Force and torque provided by a string on pulley Suppose there is a pulley of mass m and two blocks are suspended on both sides of the pulley. Now what are the forces that the string applies on the pulley?
I can visualise the normal force acting towards the centre of the pulley. I wanted to know about the forces that are tangential to the pulley. Some sites state that those are tension due to the string and friction. I can understand how friction might play a role,but I am having hard time understanding how the string can apply the tension force on the pulley.
 A: Normally questions regarding massive pulleys come with the condition of "the string is massless and does not slip", to indicate that the pulley itself must rotating in order for the masses on the atwood machine (or whatever pulley system you have) to be moving. And because of its own inertia; the rope exerts a tension force on the pulley because of whatever mass is being accelerated in the setup.
So there is a force of tension tangent to the pulley which results in a torque. In these problems you cannot assume the tension throughout the rope is the same, because pulley is itself accelerated by an imbalance in the tension of the string.
In the end you have one extra tension variable, but an extra equation of the torque given by the motion of the pulley.
Of course at the core of this force is friction; which is why questions always come with the aforementioned caveat; but this is how it we look at forces on a massive pulley.
A: Even if there is no friction between the string and the pulley, the pulley must still exert a force on the string. This is because, although the magnitude of the tension in the string is constant, its direction changes as it goes over the pulley - in fact, the direction of the tension rotates through $\pi$ radians.
Take a small length of string that is at an angle $\theta$ to the horizontal at one end and at an angle $\theta + \delta \theta$ to the horizontal at the other end. The net horizontal component of tension $T$ on this small piece of string is
$T \cos (\theta + \delta \theta) - T \cos (\theta) 
\\= T \left( \cos(\theta) \cos (\delta \theta) - \sin (\theta) \sin (\delta \theta) - \cos (\theta) \right)
\\ \approx -T \sin (\theta) \delta \theta$
since for small $\delta \theta$ we have $\cos(\delta \theta) \approx 1$ and $\sin(\delta \theta) \approx \delta \theta$. The net vertical component of $T$ on this small piece of string is
$T \sin (\theta + \delta \theta) - T \sin (\theta) 
\\= T \left( \sin(\theta) \cos (\delta \theta) + \cos (\theta) \sin (\delta \theta) - \sin (\theta) \right)
\\ \approx T \cos (\theta) \delta \theta$
If we add up all these components from $\theta=-\frac{\pi} 2$ on one side of the pulley to $\theta=\frac{\pi} 2$ on the other side and take the limit as $\delta \theta \rightarrow 0$ then we have
Total horizontal force on string $\displaystyle = -T\int^{\frac{\pi} 2}_{-\frac{\pi} 2} \sin (\theta) d\theta = 0$
as we expect. However:
Total vertical force on string $\displaystyle = T\int^{\frac{\pi} 2}_{-\frac{\pi} 2} \cos (\theta) d\theta = T \left[ \sin (\frac{\pi} 2) - \sin (-\frac{\pi} 2)\right]=2T$
This force on the string comes from the normal force (perpendicular to the pulley's surface) that the pulley exerts on each piece of string. So by Newton's third law the string must exert an equal and opposite force $2T$ vertically downwards on the pulley.
