# Lagrangian Constraint Forces

• I
Eh6794
TL;DR Summary
I am confused about the constraint I choose for my mechanics problems.

Let's say I have an inclined plane, a ball rolling off of a half hemisphere or a pendulum and I need to find a tension, friction or normal force.

After I derive the lagrangian, I need to find a constraint, but how do I choose a constraint?

I have worked through a couple of problems with solutions, but it seems like I find the coordinate/variable in the direction of the force I want? Ex. pendulum tension would be L?
ii

LCSphysicist
Summary:: I am confused about the constraint I choose for my mechanics problems.

Let's say I have an inclined plane, a ball rolling off of a half hemisphere or a pendulum and I need to find a tension, friction or normal force.

After I derive the lagrangian, I need to find a constraint, but how do I choose a constraint?

I have worked through a couple of problems with solutions, but it seems like I find the coordinate/variable in the direction of the force I want? Ex. pendulum tension would be L?

ii
"After I derive the lagrangian, I need to find a constraint, but how do I choose a constraint?" Maybe |R| = cte?
That is, ##r = a## => ##r-a = 0##, it is a good holonomic constraint. And i don't know another constraint in this cases you cited. You find a constraint studying the restrictions of the system, let's say like this.

"I have worked through a couple of problems with solutions, but it seems like I find the coordinate/variable in the direction of the force I want? Ex. pendulum tension would be L?"
I do not understand. What do you mean?

You choose the constraint that, ahem, constrains your generalized coordinates not to be just anything. For example, in the case of a block sliding down a frictionless inclined plane you can choose generalized coordinates ##x## and ##y## along the vertical and the horizontal and write $$\mathcal L=\frac{1}{2}m\dot x^2+\frac{1}{2}m\dot y^2+mgy.$$The constraint is that the block must stay on the plane so if you pick a value for one coordinate, the other is strictly specified. Here the relevant constraint is ##y=x~\tan\theta## where ##\theta## is the angle of the incline relative to the horizontal. Then the Lagrangian becomes $$\mathcal L=\frac{1}{2}m\dot x^2+\frac{1}{2}m\dot x^2\tan^2\theta+mgx\tan\theta.$$