Question

A uniform rod of mass $2\text{kg}$ is hanging from a thread attached at the midpoint $\left(\text{O}\right)$ of the rod. A block of mass $m=6\text{kg}$ hangs at the left end of the rod and a block of mass $M$ hangs at the right end at a distance of $30\text{cm}$ from the mid point $\left(\text{O}\right)$. If the system is in equilibrium, calculate the mass $\left(M\right)$, given that overall length of the rod is $100\text{cm}$. Take $g=10{\text{m/s}}^2$.

• A. $5\text{kg}$
• B. $10\text{kg}$
• C. $8\text{kg}$
• D. $12\text{kg}$

Answer 1

The correct option is B $10\text{kg}$

Given mass of rod $M^{\prime }=2\text{kg}$,
mass of left block $m=6\text{kg}$
Considering the system of mass $(m+M^{\prime }+M)$, we can see that forces $T,mg,M^{\prime }g,Mg$ produce torque.


For rotational equilibrium, let the net torque be zero about reference point $\text{O}$:
${\tau }_{\text{net}}=0...\left(i\right)$
Taking anticlockwise sense of rotation for torque as $+ve$
$\&\tau =F\times r_{\perp }$
$\because$ Forces $\left(M^{\prime }g\&T\right)$ pass through the reference point $\text{O}$, hence their torque about point $\text{O}$ is zero. $M^{\prime }g$ being weight of rod will pass through centre $\text{O}$.

Substituting the torques with proper sign in eq. $\left(i\right)$,
$\sum \tau =0$
$\Rightarrow +\left(mg\right)\left(0.5\right)-\left(Mg\right)\left(0.3\right)=0$
or, $M=\frac{m\left(0.5\right)}{0.3}=\frac{6\times 0.5}{0.3}$
$\therefore M=10\text{kg}$