# Lagrangian formalism application on a particle falling system with air resistance

I have this problem, with a first-step resolution:

$$...$$

So, I just don't know why they put the term $\frac{\partial F}{\partial \dot{z}}$ in Euler-Lagrange's equations. Why? I know that the dissipation function isn't a conservative force, but I don't know why the partial derivation. For holonomic constraints we need to partially derivate the function of constraint $f=0$ in order to $q$, the generalized coordinate: $\frac{\partial f}{\partial{q}}$. And we itroduce it in Euler-Lagrange's equation multiplied by Lagrange's multiplier $\lambda$, on RHS.

But we have here some kind of constraint with a velocity $\dot{z}$ dependence. That's why we need to put the term $\frac{\partial F}{\partial \dot{z}}$ in Euler-Lagrange's equations? But the term isn't null and they didn't had the Lagrange multiplier, so is it true that it isn't relationated to the constraints formalism? I'm a little bit condused...

• Comment to the post (v2): In a nutshell, OP's post seems to confuse (i) Lagrange's equation $\frac{d}{dt}\frac{\partial T}{\partial \dot{q}^j} - \frac{\partial T}{\partial q^j}=Q_j$ and (ii) Euler-Lagrange's equation $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}^j} - \frac{\partial L}{\partial q^j}=0$. The former does not necessarily arise from an action principle, while the latter always does. The former eq. is relevant for the Rayleigh dissipation function. – Qmechanic Apr 5 '15 at 19:47

The dissipation function is a bit like a potential. If your particle were only influenced by the gravitational force, you would say that $L=T-U$ where $U$ is the gravitational potential and the Euler-Lagrange equations would be the normal ones. But as you have a friction force, the Euler-Lagrange equations become: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{z}} - \frac{\partial L}{\partial z}=Q_j$$
where $Q_j$ is the generalized force and can be shown by its definition from the D'Alembert's principle that $Q_j=-\frac{ \partial F}{ \partial \dot{z}}$ in this particular case. Indeed, if you have a dissipation function, that's always the form of your equations I think.