Finding the value of the holonomic constraint forces So let's say I have a Lagrangian augmented with some holonomic constraints.
$$L' = L + \sum_i \lambda_i(t) f_i(q,t).\tag{i}$$
The solutions is the system of differential equations:
$$\frac{\partial L}{\partial q_k} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k} = \sum_i \lambda_i(t) \frac{\partial f_i(q,t)}{\partial q_k},\tag{ii}$$
$$f_i(q,t) = 0.\tag{iii}$$
I am not sure about the value of the constraint forces. I think they are:
$$F = \frac{\partial f_i(q,t)}{\partial q_k}.\tag{iv}$$
But I'm far from sure.
 A: Finding the value of the holonomic constraint forces is done as follows.


*

*The equations are:
\begin{align*}
 &\frac{\partial L}{\partial\vec{q}}-\frac{d}{dt}\frac{\partial L}{\partial\vec{\dot{q}}}=\sum_{i}^{n_c}\lambda_i\frac{\partial f_i(\vec{q},t)}{\partial\vec{q}}\quad,\text{($n_q$ equations )}\tag{1}\\
& f_i(\vec{q},t)=0\quad i=1\,,\ldots\,,n_c\quad\text{($n_c$ constraints equations) }\tag{2}\\\\
&\text{If we evaluate equation (1) we get:}\\
&\vec{f}_q(\ddot{\vec{q}}\,,\dot{\vec{q}}\,,\vec{q}\,,\vec{\lambda)}=0
\tag{3}\\
&f_i(\vec{q},t)=0 \tag{4}
\end{align*}
Equations (3) and (4) are $n_q+n_c$ equations for $\ddot{\vec{q}}$ and $\vec{\lambda}$ unknowns.
How to solve:


*Example: Pendulum


\begin{align*}
  &T=\frac{1}{2} m\,\dot{x}^2+\frac{1}{2} m\,\dot{y}^2 \qquad V=\,m\,g\,y\
\qquad L=T-V\\
  &\text{with:}\quad q_1=x\quad q_2=y\quad \Rightarrow\\
  &\frac{\partial L}{\partial\vec{q}}=[0\,,-m\,g]\, ,\qquad \frac{d}{dt}\frac{\partial L}{\partial\vec{\dot{q}}}=
  [m\,\ddot{q}_1\,,m\,\ddot{q}_2]\\\\
  &\text{constraint equation (where $l$ is the pendulum length) :}\\\
&f_1=x^2+y^2=l^2=q_1^2+q_2^2-l^2=0\tag{5}\\
 & \sum_{i}^{n_c}\lambda_i\frac{\partial f_i(\vec{q},t)}{\partial\vec{q}}=\lambda[2\,q_1\,,2\,q_2]\\\\
 &\Rightarrow\quad\text{Insert into equation (3)}\\
 &[0\,,m\,g]-
 [m\,\ddot{q}_1\,,m\,\ddot{q}_2]-\lambda[2\,q_1\,,2\,q_2]=0\quad &\tag{6}\\\\
 &\text{if we differentiate twice  the constraint     equation (5) we get: }\\
 &\frac{d^2}{dt^2}f_1=\frac{d^2}{dt^2}(q_1^2+q_2^2-l^2)=
 2(\dot{q}_1^2\,\ddot{q}_1+\dot{q}_2^2\ddot{q}_2)=0\tag{7}
\end{align*}
We can now solve equation (6) and (7) to get $\ddot{q}_1\,,\ddot{q}_2$ and the generalized constraint force $\lambda$:
\begin{align*}
 &\ddot{q}_1=-g\,\frac{q_1\,\dot{q}_2}{q_1\,\dot{q}_1^2+q_2\,\dot{q}_2}\tag{8}\\
 &\ddot{q}_2=-g\,\frac{q_1\,\dot{q}_1^2}{q_1\,\dot{q}_1^2+q_2\,\dot{q}_2}\tag{9}\\
 &\lambda=\frac{1}{2}\,m\,g\frac{\dot{q}_2}{q_1\,\dot{q}_1^2+q_2\,\dot{q}_2}\\\\
 &\text{The constraint force $F_x$ [N] toward the $x$-axis is:} |F_x|=|\lambda\,2\,q_1|\\
 &\text{and toward the $y$-axis: }\qquad \qquad 
\qquad \qquad \quad\, |F_y|=|\lambda\,2\,q_2|
\end{align*}
A: *

*The $i$'th holonomic constraint
$$f_i(q,t)~=~0\tag{A}$$ can be turned into an (integrable) semi-holonomic constraint
$$ \sum_{k=1}^na_{ik}(q,t)\dot{q}^k + a_{i0}(q,t)~=~0\tag{B}$$ 
by differentiation wrt. time $t$, and by using the identifications
$$ a_{ik}~=~\frac{\partial f_i}{\partial q^k}, \qquad a_{i0}~=~\frac{\partial f_i}{\partial t}. \tag{C}$$

*The generalized forces serve as sources for Lagrange equations (in the same way that forces serve as sources for Newton's second law), cf. e.g. my Phys.SE answer here. The $k$'th generalized constraint force, say $Q^{(c)}_k$, is 
$$ Q^{(c)}_k~=~\sum_i \lambda^i a_{ik}~\stackrel{(C)}{=}~\sum_i \lambda^i \frac{\partial f_i}{\partial q^k}, \tag{D}$$
where $\lambda^i$ is the $i$'th Lagrange multiplier. This answers OP's question. 
References:


*

*H. Goldstein, Classical Mechanics; Section 2.4 (Warning).

