> My grad school was basically working through F=Ma-Cv-Kx in all it's expanding varied glory, working up to full blown finite element analysis.
If I'm understanding correctly, you want another = there for a standard damped spring: F = ma = -Cv - Kx.
> My grad school was basically working through F=Ma-Cv-Kx in all it's expanding varied glory, working up to full blown finite element analysis.
If I'm understanding correctly, you want another = there for a standard damped spring: F = ma = -Cv - Kx.
F is a forcing function, not the resultant force. It’s often arranged this way with all the derivatives on one side (as opposed to having the resultant force, ma, on one side of the equality by itself) so that it matches the general form of a non-homogeneous second-order linear differential equation.
(At least I assume this is what the original commenter meant!).
Ah, that makes more sense. Thanks!