Question

One end of steel wire is fixed to a ceiling and 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 as shown in the figure. Compute the longitudinal strain produced in both the wires, if the cross-sectional area of wires is 10^-4m², (Y_steel = 20 × 10¹⁰N/m²)

Answer 1

Given: M₁ = 2.5 kg, M₂ = 2kg, A = 10^-4m²,

Y_steel = 20 × 10¹⁰ N/m², L₁ = L₂ = L

To find: Longitudinal strain of 1st wire (Strain₁).

Longitudinal strain of 2^nd wire (Strain₂)

Formula: \(Y=\frac{FL}{Al}\)

Calculation: \(Y=\frac{MgL}{Al}\)

The longitudinal strain produced in 1^st wire is 1.225 × 10^-6 and in 2nd wire is 2.205 × 10^-6.