Question

 A mild steel wire of cross-sectional area 0.60 x 10^-2 cm2 and length 2 m is stretched ( not beyond its elastic limit ) horizontally between two columns. If a 100g mass is hung at the midpoint of the wire, find the depression at the midpoint.​

Answer 1

Consider a steel wire of total length L' = 2L and a mass m is suspended at its midpoint. Now the total elongation in the wire is 2ΔL and new length is 2L + 2ΔL.A is cross sectional area of the wire and d is the depression at the midpoint.

Applying Pythagoras' Theorem in triangle QRS,

$\sf{\longrightarrow(L+\Delta L)^2=L^2+d^2}$

[ΔL is so small that (ΔL)² can be neglected on expanding the LHS.]

$\sf{\longrightarrow L^2+2L\Delta L=L^2+d^2}$

$\sf{\longrightarrow2L\Delta L=d^2}$

$\sf{\longrightarrow\Delta L=\dfrac{d^2}{2L}}$

The sine of the angle θ can be given by,

$\sf{\longrightarrow \sin\theta=\dfrac{d}{L+\Delta L}}$

$\sf{\longrightarrow \sin\theta=\dfrac{d}{L+\dfrac{d^2}{2L}}}$

$\sf{\longrightarrow \sin\theta=\dfrac{2Ld}{2L^2+d^2}}$

Here the weight of the mass m, i.e. mg, causes the elongation in the wire (F = mg) and is acting vertically downwards. So we need to consider the net vertical component of area and is given by,

$\sf{\longrightarrow A'=2A\sin\theta}$

$\sf{\longrightarrow A'=2A\cdot\dfrac{2Ld}{2L^2+d^2}}$

$\sf{\longrightarrow A'=\dfrac{4ALd}{2L^2+d^2}}$

We know Young's Modulus is given by,

$\sf{\longrightarrow Y=\dfrac{FL'}{A\,\Delta L}}$

Then, elongation will be given by,

$\sf{\longrightarrow\Delta L=\dfrac{FL'}{AY}}$

$\sf{\longrightarrow\dfrac{d^2}{2L}=\dfrac{mg\cdot 2L}{\left(\dfrac{4AYLd}{2L^2+d^2}\right)}}$

$\sf{\longrightarrow\dfrac{d^2}{2L}=\dfrac{mgL(2L^2+d^2)}{2AYLd}}$

$\sf{\longrightarrow YAd^3=mgL(2L^2+d^2)}$

$\sf{\longrightarrow YAd^3=2mgL^3+mgLd^2}$

$\sf{\longrightarrow Ad^3=\dfrac{2mgL^3}{Y}+\dfrac{mgLd^2}{Y}}$

The term $\sf{\dfrac{mgLd^2}{Y}}$ is very small since $\sf{d^2}$ is so small and $\sf{Y}$ is very large. So this term can be neglected.

Then,

$\sf{\longrightarrow Ad^3=\dfrac{2mgL^3}{Y}}$

$\sf{\longrightarrow d^3=\dfrac{2mgL^3}{AY}}$

$\longrightarrow\sf{d}=L\sqrt[\sf{3}]{\sf{\dfrac{2mg}{AY}}}$

In the question,

• $\sf{m=0.1\ kg}$

• $\sf{g=10\ m\,s^{-2}}$

• $\sf{2L=2\ m,\quad\! L=1\ m}$

• $\sf{A=0.6\times10^{-6}\ m^2}$

• Young's Modulus of Steel, $\sf{Y=2\times10^{11}\ Pa}$

Then,

$\longrightarrow\sf{d}=1\times\sqrt[\sf{3}]{\sf{\dfrac{2\times0.1\times10}{0.6\times10^{-6}\times2\times10^{11}}}}$

$\longrightarrow\sf{\underline{\underline{d=2.554\ cm}}}$

Answer 2

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