Question

Three blocks $A,B$ and $C$ of equal mass $m$ are placed on a smooth surface as shown. Coefficient of friction between any two blocks of $A,B$ and $C$ is $\mu$. The maximum value of mass $D$ so the block $A,B\&C$ move without slipping over each other is

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

Answer 1

The correct option is D $\frac{3m\mu }{(1-\mu )}$

Let mass of block $D$ be ${}^{\prime }{M}^{\prime }$
For mass $D$, from FBD we get

$Mg-T=Ma$ ... $\left(1\right)$
Now, for block $A$
$T-f=ma$
$\Rightarrow T-2\mu mg=ma$ ... $\left(2\right)$
(assuming $B\&C$ as a single block)


From FBD we can see that $B$ & $C$ combined have only friction force which provides acceleration.
So, maximum acceleration attained by single block of $B\&C$ together
$f=2ma$
$\Rightarrow 2\mu mg=2ma$
$\Rightarrow a=\mu g$
Using eq. $\left(2\right)$, we get
$T-2\mu mg=m\mu g$
$\Rightarrow T=3\mu mg$
By eq. $\left(1\right)$
$Mg-3\mu mg=Ma$
$Mg-3\mu mg=M\mu g$
$M=\frac{3\mu m}{1-\mu }$