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We know that forces of static or kinetic friction do not depend upon the area of contact of the two surfaces, is it applicable to rolling friction as well?

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If you would have a rigid wheel and surface rolling without slip, then you would have no have a line contact between the wheel and the surface parallel to the axis of the wheel, which has zero surface area. This would not yield no rolling friction forces.

In the real world nothing is completely rigid and would deform due to stresses. These deformations will lead to resistance against rolling. One example formula which can be used to approximate rolling resistance is,

$$ F_r = \frac{N b}{r}, $$

where $F_r$ is the magnitude of the force of the rolling resistance, $N$ the normal force between the wheel and the surface, $b$ the rolling resistance coefficient and $r$ the radius of the wheel. One way of intuitively understanding this equation is that the smaller the radius of the wheel, the "sharper" it is thus it would cause a larger/deeper deformation for the same normal force.

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You would think that Rolling Friction (or just friction in general) would depend greatly on the area of contact with the surface (and to some extent this is true), I mean, using logic, you could expect to think that if there are is a board of nails and you scrape your hand along it, it would hurt more to scrape your hand across more nails rather than less of it.

In it's sense it is true, and false at the same time. The microwelds that exist on both surfaces determine the coefficient of resistance/friction, and although it is not even everywhere, it could be determined that overall, it would be the same. But hey, that's just my theory.

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  • $\begingroup$ Again, I'm not a downvoter, but this is not a very careful answer. Rolling friction has no intrinsic relationship to sliding friction. "true and false at the same time"? Please try to learn from others on this site, who try to give real answers. $\endgroup$ – Mike Dunlavey Jul 9 '15 at 19:07
  • $\begingroup$ I was trying to give a real answer, it's just my knowledge and my experience on this site just isn't broad enough yet to give an explanation as in-depth as many other members of the site have given. And above all, it just seemed like a good answer to give. $\endgroup$ – phi2k Jul 9 '15 at 19:54

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