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 $10\text{N}$ 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 $110\text{N}$
• B. The normal reaction exerted by the ground on the block B is $50\text{N}$
• C. The frictional force exerted by the ground on the block B is $20\text{N}$
• 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
$\Rightarrow \mu N_1=10\Rightarrow N_1=\frac{10}{0.2}=50\text{N}$
The net force on block B in vertrical direction is zero
$\therefore N_2=50+N_1+10=110\text{N}$
$\Rightarrow$ Normal reaction exerted by ground on block B is $110\text{N}$.
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