Jacob Nie

Float orientation


Suppose you had a wooden rectangular prism with a square base that's longer than it's wide. Then you put it in water. Intuitively, it floats horizontally in the water. But does it float with a face pointing up or does it float with an edge pointing up?


Here's the key idea: whichever orientation results in the lower center of mass is the orientation that the block will assume. (Everything seeks to reduce its potential energy.)

  • Density of the block: $\rho$
  • Density of the water: $\rho_w$
  • Relative density: $\lambda = \rho/\rho_w$
  • Side length of the square: $a$
  • Length of the block's long edge: $l$
Consider scenario 1:

Using Archimede's principle, we find: $$\rho l a^2 = \rho_w l a\left(\dfrac{a}{2}+x_1\right).$$ Solving, $$\dfrac{x_1}{a} = \lambda - \dfrac{1}{2}.$$

Consider scenario 2:

(We need to assume $\lambda > 1/2$ here, since the water line is at the top part of the face.)

As before, we have the slightly more complex equation $$\rho l a^2 = \rho_w l \left(\dfrac{1}{2}a^2 + \dfrac{1}{2}a^2 - \dfrac{1}{2}(a - \sqrt{2}x_2)^2\right).$$ This simplifies to the quadratic equation: $$\left(\dfrac{x_2}{a}\right)^2 - \sqrt{2}\left(\dfrac{x_2}{a}\right) + \left(\lambda - \dfrac{1}{2}\right) = 0.$$ Solving, $$\dfrac{x_2}{a} = \dfrac{\sqrt{2}}{2} - \sqrt{1-\lambda}.$$

We do still need to address $\lambda < 1/2.$ (We don't need to do that for the first orientation though, the geometry is not significantly affected by where the water line is.)

Note that in this scenario, $x_2$ will be negative. (If $\lambda < 1/2$ in the first scenario, then $x_1$ will also be negative.)

We have $$\rho l a^2 = \rho_w l \left(\dfrac{1}{2}(a + \sqrt{2}x_2)^2\right).$$ This is the quadratic $$\left(\dfrac{x_2}{a}\right)^2 + \sqrt{2}\left(\dfrac{x_2}{a}\right) + \left(\dfrac{1}{2}-\lambda\right) = 0.$$ Solving, $$\dfrac{x_2}{a} = -\dfrac{\sqrt{2}}{2} + \sqrt{\lambda}.$$

Hence, we have to compare the following expressions:

When $\lambda > 1/2,$ which of these is smaller: $\lambda - \dfrac{1}{2}$ or $\dfrac{\sqrt{2}}{2} - \sqrt{1-\lambda}$?

When $\lambda < 1/2,$ which of these is smaller: $\lambda - \dfrac{1}{2}$ or $-\dfrac{\sqrt{2}}{2} + \sqrt{\lambda}$?

For $\lambda > 1/2$:

Here, we see that $\dfrac{\sqrt{2}}{2} - \sqrt{1-\lambda}$ is smaller for $0.5 < \lambda < \sqrt{2} - \dfrac{1}{2}$ and $\lambda - \dfrac{1}{2}$ is smaller for $\sqrt{2} - \dfrac{1}{2} < \lambda < 1.$

For $\lambda < 1/2$:

Here, we see that $-\dfrac{\sqrt{2}}{2} + \sqrt{\lambda}$ is smaller for $0 < \lambda < \dfrac{3}{2} - \sqrt{2}$ and $\lambda - \dfrac{1}{2}$ is smaller for $\dfrac{3}{2} - \sqrt{2} < \lambda < 0.5.$

So that's all the math! Let's make a table for our findings.

$0 < \lambda < \dfrac{3}{2} - \sqrt{2}$ Floats Edge Up
$\dfrac{3}{2} - \sqrt{2} < \lambda < \dfrac{1}{2}$ Floats Face Up
$\dfrac{1}{2} < \lambda < \sqrt{2} - \dfrac{1}{2}$ Floats Edge Up
$\sqrt{2} - \dfrac{1}{2} < \lambda < 1$ Floats Face Up
So that's our answer!

Further examination

We just found an equilibrium point right? Figuring out which position resulted in the lower energy. The thing is, we compared between two positions that aren't exactly "close" together. In other words, we want to see whether there's actually a local minima at the positions we determined above. And we're also curious as to whether there might possibly exist other local minima!