How to find the equation with dimensional analysis in this case?

Question

I'm doing some exercises in dimensional analysis and don't know how to proceed.

The problem is: use the dimensional analysis to find the terminal velocity $v$ of a free-faller. This speed depends on his mass $m$, his cross section $S$, air density $\rho$ and acceleration $g$.

If I try to set $ v = m^{\alpha} S^{\beta} \rho^{\gamma} g^{\delta} $, what I get by comparing the units is a system of three equations with four unknowns. Namely:

$1 = 2 \beta - 3 \gamma + \delta$

$-1 = -2 \delta$

$ 0 =\alpha + \gamma $

I can only solve for $\delta$. How to proceed? What can I do with the other unknowns?

Answer 1

I can suggest two different ways to find the answer.

First, as was mentioned in the comments, is to use the further assumption that according to the equations of motion, $g$ and $m$ always appear in the combination $mg$.

Second, which is somehow more general and is used in fluid mechanics. We introduce two different dimensions for length: One in the direction of motion which we cal $L_{||}$ and the other one in the perpendicular direction which we call $L_{\perp}$. So, the lengths in cross section are different from those that appear in $g$ and $v$. While for $\rho$ as it is proportional to inverse volume, and for volume: $$[V]=L_{||}L_{\perp}^2$$ Then, intuitively, we may write: $$[v]=L_{||}T^{-1}$$ $$[g]=L_{||}T^{-2}$$ $$[S]=L_{\perp}^2$$ $$[\rho]=ML_{||}^{-1}L_{\perp}^{-2}$$ $$[m]=M$$

And we will have an extra equation to fix the exponents.