## Main Question or Discussion Point

Please tell me if I am understanding this correctly, or please point out where im going wrong....

imagine you have an open ended u-shaped tube filled with water. The pressure on left side of the U is the same as pressure at right side, because they are both open to atmospheric pressure pushing down on the water surface, right?

OK, so then you put a rubber stopped on the right side (the bottem of the stopped is in contact with water, so theres no "air pocket)

Am I correct in saying that the pressure on the left is greater than the pressure on the right, because the left has atm pressure pushing on surface, yet the right side is not open to atm, so it doensnt have atm pressure?
so pressure on left (zero depth) is 101325Pa, and pressure on right (zero depth) is approx zero. --a vacuum. Is that correct?

if that is correct then (here is what confuse me..):

... is there a force that pushes up against the stopper? If so, then there must be force from the stopper onto the water surface (3rd law force pair), and if that is true then there actually IS pressure on the stoppered end, exactly equal (because the U-shaped tube isnt accelerating obviously) to the atm pressure that pushes down on the left side (the open ended side).

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S_Happens
Gold Member
No, that is not correct. If you imagine various situations to put your example in, then I believe the ideas will be more clear. What will happen in your situation if you were to pull a vacuum on the unstoppered side (by isolating it as well with a vapor space and removing some vapor or decreasing the whole atmosphere, who cares), what would happen if you stoppered both sides without any vapor space (did you just create a vacuum tube), etc..

Other Hints- if you stoppered one side with a certain vapor space, would you use the height of that isolated side from liquid level to bottom of stopper to determine pressure at zero depth? Also, do you believe that a simple stopper in a u-shaped tube will cause a single side to become an isolated system (meaning it does not interact with the other side and therefore the atmosphere).

exactly equal
That isn't obvious. Have you looked up how barometers work?

S_Happens
Gold Member
I believe his misunderstanding stems from his deviation from a typical manometer setup (and ignoring the inital conditions when the stopper is installed), which would have one sealed end with a vapor space. I wouldn't really fault calling the pressures equal even in a typical barometer. Sure they would be inequivalent as actual local atmospheric pressure is changing, but the difference in pressure then manifests itself as a change in height (or needle movement) as the pressures near/reach equilibrium.

Also, do you believe that a simple stopper in a u-shaped tube will cause a single side to become an isolated system (meaning it does not interact with the other side and therefore the atmosphere).
no its never isolated because its a U-shaped tube. So are you saying that the atm pressure is transmitted up the other side of tube?
thanks

OK is this correct then, (using my same example)
the pressure on right (stoppered end) is same as pressure on left (open end) because the atm pressure from left side is transmitted equally throughout the liquid (pascals law), so there is upward force on right side of 101325Pa. But this force is countered by the force from the stopper (3rd law pair). So because the system is in equilibrium, the left side pressure is still equal to the right side pressure.
Is this correct?

It sounds like you're misinterpreting Pascal's law, which does not say that the pressure is equal at all heights inside the fluid (rather than at equal heights... you understand how instead of Pa we could measure pressure in mmHg, or centimetres of water?).. at which heights are you proposing to measure the pressure?

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Pascal's Principle: "A pressure change applied to a confined fluid is transmitted undimished to every point in the fluid and the walls of the container"

there is a downward force on the surface of the water from the atmosphere.
Therefore this change is transmitted everywhere undimished in the fluid.
Is that correct?

What change?

Anyway, say your U-tube is 35m high and completely filled (and sealed) with water on the right side but the (open) left side is only filled for the first 25 metres. On the right side, from 25 to 35 metres is just air (1 atm), but as you descend the pressure increases with the weight of the water above (reaching 3.5 atm at the bottom right). On the left the pressure decreases as you go up: from 3.5atm at the bottom, 1atm 25m up (equal with the fluid level on the right), to approximately zero pressure at the very top (and if you were to extend that side to 40m then you'd have approximately 5m of vacuum/steam at the top.. in fact, you could use fluctuations in this length to measure the weather, as you ought to know by now). The unbalanced pressure on the cork causes no acceleration because is is jammed into the tube (and of course there is oppositely unbalanced external atmospheric pressure on the bottom of the U. There is also unbalanced internal pressure on the bottom of the U, and this is why the tube won't hold itself up).

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