Question

Two blocks A and B of mass $m_A=1kg$ and $m_B=3kg$ are kept on the table as shown in figure. The coefficient of friction between A and B is $0.2$ and between B and the surface of the table is also $0.2$. The maximum force $F$ (in newtons) that can be applied on B horizontally, so that the block A does not slide over the block B is [Take , $g=10m/s^2]$

Answer 1

Step 1

Acceleration a of system of blocks A and B is

$a=\frac{\text{Net force}}{\text{Total mass}}=\frac{F-f_{1_1}}{m_A+m_B}$

where, $f_{l_1}$ = limiting friction between B and the surface

$=\mu (m_A+m_B)g$

So,

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

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

Substituting the above values in Eq. (i), we have

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

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

Due to acceleration of block B, a pseudo force F' acts on A

Step 2

This force F' is given by

$F^{\prime }=m_{A}a$

where, $a$ is acceleration of A and B caused by net force acting on B.

For A to slide over B; pseudo force on A, i.e. F' must be greater than limiting friction

between A and B.

$\Rightarrow m_{A}a\ge f_{l_2}$

We consider limiting case,

$m_{A}a=f_{l_2}\Rightarrow m_{A}a=\mu \left(m_A\right)g$

$\Rightarrow a=\mu g=0.2\times 10=2m{s}^{-2}......\left(iii\right)$

Putting the value of $a$ from Eq. (iii) into Eq. (ii). we get

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

$\therefore F=16N$