Question

A metre long light rod is suspended from a rigid support horizontally by means of two vertical wires of equal length. One wire with area of cross section $10^{-3}{m}^2$ is made of steel while another with area of cross-section $2\times 10^{-3}{m}^2$ is made of brass. Find position along $2\times 10^{-3}{m}^2$ is made of brass. Find position along the rod from steel wire in $cm$ at which a weight might be hung to produce equal stresses in both the wires.

Answer 1

Answer: 66.67

Let weight $W$ be suspended at distance $x$ from steel end.
Stress in steel wire $=\frac{T_1}{a_1}$
Stress in brass wire $=\frac{T_2}{a_2}$
When stresses are equal,
$\frac{T_1}{a_1}=\frac{T_2}{a_2}$
$\therefore \frac{T_1}{T_2}=\frac{a_1}{a_2}$
$=\frac{10^{-3}}{2\times 10^{-3}}=0.5$
As system of rod and wires is in equilibrium,
$T_1\times AC=T_2\times BC$
$\therefore \frac{T_1}{T_2}=\frac{BC}{AC}=\frac{1-x}{x}$
$\therefore 0.5=\frac{1-x}{x}$
$\therefore 1.5x=1$
$\therefore x=\frac{1}{1.5}=0.6667m$ from steel wire.
$\therefore x=66.67cm$