Community Tip - Did you get called away in the middle of writing a post? Don't worry you can find your unfinished post later in the Drafts section of your profile page. X
I waited, waited for an answer, and decided to solve this problem myself
Maple
Mathcad (full solution in attach - pdf and Prime 6)
Well, it's child's play, just solve the following differential equation, obtained by deriving twice the given function:
Thanks!
But I think about it
but this is a problem of a different nature from the previous one. Now you want the tension in the belt.
There were solutions posted years ago based on the fact that the catenary tension is a constant.
yes, I vaguely remember.
The tension has to broken into two forces, horizontal and vertical. This is for each point along the catenary. The horizontal force is constant, the vertical force varies along the catenary. The solution is best solved, for starters, with the lower point being tangent to the x-axis. This is typical in ship anchor or mooring systems.
Cheers, David
in fact the horizontal component of the tension along the catenary should be constant.
The weight of the rope per meter (kg / m) is p, the tension along the catenary with equation y(x), is: T = p.y(x).
I waited, waited for an answer, and decided to solve this problem myself
Maple
Mathcad (full solution in attach - pdf and Prime 6)
with old good symbol math
New formula!!!