Question

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

Answer 1

Step 1: Draw a labelled diagram.

Step 2: Find the vertical depression
Formula used: $Y=\frac{FL}{Al}$
For equilibrium of mass 𝑀
$2T\sin \theta =mg$
From figure,
$\sin \theta \simeq \tan \theta =y/\left(L/2\right)$$=\frac{2y}{L}$

$\Delta l=2(\frac{L^2}{4}+y^2)-2\cdot \frac{L}{2}$
$=2\cdot \frac{L}{2}{\left(1\frac{y^2}{L^2/4}\right)}^{\frac{1}{2}}-L$
$=L[1+\frac{2y^2}{L^2}]-L=\frac{2y^2}{L}$

Elongation in the wire,
$\Delta L=\frac{TL}{AY}\Rightarrow \frac{2y^2}{L}=\frac{TL}{AY}$

$\frac{2y^2}{L}=\frac{Mg}{2\sin \theta }\frac{L}{AY}=\frac{Mg}{2\cdot \frac{2y^2}{L}}\frac{L}{AY}$
$8y^3=\frac{Mg}{AY}{L}^3$
$y=\frac{L}{2}{\left(\frac{Mg}{AY}\right)}^{\frac{1}{3}}$
$=\frac{1}{2}{\left(\frac{2.16\times 10}{4\times {10}^{-6}\times 2\times {10}^{11}}\right)}^{\frac{1}{3}}$

$=\frac{1}{2}{(27\times {10}^{-6})}^{\frac{1}{3}}=\frac{1}{2}\times 3\times {10}^{-2}$

$=1.5\times {10}^{-2}m=1.5cm$

Final Answer: (2)