0
$\begingroup$

While tackling an Olympiad question, it came to my mind that friction need not act in the same direction at all points on a body. I thought of using integration to evaluate the net frictional force, but was stopped by this statement.

Frictional force between 2 bodies, doesn't depend on the surface area in contact.

If that is the case, then how does the force act on a body like the following, where only the narrow strip marked is rough and hence friction can act only there? What are the individual frictional forces at the points A and B?

Top View

The image shows a hollow cylinder, mass $M$ and radius $R$ from top view. The cylinder is placed with its flat end in contact with the ground. Only the gray shaded line is rough, id est, friciton acts only there. The cylinder is rotating about the topmost point. Ignore the $v$ in the diagram.

$\endgroup$

closed as unclear what you're asking by ja72, Kyle Kanos, ACuriousMind, Neuneck, Danu Aug 26 '15 at 9:48

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Your question is unclear to me. Friction acts on each little patch of contact in a way that always opposes motion. $\endgroup$ – ja72 Aug 21 '15 at 18:06
  • $\begingroup$ Yes, I understand that. What I don't get, is this. If suppose the little patch is of area $dA$ and has height $h$ and density $\rho$, then will the friction force acting on that small patch be $F_{fr} = \mu dm g = \mu g \rho h dA$ ? If yes, then that would imply that the force IS dependent on the surface area in contact. If no, then how can the force be evaluated? $\endgroup$ – Aritra Das Aug 21 '15 at 18:20
  • $\begingroup$ @ja72 or would the force at either point A and B be dependent on the mass of the entire cylinder? $\endgroup$ – Aritra Das Aug 21 '15 at 18:44
1
$\begingroup$

As @rmhleo pointed out in his answer, the frictional force doesn't depend on the surface area, because no matter which part of the object is in contact with the other surface, the total normal force (and thus the total frictional force) is unchanged.

However, that assumes a couple of simplifying conditions: namely, that the two surfaces are consistent in their frictional coefficients. Obviously, if you have a cube with some glass sides and some rubber sides, it will slide better on some sides than others even though the area is the same for each. And the above assumption is what you've challenged by changing the average coefficient of friction without changing the contact area.

For the situation you've described, you'll have to do two calculations for frictional force: one for the majority of the cylinder, which sits on the smooth surface - I assume this will be zero since you seem to posit a frictionless main surface - and a separate one for the area which sits across the friction strip. Of course, if you want to calculate anything beyond instantaneous values, for example any properties of the path of the cylinder as it moves across the strip, you'll have to calculate the change in contact area, and in turn account for the changes in force due to the change in contact area, etc. and it could become quite complicated. Beyond my geometrical and kinematical skills, for sure, but if you happen to know any avid curlers...

$\endgroup$
1
$\begingroup$

Yes, frictional force does not depend on the area. This is clear from a simple mental experiment, think of a block resting on a surface. Assume it is a prism whose faces have different areas. Whichever face it is resting on, the friction force will be the same: when on the smaller surface, the contact is reduced compare with the case where the resting face is bigger. But the thing is that the weight is the same, so any element of area of the smaller surface receives a greater pressure than the case where all the weight is supported by the larger area. Thus, this two effects compensate each other, and finally the friction will only depend on the weight of the body.

As for the case you represent, the cylinder will perceive friction forces opposed to its rotation movement, which will be tangential to the circle, and opposed to the linear velocities in these points. So it will either stay static, or it will rotate w.r.t its center, depending on the actual values of $v$ and $\mu$.

$\endgroup$

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