# How do we prove the downwards force of a massless rope on a pulley it wraps around is $2T$?

Consider the system of the pulley, plus all of the rope directly touching the pulley. The net force on this system must be zero, but it experiences two downward forces totalling $$2T$$. This must be balanced by an upward force $$2T$$ exerted by the hand. There is absolutely no need to consider the forces between the rope and pulley here, because they are internal forces.

Now consider the system of the pulley alone. It experiences an upward force $$2T$$ from the hand, so it must also experience a downward force $$2T$$ from the rope directly touching it, by normal force.

This is completely rigorous, no integration required. But if you insist on an explicit derivation, simply note that the normal force per angle $$d\theta$$ is $$T \, d\theta$$. Furthermore, the vertical component of the normal force is $$T \sin \theta \, d\theta$$. Hence we have $$F = \int_0^\pi T \sin \theta \, d\theta = 2 T.$$