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Name: Steve
Status: educator
Age: 40s
Location: N/A
Country: N/A
Date: 1999 


Question:
When a glass is filled with water and a plate is placed on top and then it is turned upside down, what holds the plate to the glass? Is it the surrounding air pressure of about 15 psi? Does the surface tension of the water have anything to do with it? In order for the water to fall out, something would have to replace it, like air, right? With the water acting like a seal between the glass and the plate, air can't get in.


Replies:
Sounds to me that you've done a good job of explaining this phenomenon. If the glass were very tall, so that the water pressure at the bottom were higher than the surrounding air pressure, this little demonstration wouldn't work. The water pushing down would just force the plate with it. Fortunately, the glass would have to be about 32 feet deep for this to occur at sea level.

Richard Barrans Jr., Ph.D.


Hello,

Interesting question. I actually have done this years ago but never thought of why! Anyway, here is what I think.

Once your turn a half full glass upside down, the weight of water tends to increase the volume of air in the glass creating a relative "vacuum" that holds the water. The key for this to work is for water to slip down a little bit to create that "vacuum" . But how much does it have to slip? A lot or a little? If a large displacement is needed, then it will not work as water will pour out.

So, the question is this: How much displacement of water is needed to create the necessary pressure difference to hold the water. Let's see.

The pressure created by the weight of water is = p=LRg What sort of volume change in air would this change in pressure make?

Ideal gas law (applied to air): P1.V1 = P2.V2
P1.V1 = (P1+p).(V1+v)

To a first approximation:
p/P1=-v/V1
where:

P1 = atmospheric pressure
V1 = air volume in the glass without water effect
P2= pressure in the trapped air in glass (with water effect)
V2=volume of the trapped air (with water effect)
p=water column pressure
v=change in volume of the trapped air
L=water column height
R=water density
g=gravitational acceleration

Assume that the glass is cylindrical so that it has a constant cross-sectional area A

p/P1=-v/V1=Ah/AH=h/H

where H is the height of unexpended trapped air and h is the height of the expansion under water column. What we need is the slippage height h.

Let us put some number in. Let's assume that the water column in the glass is about 50 mm high.

P = Atmospheric pressure 100,000 Pa
p =height x g x density=0.05 [m]x 9.8 [m/s2]x 1000 [kg/m3]=490 Pa

Let's also assume that the air column is 20 mm. Then, h=Hp/P =0.02 [m] x 490 [Pa]/100,000 [Pa]=0.1 mm!

In order words, if water slips down by 0.1 mm, it expands the air volume enough to balance the weight of water. I think if one plays with these equations and pick the right combinations of the variable one can make interesting demonstrations and measurements. One other point: You must prevent air from being sucked into the glass. That is the purpose of the plate you use. The plate cannot be too heavy of course. If it is too heavy, the combination of surface tension and differential pressure that holds the water/plate combination can be breached. Other properties such as surface tension, viscosity, etc. will be involved but I think the essential physics is what pressure difference. And I hope I have the correct explanation!

Good luck,

AK

Dr. Ali Khounsary
Advanced Photon Source
Argonne National Laboratory


I think you've got it. And yes, the surface tension does have something to do with it. If there were no surface tension the air could easily enter into the glass. I think you also need to have a wettable surface in order to get a good seal. You might try it with a DuPont's Teflon (TM) sheet or some other surface that doesn't wet and see if it works as well.

Bradburn



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