Find the friction on B due to A when:-

(i) The wall is smooth but the surfaces of A and B in contact are rough and the system is in equilibrium

(ii) All the surfaces are rough

(p) upward

(q) downwards

(r) zero

(s) system cannot remain in equilibrium

(i) - p; (ii) - q

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(i) - q; (ii) - r

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(i) - s; (ii) - p

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(i) - r; (ii) - q

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Solution

The correct option is C
(i) - s; (ii) - p

If at all there is friction by A then its either going to be in upward or downward direction. Let's see both cases

Case I friction on B in upward direction

Since A has net force in vertically downward direction so A will move downward

Case II friction on B in upward direction

Since B has net force in vertically downward direction so B will move downward

Either case one of the blocks will move while the question is asking about friction when they are in equilibrium which is never going to happen. So (i) = d

Part - (ii)

Let's draw free body diagram of A assuming no friction between A & B.

$a_{B} = g d o w n w a r d$ $a_{B} > a_{A}$

$a_{B}$ relative to A is going to be in downward direction. So friction will oppose. So friction on B by A in upward direction (ii) - a

Alternate solution

Prove by contradiction. Assume the reverse is happening A is going down with respect to B

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Figure

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Find the friction on B due to A when:- (i) The wall is smooth but the surfaces of A and B in contact are rough and the system is in equilibrium (ii) All the surfaces are rough (p) upward (q) downwards (r) zero (s) system cannot remain in equilibrium

Find the friction on B due to A when:-

(i) The wall is smooth but the surfaces of A and B in contact are rough and the system is in equilibrium

(ii) All the surfaces are rough

(p) upward

(q) downwards

(r) zero

(s) system cannot remain in equilibrium