An aluminum cylinder with a radius of 2.5 cmcm and a height of 82 cmcm is used as one leg of a workbench. The workbench pushes down on the cylinder with a force of 3.2×104N3.2×104N. What is the compressive strain of the cylinder? Young's modulus for aluminum is 7.0×1010Pa7.0×1010Pa. Express your answer using two significant figures.

Respuesta :

To solve the problem, it will be necessary to apply the concepts related to Young's Modulus, which defines the relationship between stress and strain in a body. This mathematical relationship is explained below

[tex]\text{Young Modulus} = \frac{\text{Stress}}{\text{Strain}}[/tex]

[tex]\upsilon = \frac{\sigma}{\epsilon}[/tex]

But here,

[tex]\sigma = \frac{F}{A} = \frac{F}{\pi r^2}[/tex]

Where,

A = Area

F = Force

r = Radius

In the formula of Young modulus we have then,

[tex]\upsilon= \dfrac{\frac{F}{A}}{\epsilon}[/tex]

Replacing,

[tex]7.0*10^{10} = \dfrac{\frac{3.2*10^4}{\pi (2.5*10^{-2})^2}}{\epsilon}[/tex]

[tex]\epsilon = \dfrac{\frac{3.2*10^4}{\pi (2.5*10^{-2})^2}}{7.0*10^{10}}[/tex]

[tex]\epsilon = 2.3*10^{-4}[/tex]

Therefore the strain is [tex]2.3*10^{-4}[/tex]