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Math Help - Integration and forces

  1. #1
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    Integration and forces

    the question tells us that the buoyant force is equal to the weight of the fluid the object displaces



    po = object density
    pf = fluid density
    g = acceleration due to gravity
    A(y) = cross-sectional area of object
    h = height of object underwater
    L = total height of object
    F = buoyant force
    W = weight

    and then

    F = pf*g * integral from -h to 0 of A(y) dy

    and

    W = po*g * integral from -h to (L-h) of A(y) dy

    show that the percentage of the volume of the object above the surface of the liquid is

    100 * (pf - po) / pf
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  2. #2
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    Volumes and integration

    Hello razorfever
    Quote Originally Posted by razorfever View Post
    the question tells us that the buoyant force is equal to the weight of the fluid the object displaces



    po = object density
    pf = fluid density
    g = acceleration due to gravity
    A(y) = cross-sectional area of object
    h = height of object underwater
    L = total height of object
    F = buoyant force
    W = weight

    and then

    F = pf*g * integral from -h to 0 of A(y) dy

    and

    W = po*g * integral from -h to (L-h) of A(y) dy

    show that the percentage of the volume of the object above the surface of the liquid is

    100 * (pf - po) / pf
    First, note that, since the body is floating in equilibrium:

    F = W (1)

    Then let the volume of the body above water be V_a, and the total volume V.

    Then V = \frac{W}{\rho_og}

    and V_a = \int_0^{L-h}A(y)dy

    = \int_{-h}^{L-h}A(y)dy -\int_{-h}^0A(y)dy

    = \frac{W}{\rho_og} - \frac{F}{\rho_fg}

    = \frac{W}{g}\left(\frac{\rho_f-\rho_o}{\rho_o\rho_fg}\right), from (1)

    Can you complete it now?

    Grandad
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