Question

A steel wire of original length 1 m 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 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression ?

Y of the steel = $2.0\times {10}^{11}N{m}^{-2}.Takeg=10m{s}^{-2}$.

Answer 1

From figure $-cos\theta =\frac{X}{\sqrt{X^2+I^2}}$

$=\frac{X}{I}{\{1+\frac{X^2}{I^2}\}}^{\frac{1}{2}}$

$=\frac{X}{I}$ ....(1)

Increase in length,

$\Delta L=(AC+CB)-AB$

Here, AC $=(I^2+X^2{)}^{\frac{1}{2}}$

So, $\Delta L=2(I^2+X^2{)}^{\frac{1}{2}}$

Given that,$A=4m{m}^2$

= $4\times {10}^{-2}c{m}^2$

Substituting the values , we have

Y =$\frac{F}{A}\frac{L}{\Delta L}$

$\Rightarrow 2\times {10}^{12}$

$=\frac{T\times 1000}{(4\times {10}^{-2})\times \left[2\right({50}^2+X^2{)}^{\frac{1}{2}}-100]}$ ....(2)

From equation (i), (ii) and the freebody diagram

$2Tcos\theta =mg$

$2T\left(\frac{X}{50}\right)=2.16\times {10}^3\times 980$

$\Rightarrow \frac{2\times (2\times {10}^{1}2)\times (4\times {10}^{-2})\times [2({50}^2+X^2\frac{1}{2})-100\times X]}{100\times 50}$

= $\left(2.16\right)\times {10}^3\times 980$

Solving for X, we get X = 1.5 cm.