Question

A steel wire of mass µ per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m^–3 (Young’s modules Y = 2 × 1011 Nm^–2).

Answer 1

Consider the diagram when a small element of length dx is considered at x from the load (x = 0).

Let T(x) and T(x + dx) are tensions on the two cross-sections a distance dx apart, then – t(x + dx) + T(x) = dmg = μ dxg (where μ is the mass/length)  ......(∵ dm = μdx)

dT = μgdx  .....[∵ dT = T(x + dx) – T(x)]

⇒ T(x) = μgx + C  .....(On integrating)

At x = 0, T(0) = Mg

⇒ C = mg

∴ T(x) = μgx + Mg

Let the length dx at x increase by dr, then

Young's modulus Y = `"Stress"/"Strain"`

`((T(x))/A)/((dr)/(dx)) = Y`

⇒ `(dr)/(dx) = 1/(YA) T(x)`

⇒ `r = 1/(YA) int_0^L (μgx + Mg)dx`

= `1/(YA) [(μgx^2)/2 + Mgx]_0^L`

= `1/(YA)[(mgL^2)/2 + MgL]`  ......(m is the mass of the wire)

`A = pi xx (10^-3)^2 m^2`

`Y = 200 xx 10^9  Nm^-2`

`m = pi xx (10^-3)^2 xx 10 xx 7860` kg

∴ `r = 1/(2 xx 10^11 xx pi xx 10^-6)`  ......`[(pi xx 786 xx 10^-3 xx 10 xx 10)/2 + 25 xx 10 xx 10]`

= `[196.5 xx 10^-6 + 3.98 xx 10^-3]`

= 4 × 10^–3 m