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?

  • 2
    $\begingroup$ 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). $\endgroup$
    – user4552
    Commented Apr 28, 2013 at 16:28
  • $\begingroup$ For tires specifically friction is really complicated. $\endgroup$ Commented Nov 25, 2013 at 3:52
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/154443/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Sep 21, 2015 at 14:49

5 Answers 5


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.

  • $\begingroup$ Are both the effects equal in magnitude? If so, how? $\endgroup$
    – orionphy
    Commented Oct 26, 2011 at 11:22
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    $\begingroup$ Traction is static friction, they are to be confused. You can't have traction change and not friction. $\endgroup$
    – Ron Maimon
    Commented Oct 26, 2011 at 15:51
  • $\begingroup$ 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... $\endgroup$
    – Nic
    Commented Oct 26, 2011 at 16:23
  • $\begingroup$ 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. $\endgroup$ Commented Oct 27, 2011 at 2:28
  • $\begingroup$ 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. $\endgroup$
    – Georg
    Commented Oct 27, 2011 at 10:17

When you say underinflated tires experience more friction, do you mean static friction (i.e., resistance to slipping) or rolling resistance, which is something quite different?

Afaik the origin of the friction law is very much phenomenological, and has it's limits of applicability (especially at the static - dynamic transition). My understanding as to why drag racing vehicles have such enormous tires is spread out the shear forces and dissipate more heat. The force of friction is the same regardless, until the tires turn to liquid!


It is all about the distribution of pressure under the contact. For a block of uniform weight the pressure can be assume almost constant under the area and so when traction is broken it will happen all at once all over with a force of $\mu N$ as you stated.

But for other geometries, or for elastic parts (like tires, or marbles or billiard balls on felt) the contact pressure has various other shapes. The parts with the highest pressure (sometimes in the middle, and often at the edges) are going to stick more than the unloaded parts. The result is complicated, but in the end we call the total traction to achieve full slipping still $\mu N$, but with $\mu$ a different value that for the block above, even if the materials are the same.

A lot of scientific papers are written on the subject of how traction affects the contact properties and vice versa. Dealing with a coefficient $\mu$ for the force and not the area makes it easier to summarize the results, but in reality the pressure shape over the area is the ultimately in control here.


There are some vital considerations you are not including in your initial analysis. One is the performance and response (due to 'jiggling/vibration' of the tire at low pressure) of the tire depending on its shape.


A way that i've always (possibly naively) justified it for myself IN THIS MODEL is to consider both sides using $F=P*A$, and in the logical requirement that the area that which the pressure is applied, real or not, must be the same for both sides, and the coefficient is in fact scaling the pressures, so you can just remove the need for area algebraically.

$$ P_fA_c = {\mu} P_nA_c $$

or a similar result if you start with the definition as COF being the ratio of the normal force and the max resistive force $$ \mu = \frac{F_N}{F_{f_{max}}} $$ and in the occasion people bring up the area factor in air resistance, i assume that this comes from the fact this is a compound of elastic and fricative force components, with area coming from the former


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