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Let's say I initially have an open, empty, soda can. I then turn it over and lower it into a bowl of water, and then release it. Obviously water rises to some level in the cup and then there is air at the top of the can, which was initially the bottom.

Are there only three forces on the can once it reaches equilibrium?

  1. Buoyant force from the water (up)

  2. Weight of the can itself (down)

  3. Air pressure inside the can acting on the water surface (down)

Let me know if all of these are right, otherwise my equation for total forces is...

F_b = mg + P where m is the mass of the can, and P is the air pressure in the cup

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  • $\begingroup$ Is this not a duplication of this? $\endgroup$
    – MoonKnight
    Commented Oct 27, 2011 at 21:30
  • $\begingroup$ Yes it is. Greg seems to have asked the question, started solving the problem, and asked another question when he had doubts about his results. $\endgroup$
    – Arnoques
    Commented Dec 5, 2011 at 13:21

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Maybe you can be a bit more clear in your formulation. Unless the soda can changes into something else, do not use the word 'initially'. Are cup and bowl the same in your experiment?

In small scale experiments like this(unlike the earths atmosphere) air does not really fall anywhere, and the pressure will be the same everywhere inside the can giving a net force of zero. You do not need it here, but remember that the force from a pressure is calculated by multiplying with the area. F=P*A

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  • $\begingroup$ All the problem statement says is that we have an empty can that is slowly lowered into water with the opening of the can facing down. I then have to calculate the air pressure inside, the depth the water rises and the height of the can that becomes submerged in water. But in order to find any of these, I just wanted somebody to clarify that I have all the forces and that I have them in the right direction, because I don't want someone to flat out tell me how to do it $\endgroup$
    – Greg Harrington
    Commented Oct 25, 2011 at 11:50
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    $\begingroup$ You need Archimedes law. I do not know if the can is left to float on its own(it will displace its weight in water), or if it pressed down till fully submerged(it will displace its volume in the water). $\endgroup$
    – hpekristiansen
    Commented Oct 25, 2011 at 14:20
  • $\begingroup$ I don't believe it is fully submerged because one of the things I need to do is find the depth the water rises in the can. so there is going to be a pocket of air in the can. All I want to know is which direction the buoyant force will act and which direction the air pressure will act. then I can do the rest on my own $\endgroup$
    – Greg Harrington
    Commented Oct 25, 2011 at 19:01
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    $\begingroup$ The air pressure will act on all surfaces in all directions, giving a net force of zero. Other forces, and their directions are given from Archimedes law. $\endgroup$
    – hpekristiansen
    Commented Oct 25, 2011 at 21:35
  • $\begingroup$ Alright, thanks a lot Hans. I was drawing the air pressure acting in all directions and that's what changed everything in the end. A lot of things cancelled out and now all is well. Thanks! $\endgroup$
    – Greg Harrington
    Commented Oct 26, 2011 at 20:27

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