# What are the forces of constraint if there are multiple equivalent constraints?

Suppose a large (rigid) block is sitting on top of two smaller blocks of equal height $1$, both of which rest on the ground. We wish to find the position of the block (easy) and the forces of constraint on the block coming from the two smaller blocks.

If we write the Lagrangian along with the equations of constraint $z=1$, we can find the motion of the object using Lagrange multipliers, and we can also find the force of constraint. If there were constraints $\{G_n\}$, then the force of constraint in the $z$ direction would be $F_{n} = \lambda_n \frac{\partial G_n}{\partial z}$ on the constraint $G_n$. However, this tells us nothing about how much force each individual block is exerting.

We know from experience that the upward force from the smaller blocks must equal $mg$, where $m$ is the mass of the larger block, and furthermore, that the force is evenly distributed, so that each of the two smaller blocks exerts an equal amount of force.

It seems natural to indicate the existence of two blocks by writing down the constraint $z=1$ twice and then trying to solve the resulting system of equations. However, this doesn't seem to get us very far.

So, how is it possible to detect the fact that the force on the two smaller blocks is distributed equally?

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Can you try to explain the situation more clearly? Are you talking about statics or dynamics? You say you want to find the Lagrangian, but it sounds like you've just got one block sitting dead still on top of two other blocks; nothing will move and no Langrangians are necessary. Where are the smaller support blocks located? Unless they're distributed symmetrically under the big block it's not true that the forces on them are the same. Also, it's obvious that writing the same equation down twice won't give you any new information. What degrees of freedom do you want to constrain? –  Mark Eichenlaub May 31 '11 at 13:59
You can disambiguate the problem by adding the constraint that the angle the upper block makes with some reference is constant: $\dot{\theta} = 0$. This is equivalent to requiring that the sum of the torques on a static member be zero in Newtonian statics. –  dmckee May 31 '11 at 14:38

You can still do this with Lagrangians, even if it's the hard way around. I define $z_1$ and $z_2$ as the height of the boxes above the table, and the constraints are $f_1 = z_1-\frac{h}{2}=0$ and $f_2 = z_2-z_1-\frac{h}{2}=0$, or "the bottom box sits on the table and the top box sits on the bottom box." I assumed that the masses are equal, but you don't have to. Write the Lagrangian without assuming anything about the motion: $$L = T-V = \frac{m}{2}(\dot{z_1}^2 + \dot{z_2}^2) - mg(z_1+z_2)$$
From the Lagrange multiplier method: $$m\ddot{z_1}+mg + \lambda_1 \cdot 1 + \lambda_2 \cdot-1 = 0$$ $$m\ddot{z_2}+mg + \lambda_2 \cdot 1 = 0$$
These lead (after applying the condition that $\ddot{z_1}=\ddot{z_1}=0$) to $F_1 = \lambda_1\frac{\partial f_1}{\partial z_1}=\lambda_1\cdot1=-2mg$ and $F_2 = \lambda_2\frac{\partial f_2}{\partial z_2}=\lambda_2\cdot1=-mg$