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Magnitude of a Force on a hatch in a Vessel.

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data

    We're told the density of ever material involved here, the liquid in the vessel, called "LAB", the water on the outside, and the acryllic that the vessed is made of.
    LAB: 860 kg/m^3
    Water: 1000 kg/m^3
    Acryllic: 1185kg/m^3
    Also, the wall thickness of the vessel is 5.4 cm. The height of the neck is labeled delta h, we're also told to not take the pressure difference between the inside and the outside, due to the 5.4cm wall.

    This is a three part question, but I'm only looking for help on the first part, which asks us to find the force on the hatch, which is 1.00m^2 big, when delta h = 0.

    2. Relevant equations
    We need to use Pressure = Force/Area for this, solving for force gives us
    Force = Pressure*Area
    density * gravity * height = pressure

    3. The attempt at a solution

    My attempt to solve the question, was to solve for the value of pressure of the inside of the system, and then use that to solve for force. Since the height of liquid in the neck is 0, we have no liquid in the neck, so we only have to consider whats inside the circular area, which has a height of 12m, density of 860kg/m^3, and gravity is 9.81 m/s^2, so which we have to add the pressure of 1 atm due to the air at the top, 1.01e-5 Pa.

    Which gives us: 101239 Pa * 1.00 m^2, so the answer is 1.01e5 N?

    Edit: 1 atm is 101kPa, making it 202564 Pa * 1.00 m^2, to be 2.03e5 N.
    Last edited: Feb 10, 2009
  2. jcsd
  3. Feb 10, 2009 #2


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    Isn't air pressure about 100 kilopascals?

    The wording of the question is bothering me. Are you sure you are to ignore the force of the water pressure on the outside?
  4. Feb 10, 2009 #3
    I'm not 100% sure about the air pressure, so you're probably right, as for the second part, i'll quote it exactly from the question.

    "treat the vessel as a thin shell, ie, do not take the pressure difference between the inside and outside due to the 5.4 cm wall thickness into account"
  5. Feb 10, 2009 #4


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    I think this just means that you take the radius to be 6 m whether you are doing the inside or the outside.

    What exactly is the question? If it says find the total or net force on the hatch, you would take the difference between inside and outside forces.
  6. Feb 10, 2009 #5
    We're looking to find the magnitude of the force on the hatch. so looks like i should throw the pressure of water into the mix, too.
  7. Feb 10, 2009 #6


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    I think so. That is the force you would be interested in if you were thinking of opening the hatch!
  8. Feb 10, 2009 #7
    Alright, thanks alot delphi :D
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