A chain or rope of indefinite length passes freely over pulleys at heights y1 and y2 above the plane surface of Earth, with a horizontal distance x2 − x1 between them. If the chain or rope has a uniform linear mass density, show that the problem of finding the curve assumed between the pulleys is identical with that of the problem of minimum surface of revolution. (The transition to the Goldschmidt solution as the heights y1 and y2 are changed makes for a striking lecture demonstration. See Exercise 1.)
Exercise 1
The broken-segment solution described in the text (cf. p. 42), in which the area of revolution is only that of the end circles of radius y1 and y2, respectively, is known as the Goldschmidt solution. For the symmetric situation discussed in Exercise 2, obtain an expression for the ratio of the area generated by the catenary solutions to that given by the Goldschmidt solution. Your result should be a function only of the parameters k and α. Show that for sufficiently large values of α at least one of the catenaries gives an area below that of the Goldschmidt solution. On the other hand, show that if α = α0, the Goldschmidt solution gives a lower area than the catenary.
Exercise 2
In Example 2 of Section 2.1 we considered the problem of the minimum surface of revolution. Examine the symmetric case x1 = x2, y2 = −y1 > 0, and express the condition for the parameter a as a transcendental equation in terms of the dimension- less quantities k = x2/a, and α = y2/x2. Show that for α greater than a certain value α0 two values of k are possible, for or α = α0 only one value of k is possible, while if or α < α0 no real value of k (or a) can be found, so that no catenary solution exists in this region. Find the value of α0, numerically if necessary.
We need at least 10 more requests to produce the solution.
0 / 10 have requested this problem solution
The more requests, the faster the answer.