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I've got a simple mathematical question. I was studying the Lagrangian approach of classical mechanics and in this part I had the intention of proving that the differential of the Lagrangian is equal to $-Fr.dr$, where $Fr$ is the restriction force and $dr$ is an infinitesimal displacement orthogonal to $Fr$, this would result in the differential being 0 and the proof that the Lagrangian method works even with constraint forces, like in the section 7.4 of the book Classical Mechanics from John Taylor. But I think I am missing something, a plus or minus sign from somewhere, and not being able to find the solution. If someone could help.

part of the proof

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If you want to proof that the Lagrange method also works for constraint forces I think you're approaching the problem the wrong way. I recommend you use not L but rather L´, such that

$L' = L + \sum_{j=1}^{m} \lambda_j f_j(x,y, t) $

, where f(x,y,t) is a holonomic constraint (this wasn't specifically mentioned in your question, but I assume that these are the constraints you are talking about) and $\lambda$ is a lagragian multiplier. Lets now say that for the sake of this proof m = 1 (only one constraint force), so we get

$L' = L + \lambda f(x,y, t) $

we now need

$\delta S = \delta \int_{t1}^{t2} (L + \lambda f(x,y, t)) dt = 0 $

in order for the motion to happen. We know that if we had no constraints this equation is true iff

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

, where $q_i$ is a coordinate of the system. So what we do in order to account for the constraint force is calculate $\lambda$ as a new coordinate to the system. We hence get L' = L'(x,y,$\lambda$,t). So in addition to $q_1 = x$ and $q_2 = y$ we simply let $q_3 = \lambda$. We thus get that

$\delta S = 0 $

if additionally

$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot\lambda} \right) - \frac{\partial L}{\partial \lambda} = 0$.

So the Lagragian methods works if any such constraint Force can be written in the form

$\vec{F} = \lambda \vec{\nabla}f$

This is true because f has to describe a plane and the Force can not contribute to the acceleration in this plane, which I assume is what you mean by "Fr.dr = 0". So the Force $\vec{F}$ has to be proportional to the normal vector of the plane. It is now easy to show that this normal vector is always parallel to $\vec{\nabla}f$. So that the constraint force can be always written as stated above and this means that you get equations of motion from the lagragian method even when there are constraint forces.

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