What is the buoyancy principles?

Answer 1

The buoyancy principle states that when an object is submerged in a liquid, the object experiences an upward buoyancy force that is equal to the weight of the liquid the object displaced.

For a non-mathematical explanation of the principle go to the question "Why do some things float but not others" which is in the Related Questions section below.

In fact I suggest that you read that question before reading this one because this question is in some respects an extension of that one.

First imagine some arbitrarily shaped shell filled with water. Imagine the shell is weightless and very rigid. If we were to attach a string to it and weight it we would find it would take a certain force to hold it up, however if we were to then lower the shell of water into a basin of water the shell would neither sink to the bottom nor float to the top, it would just float around under the surface of the water held up by the buoyancy force. This is because the buoyancy force, which is up, would be exactly equal the downward gravitational force we just measured.

Now imagine if the shell surrounding the water just magically disappeared. Absolutely nothing would happen. And if a similar rigid shell just magically surrounded some other volume of the water, again nothing would happen. This is because a "pressure gradient" has developed in the water which is specifically "designed" to hold up any portion of water whether or not that portion of water is surrounded by a massless rigid shell or not.

To appreciate the nature of this pressure gradient imagine sticking a stiff drinking straw vertically down into the water. When you think about it, the only thing holding up the water in the straw is the water at the very bottom end of the straw. It is under pressure and pushing up against the water above it.

Imagine the straw has a cross sectional area A. Pressure is a measure of force per unit area, so the force with which the water at the bottom of the straw is pushing up is the pressure at that depth times the area, or...

Force = Pressure x Area ( F = P·A)

The mass of the water in the straw is the density of the water times the volume in the straw or..

mass = density x (Volume) = density x (Area x Depth)

or using symbols

m = ρ·V = ρ·(A·D)

We can see that the mass of the water in the straw depends directly on how deep the straw is. This also means that the force necessary to hold up the water in the straw directly depends on how deep the straw is. Since the area at the end of the straw is constant, the only way the upward force is going to increase with depth is if the pressure at the end of the straw increases with depth.

So again, as we lower the straw we can see how the pressure of the water needs to increase with depth if it is going to hold up the water in the straw above it.

Just as with the rigid shell, it doesn't matter whether the straw is there or not, the pressure, must increase in proportion to the depth. Think of the pressure of the water at a given depth as the amount that the water is being squeezed.

It is in fact easy to calculate what the pressure must be at a given depth.

If the mass of the water in the straw is..

m = ρ·A·D

Then the force is

F= (m)·g = (ρ·A·D)·g

where g is the acceleration of gravity.

Since the pressure is the force per unit area, dividing both sides by the area we get...

Pressure = F/A= ρ·g·D

This equation describes the pressure gradient or the way the pressure increases with depth.

To understand how this pressure gradient creates the buoyancy force, we will first think about the simplest case. Imagine if the straw we discussed above where completely submerged. We will only be thinking about the forces acting on the water in the straw. The water under pressure at the bottom of the straw is pushing up with a force that is large enough to hold up a straw of water that reaches to the surface. However the water in the straw does not reach the surface. Instead the water under pressure at the top of the straw is pushing down on the water in the straw. The upward force at the bottom is larger than the downward force at the top and the difference is exactly what is required to hold up the water in the straw. To understand why, imagine an extension to the straw that causes it to reach the surface. The force at the bottom of the extension is large enough to hold up the water in the extension. The upward force acting on the bottom of the extension is equal to the downward force acting on the top of the submerged straw. The magnitude of this downward force is exactly equal to the amount the upward force, at the bottom of the straw, is to large. Now replace the straw with an un-sharped wooden pencil. Because the wooden pencil is less dense than water the pencil will weigh less. The buoyancy forces that were acting on the water in the straw are now acting on the pencil and since the pencil weighs less it will rise to the surface.

Now imagine, submerged in water, an arbitrarily shaped shell filled with water. The buoyancy force acting on the shell is what is keeping it suspended. If you imagine the shell shape to be composed of a very large number of straws pressed together, the reasoning used to understand why the water in a single straw is suspended can now be used to understand why the water in the shell shape is suspended. Once you understand why the water in the shell shape is suspended, you completely understand the buoyancy force.