A mild steel wire of length 2l meter and cros | vedclass

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Question

(A) Let the depression at the midpoint be $x$. The original length of each half of the wire is $l$. The new length of each half is $\sqrt{l^2 + x^2}$.

Vedclass · Accepted Answer

$l \left( \frac{mg}{YA} \right)^{1/3}$

Vedclass · Answer

(A) Let the depression at the midpoint be $x$. The original length of each half of the wire is $l$. The new length of each half is $\sqrt{l^2 + x^2}$.The increase in length of the wire is $\Delta L = 2(\sqrt{l^2 + x^2} - l) = 2l(\sqrt{1 + (x/l)^2} - 1)$.Using the binomial approximation $(1 + \epsilon)^n \approx 1 + n\epsilon$ for small $x/l$,we get $\Delta L \approx 2l(1 + \frac{1}{2} \frac{x^2}{l^2} - 1) = \frac{x^2}{l}$.The strain is $\frac{\Delta L}{2l} = \frac{x^2/l}{2l} = \frac{x^2}{2l^2}$.From the force balance at the midpoint,$2T \sin 	heta = mg$,where $\sin 	heta = \frac{x}{\sqrt{l^2 + x^2}} \approx \frac{x}{l}$.Thus,$2T(x/l) = mg \implies T = \frac{mgl}{2x}$.The stress is $\frac{T}{A} = \frac{mgl}{2Ax}$.Using Young's modulus $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{mgl/2Ax}{x^2/2l^2} = \frac{mgl^3}{Ax^3}$.Solving for $x$,we get $x^3 = \frac{mgl^3}{YA} \implies x = l \left( \frac{mg}{YA} ight)^{1/3}$.

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