Recently I asked a question regarding frictional forces at math stack-exchange(because its basically part of maths syllabus) and I drew some conclusions.

  1. If A and B are in rough contact and are in limiting equilibrium, then there exist two frictional forces. One acting on A and one acting on B.

  2. The direction of frictional forces can be determined by working out the direction of motion if the friction was not present.(This is a trick to work out the direction of frictional force which was mentioned on https://physics.stackexchange.com/a/94837/128083)

  3. The direction of friction forces on A and B are opposite.

1,2 are trivial and were implied on the math stack exchange link. However, I came up with rule 3 by observing the relative velocities of each object.

Let's say A is a wall and B is a ball that is in rough contact with A. From A perspective, B is falling down hence an upward friction force should be acting on B. From B's perspective, A is moving upwards, hence a downward force should act on A.

However the problem arises in the following question


(By downward and upward, I mean upward tangential and downward tangential at point P)

The ball P is moving downwards. An upwards frictional force acts on P. Edit An upward frictional force at P imply a downward frictional force for the disc at P. Since frictional force is opposite the direction of movement, thus a downward friction imply an upward movement and hence a clockwise moment. However, it seems trivial that the disc should rotate anti-clockwise but using the rule 3, we can conclude that motion is clockwise.

Can somebody please explain the fallacies in the proposed rule 3? If there does exist a fallacy please explain some alternative that I can employ in working out the direction of frictional forces.

The link mentioned above is: Frictional forces

  • $\begingroup$ I have posted this question on physics-stackexchange but the question has remained unanswered for a while. If it is against the policy to upload similar question on different stackexchanges let me know. $\endgroup$ May 31, 2017 at 17:46
  • $\begingroup$ Your conclusions 1 and 2 are both wrong. Friction is a force, which is related to acceleration, not direction of motion. Conclusion 3 is correct, but you have the wrong reason: the reason is Newton's third law. $\endgroup$
    – David K
    May 31, 2017 at 18:09
  • $\begingroup$ @DavidK The 1,2 conclusions were obtained from physics.stackexchange.com/a/94837/128083 $\endgroup$ May 31, 2017 at 18:11
  • $\begingroup$ @DavidK Please also review the edits at 1,2. $\endgroup$ May 31, 2017 at 18:14

1 Answer 1


OK, #1 is not really incorrect as long as you realize that in some cases the force will be zero. #2 is improved by the edit, but really you should be looking at the force required to cause the motion (or lack of motion) observed, compared to what you would see if friction weren't present. That is the point that is made by https://physics.stackexchange.com/questions/94835/direction-of-force-of-friction/94837#94837. (But note that that answer really only deals with body on the incline, not with the inclined ramp itself, because we assume in that problem that the ramp is not going to move.)

In this example, without friction, the particle would slide down faster and the disk would not move at all. In order to slow down the particle and cause the disk to move, the particle must experience friction opposite its direction of motion, and the disk must experience friction at the point of contact in the same direction as its motion.

Since the particle and its point of contact on the disk are moving in the same direction, the directions of the two frictional forces are opposite to each other, as they must be.

  • $\begingroup$ Can you please elaborate on the statement "the disk must experience friction at the point of contact in the same direction as its motion"? The logic(if I don't missed it) you provided was "to slow down the particle and to move the disc". I have two confusions. 1. Is friction providing an anti clockwise moment on disc and hence instead of opposing, it is providing motion. 2. Why is rotation of disc necessary? $\endgroup$ May 31, 2017 at 18:42
  • 1
    $\begingroup$ 1. Yes, exactly so. 2. How can the particle continue to move without slipping if the disk does not move? And how can the particle slip if the tangential component of gravity on it is less than the maximum static frictional force? $\endgroup$
    – David K
    May 31, 2017 at 18:49
  • $\begingroup$ Okay perfect thanks. $\endgroup$ May 31, 2017 at 19:00

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