Mechanical Energy and Newton's Laws

[DumSpiroSpero]

Homework Statement

A block of mass $m$, attached to a rope, is dropped at the point A.
When the block reaches the point B, the tension is $T = 2 \cdot m \cdot g$ and the rope broke at that same point.
If the length of the rope is $L = 6cm$, evaluate the height $h$ where the rope broke.

Homework Equations

The solution's manual uses mechanical energy and the centripetal force ( $T - m \cdot g \cdot sin\theta = \frac{m \cdot v^{2}}{2}$). Since $h = sin \theta \cdot L$, we find the height $h = 4cm$.

The Attempt at a Solution

I understand and agree with that solution, but I can't figure out why my solution is wrong.
Since $W = m \cdot g$ is perpendicular to the horizontal line, we have a vectorial triangle composed by $\vec{T}, \vec{W}, \vec{F}$ (the green, red and purple vectors in the figure below, respectively), where F is the resultant force of T and W, i.e. $\vec{F} = \vec{T} + \vec{P}$ .
In this triangle, we have $sin \theta = \frac{W}{T} = \frac{W}{2\cdot W} = \frac{1}{2}$ and $sin \theta = \frac{h}{L}$. Therefore, $h = \frac{L}{2} = 3cm$. I couldn't find my mistake.

 

[TSny]

Welcome to PF!

You are assuming the net force, F, on the block is horizontal. If that were true in general, would the block ever have any vertical acceleration?

 

[DumSpiroSpero]

Thank you very much - for the welcome and for the solution. :)

After your comment, I noticed that in my figure the tension and the weight have similar lenghts. I redone it with more precise lenghts (T = 2W) and, indeed, it's impossible to the resultant vector to be horizontal in terms of vectors. (R is the real resultant vector, and F is the wrong one at my first attempt.)

 

[TSny]

OK, good.