Question

Two blocks $\text{A}$ and $\text{B}$ of masses $m_A=1\text{kg}$ and $m_B=3\text{kg}$ are kept on the table as shown in figure. The coefficient of friction between $\text{A}$ and $\text{B}$ is $0.2$ and between $\text{B}$ and the surface of the table is also $0.2$. The maximum force $F$ that can be applied on $\text{B}$ horizontally, so that the block $\text{A}$ does not slide over the block $\text{B}$ is

[Take, $g=10{\text{m/s}}^2$]

• A. $12\text{N}$
• B. $16\text{N}$
• C. $8\text{N}$
• D. $40\text{N}$

Answer 1

The correct option is B $16\text{N}$

Acceleration $a$ of system of blocks $\text{A}$ and $\text{B}$ is,

$a=\frac{\text{Net force}}{\text{Total masses}}=\frac{F-f_1}{m_A+m_B}$

Where, $f_1=$ friction between $\text{B}$ and the surface $=\mu (m_A+m_B)g$

So,
$a=\frac{F-\mu (m_A+m_B)g}{(m_A+m_B)}.........\left(1\right)$

Here, $\mu =0.2,m_A=1\text{kg},m_B=3\text{kg},g=10{\text{ms}}^{-2}$

Substituting the above values in Eq. $\left(1\right)$, we have

$a=\frac{F-0.2(1+3)\times 10}{1+3}$

$a=\frac{F-8}{4}.......\left(2\right)$

Due to acceleration of block $\text{B}$, a pseudo force $F^{\prime }$ acts on $\text{A}$.


This force $F^{\prime }$ is given by $F^{\prime }=m_{A}a$

Where, $a$ is acceleration of $\text{A}$ and $\text{B}$ caused by net force acting on $\text{B}$.

For $\text{A}$ to slide over $\text{B}$; pseudo force on $\text{A}$, i.e. $F^{\prime }$ must be greater than friction between $\text{A}$ and $\text{B}$.

$\Rightarrow m_{A}a\ge f_2$

We consider limiting case,
$m_{A}a=f_2$
$\Rightarrow m_{A}a=\mu \left(m_A\right)g$
$\Rightarrow a=\mu g=0.2\times 10=2{\text{ms}}^{-2}.........\left(3\right)$

Putting the value of a from Eq. $\left(3\right)$ into Eq. $\left(2\right)$ we get

$\frac{F-8}{4}=2$

$\therefore F=16\text{N}$

Hence, option $\left(\text{B}\right)$ is correct.