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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):
6gT=6a
T=6(ga) ... (1)
From FBD of 1 kg (block D):
Tg=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.


TTf=(6+3)a
TT0.2(6+3)g=9a
TT=1.8g+9a ...... (3)
From eqn. (1) and (2), we get
6g6a(g+a)=TT
i.e 6g6aga=1.8g+9a (from (3))
5g7a=1.8g+9a
a=2 m/s2

Now, for block A,
Tf=6×a


From eq. (1),
6(ga)f=6a
6×1012×2=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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