Question

A mild steel wire of length 1.0 m and cross-sectional area $0.50\times {10}^{-2}c{m}^2$ is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.

Answer 1

From the above figure,

Let $x$ be the depression at the mid point i.e. $CD=x$.

In fig.,

$AC=CB=l=0.5m$ ;

$m=100g=0.100Kg$

$AD=BD=(l^2+x^2{)}^{\frac{1}{2}}$

Increase in length, $△l=AD+DB-AB=2AD-AB$

$=2{(l^2+x^2)}^{\frac{1}{2}}-2l=2l(1+\frac{x^2}{l^2})-2l=2l[1+\frac{x^2}{2l^2}]-2l=\frac{x^2}{l}$

$\therefore$ strain $=\frac{△l}{2l}=\frac{x^2}{2l^2}$

If T is the tension in the wire, then

$2Tcos\theta =mg$

or, $T=\frac{mg}{2cos\theta }$

Here, $cos\theta =\frac{x}{(l^2+x^2{)}^{\frac{1}{2}}}=\frac{x}{l{(1+\frac{x^2}{l^2})}^{\frac{1}{2}}}=\frac{x}{(1+\frac{1}{2}\frac{x^2}{l^2})}$

As, $x<<l$, so $1>>\frac{1x^2}{2l^2}$ and $1+\frac{1}{2}\frac{x^2}{l^2}\approx 1$

Hence, $T=\frac{mg}{2\left(\frac{x}{1}\right)}=\frac{mgl}{2x}$

stress $=\frac{T}{A}=\frac{mgl}{2Ax}$

$Y=\frac{stress}{strain}=\frac{mgl}{2Ax}\times \frac{2l^2}{x^2}=\frac{mg{l}^3}{A{x}^3}$

$x=l{\left[\frac{mg}{YA}\right]}^{\frac{1}{3}}=0.5{\left[\frac{0.1\times 10}{20\times {10}^{11}\times 0.5\times {10}^{-6}}\right]}^{\frac{1}{3}}=1.074\times {10}^{-2}m$

$=1.074cm$