A flexible cable was hung across the Black Canyon between two points that were exactly 1 km apart and at the same elevation.
During the cool night the cable length was calculated to contract by .2 meters. The cable dip was actually measured to decrease by .2 meters!
What is the length of the cable after cooling?
A flexible cable always takes the shape of a catenary of the form y=Acosh(Bx). Put x=0 at the midpoint between the towers, then the "dip" in this case is the difference in y for x=+-500 vs. x=0 (units in meters). However, this alone leads to an infinite number of possible catenaries.
It's the "dip" change as a result of the length change that then defines a single solution. The length of the cable in meters is 2*integral (500 to 0) of [sqrt(1+(dy/dx)^2)] dx (calculus arc length formula), y defined above.
What needs to be done is to use these formulas twice, once with the "initial" cable length, say "a", and that "dip", and then with each of those variables decreased by 0.2m (i.e. a-0.2)
This seems a lot more difficult than a 3. I don't believe there is really any "aha" that would simplify this procedure.
Edited on March 22, 2024, 1:14 pm
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Posted by Kenny M
on 2024-03-22 13:04:26 |
Steps to a solution
