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Cup in a bathtub

  1. Nov 14, 2005 #1
    I was looking for information on calculating buoyancy of an upside-down cup in a bathtub...

    If I place a plastic cup upside-down in a bathtub and press down, the water goes up into the cup a certain distance depending on the force I press down. Does this mean that the pressure of the air in the cup is equal to the force I am pressing down with? Is there a way to calculate what the pressure of that air is depending on the force?

    I was doing some kitchen physics stuff with my nephew, and I just need a pointer to the proper equations or online tutorials. I found plenty of stuff for calculating buoyancy of floating cups (or floating anything) but nothing about air pressure.

    Thank you very much!
  2. jcsd
  3. Nov 14, 2005 #2
    the total force that u are exerting will be equal to the buoyant force + the force exerted by the pressurised air inside the cup. the buoyant force wil be propotional to volume of CUP(not the inside hollow),ie the priphery which is submerged in the liquid and the force exerted by the air inside will be propotional to its compression.
  4. Nov 15, 2005 #3


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    The pressure gradient within the air bubble will be negligible compared with the pressure gradient of the water outside the cup so, essentially, the air pressure inside the cup will be the same as the water pressure at the air-water interface.

    The force you exert will be the buoyant force minus the weight of the cup.
  5. Nov 15, 2005 #4
    Okay, just a little confused...

    I know that the buoyant force is F=dgv but I am confused as to why I would not use the air inside the cup when calculating volume.

    Let's say I use plastic wrap to seal the mouth of the cup, and perform the same experiment. Wouldn't you then use the air inside the cup when calculating average density and volume?

    For example, the cup is cylindrical and is 16cm high and has a radius of 3cm. The walls of the cup are .5cm thick and it is made out of glass. The total volume is V = pi * r^2 * h = (3.1416)(9)(16) = 452.3904cm^3

    I seal the mouth of the cup with plastic wrap, and push down on it until the entire cup is underwater and then hold it just under the surface. I am pressing down with:
    F = dgv = (1kg/m^3)(9.80m/s^2)(4.524m^3) = 44.335N

    First, do those calculations look correct? Second, why would they change if I removed the plastic seal from the cup and repeated the experiment? Would the air inside the cup offer an additional force on the water?

    Thank you all for your help so far!
  6. Nov 15, 2005 #5


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    Of course, the air inside matters. My point was that the pressure of the air inside the cup is essentially uniform.

    However, the air does displace water (providing the buoyant force) and height of the air column inside the cup will depend on how deeply immersed the cup is in the water. Air is compressible but water is not. When you place plastic wrap over the opening, you assure that the volume of the air inside will be greater and so the buoyant force will be greater. Otherwise, water rises inside the cup up to the point at which water and air pressure balance.
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