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=10⇒N1=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+10−10=0 ⇒ frictional force exerted by ground on block B is zero

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