Question

A light rod $AB$ of length $2\text{m}$ is suspended from the ceiling horizontally by means of two vertical wires as shown in Fig. One of the wires is made of steel of cross-section $0.1{\text{cm}}^2$ and the other of brass of cross-section $0.2{\text{cm}}^2$. The Young's modulus of brass is $1.0\times {10}^{11}{\text{Nm}}^{-2}$ and of steel is $2.0\times {10}^{11}{\text{Nm}}^{-2}$. A weight $W$ is hung at point $C$ at a distance $x$ from end $A$. It is found that the stress in the two wires is the same when $x=\frac{n}{3}\text{metre}$. Find the value of $n$.

Answer 1

Let $T_s$ and $T_b$ be the tension in steel and brass wire.
From the data given in the question,
Stress in steel wire =Stress in brass wire.
$\frac{T_s}{A_s}=\frac{T_b}{A_b}$ $\Rightarrow \frac{T_s}{T_b}=\frac{A_s}{A_b}=\frac{0.1{\text{cm}}^2}{0.2{\text{cm}}^2}\Rightarrow \frac{T_s}{T_b}=\frac{1}{2}......\left(1\right)$
The system is in Rotational equilibrium calculating and equating the torque about $C$.
$T_s\times \frac{n}{3}=T_b(2-\frac{n}{3})$
$\Rightarrow \frac{T_s}{T_b}=\frac{(2-\frac{n}{3})}{\frac{n}{3}}$
From equation $\left(1\right)$
$\Rightarrow \frac{1}{2}=\frac{(2-\frac{n}{3})}{\frac{n}{3}}$
$\Rightarrow \frac{n}{3}=2\times (2-\frac{n}{3})$
$\Rightarrow \frac{n}{3}=4-\frac{2n}{3}$
$\Rightarrow n=4$