Question

A mine shaft has an ore elevator hung from a single braided cable of diameter 2.5 cm. Young's modulus of the cable is 10×1010 N/m2. When the cable is fully extended, the end of the cable is 700 m below the support.

Answer 1

Answer:

The cable would stretch 14 cm when loaded with 1000 kg ore.

Explanation:

The question is incomplete.

The complete question would be.

A mine shaft has an ore elevator hung from a single braided cable of diameter 2.5 cm. Young's modulus of elasticity of the cable is $10\times 10^{10}\ N/m^2$. When the cable is fully extended, the end of the cable is 700 m below the support.

How much does the fully extended cable stretch when 1000 kg of ore is loaded into the elevator?

Given the diameter of the cable is $2.5\ cm$. The length of the cable is $700\ m$.

And the mass of the ore is $1000\ kg$. Also the Young's modulus of  elasticity of the cable is $10\times 10^{10}\ N/m^2$.

We will use Hook's law

$\sigma=E\epsilon$

Where $\sigma$ is stress. E is Young's modulus of elasticity. And $\epsilon$ is strain.

We can rewrite .

$\frac{P}{A}=E\times \frac{\delta l}{l}$

Where $P$ is the applied force, $A$ is the area of the cross-section. $\delta l$ is the change in length. $l$ is the initial length of the cable.

Also, the applied force $P$ is due to mass of the ore. That would be $P=mg\\P=1000\times 9.81\ N$

Given diameter of the cable $(d)$ $2.5\ cm$.

$d=\frac{2.5}{100}=0.025\ m\\ A=\frac{\pi}{4}d^2\\ \\A=\frac{\pi}{4}(0.025)^2=4.91\times 10^{-4}\ m^2$

$E=10\times 10^{10}\ N/m^2$

$l=700\ m$

Plugging these values

$\frac{P}{A}=E\times \frac{\delta l}{l}$

$\frac{P}{A}\times \frac{l}{E}=\delta l \\\\ \delta l =\frac{1000\times 9.81\times 700}{4.91\times 10^{-4}\times 10\times 10^{10}} \\\delta l=.139\ m\\\delta l=14\ cm$.

So, the cable would stretch 14 cm when loaded with 1000 kg ore.