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Math Help - hydrostatic pressure

  1. #1
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    hydrostatic pressure

    An aquarium 8 m long, 5 m wide, and 4 m deep is full of water. Find the following:
    the hydrostatic pressure on the bottom of the aquarium 39200,
    the hydrostatic force on the bottom of the aquarium 568000 ,
    the hydrostatic force on one end of the aquarium?

    i got the first 2 but i cant get the force on 1 end

    the equation i got is integral from depth1 to d2of (density)(gravity)(length)x
    what i tried was integral for 0 to 4 of (1000)(9.8)(4)x
    but it did not work for me
    Last edited by dat1611; August 19th 2009 at 01:54 PM.
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  2. #2
    Member Mauritzvdworm's Avatar
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    try
    \int^{5}_{0}\int^{4}_{0}\rho gydydx
    because \rho gy is the pressure and to find the force you will need to consider one of the ends of the equarium and calculate the force and each differential area
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  3. #3
    Junior Member enjam's Avatar
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    I don't think you calculated the hydrostatic force at the bottom of the aquarium correctly.

    Water is 1000kg/m^3 = 9800N/m^3
    So the weight of the water in the tank is 9800N/m^3 x (160m^3)
    So F = 1568x10^3 N, this is the force that the bottom of the tank must oppose.
    Or, you could say P at the bottom = 39,200Pa
    Since P x A = F, we have 39,200 N / m^2 x 40 m^2 = 1568kN

    By 'one end', do you mean the side of the tank, or one of the corners?
    If you mean the side, it's simply F = rho x g x height to centroid
    = 1000 kg/m^3 x 9.8 m/s^2 x 2 m x (1 N / kg.m/s^2)
    = 19,600 N/m^2
    Multiply this by the area of the side, and you have
    19,600N/m^2 x either (4m x 5m) or (4m x 8m)
    and you will get a value for force.
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