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Suppose we hang a heavy ball with a mass 13 kg (so the weight is ) from a steel wire 3.9 m long that is 3.1 mm in diameter (radius = 1.55×10-3 m). Steel is very stiff, and Young's modulus for steel is unusually large, 2 × 1011 N/m2. Calculate the stretch of the steel wire.

Respuesta :

abidemiokin

Answer:

1.635×10^-3m

Explanation:

Young modulus is the ratio of the tensile stress of a material to its tensile strain.

Young modulus = Tensile stress/tensile strain

Tensile stress = Force/Area

Given force = 130N

Area = Πr² = Π×(1.55×10^-3)²

Area = 4.87×10^-6m²

Tensile stress = 130/4.87×10^-6 = 8.39×10^7N/m²

Tensile strain = extension/original length

Tensile strain = e/3.9

Substituting in the young modulus formula given young modulus to be 2×10¹¹N/m²

2×10¹¹N/m² = 8.39×10^7/{e/3.9)}

2×10¹¹ = (8.39×10^7×3.9)/e

2×10¹¹e = 3.27×10^8

e = 3.27×10^8/2×10¹¹

e = 1.635×10^-3m

The stretch of the steel wire will be

1.635×10^-3m

boffeemadrid

The extension of the steel wire is required.

The stretch of the wire is 0.00033 m.

Young's modulus

m = Mass of ball = 13 kg

L = Initial length = 3.9 m

r = Radius = [tex]1.55\times 10^{-3}\ \text{m}[/tex]

Y = Young's modulus = [tex]2\times 10^{11}\ \text{N/m}^2[/tex]

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

A = Area = [tex]\pi r^2[/tex]

The elongation of a material is derived from the Hooke's law.

Elongation is given by

[tex]\Delta L=\dfrac{FL}{AY}\\ =\dfrac{mgL}{\pi r^2Y}\\ =\dfrac{13\times 9.81\times 3.9}{\pi (1.55\times 10^{-3})^2\times 2\times 10^{11}}\\ =0.00033\ \text{m}[/tex]

Learn more about Young's modulus:

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