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Question

All the blocks shown in the figure are at rest. The pulley is smooth and the strings are light. Coefficient friction at all the contacts is 0.2. A frictional force of 10 N acts between A and B. The block A is just about to slide on block B. Then

A
The normal reaction exerted by the ground on the block B is 110 N
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B
The normal reaction exerted by the ground on the block B is 50 N
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C
The frictional force exerted by the ground on the block B is 20 N
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D
The frictional force exerted by the ground on the block B is zero.
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Solution

The correct options are
A The normal reaction exerted by the ground on the block B is 110 N
D The frictional force exerted by the ground on the block B is zero.

The frictional force on block A is
μN1=10N1=100.2=50 N
The net force on block B in vertrical direction is zero
N2=50+N1+10=110 N
Normal reaction exerted by ground on block B is 110 N.
The net force on block B in horizontal direction is zero
f+1010=0
frictional force exerted by ground on block B is zero

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