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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.

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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