Question

Consider the situation shown in figure. The wall is smooth but the surface of A and B in contact are rough. The friction on B due to A in equilibrium-
(a) is upward
(b) is downward
(c) is zero
(d) the system can't remain in equilibrium

A. Only a
B. Only b
C. Only d
D. c and d both

Answer 1

Hint: we are given two blocks attached together which is attached to a wall. By drawing the free body diagram of the given situation we can easily find all the forces acting on the bodies and their direction. From the figure we can find the friction acting on ‘B’ due to ‘A’. thus we will get the solution.

Complete answer:
In the question we are given two blocks A and B who are attached together.
To solve the question, let us draw the free body diagram of the given case.

From the figure we can see the different forces acting on both the blocks.
The weight of the two bodies A and B, ‘\[{{m}_{A}}g\]’ and ‘${{m}_{B}}g$’ respectively will be vertically downwards.
The normal force acting on ‘B’ by ‘A’ and on ‘A’ by ‘B’ will be equal and opposite, and is represented as N.
The force acting on the block ‘A’ due the wall will also be ‘N’.
The frictional force exerted by ‘A’ on ‘B’, ${{f}_{AB}}$ will be the same as the frictional force exerted by ‘B’ on ‘A’, ‘${{f}_{BA}}$’.
We are asked about the frictional force on B due to A.
From the figure we can see that,
${{f}_{AB}}=\mu N+mg$, were ‘$\mu $’ is the coefficient of friction between the two surfaces.
Since the frictional force is not equal to zero, we know that the system will not remain in equilibrium.

So, the correct answer is “Option C”.

Note:
Here it is given that the wall is smooth and the surfaces between the blocks are rough. Since the surface of the wall is smooth, we know that there is no friction acting. Hence due to the gravitational force both the blocks will fall downward with acceleration due to gravity.