# Why should water fall down from an upside down glass of water?

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.

• Did you mean for your cup to be cylindrical and not tapered? Because reducing V would decrease A. – DKNguyen May 28 at 5:19
• @DKNguyen Yes that or the angle of slope can be considered very small. – FullBridge May 28 at 6:08
• Is the glass allowed to free with the water or will you hold it? In case you hold it, there won't be any F_g – Sidarth May 28 at 10:24
• @Sidarth Why would there not be any $F_g$ if I hold it? – FullBridge May 28 at 10:40
• I have to agree that pressure from the air is only on one side of the water. I started thinking of your water as solid ice, making it better to think about. In case the glass was a cylinder then as the ice block block falls, the vaccum suction in the top will prevent the ice from falling down. That is certainly true. – Sidarth May 28 at 10:53

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.

• Here's a very interesting demonstration of this very phenomenon, including the weird effects of artifically stabilizing that equilibrium: youtube.com/watch?v=gMAKamGIiMc – João Mendes May 28 at 15:09
• @JoãoMendes yes, studying the paper about that video is exactly where I first learned about the Rayleigh Taylor instability. What a bizarre phenomenon! See: physics.stackexchange.com/questions/580651/… – Dale May 28 at 15:16
• "Surface tension can actually stabilize very small deviations in some fluid interfaces." Readily seen when the entire setup is much smaller, such as in a drinking straw with the top opening sealed with a finger. – Arthur May 28 at 19:30
• I've recently 3D-printed an animal-watering station for my garden, using a soda-bottle as the reservoir. I was amazed at how large I had to make the opening of the nozzle to stop surface tension winning (about 20mm, if anyone is wondering) – SiHa May 31 at 10:18

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:

• I think the question is why does this not happen without a card . Isnt the pressure on the card same as the pressure on the surface of water ? – silverrahul May 28 at 6:22
• @silverrahul Even if it worked the surface of the water is really malleable and imperfect compared to the card and I would imagine pressure variations and imperfections in the water surface would quickly cause concaves in the water which would just get worse and get behind some water, isolating making it fall away and it would just repeat until everything had fallen away. And supposing it did work, I don't see how it being water is any different than say, a block of ice which we know just falls out. – DKNguyen May 28 at 6:35
• My guess about ice vs water would be that, ice is more likely to have small gaps along walls, whereas water is more airtight. I dont think, the bill nye trick would have worked if it was ice instead of water. Unless, it was due to ice being stuck frozen, in which case obviously, no card would have been needed because air pressure would not be what is causing it to stick – silverrahul May 28 at 7:00
• @DKNguyen: If you turn an ice-cube tray upside down, the ice doesn't fall out even if you shake it. Because it's frozen to the tray. If you made a plug of ice with an air-tight seal in a test tube, I'd certainly expect it to be held in by air pressure. The seal could be a layer of melt-water. (Assuming the ring of water is small enough for surface tension to stably seal against the weight of the ice plug). Obviously if you just put some loose ice cubes in a tube, you can dump them out easily, so the question is how tight a fit it is. An airtight metal plug would equally stay in place. – Peter Cordes May 29 at 5:16
• @PeterCordes Now that you mention it, if you plug up a syringe with your finger and pull it on the handle it really won't want to come out. Even though the friction in the syringe is more than enough to overcome gravity, you pull it on it much harder than gravity would. – DKNguyen May 29 at 5:58

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.

• What do you mean by "equal pressure in the water"? – FullBridge May 28 at 6:10
• @FullBridge Take a tennis ball in your hand. Squeeze down on it. Your hand is the "air", pressing down on the ball which is the "water". Feel how the ball is pressing BACK at you? Same thing. Newton's laws.. If you apply pressure on something, and it cannot move away, it simply has to press back with equal force. This produces pressure in the object. – PcMan May 28 at 6:15
• I agree, and what I'm saying is that the water pushes on the glass with a force $F_g$ and the water is pushed back with the same force. – FullBridge May 28 at 6:53

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...?

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.

• But that pressure would be exerted on the glass right? What if the glass is fixed through some rigid support? – FullBridge May 28 at 6:05
• How the glass is supported is irrelevant. I am assuming the glass is stationary the way your case is stated. Think about it. Why would you think there is zero pressure at the top of the liquid? Consider a small particle of liquid in the liquid volume somewhere. Pressure acts on it in every direction and the net effect is zero force. The only net force on that particle is gravity. – Bill Watts May 28 at 6:19
• I didn't assume zero pressure at the top, in that case $F_g$ would be $0$. I think you're saying that the pressure at the top should be atmospheric pressure, but why should that be the case? Why can't it be $\frac{P_aA-\rho Vg}{A}$? – FullBridge May 28 at 6:42
• But then the drop in pressure associated with some height $\Delta h$ would be exactly equal to the amount required to hold the water above it and maintain equilibrium. – FullBridge May 28 at 8:36
• Actually the variation with depth is for the liquid sitting on the bottom where the pressure is determined by the weight of the fluid above it. In the case where the fluid is at the top, it will be in free fall at first, and the pressure will be atmospheric all the way around it. As it accelerates, air drag will become significant, causing a pressure difference between top and bottom, but the liquid will still fall. – Bill Watts May 28 at 17:51

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.