Question

A mild steel wire of length 2l meter cross sectional area A square meter is horizontally between two pillars.. A small mass m kg is suspended from themild point of the wire. If extension in wire are within rlastic limit then depression at the mid point of the wire will be

Answer 1

Given that,

Length of wire = 2l

Area = A

Mass = m kg

According to figure,

Let x be the depression at the mid point of the wire.

$SR=x$

$PQ=QR=\sqrt{l^2+x^2}$

We need to calculate the strain

Using formula of change in length

$\Delta l=(PR+RQ)-PQ$

$\Delta l=2PR-PQ$

Put the value into the formula

$\Delta l=2(l^2+x^2)^{\frac{1}{2}}-2l$

$\Delta l=2l(1+\dfrac{x^2}{2l^2})-2l$

$\Delta l=\dfrac{x^2}{l}$

Using formula of strain

$strain=\dfrac{\Delta l}{2l}$

Put the value into the formula

$strain=\dfrac{x^2}{2l^2}$

We need to calculate the tension in the wire

Using balance equation

$2T\cos\theta=mg$

$T=\dfrac{mg}{2\cos\theta}$.....(I)

We need to calculate the value of cos θ

Using figure

$\cos\theta=\dfrac{x}{(l^2+x^2)^{\frac{1}{2}}}$

$\cos\theta=\dfrac{x}{l(1+\dfrac{x^2}{l^2})^{\frac{1}{2}}}$

$\cos\theta =\dfrac{x}{l(1+\dfrac{x^2}{2l^2})}$

As, x < < l,

So, $1 > >\dfrac{x^2}{2l^2}$

$1+\dfrac{x^2}{2l^2}\approx 1$

Then, $\cos\theta=\dfrac{x}{l}$

Put the value in the equation (I)

$T=\dfrac{mg}{2\times\dfrac{x}{l}}$

$T=\dfrac{mgl}{2x}$

We need to calculate the stress

Using formula of stress

$stress=\dfrac{T}{A}$

Where, T = tension

A = area

Put the value into the formula

$stress =\dfrac{mgl}{2Ax}$

We need to calculate the depression at the mid point of the wire

Using formula of young's modulus

$Y=\dfrac{stress}{strain}$

Put the value into the formula

$Y=\dfrac{mgl}{2Ax}\times\dfrac{2l^2}{x^2}$

$Y=\dfrac{mgl^3}{Ax^3}$

$x = l(\dfrac{mg}{YA})^{\frac{1}{3}}$

Hence, The depression at the mid point of the wire is $l(\dfrac{mg}{YA})^{\frac{1}{3}}$.