The classic falling rope problem 
A rope was on top of a table with a part of it hanging off the edge, without any friction.
The equation of the motion is:
\begin{equation}
\frac{mgx}{L}=m \frac{dx^2}{dt^2} \to \ddot{x}=\frac{gx}{L}
\end{equation}
In which $x$ is the length of rope hanging over the edge of the table, and the $L$ is the total length of the rope.
My question is: why is the tension force not considered? Since, the rope that is still on top of the table, needs a force to start it's movement, and I was thinking that this force should be considered on the vertical part of the rope as well.
If we think in a similar problem, in which there is two cubes connected by an ideal rope and a pulley, and one of them is hanging off the table, the tension should be considered. But, in the case that is just the rope (with mass), there is no traction, and I want to know why?
 A: The tension force is not relevant because we are considering the motion of the chain (mass $m$) as a whole, but the tension is internal to the chain, that is if one part, A, of the chain exerts a force $\mathbf F$ on another part, B, then B exerts a force $-\mathbf F$ on A so there is no resultant force on the chain as a whole due to tension.
Things are a little more complicated at the edge of the table, where the chain bends through a right angle. [The edge of the table may be taken as shaped like a quarter-circle in section.] While it is true that the vertical and horizontal parts of the chain exert forces of equal magnitude on each other, we cannot justify this from Newton's third law, as the forces are at right angles to each other. A full analysis would include the normal contact force of the table-edge on the chain, but would lead to the same result: the vertical and horizontal parts of the chain exert forces of equal magnitude on each other.
A: This problem is discussed here and here, along with the various incorrect solutions, including the one you suggest.
