Here is another way to look at where $L$. For a dissipative system it is more natural to write the Lagrange just in terms of the kinetic energy $T$ instead of $L$:

$$ \frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial T}{\partial q_i} = F_i = \mathbf{F} \cdot \frac{\partial \mathbf{r}}{\partial q^i} $$

This is the equation that the Beers actually derives.

It is the general form. The conservative case is obtained by setting $ \mathbf{F} = -\nabla V$.

$$ \begin{align}

\frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial T}{\partial q_i} &= - \frac{\partial V}{\partial q_i}, \text{ since } \frac{\partial V}{\partial q_i} = \nabla V \cdot \frac{\partial \mathbf{r}}{\partial q^i} \\

\therefore \frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial (T - V) }{\partial q_i} &= 0 \\

\frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial L }{\partial q_i} &= 0

\end{align} $$

Now $\frac{\partial V}{\partial \dot{q}_i} = 0 $ since by definition $V$ is a function of only the $q_i$s and thus independent of $\dot{q}_i$, so

$$ \begin{align}

\frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial V}{\partial \dot{q}_i} \right) - \frac{\partial L }{\partial q_i} &= 0 \\

\frac{d}{dt} \frac{\partial (T - V)}{\partial \dot{q}_i} - \frac{\partial L }{\partial q_i} &= 0 \\

\frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L }{\partial q_i} &= 0 \\

\end{align} $$

Seen this way $L$ looks like a bit of a hack.

It is really an artifact of the system being conservative and even then replacing $T$ by $L$ in the first term of the equation is kind of redundant.

$$ \frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial T}{\partial q_i} = F_i = \mathbf{F} \cdot \frac{\partial \mathbf{r}}{\partial q^i} $$

This is the equation that the Beers actually derives.

It is the general form. The conservative case is obtained by setting $ \mathbf{F} = -\nabla V$.

$$ \begin{align}

\frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial T}{\partial q_i} &= - \frac{\partial V}{\partial q_i}, \text{ since } \frac{\partial V}{\partial q_i} = \nabla V \cdot \frac{\partial \mathbf{r}}{\partial q^i} \\

\therefore \frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial (T - V) }{\partial q_i} &= 0 \\

\frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial L }{\partial q_i} &= 0

\end{align} $$

Now $\frac{\partial V}{\partial \dot{q}_i} = 0 $ since by definition $V$ is a function of only the $q_i$s and thus independent of $\dot{q}_i$, so

$$ \begin{align}

\frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial V}{\partial \dot{q}_i} \right) - \frac{\partial L }{\partial q_i} &= 0 \\

\frac{d}{dt} \frac{\partial (T - V)}{\partial \dot{q}_i} - \frac{\partial L }{\partial q_i} &= 0 \\

\frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L }{\partial q_i} &= 0 \\

\end{align} $$

Seen this way $L$ looks like a bit of a hack.

It is really an artifact of the system being conservative and even then replacing $T$ by $L$ in the first term of the equation is kind of redundant.