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Is friction really independent of area? The friction force, $f_s = \mu_s N$. The equation says that friction only depends on the normal force, which is $ N = W = mg$, and nature of sliding surface, due to $\mu_S$.

Now, less inflated tires experiences more friction compared to well inflated tire. Can someone give clear explanation, why friction does not depend on area, as the textbooks says?

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You need to distinguish between reality and models of reality. The textbook model of friction that you're referring to is called the Amonton model. It's only one of many models. It's fairly accurate in certain situations but not in other (e.g., a lubricated bearing). – Ben Crowell Apr 28 at 16:28

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The increased 'resistance' of an underinflated tyre is due to mechanical deformation, friction is independent of area as suggested. The simplest explanation for me is that: as area increases the applied force per unit area decreases, but there is more contact surface to resist motion.

Added as per Zass' suggestion below:

$$\rm{Friction}= \rm{Material\ Coefficient} \times \rm{Pressure} \times \rm{Contact Area}$$

Where the material coefficient is a measure of the 'grippiness' of the material, the pressure applied to the surface and the area of the surfaces in contact. So we can see the area in the pressure term cancels with the third term.

This is not to be confused with traction, where spreading the motive force over a larger area can help.

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Are both the effects equal in magnitude? If so, how? – orion Oct 26 '11 at 11:22
Traction is static friction, they are to be confused. You can't have traction change and not friction. – Ron Maimon Oct 26 '11 at 15:51
I use traction to include the mechanical strength of the 'ground'. Using soft wide tyres in a muddy field will get you further than hard thin tyres. This isn't a friction issue... – Nic Oct 26 '11 at 16:23
This is the correct answer to the question (which I think most students have). You could, by all means, introduce another coefficient specific only to the surface material that, multiplied by the area, gives the total coefficient of friction. Then (material coefficient)x(pressure)x(area) will show the same behavior. So it's not that area doesn't matter, but it cancels out in this view. – AlanSE Oct 27 '11 at 2:28
After this correction, Your equation is dealing with traction (friction at rest) predominantly! The bigger "friction" of a underinflated tire is a question of deformation work within the rubber, which is not reversible entirely. – Georg Oct 27 '11 at 10:17
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