A Zip Line Problem

  The section on finding the minimum of a function in a mechanical problem is a little difficult to follow in Lagrange's Analytical Mechanics so an example to illustrate the procedure may be helpful. Let's consider a zip line with a weight suspended on it and see if we can find the equilibrium position. We can set up a coordinate system with the origin at the lowest anchor for the line and the coordinates for the higher anchor will be (L,h). The length of the cable is ℓ=a+b and a point on the line can be designated by the coordinates (x,y) as shown.

Two unit vectors e₁ and e₂ are needed to simplify the specification of the problem and that the sum of the two vectors a and b will always equal the position of the higher anchor. The function that we want to minimize is the potential V. We can use the length of segment b to show that length of segment a is a linear function of the coordinates.

We can also show that the weight moves along an elliptical path by substituting the formulas for x and y as a function of a and θ into the linear equation for a and then solve the result for a.

The variation of the potential function V is subject to two constraints which are the components of the equation for the sum of the two segments. Multiplying the constraints by two undetermined multipliers assumed to be known constants and adding to the potential V we get a modified potential function V' for which we want the minimum. Taking the derivative of V' an setting it equal to zero we get three equations which we need to solve for the three unknowns a, θ, and φ.

The last equation tells us that the dot product of the second unit vector with a vector formed from the Lagrange multipliers is zero so we know it is perpendicular to this vector. Dividing by the square root of the sum of the squares of the multipliers gives us the second unit vector. The same trick works for the second equation and we get the components of the first unit vector in terms of the multipliers. Substituting these results into the first equation allows us to determine the value of μ which in turn allows us to show that the two angles are equal at the equilibrium position.

This result also allows us to simplify the equations for sum of the segments and solve for the unknowns in terms of the known quantities.

We are now in a position to do a numerical problem and find the values of x and y for the equilibrium position,

and plot the results.