Question

Block $\text{A}$ is kept on top of block $\text{B}$. Both of them have a mass of $1\text{kg}$. The co-efficient of friction between the blocks are ${\mu }_s=0.75$ and ${\mu }_k=0.60$. The table over which block $\text{B}$ is kept is frictionless. A force $\frac{P}{2}$ is applied in block $\text{A}$ to the left and force $\text{P}$ on block $\text{B}$ to the right. The minimum value of $\text{P}\left(\text{in N}\right)$ such that sliding occures between the two block is

Answer 1

Given,
$\text{Mass of block}\left(m_A\right)=\text{Mass of block}\left(m_B\right)=1\text{kg}$

Considering $f$ as static frictional force and $a$ acceleration, free body diagram of the blocks:

From FBD,

$P-f=m_{B}a$
$\Rightarrow P-f=1\times a=a...\left(1\right)$

And $f-\frac{P}{2}=m_{A}a$
$\Rightarrow f-\frac{P}{2}=1\times a=a...\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$,

$\frac{P}{2}=2a$ and $f=\frac{3P}{4}$

As $f$ is static frictional force, so

$f=\frac{3P}{4}\ge {\mu }_{s}mg$

$\Rightarrow P\ge \frac{4}{3}\times 0.75\times 1\times 10$

$\Rightarrow P\ge 10\text{N}$

Therefore, the minimum value of $P$ is $10\text{N}$ such that sliding occurs between the two blocks.

Accepted answer : $10$