Grashof number as a ratio of buoyant and viscous forces The Grashof number is supposed to be a ratio of buoyant forces to viscous forces. 
I find this hard to believe, since if
$$F_b=\beta g \rho \Delta T$$
is the buoyancy force, the definition of the Grashof number,
$$\text{Gr}=\frac{\beta g\Delta T L^3}{\nu^2},$$
implies that the viscous force is something like $\frac{\rho}{L^3}\nu^2$, instead of something linear in $\nu$. How is this supposed to be the viscous force?
 A: I don't agree with @Pirx that it is to be understood as vague metaphors although I admit it sometimes is a little bit difficult to understand exactly how they are ratio of scales as you have clearly found out.
What makes it a bit difficult is that dimensionless numbers are sometimes themselves ratios of other dimensionless numbers. For example the definition of $Gr$ can be rewritten as:
$$Gr=\frac{\beta g\Delta TL^{2}}{\nu U}\frac{UL}{\nu}$$
clearly we see a role for the Reynolds number here:
$$Re=\frac{UL}{\nu}=\frac{\rho U^2/L}{\mu U/L^2}=\frac{inertial}{viscous}$$
The other term in $Gr$ is easily decomposed:
$$\frac{\beta g\Delta TL^{2}}{\nu U}=\frac{\beta \rho g\Delta T}{\mu U/L^2}=\frac{bouyancy}{viscous}$$
A: Don't take those intuitive notions of dimensionless numbers as ratios of forces too seriously. Those kinds of statements are to be understood as vague metaphors more than anything else. 
But, clearly the expression $\frac{\rho}{L^3}\nu^2$ has the dimension of a force, and clearly this force depends on viscosity. That's pretty much all there is to say about this. How exactly viscous and buoyancy forces arise in convection problems depends on the boundary conditions and will be complex in general.
A: you both are right. I think the closure point is that the velocity $U$ which is normally the velocity outside the boundary layer (BL) in the forced convection problem, here is defined in terms of the buoyancy force. This comes in when going from the dimensional form of the BL equations for natural convection to the non-dimensional form. Here,
$$u_o=[gβ(T_s-T_∞)L]^{1/2}$$
In order to clean the first term of the RHS of the momentum equation, see Page 565 of
Incropera F.P., Dewitt D. P., Bergman T. L., Lavine A. S., Fundamentals of Heat and Mass Transfer, John Wiley & Sons, 6th Ed, 2007.
So at the end of the day, although $U$ defines $Re$, true, this is a $Re$ number based on a $U$ velocity promoted by the buoyancy force.
Lastly, thank you guys. You help me to explain in detail this to my students.
