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

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 } $
$\therefore 1.5 x =1$
$\therefore x =\frac{1}{1.5}=0.6667\, m$ from steel wire.
$\therefore x=66.67\, cm$