A catenary curve is the curve followed by a rope suspended at both ends in uniform gravity.
I thought I would try to solve it myself because it seemed like a good challenge, but almost immediately I got stuck on this question.
Given that nothing is moving, we can say
$$T_{left} + T_{right} + F_g = 0$$
where $T_{left}$ is the tension along the rope to the left and $T_{right}$ is the tension along the rope to the right. Fine and dandy.
Now given that the catenary curve is smooth between (but not including) its endpoints, $T_{left}$ should then always be colinear with $T_{right}$. So then the x-direction components cancel out and the net of the y-direction components cancels with gravity.
But what about at the very lowest point on the catenary curve? Intuitively, this should be the point where the tangent is perfectly orthogonal to gravity, and as such $\hat{T}_{right} = - \hat{T}_{left} = \hat{i}$.
Since both $T_{left}$ and $T_{right}$ are orthogonal to $F_g$...
what holds up the lowest point of a rope following a catenary curve?