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I recall studying a law of friction some years ago, in engineering school. All I remember is that when first approximation was taken, the popular $f=k \times N$ was derived.

What could that be? Or if this does not exist, what is the proof that contact area does not affect friction force?

EDIT:
So, the book is about designing a friction clutch. This device consists of a number z of disc pair, one connected to the input shaft and the other disc from the pair connected to the output shaft.

If I understand the formula and meanings of the variables correctly, the maximum torque, that the device cn transmit is proportional to $F \times z$. F is the force, pushing the disc package together.

Does this change anything?

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  • $\begingroup$ The frictional force is proportional to the real, not apparent, area of contact, and the real area of contact is proportional to the load. Therefore the area cancels and the frictional force is proportional to the load. $\endgroup$ – John Rennie Oct 31 '13 at 15:51
  • $\begingroup$ @Thank you for the response. However, I am not aware what is real contact area and apparent contact area? Does it have to do something with deforming of soft bodies? With micro-roughness? We were assured that higher contact area increases the friction (designing a friction clutch), but the formulas were never proven. $\endgroup$ – Vorac Oct 31 '13 at 15:54
  • $\begingroup$ Yes, it's the atomic scale roughness. When you touch the surfaces it's only the highest points that touch. As you increase the load these asperities deform and spread so the area of contact increases. The force divided by the real area is roughly the yield pressure, which is a material property and roughly constant, so you end up with the real area of contact roughly proportional to the load. I'm sure Google can find you a more rigorous treatment! $\endgroup$ – John Rennie Oct 31 '13 at 16:00
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    $\begingroup$ Cite the formula, indicating a source for it, and I'll have a look. I'm not familiar with clutch design so I don't know the answer offhand. $\endgroup$ – John Rennie Oct 31 '13 at 16:08
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/16213/2451 , physics.stackexchange.com/q/154443/2451 and links therein. $\endgroup$ – Qmechanic Nov 26 '13 at 12:14
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Of course the simple linear relationship F=k*N is just a crude approximation.

The major force in friction of mainly flat areas is the van der Waals force between the molecules in the two layers being close to each others, which also are affected by the deeper layers in the materials.

The amount of friction also depends on how long the two areas have been in contact, their temperature and a myriad of other factors, with plenty of ongoing research.

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Well I'm not aware of any well publicized change in the standard teaching that dry static friction is area independent, and due to the microscopic contacting as described above by John Rennie. However when it comes to clutch design, there are other factors to consider. It is desirable to have very smooth increase in friction with increased contact pressure (macro), and this dictates a macro surface area, that is hugely greater than the real micro contact area. So that the making and breaking of these contacts is not detectible at the individual contact level. The actual contacts tend to weld, in the case of metal on metal, and the friction force itself is the shearing of these micro welds (when sliding friction occurs.) The other major concern as regards clutch design, is the removal of heat; and that feeds into the calculation of suitable macro areas, so that temperature rise is minimized. This is even more important, in the case of a brake shoe, where energy dissipation as heat, is the end aim of the system.

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It affects and it does not in different cases

Case 1: When contact area do not affects the frictional force

You know that frictional force $f$ is $$f = N*\mu$$ where $\mu$ is the frictional coefficient of the surface and $N$ is the normal force acting on the surface. Now look at the image below

enter image description here

both the blocks have same mass in $fig(1)$ and $fig(2)$ but the contact area is different let the block 1 have the are $Am^2$ and block 2 have the are $2Am^2$, in the both the images there is equal frictional coefficient that is $\mu$

Now force exerted by the first block per square meter is $$N_1 =\frac{mg}{Am^2}$$and the force exerted by the second block is $$N_2 = \frac{mg}{2Am^2}$$so since both the blocks have same frictional coefficient $\mu$ so frictional force between block 1 and the bottom plank is$$f_1 = N_1*\mu*a_1$$ here $a_1$ is the area of the contact between the blocks, so$$f_1=\frac{mg}{Am^2}*\mu*Am^2\implies mg\mu$$ and the contact force between block 2 and bottom plank is$$f_2 = N_2*\mu*a_2$$ here $a_2$ is the area of the contact between the blocks, so$$f_2=\frac{mg}{2Am^2}*\mu*2Am^2\implies mg\mu$$$$\implies f_1=f_2$$ so you see here there is no effect of contact area on the frictional force.

Case 2:When contact area do affects the frictional force

i don't if this is a correct example or not but here is goes

take example of books kept in shelf standing vertically side by side and are tightly packed so if you will try to remove a big book with a greater surface are you will have to apply more force as compared to in removing smaller books with smaller surface area

Hope you got it

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  • $\begingroup$ So here is the edit. I suspect that either I am misunderstanding the meanings of the variables, or am missing something sismple in that the discs are separate units, not one large surface area. $\endgroup$ – Vorac Jan 6 '14 at 20:19
  • $\begingroup$ i didn't got your wuestion correctly but according to what i was acle to understand disc are separate units but at last they are all connected to same shaft so they all apply combine force on ut $\endgroup$ – Dimensionless Jan 7 '14 at 7:42
  • $\begingroup$ Imagine 10 discs. The first is connected to a shaft. The second is stationary, connected to the casing of the machine. The third is connected to the same shaft etc. F is the force, that is pushing those discs together. $\endgroup$ – Vorac Jan 7 '14 at 8:34
  • $\begingroup$ The force acts on each disc which is then transferred to the shaft as the shaft is connected to the disc and we get a combined force of every disc on the shaft $\endgroup$ – Dimensionless Jan 8 '14 at 3:37

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