Question

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

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

Answer 1

Answer: (A), (D)

The frictional force exerted by the ground on the block B is zero

The frictional force on block A is
$\Rightarrow \mu N_1=10\Rightarrow N_1=\frac{10}{0.2}=50N$
The net force on block B in vertical direction is zero
$\therefore N_2=50+N_1+10=110N$
$\Rightarrow$ Normal reaction exerted by ground on block B is 110 N.
The net force on block B in horizontal direction is zero
$\therefore$ f + 10 - 10 = 0
$\Rightarrow$ frictional force exerted by ground on block B is zero