Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to make a home experiment where I roughly explain the phenomenon of water remaining in a straw if you close one end of the straw.

So I'm thinking that the weight of the water is pulling it down, but the pressure underneath the straw is keeping it up. If there was no finger on the top, the pressure would also be pushing it down so it would fall.

Would this be a valid way to show this?

Mass of water: volume*density: $\pi r^2*H*1000$

Weight of water: $\pi r^2*H*1000*9.8$

Pressure*area = force

force= $101325*\pi r^2$ (weight of water)

And I somehow want to combine this with the upward force of pressure in the room.

Am I on the right track?

share|cite|improve this question
You need to take into account viscosity as well. – Jiminion Mar 3 '15 at 3:31
up vote 1 down vote accepted

Yes, what you have formulated is fine. The pressure acting on the water from the bottom of the straw will be equal to the weight of the water times the cross section area. So $101325\pi r^2$ is the force acting from below also. That is precisely why there is an equilibrium and the water is not falling.

share|cite|improve this answer
That is because the cross section area increases, which is inversely proportional to pressure. So basically, the pressure from below is no longer strong enough to hold the pressure from the weight of the water. – user42733 Mar 31 '14 at 13:01
Ahm, this is wrong. The pressure from the top is not $0$, why would it be? Indeed for a short straw the pressure from top will be almost equal to the pressure from bottom, only slightly less to counter the weight of the water. (Of course, you can argue with relative pressures alone, simply gauging the pressure on top to $0$, but then please make that explicit.) – leftaroundabout Mar 31 '14 at 13:08
The pressure from the water is constant with an increase in diameter, it stays at $H*1000*9.8$and the pressure from the air stays at $101325$ in this example. The water does not fall out when the diameter increases because the air cannot support the water. It falls out because the adhesion of water to the straw allows it to creep downwards along the sides, and a wider diameter lets it get to a point where bubbles can then propagate through the water in the straw. Once this starts happening, the pressure above equalizes with that below and the water falls – Jim Mar 31 '14 at 13:21
additionally, with larger diameter, you can have more fluctuations over the air-water surface, if water ripples and starts to clump somewhere, that increases the pressure in one spot while decreasing it somewhere else. This is a positive feedback system, the result being that the water can pool to the center (or somewhere else) and the air sidles around it to the top of the straw... Water falls – Jim Mar 31 '14 at 13:27
@Parth: Forces are not equal at top and bottom, you have the weight of the water column as a difference. I maintain that your explanation is misleading. – Joce Apr 2 '14 at 9:44

This is not altogether correct. In particular, it does not matter what the pressure of the room is (unless you reach extreme values), because actually what is holding the water column is not some "high" value of the pressure at the bottom of it, it is the fact that the pressure at the top of the water column (at the water-finger interface, or in the air pocket between water and finger) is lower than the atmospheric pressure. It is exactly the room pressure minus $\rho g h$, where $\rho$ is the density of water, $g$ gravitational constant and $h$ the height of the column.

This is true whatever the radius of the straw. However, you need two more ingredients to keep the system stable :

First, the air pocket should remain incompressible : this is true for small enough water columns, but there'll be some height $h$ above which it will expand under the weight of water. And even if there is no air, if you get to pressures lower than 0 at the top, the phenomenon of cavitation will nucleate a bubble of void that will expand.

Second, if the straw is too wide, the bottom surface will grow unstable and water will flow e.g. on one side, allowing bubbles to rise on the other. What actually maintains the stability there is a force called surface tension. You can lower it by adding a surfactant (dish washing soap e.g.) to the water, you'll see that the biggest straw possible is then smaller!

share|cite|improve this answer

Weight of water is too more to get hold from bottom just by h$\rho$g pressure but actually when you put thumb you cutoff above pressure and then the pressure at bottom then dominates and hold water in straw.You can find the force now using formula above and it will be right to say that the force just due to pressure on bottom surface is $1.01325*10^5*\pi r^2$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.