Question

A thin, light wire 75.1 cm long having a circular cross section 0.555 mm in diameter has a 25.4 kg weight attached to it, causing it to stretch by 1.10 mm.
A) what is the stress of the wire.
B) what is the strain in the wire.
C) find the Young Modulus for the material of the wire.

Answer 1

Answer:

(a) 3.23×10⁸ N/m²

(b)  1.46×10⁻³

(c) 2.21×10¹¹ N/m²

Explanation:

(a) Stress = Force/Area.

Stress = F/A................ Equation 1

But,

F = mg................. Equation 2

Where m = mass, and g = acceleration due to gravity

A = πd²/4................. Equation 3

d = diameter of the circular cross section.

Substitute equation 2 and equation 3 into equation 1

Stress = 4mg/πd²............. Equation 4

Given: m = 25.4 kg, d = 0.555 mm = 0.000555 m

Constant: g = 9.8 m/s², π = 3.142

Substitute these values into equation 4

Stress = 4(25.4)(9.8)/(3.142×0.00555²)

Stress = 995.68/(3.08×10⁻⁶)

Stress = 3.23×10⁸ N/m²

(b)

Strain = ΔL/L.............. Equation 5

Where ΔL = extension, L = Length.

Given: ΔL = 1.1 mm = 0.0011 m, L = 75.1 cm = 0.751 m

Substitute into equation 5

Strain = 0.0011/0.751

Strain = 1.46×10⁻³.

(c)

Young modulus = Stress/Strain

Young modulus = 3.23×10⁸/ 1.46×10⁻³

Young modulus = 2.21×10¹¹ N/m²