Force analysis in a compound pulley system At 4:56 of this video on compound pulley systems with the following schematic set up:

The professor does a force analysis of the forces on the pulley A , and strangely (for me) , he finds that there is tension acting vertically at two diametrically opposite ends of the pulley.. but why does it specifically act there?  Why does tension not push the pulley down for the region where the rope curves around the pulley ?
 A: When analyzing pulley systems we do not normally take into account the force of the loop on top of the pulley. We would use the tension of the rope on each side of the pulley. Discounting small frictions in the pulley and rope, the tension is the same throughout the rope. So pulley A has rope tension on each side for a total downward force of 2T.
A: There are no frictions between the rope and the wheel of pulley. Therefore, the wheel will not rotate, and the rope slips over the wheel frictionlessly.
Consider a finite segment of the rope between $\theta_1$ and $\theta_2$, the total force acting on this segment of the rope (see the Figure):

$$
  \vec{F}_{total}(\theta_1 \to \theta_2) = \vec{T}_1 + \vec{T}_2 + \int_{\theta_1}^{\theta_2} \vec{N}(\theta) d \theta = 0; 
$$
where the vectors $\vec{T}_1$ and $\vec{T}_2$ are the draging forces from th rest of the rope ( in the tangential direction), and $\vec{N}(\theta)$ is the normal force per angle exerted from the wheel to this arc of rope. The normal force may be a function of angle. Since the mass of rope is neglected, the total foce has to be zero.
Then, we focus on the infinitesmal segment of rope betweem angle $\theta_2$ and $\theta_2 + d \theta$:
$$
   d\vec{F}({\theta_2}) =  \vec{F}_{total} (\theta_1 \to \theta_2 + d\theta) - \vec{F}_{total}(\theta_1 \to \theta_2) = \frac{\partial \vec{F}_{total}(\theta_1 \to \theta_2) }{\partial \theta_2} d \theta = 0.
$$
The force on the infinitesmal segment should also vanish, as detailed balance of the force.
The force $\vec{T}_1$ depends only one $\theta_1$, thus vanishes as derived by  $\theta_2$ :
$$
  \frac{d \vec{T}_2 (\theta_2) }{d \theta_2} + \vec{N}(\theta_2) = 0.
$$
Since $ \vec{T}_2$ is in the tangential direction, $  \vec{T}_2 = T \hat{\theta}_2$. The derivative of a unit tangential vector renders a unit vector in negative $r$-direction, $-\hat{r}$.  This render the normal force at angle $\theta_2$ to be:
$$
   \vec{N}(\theta_2) = - T \frac{d \hat{\theta_2}}{d\theta_2} = T \hat{r}. 
$$
Surprisingly, the magnitude of the normal force per angle is a constant independent of the angle, equals to tension of the rope.
