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Suppose I have an upside down glass of water that I somehow brought in the configuration shown below (without any air between the glass and water). Now the water will obviously fall down but my question is why exactly?

The forces on the water column are $\rho Vg$ downwards due to gravity, $P_aA$ upwards due to the atmosphere and $F_g$ downwards due to the glass. If $P_aA>\rho Vg$ (which can be easily achieved by reducing $V$), $F_g$ could simply be equal to $P_aA-\rho Vg$ and the water would be in equilibrium and would have no reason to fall down.

enter image description here

Dale
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FullBridge
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6 Answers6

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You are absolutely correct that the picture shown is in equilibrium. The pressure at the top and sides need not be 0, but it will be less than the pressure on the bottom by the amount of the weight of the fluid. So the net force and net torque on the fluid is zero and there is no tendency to accelerate or rotate. This indeed means that the fluid in this configuration is in equilibrium.

There are two types of equilibrium: stable and unstable. Although this configuration is an equilibrium it is an unstable equilibrium. Specifically, this configuration is subject to Rayleigh-Taylor instability

Basically, if a small parcel of the water descends and is replaced by an equal volume of air going up, the potential energy of the system is reduced. This means that the system will not tend to return to the original configuration. So any deviation from the perfect equilibrium state will grow exponentially, regardless of how minuscule* it is initially.

Since there is always some small deviation, the fluid deforms, forms drops, and falls down as expected from common experience.

*Surface tension can actually stabilize very small deviations in some fluid interfaces.

Community
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Dale
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There is a variation of this experiment that demonstrates the very principle that you are asking about. A playing card is placed over the top of a filled test tube, which is then inverted. The atmospheric pressure against the card on the bottom provides enough force to push the card against the test tube, in spite of the force of gravity pulling in the downwards direction.

See this Bill Nye video demonstrating this at about 1 minute:

https://www.youtube.com/watch?v=QeAp3CuGjk8

lamplamp
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Because your diagram is incomplete, there is equal pressure in the water too. So pressure neither supports nor pulls the water from the glass.

Effectively, the only unbalanced forces are gravity, the water's own cohesion to itself(manifesting partly as surface tension), the water's adhesion to the surface of the glass, and the water's own structural rigidity.

Unfortunately that last item is almost nonexistent, as the water is a liquid. It deforms, and and slumps down under the force of gravity, thus falling out of the glass.

If you alter the parameters just a bit and greatly increase the structural strength of the water by freezing it, it will not slump. Then the fight is between the adhesion of the water to the glass and gravity, which is a more balanced fight.

PcMan
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Water *doesn't fall down if you use a very narrow glass. This principle is used in sucking up water with a pipette. so depending on the width of the glass, it water will fall down or stay inside. If the $A$ of your glass was very small it would stay inside. In this case, the surface tension of the water will keep it inside, because it's stronger than the force of gravity trying to pull it down. It would be interesting to see what would happen if liquid helium moved up a small tube. The helium would collect on the top. And then...?

Deschele Schilder
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Don't forget the pressure contribution at the top and the sides of the liquid. The only net force in your case is gravity.

The Bill Nye experiment where he pressures up the vessel and relieves the pressure at the top causing flow upward, is a very different case.

Bill Watts
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Assuming you hold the glass:

If the glass was cylindrical and if your water was actually ice, as the ice falls a tiny bit, ( assuming no air escapes from the sides) , a good vaccum inside and the atmospheric pressure on the outside will simply stop the ice from falling.

If the ice was inside the glass shaped as in the question, if ice falls a tiny bit, a gap gets created between the glass wall and the ice and air can go inside and stabilise the pressure, letting the ice fall only due to gravity.

Now replacing the ice with the water again in your question, the water has the shape of the trapezoid initially and if it falls a tiny bit due to gravity even, the vaccum temporarily created above will hold some layer of the water. Now, water has no ability to withstand shear forces and "breaks" along some plane and the remaining will fall.

The layer stuck above,I guess it will be more chaotic but finally will fall down from a repetetive process of vacuum creation, shearing and breaking.

Sidarth
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