Buoyancy is a familiar concept to most people. We all have experience with floating – a floating cork, boats, floating in a pool, and so on. Consider two examples:

1. People with joint injuries often exercise in shallow (say chest-high) water in swimming pools. They do this because there is less force on their joints while submerged in water. They *weigh* less. So there must be some upward force due to the water that counteracts the downward pull of gravity, the force that might otherwise cause aching joints to grind together.

2. Boats float on top of water. Heavy boats will lie lower in the water than light boats, but all except overloaded boats stay on the surface. Therefore, there must be some upward force due to the presence of water that counteracts the force of gravity, which would otherwise pull the boat to the bottom, just as it might pull a marble down.

We call the buoyant force the difference between the force of gravity on an object and the upward force caused by the pressure of the surrounding fluid. Here's a diagram:

The upward force is not a direct upward force from the liquid, *per se*. It arises from the difference in pressure (ultimately caused by the gravitational force on the liquid) at the top of a submerged object compared to that at the bottom. It's simply due to the difference in water pressure at the top and bottom of an object.

In our figure, the pressure at the top is

$$P = \rho g A h_1,$$

where $\rho$ is the density of the fluid (usually Kg/m^{3} or g/cm^{3}), $A$ is the area of the top or bottom of the object, $h$ (h_{1} and h_{2} in the figure) is the depth of the liquid, and $g$ is the acceleration of gravity. Likewise, the pressure at the bottom is

$$P = \rho g A h_2,$$

Now the difference in the pressure between the top and the bottom of our submerged object is

$$\Delta F = \rho A g(h_2 - h_1)$$

Because $h_2 \gt h_1,$ the pressure is greater at the bottom and less at the top, causing a net force, the buoyant force, $F_b,$ upward. When $F_b = F_g,$ the object is said to be **neutrally buoyant**. Divers strive for neutral buoyancy by balancing the effects of a weight belt with a buoyancy compensator, a vest they can blow up with air from diving tanks to counteract the downward pull of the weights.

Now notice that $h_2 - h_1$ is the height of our cylinder, $h,$ so we have

$$\Delta F = \rho g A h, $$

where the area multiplied by the height is the volume:

$$\Delta F = \rho g V$$

Now $ \rho V$ is a mass. You can see that from this unit analysis:

$$ \require{cancel} \rho V \; \rightarrow \; \frac{Kg}{\cancel{m^3}} \cdot \cancel{m^3} = Kg$$

But because the density we're using here is the density of the *fluid*, this mass is the mass that a volume of the fluid equal to the volume of the cylinder would occupy. This leads us to a statement of **Archimedes' principle**:

The upward, or buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced (moved out of the way) by that object.

The buoyant (upward) force on a body immersed in a fluid is equal to the weight of the fluid displaced by the body.

An object that sinks displaces its entire volume of fluid. An object that floats displaces only the volume of fluid equal to the volume of the object lying below the level line of the fluid.

We all have experience with things that float and things that don't. We know that if we place a nail, a bolt or a rock in water, it will sink. On the other hand, we build giant ships out of thick steel and those float. This must be related to the buoyant force. What's the difference?

Consider dropping a solid marble of mass 10 g and radius 0.5 cm into water $(\rho_{water} = 1 \; g/cm^3).$ The volume of the marble is $V = \frac{4}{3} \pi r^3,$ so $V = 0.5236 \; cm^3.$ The mass of the water displaced (moved aside) by the marble is thus 0.5236 g. The masses of the marble and the water are directly proportional (to within the acceleration, g) to the downward and upward forces, respectively, so it's easy to see that the buoyant force is nowhere near the downward force on the marble. It sinks.

Now consider taking the same amount of marble material and constructing a hollow sphere 3 cm in radius. The volume of such a sphere is

$$ \begin{align} V = \frac{4}{3}\pi r^3 &= \frac{4 \pi 3^3}{3} \\[5pt] &= 37.7 \; cm^3 \end{align}$$

The gravitational force on the sphere is

$$F = mg = 0.01 \, Kg \cdot 9.8 \, \frac{m}{s^2} = 0.098 \, N.$$

The buoyant force is the weight of the displaced water:

$$F = mg = 0.0377 \, Kg \cdot 9.8 \, \frac{m}{s^2} = 0.377 \, N.$$

The buoyant force is greater than the pull of gravity on the sphere, so it will float.

A cubic box without a top has dimensions of 20 cm and a mass of 1.0 Kg. Will the box float on pure water (density = 1 Kg/m^{3}), and if so, how far below the surface of pure water will the bottom be?

**Solution**

Let's assume the box will float, and calculate how much of it will be below the water level. If this box *won't* float, that depth will exceed the box height of 20 cm.

In order for the box to float, the mass of water displaced must equal the mass of the water. The mass of the water displaced will be

$$m = (20 \, cm)(20 \, cm) \cdot h \cdot \rho_{water}$$

So setting this equal to the mass of the box gives us our equation to solve for h:

$$ \require{cancel} \begin{align} 400 \cancel{cm^2} \cdot h \cdot \frac{1 \, \cancel{g}}{cm^{\cancel{3}}} &= 1000 \cancel{g} \\[5pt] h &= \frac{1000}{400} = 2.5 \, cm \end{align}$$

So our box will float with 7.5 cm of its height above the water and 2.5 cm below it. Ship builders obviously have to worry about calculations like this. An unloaded container ship will "sit high" in the water. Its displacement (depth below the water line) will be greater once it's piled high with freight containers.

The calculation for ship builders and loaders is the same, the geometry is just a little trickier.

A 10 Kg block of lithium (Li) metal is encased in plastic* (assume negligible volume of plastic) and submerged in water. Lithium is a low-density metal, $\rho_{Li} = 0.534 \; g/cm^3$. How much force would be needed to lift the Li block while it is still under water?

*_{2}), which ignites.

**Solution**

$$\rho_{Li} = \frac{m}{V},$$

where m and V are the mass and volume. Rearranging to solve for volume gives

$$ \begin{align} V &= \frac{m}{\rho_{Li}} \\[5pt] &= \frac{1000 \, \cancel{g}}{0.534 \cancel{g}/cm^3} \\[5pt] &= 1873 \, cm^3 \end{align}$$

Now the buoyant force on that volume of water is the mass of the water $(\rho_{water} = 1 \, g/cm^3)$ multiplied by the volume and the acceleration of gravity:

$$ \begin{align} F_b = \rho_{water} V g &= \frac{0.001 \, Kg}{\cancel{cm^3}}(1873 \, \cancel{cm^3})\frac{9.8 \, m}{s^2} \\[5pt] &= 18.4 \, N \end{align}$$

Now the downward pull of gravity on a 10 Kg mass is

$$ \begin{align} F_g &= mg \\[5pt] &= 10 \, Kg \, (9.8 \frac{m}{s^2}) \\[5pt] &= 98 \; N \end{align}$$

When the lithium is under water, the buoyancy force assists us in lifting the block. The force needed will be

$$ \begin{align} F &= F_g - F_b \\[5pt] &= 98 \, N - 18.4 \, N = 79.6 \, N \end{align}$$

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