A flexible chain of length L is suspended between two poles of equal height separated by a distance 2M (Figure 10). By Newton’s laws, the chain describes a curve (called a catenary) with equation y = a cosh(x/a) + C. The constant C is arbitrary and a is the number such that L = 2a sinh(M/a). The sag s is the vertical distance from the highest to the lowest point on the chain.

Assume that M = 50 and L = 160. In this case, a CAS can be used to show that a ≈ 28.46.
(a) Use Eq. (6) and the Linear Approximation to estimate the increase in sag if L is increased from L = 160 to L = 161 and from L = 160 to L = 165.

(b) If you have a CAS, compute s(161) − s(160) and s(165) − s(160) directly and compare with your estimates in (a).

Transcribed Image Text:

y 1 1_ 2 M y = a cosh(x/a) S 11 -x