Respuesta :
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 [tex]10\times 10^{10}\ N/m^2[/tex]. 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 [tex]2.5\ cm[/tex]. The length of the cable is [tex]700\ m[/tex].
And the mass of the ore is [tex]1000\ kg[/tex]. Also the Young's modulus of elasticity of the cable is [tex]10\times 10^{10}\ N/m^2[/tex].
We will use Hook's law
[tex]\sigma=E\epsilon[/tex]
Where [tex]\sigma[/tex] is stress. E is Young's modulus of elasticity. And [tex]\epsilon[/tex] is strain.
We can rewrite .
[tex]\frac{P}{A}=E\times \frac{\delta l}{l}[/tex]
Where [tex]P[/tex] is the applied force, [tex]A[/tex] is the area of the cross-section. [tex]\delta l[/tex] is the change in length. [tex]l[/tex] is the initial length of the cable.
Also, the applied force [tex]P[/tex] is due to mass of the ore. That would be [tex]P=mg\\P=1000\times 9.81\ N[/tex]
Given diameter of the cable [tex](d)[/tex] [tex]2.5\ cm[/tex].
[tex]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[/tex]
[tex]E=10\times 10^{10}\ N/m^2[/tex]
[tex]l=700\ m[/tex]
Plugging these values
[tex]\frac{P}{A}=E\times \frac{\delta l}{l}[/tex]
[tex]\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[/tex].
So, the cable would stretch 14 cm when loaded with 1000 kg ore.
