Question

Consider the situation shown in figure. The horizontal surface below the bigger block is smooth. The coefficient of friction between the blocks is $\mu$. Find the minimum force F that can be applied in order to keep the smaller block at rest with respect to the bigger block

• A. $\frac{1+\mu }{1-\mu }(2M+m)g$
• B. $\frac{1+\mu }{1-\mu }(M+2m)g$
• C. $\frac{1+\mu }{1-\mu }(3M+2m)g$
• D. $\frac{1-\mu }{1+\mu }(M+2m)g$

Answer 1

The correct option is D

$\frac{1-\mu }{1+\mu }(M+2m)g$

If no force is applied, the block A will slip on C towards right and the block B will move downward. Suppose the minimum force needed to prevent slipping is F. Taking A + B + C as the system, the only external horizontal force on the system is F. Hence the acceleration of the system is
$a=\frac{F}{M+2m}$ ...... (i)
Now take the block A as the system. The forces on A are

(i) tension T by the string towards right,

(ii) friction f by the block C towards left,

(iii) weight mag downward and

(iv) normal force N upward.

For vertical equilibrium N = mg.

As the minimum force needed to prevent slipping is applied, the friction is limiting. Thus,
$f=\mu N=\mu mg$
As the block moves towards right with an acceleration

(i) tension T upward,

(ii) weight mg downward,

(iii) normal force N’ = ma.

(iv) friction f ‘ upward.

As the block moves towards the right with an acceleration a,

N’ = ma.

As the friction is limiting.
$f^{\prime }=\mu N^{\prime }=\mu ma$

For horizontal equilibrium of Block A

$T-\mu mg=ma$ ...(ii)

For vertical equilibrium

T + f’ = mg

Or, $T+\mu ma=mg$. …(iii)

Elimination T from (ii) and (iii)
$a_{min}=\frac{1-\mu }{1+\mu }g$
When a large force is applied, the block A slips on C towards left and the block B slips besides C in the upward direction. The friction on A is towards right and that on B is downwards. Solving as above, the acceleration in this case is
$a_{max}=\frac{1+\mu }{1-\mu }g$
thus, a lies between $\frac{1-\mu }{1+\mu }g$ and $\frac{1+\mu }{1-\mu }g$
From (i) the force F should be between
$\frac{1-\mu }{1+\mu }(M+2m)g$ and $\frac{1+\mu }{1-\mu }(M+2m)g$