Why is the self-weight of a cable not uniformly distributed? I have learned that a free cable that is hanging with only its self-weight to consider will form a caternary while a cable with a uniformly distributed load forms a parabola. Why is the self-weight of the cable not considered to be uniformly distributed? Is it because "uniformly distributed" is defined with respect to the span?
 A: Yes, because when they say "uniformly distributed" they mean uniformly distributed along the horizontal direction.
So if the cable (in the center of the span) is horizontal it takes 1 m of cable to span 1 m of horizontal distance.
But if the cable is angled at 45 degrees (near a support) it takes 1.4 m of cable to span 1 m of horizontal distance, and 1.4 m of cable weighs 1.4 times as much as 1 m of cable.
A: Yes, uniform load means constant-valued $F(x)$ pointing down. I guess the easiest way to think about it is that when you consider self-weight, the mass is distributed along an arclength $dm = \lambda ds$ (where $\lambda$ is some constant) and therefore the weight is distributed as $dF = g \lambda ds$ but $ds = \sqrt{dx^2 + dy^2} = \sqrt{1 + [y'(x)]^2}dx$ after some very useful abuse of notation. $y'(x)$ is the derivative of the shape of the cable $y(x)$ and if there is any curvature, that is, if $y'(x)$ is not constant and there is some value $y''(x) \neq 0$, then the weight varies with $x$.
