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

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
$a=\frac{F-02(1+3)\times 10}{1+3}$
$a=\frac{F-8}{4}.............\left(ii\right)$
Due to acceleration of block B, a pseudo force F' acts on A
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
$\frac{F-8}{4}=2$
$\therefore F=16N$