Question

A steel wire of original length 1 m and cross-sectional area $4.00{mm}^2$ is clamped at the two ends so that it lies horizontally and without tension. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression?Y of the steel $=2.0\times {10}^{11}{Nm}^{-2}$.

Take g $=10{ms}^{-2}$.

Answer 1

let T be the tension in string after load is suspended

$cos\theta =\frac{x}{\surd x^2+l^2}$ = $\frac{x}{l}$ {$1+\frac{x^2}{l^2}$} ${}^{-1/2}$

expanding above equation using the binomial theorem

$cos\theta =\frac{x}{l}$ ($-1\frac{1\times x^2}{2\times l^2}$)

since $x<<<l$

Increase in length

$\Delta L=AC+BC-AB$

$AB=(l^2+x^2{)}^{1/2}$

$\Delta L=2(l^2=x^2{)}^{1/2}-2l$

we know that

$2Tcos\theta =mg$

put value from given data

we get

$x=1.5cm$

hence the verticle depressipn is 1.5cm