Dimensional analysis: a particular problem I don't know how to solve I have the following configuration:

in which a viscous fluid with dynamical viscosity $\mu$ and density $\rho$ slides down the inclined plane due to gravity $g$. 
After having solved the Navier-Stokes equation for this extremely symmetric and idealized situation I am asked to find the force per unit area the inclined surfaces feels using dimensional analysis, and then to compare the result with the analytical one.
What I did was to build a "table of units" as follows:

where the leftmost column L, M and T stand for units of length, mass and time and the uppermost row indicates each of the parameters of the problem. $F$ is the net force, not per unit area. And $\nu = \mu / \rho$ is the kinematic viscosity. Watching at this table it seems obvious to propose that $F \propto p_0$. Thus $$ F = p_0 d^\alpha g^\beta \nu^\gamma $$ from where I obtain the following equations for each unit

This tuns to be an undetermined system so one of the exponents in $ F = p_0 d^\alpha g^\beta \nu^\gamma $ is free. 
How do I know which one is free? Is all this ok?
Thanks!
 A: Dimensional analysis usually requires you to think about which physics is important to the problem.
In this case you are asked to find the force per area on the ramp, and I assume that means the force in the x direction due to viscosity. That force is equal to the force exerted by the ramp on the fluid.
Now the fluid is in steady state so the sum of all external forces is zero. The external pressure $p_0$ produces a force in the y direction not the $x$, so it's not important here. (By the way you could know that $F_x \propto p_0$ is wrong because even if $p_0=0$ the force on the ramp definitely should not be zero). The only other external force is gravity which depends on the mass and $g$. $\mu$ should not appear in your result since that describes the internal forces.
Now you only have $\rho, d, g$ to play with and dimensional analysis leads to a unique answer up to a dimensionless constant.
A: A force per unit area has dimension $L^{-1}MT^{-2}$, as does $\rho(\nu/d)^2$. Since $d^3g/\nu^2$ is dimensionless, $\rho d g$ has the right units too. As octonion has noted, we expect a result $\propto dg$.
