Question

A steel wire of diameter $d$, area of cross-section $A$ and length $2l$ is clamped firmly at two points A and B, which are $2lm$ apart and in the same plane. A body of mass $m$ is hung from the middle point of the wire such that the middle point sags by x lower from the original position. If Young's modulus is $Y$, then $m$ is given by:

• A. $\frac{1}{2}\frac{YA{x}^2}{g{l}^2}$
• B. $\frac{1}{2}\frac{YA{l}^2}{g{x}^2}$
• C. $\frac{YA{x}^3}{g{l}^3}$
• D. $\frac{YA{l}^2}{g{x}^2}$

Answer 1

Answer: (C)

$\frac{YA{x}^3}{g{l}^3}$

We have $mg=2Tsin\theta$
Also for small values of $\theta$ we have $sin\theta =tan\theta =\frac{x}{l}$
$T=\frac{YA}{l}\Delta l$
or
$T=\frac{YA}{l}\left(\right(x^2+l^2{)}^{1/2}-l)$
As $x<<l$ we have
$T=\frac{YA}{l}\left(l\right((1+\frac{x^2}{l^2}{)}^{1/2}-1))$
or
$T=YA\left(\right(1+\frac{x^2}{l^2}{)}^{1/2}-1)$
or
$T=YA\left(\right(1+\frac{1}{2}\frac{x^2}{l^2})-1)$
$=\frac{YA{x}^2}{2l^2}$
Thus we get
$mg=2\frac{YA{x}^2}{2l^2}\frac{x}{l}$
or
$m=\frac{YA{x}^3}{g{l}^3}$