Consider a circular horizontal plane (like a round tabletop) rotating around its center.

Consider a body A resting on this tabletop. Since the tabletop is rotating around its center, the body is moving along with it in a circular motion around the center of the circle.

I understand that in circular motion, there's always a centripetal force, pushing the object towards the center of the circle. And in this case, this force is the static friction force, operating on the radial axis towards the center.

I have two questions regarding this:

  1. I was under the impression that a static friction force, or any friction force, always operates in the opposite direction of some other force which operates on the same body. Is this correct? If so, what is the additional force (opposite to the friction force) that I'm missing in the above scenario?

  2. Is the static friction force operating only on the radial axis towards the center? Or is there also static friction along another axis?

  • $\begingroup$ Every force is opposite to some other force. "Actio es reactio." Perhaps you can be a bit more specific? ;) $\endgroup$
    – kricheli
    Aug 31, 2023 at 7:04
  • $\begingroup$ @kricheli I edited to clarify: "I was under the impression that a static friction force, or any friction force, always operates in the opposite direction of some other force which operates on the same body". Is this false? $\endgroup$
    – Aviv Cohn
    Aug 31, 2023 at 7:29

2 Answers 2


For your question 1: In the rest frame of the body A (which is an accelerated frame!) there is a fictitious force, the centrifugal force, which acts on body A opposite to the friction force (the centrifugal force points outwards in a radial direction, the friction force inwards). Both forces cancel and thus the body is at rest in that frame. Due to Newton's "actio est reactio" the friction force causes and equal (in magnitude) and opposite force on the disc at the point where A rests on.

In the rest frame of the rotating disk/the lab system the body A is in motion. It is accelerated according to $F=ma$ wherein $F$ is the centripetal force/friction force acting on the body A. There is no opposite force on it, otherwise it would not be on its circular trajectory. The "actio est reactio"-outward-pointing friction force on the disc is the same in the lab frame.

For question 2: If A is a mass point then yes, there is only a radial force. However, for an extended body this is not as straightforward, you'd have to discuss a stress distribution at the contact interface... And to see that there is not only a radial force in that case, take a rod of length $L$ which you place tangentially at a radius $R$. For $L \ll R$ the force/stress is mainly in the radial direction, for $R \ll L$ not.


Your impression about frictional force requiring some other force for it to take effect is incorrect. Also, there is no such thing as a centripetal force; there is only centripetal acceleration, which is the component of acceleration toward the center of rotation.

In the laboratory form of reference, only the frictional force F is acting on the body, and it is directed toward the center of rotation. So we have $$ma=F$$where $a$ is the centripetal acceleration.

In the (non-inertial) frame of reference of an observer rotating with the table, we envision an imaginary body force acting in the positive radial direction, of magnitude ma pushing the body outward. We call this fictitious force the "centrifugal force" and, since the body is stationary in our frame of reference, we have as our static force balance $$F-ma=0$$We say that the frictional force F balances the fictitious centrifugal force ma.

Both these alternate approaches are valid and give the same answer.

  • $\begingroup$ Cf. my almost identical earlier answer. One of us could have saved themselves some time and effort. $\endgroup$
    – kricheli
    Aug 31, 2023 at 13:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.