Pressure due to depth in a liquid; buoyant force

Pressure is force perpendicular to a surface/ area of the surface. In a static fluid (liquid or gas), the pressure increases with depth due to the weight of the fluid. It is easy to show (see any beginning physics book) that the pressure at depth h is given by

P = P_o + r gh,

where r = density (=mass/volume), g = gravitational acceleration, and P_o = pressure at the surface. The same equation would work for pressure in a stack of newspapers. The difference is that newspapers do not push sideways.

If a tank of liquid is open to the atmosphere, P_o is atmospheric pressure. You should also be aware that if you travel downward a distance D h, the pressure increase is
D P = r gD h. This is true in a gas, also. In air for example: Go to the top floor of a skyscraper in a fast elevator, and you can detect the change of pressure by a feeling in your ears (in this case a decrease). Now you can calculate it. Do not try this with too large a D h though, because r changes. In a liquid, r is almost exactly constant.

This shows a tank with a plug in it. If the area of the hole is A, what is the net outward force due to the pressures on the inside and outside? To prevent its popping out, there needs to be an equal and opposite force on it, the static friction force on the plug by the hole edge.

Did you answer the question in the previous paragraph? It is r ghA, if h is the depth. Atmospheric pressure drops out of the picture. It contributes to the force to the right and to the left, so zero net effect.

The buoyant force (usually pronounced boy-ant even though a buoy is usually pronounced booee) is the upward force on an object due to the fact that the pressure on the bottom of it is greater than the pressure on top. Consider an unopened can of beans sitting on the table. If the air pressure on the top of it is P, then on the bottom it is
P + r gD h, where D h is the height of the can. So the net upward force due to this difference is
r gD hA. But D hA is the volume of the can, and r g times the volume is the weight of the air displaced by the can. This is known as Archimedes' principle: buoyant force equals weight of the fluid displaced. Armed with this principal principle, Archimedes was able to determine that the king's crown was not as dense as pure gold, exposing the crook who crafted it. (How do you find density from buoyant force? Find V from the buoyant force then r = m/V.)

So now you know that when you step on a scale, the reading you see is not your true weight, it is your weight minus the buoyant force on you. For the typical adult, this is 0.1 to 0.3 pounds, not enough to be concerned about for most purposes, but detectible. For a gas-filled balloon, the buoyant force is a little more or less than the weight, depending on what it is filled with. Helium, like any mass, is gravitationally attracted to the Earth, but being less dense than air, a helium filled balloon can weigh less than the buoyant force on it, and therefore can float upward in the atmosphere.

If you weigh yourself in a swimming pool, you would see a much more dramatic reduction in your apparent weight, because you have approximately the same average density as water. But don't use a scale that you stand on (they are not designed for immersion), rig up a pulley overhead and see what counterweight is needed for balance. Of course it depends on how much of you is out of water.

A floating object has a buoyant force on it equal to its weight. Mountains float in the Earth, because a mountain is so heavy that it can cause flow. If it weren’t supported by a buoyant force, a mountain would slowly sink into the Earth. So there must be a region below the mountain with a lower density than normal.

Swim back to the main page on fluids, heat, and E&M. Or check out this simple experiment on buoyancy.

My other main pages:

mechanics
vibrations and waves
quantum
index has stuff alphabetized.

Comments, questions: fredrick.gram at tri-c.edu (but use @, not at, and remove spaces)