Question

An uniform rod of mass $3\text{kg}$ is hinged at the wall and connected through a light string as shown in the figure. Find the tension in the string.(Take $g=10{\text{m/s}}^2$).

• A. $30\text{N}$
• B. $15\text{N}$
• C. $5\text{N}$
• D. $10\text{N}$

Answer 1

The correct option is B $15\text{N}$

Given, mass of the rod, $m=3\text{kg}$
Let $T$ be the tension in the string and $L$ be the length of the rod.


Let the reaction force on rod from the hinge be $N_x$ and $N_y$.
As rod is in equilibrium, thus net torque about point $\text{O}$ should be zero.
$mg$ will act at the geometric centre of rod, so the $\perp$ distance for $mg$ from $\text{O}$ is $\frac{L}{2}\sin \theta$.
Similarly, $\perp$ distance for $T$ from $O$ is $L\sin \theta$

${\tau }_{net}=0...\left(i\right)$

Considering anticlockwise sense of rotation due to torque as $+ve$, torque about point $\text{O}$ is given as:
${\tau }_{mg}=-mg\left(\frac{L}{2}\sin \theta \right)$
${\tau }_T=+TL\sin \theta$
$\because N_x\&N_y$ passes through point $\text{O}$ hence their torque about $\text{O}$ is zero.

Substituting in Eq. $\left(i\right)$,
${\tau }_{net}=0$
$\Rightarrow {\tau }_T+{\tau }_{mg}+{\tau }_{N_x}+{\tau }_{N_y}=0$
$\Rightarrow +T\left(L\sin \theta \right)-mg\left(\frac{L}{2}\sin \theta \right)+0+0=0$
$\Rightarrow T=\frac{mg}{2}$
$\therefore T=\frac{3\times 10}{2}=15\text{N}$