## Today’s Fact: Foot On Water

Andi came up with an insane question: How wide does a human foot have to be for a human to stand on water? Of course, he came up with something insane, reproduced here.

## Foot On Water, Andi Gu

### Relevant Parameters

Let $$m$$ be the mass of some person, let $$R$$ be the radius of his foot (which we model as a circle), and let $$\gamma$$ be the surface energy of water (i.e. surface tension).

We make a simplistic model in which the water deforms in a circularly symmetric pattern around the foot, that when viewed at a cross-section appears as follows: ### Derivation of Foot Size

The increase in the water surface area is

\begin{align*} \Delta A &= \int_{R}^{R+\frac{h}{\tan \theta}} 2\pi r (\frac{1}{\cos \theta}-1) d{r} \\ &= \pi (\frac{h^2}{\tan^2 \theta} + \frac{2Rh}{\tan \theta}) \frac{1-\cos \theta}{\cos \theta} \end{align*}

Since $$\Delta E=\gamma \Delta A$$, we have:

\begin{align*} F_{up} &= \frac{\partial E}{\partial h} \\ &= \gamma \pi (\frac{1-\cos \theta}{\cos \theta}) (\frac{2h}{\tan^2 \theta}+\frac{2R}{\tan \theta}) \\ &= 2\gamma\pi (\frac{1-\cos \theta}{\cos \theta}) (\frac{h+R \tan \theta}{\tan^2 \theta}) \\ &\approx 2\gamma \pi R \frac{1-\cos \theta}{\sin \theta} \end{align*}

with the last step following since it is reasonable to assume $$R \gg h$$

The only remaining variable is $$\theta$$ – it is reasonable to assume a small angle $$\theta$$ (i.e. $$\theta \approx \frac{\pi}{10}$$, so that this adds a factor of approximately 0.1). With $$\gamma=72.8$$ dynes per centimeter and $$m=80$$ kilograms, $$R \approx 13 \text{km}$$.