perplexus.info :: Tricks : Cables II

Cables II (Posted on 2024-03-27)

In the problem Cables provided by K Sengupta, the solution required the knowledge that a hanging flexible cable takes the form of a catenary, the general equation of which is y=a*cosh(x/a), combined with calculating the overall length of the catenary between the two fixed points of the cable.

This time we have a cable strung between two towers on a flat level plane. The low point of the cable is 10m above the plane. Later, the cable heats up and expands to a length of 80m. The towers also expand and gain 0.2m in height – they are now exactly 50m high. The low point of the cable is still 10m above the plane. How far apart are the towers?

Submitted by Kenny M

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Solution: (Hide)

This problem is full of red herrings. The only possible solution is that the towers are zero distance apart. After heating, the cable drops 40m from the top of one tower and rises 40m back to the top of the other, which is 80m in total length. The low point is 10m above the plane (50-40=10). Whatever happens before heating does not impact the solution because not enough info is given to constrain the problem any further.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
(spoiler)	Larry	2024-03-27 14:21:04
solution	Charlie	2024-03-27 14:20:36

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