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(Edited question, 07/04/23)

For someone standing on a carousel, there is a frictional force at their feet of magnitude $|F| \leq \mu N = \mu mg$ that is directed towards the centre of the carousel, where $N$ is the normal force, $m$ is the person's mass, and $\mu$ is the coefficient of friction. If $F \geq mr \omega^2$, where $r$ is the distance from the person to the centre and $\omega$ is the angular velocity of the carousel, then the frictional force is large enough that the person undergoes circular motion.

However, what happens if the person is seated facing the centre, leaning forward such that their back is not in contact with the back of the seat? Their feet are in contact with the carousel as before, but now their body is also in contact with the seat (with a different coefficient of friction), since they are sitting down. In this case, how is the frictional force recalculated, if it is no longer possible to simply use $|F| \leq \mu N$?

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  • $\begingroup$ Do you know how to work with force vectors and do you understand the notion of torque already? In a scenario like this there will be three different forces involved and if they are not all acting on the same point, then there will be a torque that will try to turn your body. The problem is not fully specified until we know the possible degrees of freedom of the motion (in a plane, in all three dimensions etc.). $\endgroup$
    – FlatterMann
    Commented Mar 25, 2023 at 21:06
  • $\begingroup$ I can work with force vectors and have some understanding of torque. I agree that the person should turn because the friction force at one foot will be much smaller than at the other. I suppose my problem is trying to work out how to quantify the forces? $\endgroup$
    – Thomas
    Commented Mar 26, 2023 at 8:45
  • $\begingroup$ @Thomas Also the frictional force is not always $muN$, it depends on the situation. $\endgroup$
    – Tanmay Gupta
    Commented Mar 26, 2023 at 13:35
  • $\begingroup$ Yes, for an object at rest the force of friction is just sufficient to counter impending motion. $\endgroup$
    – John Darby
    Commented Mar 26, 2023 at 20:44
  • $\begingroup$ General tip: Let's not have posts look like revision histories. $\endgroup$
    – Qmechanic ♦
    Commented Apr 7, 2023 at 9:41

2 Answers 2

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There will be one overall net effective coefficient of friction determined by how the person distributes their weight. If the sitting person puts all their weight into the seat, the centripetal acceleration occurs according to the coefficient of friction with the seat alone. If the person puts all their weight on their feet instead, the centripetal acceleration will occur according to the coefficient of friction with the floor alone. If they evenly distribute their weight, then the effective friction will be an average of the two values.

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  • $\begingroup$ Thanks very much for your answer, @NuclearHoagie, that's very helpful. $\endgroup$
    – Thomas
    Commented Apr 12, 2023 at 9:40
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You need more information. For example, the point at which the force for the push acts may cause motion in three dimensions and can cause a torque about the center of mass. In general, you need to consider the changes in linear and angular momentum for the total system and for each foot on the different surfaces. For the latter, you need to know the forces the person applies to each foot (forces internal to the total system). See the comment by @FlatterMann.

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