Question

A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10^–2 cm² 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. 

Answer 1

Length of the steel wire = 1.0 m

Area of cross-section, A = 0.50 x 10^-2 cm² = 0.50 x 10^-6 m² 

A mass 100 g is suspended from its maidpoint

m = 100 g = 0.1 kg

Hence, the wire dips, as shwn 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:

Δl = (XO + OZ) – XZ

Where,

Expanding the expression and eliminating the higher terms:

Hence, the depression at the midpoint is 0.0106 m.