Question

In the arrangement shown in the figure, mass of the block $B$ and $A$ are $2m$, $8m$ respectively. Surface between $B$ and the floor is smooth. The block $B$ is connected to block $C$ by means of a pulley. If the whole system is released from rest, then find the minimum value of mass of block $C$ so that the block $A$ remains stationary with respect to $B$. Given coefficient of static friction between $A$ and $B$ is $\mu$ and pulley is ideal.

• A. $\frac{m}{\mu }$
• B. $\frac{2m}{\mu +1}$
• C. $\frac{10m}{\mu }$
• D. $\frac{10m}{\mu -1}$

Answer 1

The correct option is D $\frac{10m}{\mu -1}$

Assume the acceleration of system is $a$. Hence to maintain string constraint, $a_B=a_C=a$ and if block $A$ remains stationary w.r.t block B, then $a_A=a_B=a$
FBD of block $A$:


Applying equilibrium condition on $A$ from frame of block $B$ in horizontal direction.
$N=8ma...\left(i\right)$
Equilibrium condition in vertical direction,
$f_s=8mg....\left(ii\right)$
But $f_s=\mu N...\left(iii\right)$ (limiting condition for minimum value of block $C$)
From Eq $\left(i\right),\left(ii\right)\&\left(iii\right)$
$8mg=\mu \times \left(8ma\right)$
$\therefore a=\frac{g}{\mu }...\left(iv\right)$

Net pulling force on the system of $A,B,C$ is the weight of block $C$ i.e $m_{1}g$ [assume $m_1$ is the minimum mass of $C$]


Hence, acceleration $a$ of system can be written as:
$F=ma$
$m_{1}g=(m_1+2m+8m)a$
From eqn. $\left(iv\right)$,
$m_{1}g=(10m+m_1)\times \left(\frac{g}{\mu }\right)$
$\mu m_1=10m+m_1$
$\therefore m_1=\frac{10m}{\mu -1}$ is the minimum mass of $C$ for $A$ to remain stationary w.r.t $B$.