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?
The towers are 50 m tall
The cable is 80 m with its low point 10 m above the ground.
That's 40 down, and 40 up.
The towers are zero meters distance from each other.
(The category is Tricks)
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Posted by Larry
on 2024-03-27 14:21:04 |
(spoiler)
