Caternary Action

Factual Questions
1

This is not a question from a textbook. This is a question about the textbook method! An iterative method is provided to find the tension (T) in a caternary cable. Following the example, a spreadsheet was built to emulate the example thus validating the spreadsheet. Yet, as sag increases to a ridiculous amount (shown below), the resulting tension seems ridiculous. Consider:

Case 1) A 2 lb/ft cable spans a 10-ft gap with a 1-ft sag, and T = 27 lbs.
Case 2) A 2 lb/ft cable spans a 10-ft gap with a 100-ft sag, and T = 225 lbs? (i.e., I picture a large, slack “U” shape)

For Case 2, I would expect an abundance of slack! Thus, T might drop to zero! WTF?
Any civil engineers that may be able to explain what is happening?

2

You’re forgetting that the 100-ft sag means that the rope must be in excess of 200 ft long (but no more than 210 ft). Each side has to support half the weight, which is somewhere between 200 and 210 lbs. Add a bit for the lateral component of the tension, and I can easily imagine 225 lbs.