An aluminum cylinder with a radius of 2.5 cmcm and a height of 82 cmcm is used as one leg of a workbench. The workbench pushes down on the c

Question

An aluminum cylinder with a radius of 2.5 cmcm and a height of 82 cmcm is used as one leg of a workbench. The workbench pushes down on the cylinder with a force of 3.2×104N3.2×104N. What is the compressive strain of the cylinder? Young’s modulus for aluminum is 7.0×1010Pa7.0×1010Pa. Express your answer using two significant figures.

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Nho 5 years 2021-07-14T14:42:06+00:00 1 Answers 17 views 0

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  1. minhkhang
    0
    2021-07-14T14:43:21+00:00
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    To solve the problem, it will be necessary to apply the concepts related to Young’s Modulus, which defines the relationship between stress and strain in a body. This mathematical relationship is explained below

    \text{Young Modulus} = \frac{\text{Stress}}{\text{Strain}}

    \upsilon = \frac{\sigma}{\epsilon}

    But here,

    \sigma = \frac{F}{A} = \frac{F}{\pi r^2}

    Where,

    A = Area

    F = Force

    r = Radius

    In the formula of Young modulus we have then,

    \upsilon= \dfrac{\frac{F}{A}}{\epsilon}

    Replacing,

    7.0*10^{10} = \dfrac{\frac{3.2*10^4}{\pi (2.5*10^{-2})^2}}{\epsilon}

    \epsilon = \dfrac{\frac{3.2*10^4}{\pi (2.5*10^{-2})^2}}{7.0*10^{10}}

    \epsilon = 2.3*10^{-4}

    Therefore the strain is 2.3*10^{-4}

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