Consider the following setup:

A (homogeneous) pulley of mass $M>0$ is placed in the vertical plane, where it is fixed at its center but can freely rotate. An ideal rope is placed around the pulley and two point masses are attached to the ends of the rope. Assume, moreover, that the rope does not slip over the pulley.

Now, what I want to do is to find all the forces acting on the pulley. Surely, we have the gravitational force $Mg$, pointing downwards and acting on the center of mass of the pulley, and a constraint force $V$ which prevents it from falling.

However, these cannnot be all the forces. In fact, there is a force of static friction between the rope and the pulley which allows the former not to slip and the latter to rotate. (I imagine one should also take into account the "weight" of the rope "pressing" on the pulley, but since the rope is ideal, hence massless, this force should be null. Correct me if I'm wrong.)

My question, now, is

How do we find the magnitude, the direction and the point of application of this friction force so that we can write down the equations of motion?

Perhaps a line integral is required? This seems overkill...but still, how do we find the remaining forces (i.e. the forces beside gravity and the constraint force) acting on the pulley in a strictly rigorous way?

Please, note that I'm not just interested in the answer itself, but on how one arrives at the answer as well!

  • $\begingroup$ If you simplify/idealize the problem from the start by saying you have a non-slip massless rope, I don't see much point in trying to consider things like friction afterwards. In the end it depends on what you want to calculate, but from the text it seems that you should be more concerned about the forces from the point masses than friction or the constraint force. $\endgroup$ Commented Aug 12, 2017 at 15:49
  • $\begingroup$ @user1583209 there are several problems which involve such a setup, or a similar one, in introductory classical mechanics, and in practically all cases it is needed to find the forces which act on the pulley, or at least those which generate momentum. So yes, there is a point in the question. $\endgroup$
    – Adrian
    Commented Aug 12, 2017 at 18:20
  • $\begingroup$ Yes, I am aware of these kind of problems. My point was, that with the initial assumptions, you basically get rid of any discussion of how the rope interacts with the pulley (i.e. friction). So it is kind of strange to me trying to consider friction afterwards. To give you another example: Implicitly you assume that the pulley is rigid (cannot be deformed). That's why you are not considering forces leading to deformation either. $\endgroup$ Commented Aug 12, 2017 at 20:00

1 Answer 1


Interesting question. I think your conceptual difficulty comes from the fact that the rope is in contact with the pulley over an extended distance. Complicating matters is the fact that the rope has a different force being applied to it from each end. As you have realized, a complete description of the system is difficult. But you can simplify your model of the system. The fact that the rope is massless and does not slip makes that fairly easy to do in this case.

You can simply think of the portion of the rope in contact with the pulley rotor as part of the pulley rotor itself. Think of the point where the rope meets the pulley to be the point of contact between the rope and the pulley. Now the extended friction force becomes two point-contact tension forces.

With this model of the situation it should be clear what directions those tension forces act. The magnitude of those forces is not obvious; they have to be computed from the dynamics of the situation. Hint: the magnitudes of the two tension forces are not necessarily equal to each other.

  • $\begingroup$ Thanks, I actually get it now! This is pretty much the answer I was looking for. One question though: what changes if the rope slips over the rotor? Of course the whole dynamics of the system changes and it won't be so easy to compute the magnitudes of the two forces, but are the two forces still acting on the points of contact and still equal to the tensions? $\endgroup$
    – Adrian
    Commented Aug 12, 2017 at 18:50

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