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Norma-Jean [14]
3 years ago
12

An aluminum cylinder with a radius of 2.7 cm and a height of 67 cm is used as one leg of a workbench. The workbench pushes down

on the cylinder with a force of 3.2×104N. What is the compressive strain of the cylinder? Young's modulus for aluminum is 7.0×1010Pa. Express your answer using two significant figures. Compressive strain = nothing
By what distance does the cylinder's height decrease as a result of the forces on it?
Physics
1 answer:
soldier1979 [14.2K]3 years ago
3 0

Answer:

1.9\times 10^{-4}

1.2\times 10^{-4}\ m

Explanation:

r = Radius = 2.7 cm

F = Force = 3.2\times 10^4\ N

A = Area = \pi r^2

\sigma = Stress = \frac{F}{A}

E = Young's modulus = 7\times 10^{10}\ Pa

\epsilon = Strain

L_0 = Original length = 67 cm

\Delta L = Change in length

Young's modulus is given by

E=\frac{\sigma}{\epsilon}\\\Rightarrow \epsilon=\frac{\sigma}{E}\\\Rightarrow \epsilon=\frac{\frac{3.2\times 10^4}{\pi 0.027^2}}{7\times 10^{10}}\\\Rightarrow \epsilon=0.0001996=1.9\times 10^{-4}

Strain is 1.9\times 10^{-4}

Strain is given by

\epsilon=\frac{\Delta L}{L_0}\\\Rightarrow \Delta L=\epsilon\times L_0\\\Rightarrow \Delta L=1.9\times 10^{-4}\times 0.67\\\Rightarrow \Delta L=0.0001273\\\Rightarrow \Delta L=1.2\times 10^{-4}\ m

The cylinder height decreases by 1.2\times 10^{-4}\ m

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