# Atmospheric pressure In FBD doubt

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

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.

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.

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?

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.
• 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 $m_{water}g$ – user118752 Oct 12 '16 at 14:55