Question

Minimum horizontal force $F$ required to keep the smaller block stationary with respect to the bigger block as shown in the figure is :

• A. $\frac{(M-m)g}{\mu }$
• B. $\frac{(M+m)g}{\mu }$
• C. $\frac{Mg}{\mu }$
• D. $\frac{mg}{\mu }$

Answer 1

The correct option is B $\frac{(M+m)g}{\mu }$

Let us suppose that, acceleration of the system is ${}^{\prime }{a}^{\prime }$, then
$F_{\text{net}}=m_{\text{total}}a$
$a=\frac{F}{M+m}$ ...... (1)


Taking ground as the frame of reference.
F.B.D of smaller block:


On applying $\sum F=ma$ along horizontal direction,
$N=ma=\frac{mF}{M+m}$... (2) [from (1)]
Now, on applying $\sum F=ma$ along vertical direction,
$f-mg=0$ [No acceleration along vertical direction]
$\Rightarrow f=mg$
In limiting case,
Friction $\left(f\right)=\mu N$
$\Rightarrow \mu N=mg$
$\Rightarrow \frac{\mu mF}{M+m}=mg$ [from (2)]
$\Rightarrow F=\frac{(M+m)g}{\mu }$
Minimum horizontal force required is $\frac{(M+m)g}{\mu }$ to keep the smaller block stationary with respect to the bigger block.