A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10^–2 cm²is 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.

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Solution

Length of the steel wire = 1.0 m

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

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:

Δl = (XO + OZ) – XZ

Where,

XO = OZ =

Let T be the tension in the wire.

∴mg = 2T cosθ

Using the figure, it can be written as:

Expanding the expression and eliminating the higher terms:

Young’s modulus of steel, Y =

Hence, the depression at the midpoint is 0.0106 m.

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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.

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