# About the real reason why paper doesn't fall in HALF-FILLED inverted cup [duplicate]

I just happen to see a show about air pressure host by Brian Cox. As the link:http://imgur.com/EsRadFi shows, he flipped a half-filled water cup upside down then the paper on cup doesn't drop:

It puzzled me, because the cup is half-filled, the pressure shouldn't balance since inside water pressure+atmosphere pressure should be bigger than outside atmosphere pressure only.

All the following answers seems not right, now I came the answer with myself: Treat the air inside as gas of a certain density. It's then obvious that the pressure of inside of the paper always equal to the outside unless the pressure exerted by gravity of {water+gas} system beyond the outside atmosphere pressure. The key is that inside is not {water pressure+atmosphere pressure} but {1: water pressure+ 2: small air pressure by inside air's own gravity+ 3: big atmosphere pressure-(1.+2.)}=atmosphere pressure. The third force came from the top inside of the flipped cup. So the paper wouldn't drop until (1.+2.)> atmosphere pressure, even when the paper is rigid body, in constrast to Steeven or others' result.

## marked as duplicate by Bill N, John Duffield, user10851, Gert, ACuriousMind♦Nov 13 '15 at 2:22

• Please consider choosing a title that actually reflects what the question is about. See also our meta post on good question titles. – ACuriousMind Nov 12 '15 at 16:59
• Why would there be atmospheric pressure inside the cup? – Steeven Nov 12 '15 at 17:20
• @Steeven because the glass is initially half filled, so the air inside just moves to the top of the glass, I do not see how a vacuum could be generated – user83548 Nov 12 '15 at 17:28
• I did a ball park estimate and got that a change in h (a bulge at the bottom) six orders of magnitude smaller than h results in a change in pressure enough to counter the weight (I did not post it because I did not check if the result is correct) – user83548 Nov 12 '15 at 18:31

When the cup is turned upside down, the water wants to fall out. The air-filled cavity is therefore stretched a bit as the gravity pulls down the water. The air expands a bit. This reduces the air pressure inside the cup, since increasing volume reduces pressure. This is hinted in the ideal gas equation:

$$pV=nRT$$

Soon this lower pressure pulls upwards with the same force as the weight of the water pulls downwards. The water is now kept in place and the pressure inside is lower than atmospheric pressure outside.

If you disturb the water by shaking the cup, suddenly different forces work on different parts of the water surface as the water level is not constant. If the water amount at any point is displaced enough for a breach to appear so outside air comes in, then atmospharic pressure is restored inside the cup, the water is not kept up anymore and it falls out.

• so if you use a rigid material instead of paper it will not work because the volume of water, and thus of the air will not change? – user83548 Nov 12 '15 at 17:43
• @brucesmitherson It should be possible - it doesn't move much. Also the water doesn't necessarily need the paper under it. That is simply to avoid disturbance in the water. So if the gab made from putting some other plane but rigid material under it, just has to be small enough to still avoid the water from "changing shape" – Steeven Nov 12 '15 at 17:51
• Please see my answers in my modified post. Now I think the reason is clear. The changing shape of the paper doens't matter. – Curio Nov 13 '15 at 3:46
• Who talked about a change of paper shape? In order to prevent the volume to increase, the pressure is falling according the the ideal-gas equation. This is the case without any air in the cup and it is the case with air in the cup. Also I do not clearly understand your solution that you wrote in your question - could you write it out more clearly? – Steeven Nov 13 '15 at 8:28

How does air pressure on a paper sheet hold water in an half-filled inverted cup?

A simplified model of the experiment you have seen is illustrated in the attached picture. You initially have a piston with a certain quantity of air at 1 atm inside and a piston of mass m and surface S somewhere in the cylinder. Then you let the piston fall. It will reach a static equilibrium when:

$P_{AirCylinder}*S + mg=1 atm * S$

Solving the equation you get:

$P_{AirCylinder} = 1 atm - mg/S$

• It's different case. In your case the air is in top so the volume can be stretched which decrease the air pressure. However in the water cup case the air is in bottom so if the paper is rigid as piston, the air pressure will increase thus would cause falling. – Curio Nov 13 '15 at 0:13
• In my case the black piston symbolizes the mass of water not the paper. The paper can be taken as the lower side (the base) of the piston. Regarding the second part of your comment, the water has to be in contact with the paper. It is not possible to have paper - air - water when the glass is turned upside down. – Energizer777 Nov 13 '15 at 0:31
• the paper-air-water setup is possible, the picture imgur.com/EsRadFi already showed this. For the reasoning please see Steeven's and my comments. – Curio Nov 13 '15 at 1:35
• Curio, I magnified your pictures and the air is clearly above the water when the glass is turned upside down. You simply have not studied the pictures carefully. – Energizer777 Nov 13 '15 at 3:36
• Please see my answers in my modified post. Now I think the reason is clear. Just think the air inside as certain fluid or gas, even can be with somewhat higher density than air the paper wouldn't drop. The the paper-air-water setup is possible. – Curio Nov 13 '15 at 3:49