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.

Question

Philoid · Accepted Answer

Now here when a mass is suspended at midpoint of the stretched wire, the wire will elongate i.e. its size will increase due to which a depression will be there on the midpoint, and originally horizontal wire will make some angle with the horizontal

let us say the original length of wire is

l = 1m

Mass suspended is

m = 100g = 100 × 10^-3 Kg = 0.1 Kg

area of cross section of wire is

A = 0.50 × 10^-2 cm² = 0.50 × 10^-6 m² (since 1 cm² = 10^-4 m²)

Let the depression in the middle of wire be x

After elongation let the new length of wire be l’

And angle made by wire with horizontal on both sides be

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.

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.