Lagrangian formalism application on a particle falling system with air resistance I have this problem, with a first-step resolution:


Obtain the equation of motion for a particle falling vertically under the influence of gravity when frictional forces obtainable from a dissipation function $\frac12kv^2$ are present. Integrate the equation to obtain the velocity as a function of time and show that the maximum possible velocity for a fall from rest is $v+mg/k$.

Answer:
Work in one dimension, and use the most simple Lagrangian possible:
$$ L = \frac 12 m \dot z^2 - mgz$$
With dissipation function:
$$ F=\frac 12 k \dot z^2 $$
The lagrangian formulation is now:
$$ \frac{d}{dt} \frac{\partial L}{\partial \dot z} - \frac{\partial L}{\partial z} + \frac{\partial F}{\partial \dot z} = 0 $$

So, I just don't know why they put the term $\frac{\partial F}{\partial \dot{z}}$ in Lagrange's equations. Why? I know that the Rayleigh 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 introduce it in 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 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?
 A: I would say that this exercise in particular has nothing to do with constraints, because the existence of a dissipation force is not giving you any type of constraint in the coordinates - I mean, it certainly isn't holonomic since it doesn't eliminate any degree of freedom from the problem. It is just more information on how the system behaves along the only coordinate that describes it - z, in this case.
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:
\begin{equation}
\frac{d}{dt}\frac{\partial L}{\partial \dot{z}} - \frac{\partial L}{\partial z}=Q_j
\end{equation}
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.
