Question

A mild steel wire of length 1.0 m and cross-sectional area 0.50 x 10^-2 cm² is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the midpoint of the wire. Calculate the depression at the mid-point.

Answer 1

Take, g = 10 ms^-2; Y = 2 x 10¹¹ Nm^-2

In the figure, let x be the depression at the mid point, i.e., CD = x

AC = CB = l = 0.5 m;

m = 100 g = 0.1 kg

AD = BD = (l² + x²)^1/2

Increase in length, Δl = AD + DB - AB

= 2AD - AB

= 2 (l² + x²)^1/2 - 2l

= 2l(l + x²/2l²) - 2l = x²/l

Strain = Δl/2l = x²/2l²

If T be the tension in the wire, then 2T cosθ

= mg or T = mg/2cosθ

= 1.074 x 10^-2 m = 1.074 cm = 1.074/100 = 0.01074 m