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Let us say I have a beaker floating in a water tank. Suddenly I put some water in the beaker also. The figure here shows the exact situation enter image description here

Now I want to draw the FBD of the beaker. I am somehow confused with the role of the atmospheric pressure in drawing the FBD. I thought of two ways of drawing this and I am just unable to figure out which one is right.

enter image description here

Here since the beaker is floating an $F_{buoyant}$ is there in the upward direction which the net mass (beaker+water in beaker) would put its weight downward.

But while solving one question I came to know that total force that a liquid exerts on the bottom of the container is $P_0+m_{liquid}g$ ($P_0$ is the atmospheric pressure). So should the downward force on the beaker be $(M_{beaker}+M_{water in beaker})g+P_0 S$ (S being the base area of the beaker) as in the second diagram down here.enter image description here

enter image description here

Here the $P_0S$ due to atmospher on the tube gets cancelled by the $P_0S$ exerted by the water in tank on the tube. But what about the $P_0S$ from the water present in the tube? Do the FBD's apply like the one in the new image?

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You get the same answer either way. In the second case, you get$$(M_{beaker}+M_{water})g+P_0S=F_B+P_0S$$In the first case, you get $$(M_{beaker}+M_{water})g=F_B$$The key to this is that the buoyant force $F_B$ omits the effect of atmospheric pressure because it always cancels.

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  • $\begingroup$ And what about the total force that the water in the beaker exerts on the beaker. Is it $P_0S+m_{water}g$ or is it just $m_{water}g$ $\endgroup$ – user118752 Oct 12 '16 at 14:55
  • $\begingroup$ I have edited my question to clarify my doubt. $\endgroup$ – user118752 Oct 12 '16 at 16:32
  • $\begingroup$ What we are talking about here is the difference between using "gauge pressure" and using "absolute pressure." For a problem like this, it doesn't matter whether you include the gas pressure or not. Both methods give exactly the same result, as you can see from the equations I wrote. The only time it is necessary to work with absolute pressure is when some of the gas is being compressed within the system. Google "gauge pressure" and see the discussion. $\endgroup$ – Chet Miller Oct 12 '16 at 17:45

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