A problem I encountered is as follows: A pulley consists of a circle of radius $R$ is pivoted about a point a distance $R/2$ from the circle’s
center. A string attaches to a block hanging from the pulley as shown. The coefficient of friction
between the string and pulley is infinite. Find the tension in the horizontal part of the string if the
system is at rest.
The solution is simple at first glance: the vertical part of the string must have a tension of $mg$ to hold the weight, so to balance the torque on the pulley about its pivot, the horizontal part of the string must have a tension of $2mg$.
However, how can the torque from the string be reduced to the torque of the two forces of tension at the top and on the side of the pulley? From my understanding, it's not really "tension" that is pulling on the sides of the pulley; rather, it is friction and normal force from each segment of the string that are truly acting on the pulley. It's not obvious to me how summing the torques from each of the segments' friction and normal forces add up to the torque of the two forces I described earlier, especially when taken about a pivot that is not the center of the pulley.
Best Answer
[answer edited to reflect comments]
The question is why the problem is equivalent to replacing the pulley with two rigid, joined, rods leading from the pivot point to the first and last points of contact respectively, as shown (such a system is easily solvable by equating the clockwise and anti-clockwise moments).
The reason that the other parts of the pulley can be ignored for the purposes of the calculation, is because we are told that the coefficient of friction is infinite and the string is inextensible. This means that in the region between the two contact points, the string might as well be welded to the pulley, or absent altogether. Either way, elements of the string in that region cannot exert any torque about the pivot.