You can show that it has to be that way if you reduce it to dimensional analysis. All units resolve to one of three dimensions - kilogrammes (mass), metres (distance) or seconds (time).
So lift force has dimensions of N. One Newton is the force to accelerate 1kg at 1m/s/s. So the dimension is kg.m.s^-2
Now look at the basic equation of lift, L=0.5 Rho V^2 S Cl
Rho, density, is kg.m^-3
V, speed, is m.s^-1
S, area is m^2
So take the right hand side of the equation and you get
0.5 x kg.m^-3 x (m.s^-1)^2 x m^2 x Cl
You can actually ignore the 0.5 and the Cl at the moment as we're only interested in dimensions. So, ignore those, and multiply out, we get the dimensions for the right hand side of the equation as...
kg.m^-3.m^2.s^-2.m^2
Which is kg.m.s^-2
Which has the same dimensions as force, hence the dimensions of the equation work out. If it was ony V on the right hand side, it wouldn't.
Not for the first time, I wish that Pprune would let me type equations like I can in MS Word !
This is actually a pretty standard physicists trick. If you know that one term must be a function of some other terms, in this case...
Lift =F(speed, air density, aeroplane size, something to do with shape)
You reduce speed, density and size to dimensions, and throw in a coefficient, which here we call coefficient of lift nad has no dimensions. Then doing it this way you can work out what you *think* the dimensions look like, and from there you throw in a load of experimental data and start trying to see (a) if you are right, or need to go back and re-think the whole thing, and (b) what values your coefficient has. The same basic approach was used to work out the dimensions, and thus units, for things like Reynolds Number, viscosity, Strouhall number and dozens of others.
G
Last edited by Genghis the Engineer; 4th Nov 2017 at 11:17.