Video of the “Atmospheric pressure, hydrostatics, and upside down water glasses” demonstration

Here are videos of the “Atmospheric pressure, hydrostatics, and upside down water glasses” demonstration. The full videos are linked form the GIF titles

Demonstration with the cup full

Demonstration with the cup only partially full

Again thanks to Henry Burridge for this demonstration.

An index of all the demonstrations posted on this blog can be found here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (nbkaye@clemson.edu). I also welcome comments (through the comments section or via email) on improving the demonstrations.

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Atmospheric pressure, hydrostatics, and upside down water glasses

Courtesy of Henry Burridge

Background

The demo was inspired by a suggestion in the Cambridge engineering supervisions for first years and a question that Colm Caulfield asked me during a summer school. The question was “Given that the atmospheric air pressure is roughly equivalent to a 10m head of water, why can’t we build 10m deep swimming pools and turn them upside down?”.

This question works well with engineers when being introduced to hydrostatics, who (typically) are quick to appreciate that it is relatively simple to create a structure that can support a pressure drop across it of 1 atmosphere. Hence, one can imagine building a 10m deep swimming upside down, the ‘ceiling’ of which is exposed to atmospheric air pressure above and near zero pressure within the water just below.

So if we could fill the upside-down swimming pool with water to a depth of 10m then the water at the ceiling could be at near zero pressure whilst the water 10m below would be roughly at atmospheric pressure.

Equipment:

1. A pint glass.
2. A piece of thin waxy card, just a bit bigger than the size of the rim of the pint glass (a postcard, photographic paper, or the cover of a CD case should work).
3. A piece of fine metal gauze or mesh, just a bit bigger than the size of the rim of the pint glass. I have used wire mesh with mesh gaps of about 0.5mm to 1mm and a metal plate perforated with holes of about 3mm diameter.
4. Some Blu-Tack which can be used to seal the gauze or mesh to the rim of the glass.
5. A bucket – just to catch any spills.

Equipment required for the first part of the demonstration (no gauze)

Aim

To encourage consideration of hydrostatic pressures and the role of atmospheric pressure (useful for engineers and the concept of gauge pressures).

Demonstration

After posing the above question to them and having a discussion around it, I usually tell the students to think about it whilst we test the ideas on a smaller scale. I then tell the students that I’m going to fill the pint glass with water and, whilst holding the card in place on the top of the glass, turn the glass upside down. I ask them to think about what will happen, and why, when I remove my support from the card. The answer is that nothing happens (except perhaps the card bowing slightly) – I suggest that the reason nothing happens is that the upside-down glass can support a lower than atmospheric pressure in the water at the top and hence the water at the bottom of the glass can be at atmospheric pressure and hence there is no net pressure forcing the water out of the glass. Thus, the weight of the card can be supported by the surface tension in the interface that forms between the glass and the card.

I then ask them to think about what will happen if we only partially fill the glass with water – the results are the same (since any pressure drop within the air contained inside the glass, required to attain atmospheric pressure within the water at the base of glass, is slight and any expansion of the air typically is small and hence the interface between the card and glass remains small). Finally, to convince them that there really is no pressure force ‘pushing’ the water out of the glass, I repeat the demonstration with the card replaced with the wire mesh or perforated plate (note that until the glass is inverted it is wise to cover the mesh or plate with the card!).

Having convinced them (hopefully) that there really is no force driving the water from the glass, I then return to the question of the upside-down 10m swimming pool and again emphasize that such a situation is in equilibrium but the equilibrium is unstable. It is worth reminding them of more familiar examples of unstable equilibrium, e.g. a pencil balancing vertically on your finger, a ball on the apex of a convex surface. The discussion then typically follows onto to the fact that the density difference across the interface makes the interface unstable and thus any tiny perturbation (or vibration) will take the system away from the unstable equilibrium. However, surface tension is able to stabilize the interface if length of the interface is small – hence since the holes in the mesh or the plate are small enough, surface tension is able to stabilize the interface to small perturbations.

An index of all the demonstrations posted on this blog can be found here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (nbkaye@clemson.edu). I also welcome comments (through the comments section or via email) on improving the demonstrations.

Atmospheric pressure – plunger tug-o-war

One of the difficulties some students have with any engineering course is to have a feel for the scale of particular forces or the scale of different parameters. This can be hard to communicate. Here is an easy demonstration to illustrate the scale of atmospheric pressure.

Equipment

1. Two suction cup plungers
2. A chap stick (or some petroleum jelly)
3. Two student volunteers

Demonstration

1. Rub the chap stick or petroleum jelly along the rim of the two plungers
2. Align the two rims and push them together pushing out as much of the air between them as possible
3. Have the two student volunteers try and pull the plungers apart. This will be quite difficult if the seal is good enough and the bulk of the air is removed from between the plungers.

Analysis

It is not possible to easily calculate or measure the force required to pull the plungers apart. The answer mainly depends on the air pressure and volume in the gap between the plungers before you start to pull. The easiest way to get a rough (over) estimate of the force is to assume that there is a vacuum between the plungers and that the chap stick and the plungers do not provide any significant mechanical resistance. In that case one can draw a free body diagram of one plunger and sum forces along the direction of the axis of symmetry.

∑Fx=0=Patm πD2/4 – Fstudent

Substituting in values for atmospheric pressure (14.7 psi) and the plunger diameter (5½” in my case) then each student will need to provide approximately 350 lb of force to pull the plungers apart. In reality the plungers are unlikely to be perfectly aligned and the seal will also be less than perfect so this will be an upper estimate of the required force. Even if the force actually required is a bit less than this the students will still get a sense that atmospheric pressure is quite large.

An index of all the demonstrations posted on this blog can be found here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (nbkaye@clemson.edu). I also welcome comments (through the comments section or via email) on improving the demonstrations.