Question

In the arrangement shown in the figure, masses of the blocks $A$ and $B$ are $\text{m}$ and $2\text{m}$ respectively. Surface between block $B$ and floor is smooth. The block $B$ is connected to the block $C$ by means of a string pulley system. If the whole system is released, then find the minimum value of mass of block $C$ so that block $A$ remains stationary w.r.t. $B$. Coefficient of friction between $A$ and $B$ is $\mu$.

• A. $\frac{m}{\mu }$
• B. $\frac{2m+1}{\mu +1}$
• C. $\frac{3m}{\mu -1}$
• D. $\frac{6m}{\mu +1}$

Answer 1

The correct option is C $\frac{3m}{\mu -1}$


As a system, both $A$ and $B$ move together with an acceleration $a$.
$F=ma\Rightarrow T=3ma$
$a=\frac{T}{3m}$ ... (1)
Applying FBD for $C$
$Mg-T=Ma$ ... (2)
Applying FBD for $A$, to keep that at rest on $B$
$R=ma;f_s=mg\Rightarrow \mu R=mg$ ... (3)
Putting value of $a$ from (1)
$\frac{\mu T}{3m}=g\Rightarrow T=\frac{3mg}{\mu }$ ... (4)
$\Rightarrow a=\frac{T}{3m}=\frac{g}{\mu }$
Substituting (4) in (2) we get,
$Mg-\frac{3mg}{\mu }=Ma$
$M(g-\frac{g}{\mu })=\frac{3mg}{\mu }$
$Mg\left(\frac{\mu -1}{\mu }\right)=\frac{3mg}{\mu }$
$\Rightarrow M=\frac{3m}{\mu -1}$