One can modify the zip line problem is the previous post by adding a counterweight m. We learned that the two angles were equal for equilibrium so we can simplify the solution making same assumption here.
The length of the line is now ℓ=a+b+c so we have eliminated one variable and replaced it with another, c. The potential has an additional term for the height of m as well.
We have to modify the potential again to allow an unconstrained variation of the unknown parameters. The constraints here are constants of the variation and we can write them in a form that doesn't change the potential. We again take the derivative of the modified potential, V', and set it equal to zero to find the minimum of V'. If this is true for arbitrary changes in the unknowns their coefficients have to be equal to zero. The first of the three resulting equations gives us the same value for μ as before which allows us to simplify the second equation. We now have two simultaneous equations which allow us to solve for the undetermined multipliers λ and μ. The value of μ allows us to solve for sinθ.
We now have enough information to solve for the values of a and c.
The values of x, y, and sinθ from the previous problem allows us to find the mass m needed to maintain the equilibrium there. The formulas above allow us to solve for the remaining unknowns.
We can plot the data to help visualize the results.
The counterweight allows us to determine that the tension of the line is T=mg for equilibrium. The action of simple machines was all that Lagrange would have needed to develop general methods for mechanics.
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