1.8g+9a (from (3)) ⇒5g−7a=1.8g+9a ⇒a=2 m/s2 Now, for block A, T−f=6×a From eq. (1), 6(g−a)−f=6a 6×10−12×2=f ⇒f=36 N Maximum friction force that can act between block A and block B =μ(mg)=0.7×(6×10)=42 N=fmax Since, f<fmax, so both the blocks will move together without slipping. Hence our assumption was correct.

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