Question

A steel wire of diameter $d=1.0mm$ is stretched horizontally between two clamps located at the distance $l=2.0m$ from each other. A weight of mass $m=0.25kg$ is suspended from the midpoint $O$ of the wire. What will the resulting descent of the point $O$ be in millimetres?E=$2\times {10}^{11}N/m^2$

Answer 1

Let the point $O$ descend by the distance $x$ (figure shown below). From the condition of equilibrium of point $O$.
$2T\sin \theta =mg$ or $T=\frac{mg}{2\sin \theta }=\frac{mg}{2x}\sqrt{{\left(\frac{l}{2}\right)}^2+x^2}..........................................\left(1\right)$
Now,
$\frac{T}{\pi {\left(\frac{d}{2}\right)}^2}=\sigma =\epsilon E$ or $T=\epsilon E\pi \frac{d^2}{4}................\left(2\right)$
$(\sigma$ here is stress and $\epsilon$ here is strain.$)$
In addition to it,
$\epsilon =\frac{\sqrt{{\left(\frac{l}{2}\right)}^2+x^2}-\frac{l}{2}}{\frac{l}{2}}=\sqrt{1+{\left(\frac{2x}{l}\right)}^2}-1...........................................\left(3\right)$
From equations $\left(1\right),\left(2\right)$ and $\left(3\right),$
$x-\frac{x}{\sqrt{1+{\left(\frac{2x}{l}\right)}^2}}=\frac{mgl}{\pi E{d}^2}$

Using Binomial Expansion,

$x(1-(1-4x^2/l^2+6x^4/l^4+....))=\frac{mgl}{\pi E{d}^2}$

As $x\ll l$, higher order terms can be ignored
So, $\frac{4x^3}{2l^2}\approx \frac{mgl}{\pi E{d}^2}$
or, $x=l{\left(\frac{mg}{2\pi E{d}^2}\right)}^{\frac{1}{3}}=2.5cm=25mm$