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

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

Step-by-step 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* 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* 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 .

`(P)/(A)=E* (\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* 9.81\ N`

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

`d=(2.5)/(100)=0.025\ m\\ A=(\pi)/(4)d^2\\ \\A=(\pi)/(4)(0.025)^2=4.91* 10^(-4)\ m^2`

`E=10* 10^(10)\ N/m^2`

`l=700\ m`

Plugging these values

`(P)/(A)=E* (\delta l)/(l)`

`(P)/(A)* (l)/(E)=\delta l \\\\ \delta l =(1000* 9.81* 700)/(4.91* 10^(-4)* 10* 10^(10)) \\\delta l=.139\ m\\\delta l=14\ cm`.

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