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.

  • $\begingroup$ If friction is infinite, the vertical part of the string could support the whole weight, meaning that a tension of 0 in the horizontal string still satisfies the constraints of the problem. In my opinion, this problem is indeterminate without more or different information. $\endgroup$ Jan 9, 2022 at 22:01
  • $\begingroup$ @DavidWhite if there is no tension in the horizontal part, the pulley would rotate clockwise. If that is not happening, where is the torque to prevent that rotation? $\endgroup$
    – BowlOfRed
    Jan 9, 2022 at 22:28
  • $\begingroup$ @DavidWhite For what it's worth, my initial reaction was the same as yours, until I realized the pulley can still rotate about its pivot if there is a net torque. Certainly if the pulley was fixed (unable to rotate) the tension in the horizontal string would be zero. $\endgroup$
    – Bob D
    Jan 10, 2022 at 0:05
  • $\begingroup$ @BobD, even under rotation, if there is any part of the string still touching the pulley, the weight will not fall. In my opinion, this is an ambiguous question. $\endgroup$ Jan 10, 2022 at 18:24
  • $\begingroup$ @DavidWhite It's not a question of whether or not the weight will fall. The question is. whether or not the system, as shown, will be in equilibrium with no tension in the horizontal string, which it will not. Without tension in the horizontal string, the pulley will rotate about the pivot point until the string attached to the weight is be perpendicular to the surface of the pulley. The weight will shift to the left. $\endgroup$
    – Bob D
    Jan 10, 2022 at 21:33

1 Answer 1


[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.

diagram showing the position of the two notional 'rods'

  • $\begingroup$ Sorry, this doesn't exactly answer my question. I'm aware that once you boil down the torques from the string to two forces exerted at the first and last points of contact, you can use the method you just described. My question was concerning why you can consider the string to only be acting at those two locations. $\endgroup$
    – wheelix
    Jan 9, 2022 at 22:13
  • 1
    $\begingroup$ Generally these sorts of toy or exam questions assume "no-slip" so the dynamics (or statics) of the tension along the chord contact points really doesn't enter into the problem. $\endgroup$
    – DWin
    Jan 9, 2022 at 22:20
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    $\begingroup$ The reason you can reduce the problem to a 'two levers' solution is that we are told to consider the friction to be infinite, and the string inextensible. This implies that the torque generated at any other contact point must be zero. Consider, for example, what would happen if the part of the pulley between (but not at) the two contact points were to be coated in super-slippy zero friction teflon. Would you expect anything to change? $\endgroup$
    – Martin CR
    Jan 9, 2022 at 22:22
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    $\begingroup$ @wheelix between the first and last point of contact you can consider the string and pulley to be as one (the pulley only) due to the infinite coefficient of static friction. $\endgroup$
    – Bob D
    Jan 9, 2022 at 22:22
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    $\begingroup$ @Martin CR nice answer. $\endgroup$
    – Bob D
    Jan 9, 2022 at 22:24

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