Question

One end of steel wire is fixed to a ceiling and a load of 2.5 kg is attached to the free end of the wire. Another identical wire is attached to the bottom of load and another load of 2.0 kg., is attached to the lower end of this wire. Compute the longitudinal strain produced in both the wires, if the cross-sectional area of wires is 10⁻⁴ m².

Answer 1

Given :

m1 = 2.5 kg,

m2 = 2 kg

A = 10⁻⁴ m²

Ysteel = 20 × 10¹⁰N/m²

Formula : Y = FL/ Al

For 1st wire, m = 2.5 + 2 = 4.5 kg

Y = FL/ Al = mgL/ Al [F = mg]  

∴  l/L = mg /YA

∴ Strain1 =  4.5x 9.8/ 20x 10 ¹⁰ x10⁻⁴

∴ Strain1 = 2.205 × 10⁻⁶

For 2nd wire, m2 = 2 kg

∴ Strain2 =  l2/L = m2 g /YA  

=  2.0x 9.8/ 20 x10¹⁰x 10⁻⁴

∴ Strain2 = 9.8 × 10⁻⁷

Answer 2

Answer:

A steel wire is loaded at one end and fixed with other end to a rigid roof of a house Draw a neat labelled diagram showing force acting on the load and wire