If you are assuming a parabola, then a little work will show that the equation is
y = ax^2+bx+c
2500a-50b+c = 50
2500a+50b+c = 70
a(-b/2a)^2 + b(-b/2a) + c = 47
y = 1/200 x^2 - 1/5 x + 47
and the vertex is at (20,47)
Now, the length of cable is thus
s = ∫[-50,50] √(1+(x/100-1/5)^2) dx
= 105.8
The curve of a freely hanging cable of uniform density, however, is not a parabola, but a catenary of the form
y = a cosh((x-h)/a)
This gets rather complicated, and the parabola is a good approximation anyway, so let's stop here.
A cable is suspended from two towers with a distance of 100m between them. The 1st tower has
a height of 50m and the 2nd tower has a height of 70m from the ground. The lowest part of the
cable has a distance of 47m from the ground. Find the length of the cable
1 answer