SI vs. Imperial: Can lbf be broken down into simpler units?

[Philly]

So we're working with the force of lift generated by airfoils and I ran into a bit of an issue with units:

The equation we are working with is

, except solved for F(lift).

In the SI system, after all the simplifying, your units will break down to (kg x m)/(s^s) which is also equal to one Newton. This makes sense as the Newton is the basic unit of force in the SI system, and F(lift) will often times be given in Newtons.

In the Imperial system, running the equation will produce the units (lb x ft)/(s^2), the same dimensions as the SI system, but the units for force in the Imperial system are in pounds, which are already in the result above. Are these even equal? This just seems strange.

Or, upon further looking, do I need to go to pounds mass to get to ([M][L])/([T]^2) dimensions? How would I go about such a conversion?

The use of the Imperial system in Engineering here after years of SI in sciences is really getting rather frustrating....

 

[Azuremen]

Trust me, you aren't alone. Pounds mass versus pounds weight is about the most annoying thing to deal with in engineering. In this context, it looks like pounds mass, though working with slugs instead of the annoyingly vague pound.

If I recall correctly, a pound mass = a pound force, just pound force represents(m * a) and pound mass (m), assuming you are on earth and working with standard gravity constants, so 9.81 m/s^2 (like hell I remember it in imperial )

 

[Kylehnat]

Whoever decided that lbs should be force AND mass is probably the same lazy-ass who decided that ounces were good enough to describe both a mass and a volume.

As I recall, lbm and lbf are equivalent on Earth (g = 32.2 ft/s^2), and anywhere else in the universe, you have to use a ratio of g values to convert. It's incredibly pointless and unnecessary.