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

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.

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Gia Bảo 3 years 2021-09-03T08:59:25+00:00 1 Answers 170 views 0

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  1. thuthao
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    2021-09-03T09:01:19+00:00
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    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.

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