Part III: Buoyancy

With a decent understanding of hydrostatics and the way pressure and forces interact under our belt, we can then begin to ask some important questions about how objects react to the presence of fluids. For example, how do I know if a boat will sink or float? Is thinking about sinking and floating the only thing I need to thinking about when I’m designing a boat?

To be able to answer those questions, we need to recall the example we saw before of the fluid point at the bottom of the cup. There, we saw that a surface force \vec{f}_s of the same magnitude as the pressure within the fluid point ensures that the fluid point remained stationary. This fact turns out to generalize for any interface between two continuum objects, be they solid or fluid; at every point of the interface between two continua, there is a traction with magnitude equal to the pressure at that point, pointing perpendicular to the interface in the direction of the continua of interest. This is what we saw in the cup example— in that scenario, our continuum of interest was actually the water, and the cup was exerting a traction on the fluid point proportional to its local pressure |\vec{t}_s| = p.

A solid sphere submerged in a cup of water experiences a traction at every point on its surface, which is pointing perpendicular to the surface at that point, and with magnitude equal to the fluid pressure at that point.

This concept allows us to understand & calculate the hydrostatic forces acting on an object embedded in a fluid through the following analysis:

  1. Determine the pressure distribution of the fluid.
  2. Determine the interface between the continuum of interest and the fluid.
  3. Determine the pressure of the fluid at the interface.
  4. Calculate the pressure-induced traction at each point on the interface.
  5. “Add up” (integrate) those tractions to get the total hydrostatic force on the object.

We can express this last step of the process succinctly using mathematical language:

\displaystyle \vec{F}_b = \int_{A} \vec{t}_s\ dA

This procedure involves applying tools from vector calculus, which beginning practitioners of fluid mechanics might find daunting. Luckily, we can obtain the total buoyancy force on any object embedded in a static liquid on Earth without having to do so. In short, the buoyancy force on an object in a fluid is always equal to the weight of the fluid it displaces, and always points up. This is mathematically represented as Archimedes’ law:

\vec{F}_{\text{b}} = -\rho V_{\text{displaced}} \vec{g}

For the sake of ideological consistency, we can show how Archimedes’ simple law is derived from the more complicated process of adding up tractions described above. This relies chiefly on the fact that the pressure distribution in a given static fluid on Earth is essentially always the same, stemming from the solution to \nabla p = \rho \vec{g}. Since the gravitational force points strictly downwards, we can simply integrate in the “depth” direction z to find that:

p = p_{0} + \rho|\vec{g}|z

where p_0 represents the pressure at the surface of the fluid and z represents the depth of the fluid at which the fluid point is located. Clearly, the pressure of a fluid point here only depends on its depth within the fluid. As a result, horizontal pressure-induced tractions that act on the surface of an object in such a fluid must cancel out, since a pressure-induced tractions on the “left” side of the object will inevitably be canceled by tractions on the “right” side with the same net magnitude.

Now consider a cube of side length a within the fluid. Its interface is determined by six distinct surfaces; the top, bottom, and four sides. Because the traction on the side surfaces must necessarily cancel by the argument above, the only contributions to the buoyancy force are going to come from the top and bottom, in which the tractions are uniform but distinct.

Frontal view of a submerged cube with side length a and the pressure-induced tractions it feels. From symmetry, the tractions on the sides are equal and opposite, and the net buoyancy force comes from the difference in tractions between the top and bottom of the cube.

Therefore, the total traction on the cube is just the difference in tractions between the top and bottom of the cube, multiplied by the area over which they act:

\vec{F}_{\text{b}} =   \int_{A} \vec{t}_s\ dA = a^2 \vec{t}_{\text{top}} + a^2 \vec{t}_{\text{bottom}}

\vec{F}_{\text{b}} \cdot \hat{y}  =  a^2(p_{0} + \rho|\vec{g}|z_{\text{bottom}}) - a^2(p_{0} + \rho|\vec{g}|z_{\text{top}})

\vec{F}_{\text{b}} \cdot \hat{y}  =  a^2 \rho|\vec{g}|(z_{\text{bottom}} - z_{\text{top}})

\vec{F}_{\text{b}} \cdot \hat{y}  =  a^3 \rho|\vec{g}|=V_{\text{cube}} \rho|\vec{g}|

Because this force is additive in the volume, and because every solid body can be approximated to arbitrary accuracy as a combination of sufficiently small cubes, we have by consequence derived Archimedes’ law for arbitrarily shaped solid objects. Voilà!

This begs the question; why bother thinking about this in a way that requires vector calculus if we don’t need it to calculate the buoyancy force? Luckily, the answer is simple—rotation!

Think of a cylinder wrapped in string. If you pull the string on both sides with equal and opposite force, the center of mass will surely remain stationary by virtue of Newton’s law, but the cylinder as a whole will spin about its axis. Clearly, just the total force on an object doesn’t paint the whole picture of how the cylinder moves; the location of those forces on the object also matters! And since we don’t want our boats to spontaneously capsize as we sail on the ocean, understanding this phenomenon is imperative for any practical application of buoyancy.

A cylinder wrapped in string is pulled on opposite ends by equal and opposite forces. Although the cylinder’s center of mass doesn’t move due to zero net force, the cylinder spins due to a torque equal to the sum of the moments induced by the forces.

In essence, hydrostatic pressure-induced tractions don’t only induce a net buoyancy force \vec{F}_b on an immersed object, but a buoyancy torque \vec{\tau}_b as well. And unlike the buoyancy force, we don’t really have a neat torque version of Archimedes’ law that lets us calculate this buoyancy torque without using vector calculus. As a result, we find the process of calculating it nearly identical to the original process of determining the net buoyancy force:

  1. Determine the pressure distribution of the fluid.
  2. Determine the interface between the continuum of interest and the fluid.
  3. Determine the pressure of the fluid at the interface.
  4. Calculate the pressure-induced traction torques (\vec{r}\times \vec{t}_s) at each point on the interface.
  5. “Add up” (integrate) those traction torques to get the total hydrostatic torque on the object.

We can again list out this last step succinctly using mathematical language:

\displaystyle \vec{\tau}_b = \int_{A} \vec{r}_{cm} \times \vec{t}_s\ dA

where \vec{r}_{cm} represents the position of the point being analyzed relative to the object’s center of mass.

As it turns out, this process of summing up small contributions through integration is a critical tool in all areas of fluid mechanics, which we shall soon observe.

Things to Think About

1. How would you determine if an object will sink or float? How would you determine if an object will spin when it either sinks or floats, and in which direction?

2. A static, floating object is always partially submerged in the liquid below it. Can you determine the volume of the submerged part of the object using the techniques above? (Ignore air, then ask yourself why you could do that.)

3. Think of the pressure-induced traction distribution on a triangle (◣). Convince yourself that the horizontal tractions cancel out. Which way would the triangle spin if it floats? Is it the same direction if it sinks?

4. Use the result above to justify why boat hulls look the way they do. Try to think of reasons for the design of boat hulls in general (shape, length, width, depth, etcetra).

5. If an object possesses a uniform identical density to water, it will neither sink nor float—its center of mass simply remaining stationary. In this scenario, could it rotate due to hydrostatic forces?

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