- Buoyancy
is the force that causes objects to float in fluids.
- The simple rule to remember is the
buoyancy force equals the
weight of the
fluid:
### F = m

where m_f is the mass of displaced fluid and g is Earth's surface gravitational field (fiducial value g = 9.8 N/kg)._{f}*g , - For simple understanding, note that an object denser/(less denser) than the
fluid sinks/rises because the
buoyancy force is less/greater than its
weight.
If object and fluid are equal in density, the object is neutrally floating, neither sinking nor rising.

Humans in water are close to neutral floaters---sometimes we sink, sometimes we float up to the surface where their is a much less dense fluid air in which we sink.

- You've all known buoyancy since forever.
You all tried to push a beach ball under the water once---don't deny it---and it was great fun trying, but you never succeeded.

- We now proceed to give a fuller explication.
We restrict our discussion to incompressible fluids to avoid the difficulties of generality. An incompressible fluid has constant density even though the pressure. It is an ideal case that most terrestrial liquids approximate very well. Even air can be treated as incompressible for many cases.

- If you fully immerse an object in a fluid,
it replaces a clump of
fluid
that has its shape.
- The net pressure force on the object is the same
as on the replaced clump.
- This net pressure force
is the buoyancy force itself.
- Calculating the buoyancy force
by integrating the
pressure all over an object
is difficult.
But there is a clever trick. Assuming a static fluid, the clump of fluid replaced by the object is floating neutrally.

Thus, applying Newton's 2nd law of motion (AKA F_net=ma) (with upward positive) to the clump (which has zero acceleration), we get

### F

_{net}= -m_{f}*g + F_{f}= -ρ_{f}Vg + F_{f}= 0 ,where -m_f*g is the gravity force on the clump, m_f is the clump's mass, g is (as above) the Earth's surface gravitational field (fiducial value g = 9.8 N/kg), ρ_f is the

**MEAN**fluid density, V is the clump volume, and F_f is the buoyancy force.Thus we find

### F

_{f}= ρ_{f}Vg .So we solved for the buoyancy force from Newton's 2nd law and the gravitational force rather than directly from the pressure force.

Since the net pressure force on the clump only depends on the shape of the clump, it must be the same on the object that replaces the clump.

So the formula above is the general buoyancy force formula.

- If an object of mass m is immersed in a single fluid
there are three cases:
- F_f < mg: In this case, the net force is downward and the object sinks.
- F_f = mg: In this case, the net force is zero and the object stays at rest or, in the buoyancy jargon, has neutral buoyancy.
- F > mg: In this case, the object floats upward.

- What if an ojbect of mass m is at rest on a solid bottom of a
fluid layer?
Well

### F

_{net}= -mg + F_{p}+ F_{n}= 0 ,where F_p is the sum of the pressure force from the fluid and the solid bottom and F_n is normal force that the solid bottom must supply in addition to supplying some of the pressure force.

The pressure force from solid bottom is just what the fluid would supply if it replaced the solid bottom. This pressure force from solid bottom is, of course, what supports all the fluid in all the layers above the solid bottom.

Obviously,

### F

_{n}= mg - ρ_{f}Vg ,just the force needed to balance the extra weight of the object beyond that of the clump of fluid it has replaced. The ideal normal force (which is what we are considering here) supplies just the force need to to satisfy the condition of no motion through the solid bottom.

Now we find that

### F

_{p}= ρ_{f}Vg = F_{f},the buoyancy force all over again. However, we note that this is supplied both the fluid and the solid bottom.

Finding the separate combributions to F_f is tricky in general. There is one ideal case where it is easy. In the limit that the object touches the bottom only a true points, the pressure force from solid bottom is zero and all the buoyancy force is supplied by the fluid.

Some very special real cases that one can think of approximate the ideal case: e.g., an object resting on narrow legs.

- What of the case of object at rest straddling an
interface between
a lower layer of fluid 1
and an upper layer of fluid 2?
Note that stability requires ρ_1 > ρ_2 where ρ_i is density of fluid i. Note also the two fluids must be immiscible for stability (see Wikipedia: Miscibility).

In this case, Newton's 2nd law gives

### F

_{net}= -mg + F_{f}= -ρVg + ρ_{f}Vg = 0 ,where ρ is the object density.

Recall ρ_f is the mean density of the clump of fluid. Thus

### ρ

_{f}= m_{f}/V = (m_{1}+m_{2})/V = (ρ_{1}V_{1}+ ρ_{2}V_{2})/V ,where m_i is the mass of fluid i displaced, V_i is the volume of fluid i displaced, and ρ_i is the density of fluid i.

Substituting the last formula into the second to last formula and canceling g out gives

### ρV = ρ

_{1}V_{1}+ ρ_{2}V_{2}or ρ = ρ_{1}f_{1}+ ρ_{2}f_{2},where f

_{i}is the volume fraction of the object in fluid i.We have, of course, another equation f_1 + f_2 = 1. Sovling for f_1 and f_2, we obtain the formulae

### f

_{1}= (ρ - ρ_{2})/(ρ_{1}- ρ_{2}) and f_{2}= (ρ_{1}- ρ)/(ρ_{1}- ρ_{2}) .We can consider some special cases:

- ρ_1 = ρ_2: This is the effectively the single
fluid
case with neutral buoyancy that
we considered above.
The derivation of our formulae assumed ρ_1 > ρ_2, in fact.
If we allow ρ_1 = ρ_2, then we get ρ = ρ(f_1 + f_2) and f_1 and f_2
are indeterminate other than f_1 + f_2 = 1 which we already knew.
The object could float neutrally in either fluid or at the
interface in this case.
- ρ > ρ_1: This is the case where the object will actually sink in
fluid 1 and our analysis does
**NOT**apply. - ρ > ρ_2: This is the case where the object will actually rise in
fluid 2 and our analysis does
**NOT**apply. - The zeroth order approximations
for ρ_2/ρ_1 and ρ_2/ρ are
### f

_{1}= ρ/ρ_{1}and f_{2}= (1 - ρ)/ρ_{1}.The zeroth order approximations are usually good approximations for fluid 1 being any liquid and fluid 2 being air since ρ_2/ρ_1 and ρ_2/ρ are of order 1/1000.

For example, consider an object of ice in water (i.e., liquid water). The fiducial densities of ice and water are, respectively 0.92 g/cm**3 and 1.00 g/cm**3 (see Wikipedia: Properties of Water: Density of water and ice). We obtain f_1 = 0.92 ≅ 0.9 and f_2 = 0.08 ≅ 0.1. Hence, the oft quoted remark that icebergs are 90 % below the surface. For a more accurate result, one should use accurate densities for ice and water for the temperature, pressure, and impurities of the particular case.

- ρ_1 = ρ_2: This is the effectively the single
fluid
case with neutral buoyancy that
we considered above.
The derivation of our formulae assumed ρ_1 > ρ_2, in fact.
If we allow ρ_1 = ρ_2, then we get ρ = ρ(f_1 + f_2) and f_1 and f_2
are indeterminate other than f_1 + f_2 = 1 which we already knew.
The object could float neutrally in either fluid or at the
interface in this case.

Caption: Buoyancy illustrated and explicated.

Features:

Image link: Wikipedia: File:Buoyancy.svg.

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