A mild steel wire of length $1.0 \;m$ and cross-sectional area $0.50 \times 10^{-2} \;cm ^{2}$ 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 midpoint.
A mild steel wire of length $1.0 \;m$ and cross-sectional area $0.50 \times 10^{-2} \;cm ^{2}$ 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 midpoint.
Length of the steel wire $=1.0 m$
Area of cross-section, $A=0.50 \times 10^{-2} cm ^{2}-0.50 \times 10^{-6} m ^{2}$
A mass $100 g$ is suspended from its midpoint.
$m=100 g =0.1 kg$
Hence, the wire dips, as shown in the given figure.
Original length $= XZ$
Depression $=l$
The length after mass $m$, is attached to the wire $= XO + OZ$
Increase in the length of the wire:
$\Delta l=( XO + OZ )- XZ$
$XO = OZ =\left[(0.5)^{2}+l^{2}\right]^{\frac{1}{2}}$
$\therefore \Delta l=2\left[(0.5)^{2}+(l)^{2}\right]^{\frac{1}{2}}-1.0$
$=2 \times 0.5\left[1+\left(\frac{l}{0.5}\right)^{2}\right]^{\frac{1}{2}}-1.0$
Expanding and neglecting higher terms, we get:
$\Delta l=\frac{l^{2}}{0.5}$
Strain $=\frac{\text { Increase in length }}{\text { Original length }}$
Let $T$ be the tension in the wire.
$\therefore m g=2 T \cos \theta$
Using the figure, it can be written as
$\cos \theta=\frac{1}{\left((0.5)^{2}+l^{2}\right)^{\frac{1}{2}}}$
$=\frac{1}{(0.5)\left(1+\left(\frac{l}{0.5}\right)^{2}\right)^{\frac{1}{2}}}$
Expanding the expression and eliminating the higher terms
$\cos \theta=\frac{1}{(0.5)\left(1+\frac{l^{2}}{2(0.5)^{2}}\right)}$
$\left(1+\frac{l^{2}}{0.5}\right)=1$ for small $l$
$\therefore \cos \theta=\frac{l}{0.5}$
$\therefore T=\frac{m g}{2\left(\frac{l}{0.5}\right)}=\frac{m g \times 0.5}{2 l}=\frac{m g}{4 l}$
Stress $=\frac{\text { Tension }}{\text { Area }}=\frac{m g}{4 l \times A}$
Young's modulus $=\frac{\text { Stress }}{\text { Strain }}$
$Y=\frac{m g \times 0.5}{4 l \times A \times l^{2}}$
$I=\sqrt[3]{\frac{m g \times 0.5}{4 Y A}}$
Young's modulus of steel, $Y=2 \times 10^{11} Pa$
$\therefore l=\sqrt{\frac{0.1 \times 9.8 \times 0.5}{4 \times 2 \times 10^{11} \times 0.50 \times 10^{-6}}}$
$=0.0106 m$
Hence, the depression at the midpoint is $0.0106 m$
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