Question

Three blocks $A,B$ and $C$ of equal masses $‘m^{\prime }$ each are placed one over the other on a frictionless table. The coefficient of friction between any two blocks is $\mu$. Find the maximum value of mass of block $D$ so that the blocks $A,B$ and $C$ move without slipping over each other.

• 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 $\frac{3m\mu }{(1-\mu )}$

Taking blocks $B$ and $C$ as a system, and applying the condition of no slip between $A$ and $B$.
The FBD of the system is as shown

Let the mass of block $D$ be $m_D$
from the FBD we have
$N_1=2mg$
$f=\mu N_1=\mu \left(2mg\right)$

Since the system should move without slipping over each other, then the common acceleration of the system is given by
$a_c=\frac{\text{(Supporting force - Opposing Force)}}{\text{Total mass}}$
$\Rightarrow a_c=\frac{m_{D}g}{3m+m_D}$........(i)

Also, from equation of force we have
$f=2m{a}_c$
$\Rightarrow 2m{a}_c=\mu 2mg\Rightarrow a_c=\mu g$....(ii)
From eq (i) & (ii) we get
$\mu g=\frac{m_{D}g}{3m+m_D}$
$\Rightarrow m_D(1-\mu )=3m\mu$
$\Rightarrow m_D=\frac{3m\mu }{(1-\mu )}$