# Dimensional analysis: a particular problem I don't know how to solve [closed]

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!

• If one of the variables is free, you just sum over all possible configurations of the variable with undetermined coefficients. physics.stackexchange.com/questions/389586/… might help you Mar 8, 2018 at 4:14
• I didn't see $\rho$ in your table.
– JEB
Mar 8, 2018 at 4:37
• Are you trying to find the normal component of the force or the shear component? Mar 8, 2018 at 12:24

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 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 $\propto dg$.