Question

A light rod of length $2.00\text{m}$ is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel of cross-section ${10}^{-3}{\text{m}}^2$ and the other is of brass of cross-section $2\times {10}^{-3}{\text{m}}^2$. Find out the position along the rod at which a weight may be hung to produce equal stress in both the wires.

Given:
$Y_{\text{steel}}=2\times {10}^{11}{\text{N/m}}^2$
$Y_{\text{brass}}={10}^{11}{\text{N/m}}^2$

• A. $1.33\text{m}$ from end $B$
• B. $1.33\text{m}$ from end $A$
• C. $1.57\text{m}$ from end $B$
• D. $1.57\text{m}$ from end $A$

Answer 1

The correct option is B $1.33\text{m}$ from end $A$

Let the distance of the weight ${}^{\prime }{W}^{\prime }$ from the end $A$ be ${}^{\prime }{x}^{\prime }$.

Given:

Stress in steel $=$ stress in brass

$\Rightarrow \frac{T_A}{A_A}=\frac{T_B}{A_B}$

$\Rightarrow \frac{T_A}{T_B}=\frac{A_A}{A_B}=\frac{{10}^{-3}}{2\times {10}^{-3}}=\frac{1}{2}...\left(i\right)$

As the system is in equilibrium, taking moments about $C$, we have,

$T_A.x=T_B(2-x)$

$\Rightarrow \frac{T_A}{T_B}=\frac{2-x}{x}...\left(ii\right)$

On solving $\left(i\right)$ and $\left(ii\right)$, we get,

$x=1.33\text{m}$