A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 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.

#### Solution 1

Let AB be a mild steel wire of length 2L = lm and its cross-section area A = 0.50 x 10^{-2} cm^{2}. A mass m = 100 g = 0.1 kg is suspended at mid-point C of wire as shown in figure. Let x be the depression at mid-point i.e., CD = x

`:. AD = DB = sqrt((AC^2 + CD^2)) = sqrt(L^2+x^2)`

:. Increase in length `triangleL = (AD+DB) - AB = 2sqrt(L^2+x^2) - 2L`

`= 2L[(1+x^2/L^2)^(1/2) - 1] = 2L.x^2/(2L^2) = x^2/L`

:. Longitudinal strain = `(triangleL)/(2L) = x^2/(2L^2)`

If T be the tension in the wire as shown figure then in equilibrium `2T cos theta = mg`

or `T = "mg"/(2costheta)`

`= "mg"/(2 x/(sqrt(x^2+L^2))) = (mgsqrt(x^2+L^2))/(2x) = "mgL"/(2x)`

:. Stress = `T/A = "mgL"/(2 x A)`

As Young's modulus Y = `"stress"/"strain"`

`=((mgL)/(2 x A))/((x^2)/(2L^2)) = (mgL)/(2 x A) xx (2L^2)/x^2 = (mgL^3)/(Ax^3)`

`=> x = [(mgL^3)/(YA)]^(1/3) = L[(mg)/(YA)]^(1/3)`

`= 1/2[(0.1xx9.8)/(2xx10^11xx0.50xx10^(-2)xx10^(-4))]^(1/3) = 1.074 xx 10^(-2) m`

= 1.074 cm ≈ 1.07 cm or 0.01 m

#### Solution 2

Length of the steel wire = 1.0 m

Area of cross-section, *A* = 0.50 × 10^{–2} cm^{2 }= 0.50 × 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:

Δ*l *= (XO + OZ) – XZ

Where,

XO = OZ = `[(0.5)^2 + l^2]^(1/2)`

`:.triangle l = 2[(0.5)^2 + (l)^2]^(1/2) - 1.0`

`= 2 xx 0.5 [1+(l/0.5)^2]^(1/2) - 1.0`

Expanding and neglecting higher terms, we get

`triangle l = l^2/0.5`

`"Strain" = "Increase in length"/"Original length"`

Let *T* be the tension in the wire.

∴*m*g = 2*T* cos*θ*

Using the figure, it can be written as:

`cos theta = l/((0.5)^2 + l^2)^(1/2)`

`= l/ ((0.5)(1+(l/0.5)^2)^(1/2))`

Expanding the expression and eliminating the higher terms:

`cos theta = l/((0.5)(1+l^2/(2(0.5)^2)))`

`(1+l^2/(0.5))=~1` for small l

`:. cos theta = l/(0.5)`

`:. T = (mg)/(2(l/(0.5))) = (mgxx 0.5)/(2l) = (mg)/(4l)`

`"Stress" = "Tension"/"Area" = (mg)/(4lxxA)`

`Y = (mg xx 0.5)/(4lxxAxxl^2)`

`l = sqrt((mgxx0.5)/(4YA)) `

Young’s modulus of steel, *Y* = 2 x 10^{11} Pa

`:.l = sqrt((0.1xx9.8xx0.5)/(4xx2xx10^11xx0.50xx10^(-6)))`

= 0.0106 m

Hence, the depression at the midpoint is 0.0106 m.