# Force of constraint in Lagrangian formation

• BearY
In summary, by using the Euler-Lagrange equation and considering a constraining potential, we can show that the normal force from the plane on the mass is equal to mg cos(θ). This is done by setting up a Lagrangian with two equations of motion and two additional equations representing the constraint and the direction of the normal force.

## Homework Statement

A mass m slides down a frictionless plane that is inclined at angle θ. Show, by considering the force of constraint in the Lagrangian formulation, that the normal force from the plane on the mass is the familiar mg cos(θ).
Hint: Consider the Normal force to be the result of a steep constraining potential V(z) confining the mass to the surface of the plane.

## The Attempt at a Solution

This question itself can be solved by using the Euler-Lagrange equation to get force along the plane is ##mgsin(\theta)## and simply knowing the total force is mg. But I don't really know what the hint part means. Assuming the normal force is to be the result of a potential, and then add another coordinate that is perpendicular to the surface and have another Lagrangian for it?

It's possible that what is meant is this:

Use a Lagrangian: ##L = \frac{1}{2} m (\dot{x})^2 + \frac{1}{2} m (\dot{z})^2 - mgz - V(x,z)##

Then the Lagrangian equations of motion give you 2 equations:
1. One involving ##\ddot{x}## and ##F_x = - \frac{\partial V}{\partial x}##
2. One involving ##\ddot{z}## and ##F_z = - \frac{\partial V}{\partial z}##
In addition, we get two more equations:
• The constraint equation: ##\dot{x} = \dot{z} cot(\theta)## (the velocity must be directed down the slide)
• ##F_x = F_z tan(\theta)## (the normal force must be directed normal to the slide)

• BearY
stevendaryl said:
It's possible that what is meant is this:

Use a Lagrangian: ##L = \frac{1}{2} m (\dot{x})^2 + \frac{1}{2} m (\dot{z})^2 - mgz - V(x,z)##

Then the Lagrangian equations of motion give you 2 equations:
1. One involving ##\ddot{x}## and ##F_x = - \frac{\partial V}{\partial x}##
2. One involving ##\ddot{z}## and ##F_z = - \frac{\partial V}{\partial z}##
In addition, we get two more equations:
• The constraint equation: ##\dot{x} = \dot{z} cot(\theta)## (the velocity must be directed down the slide)
• ##F_x = F_z tan(\theta)## (the normal force must be directed normal to the slide)

Actually, since ##\dot{z}## is negative, it should be

##\dot{x} = -\dot{z} cot(\theta)##