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.
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
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