I think because that the contact occurs only between peaks which have irregular distribution in contacting surfaces, therefore the contact area is much smaller than the projected area.

In addition to the above, you can see the books "Static and dynamic by Hibbeler" and "Engineering Mechanics,Static by Meriam"

Because both the friction force and the normal force (which are abstractions, there're no concentrated forces in reality) are proportional to the area of contact, say f = ηA, N = νA. The ratio of these integral coefficients gives friction coefficient μ = η/ν

Thanks Mustafa Kadhim, Vadique Myself and Mohammad Amin Motamedi for your answers. I think, Mohammad Amin Motamedi made a convincing argument. Now it make sense to me.

Maybe it is more intuitive to think about it on a stress-level.

Think about a body with weight F [N] that is placed on a surface with friction. Let the contact area be A [m^2].

The normal stress in the contact area then follows to:

sigma [N/m^2] = F / A (1)

This leads to a resistance against movement due to friction (mu [-]) of:

tau [N/m^2] = mu * sigma (2)

So every part of the contact area contributes to the resistance against movement. However, for the resisting force V [N] you have to sum up the stresses.

V = tau * A = mu * F (3)

So I think you could say that the contact area matters in a way. But what really matters is the product of the contact area and the normal stresses and that product stays the same for different geometries as long as your body has the same weight.

Hope that helps :)

Regards,

Stefan

To Bhanuday Sharma

Early (19 August 2019) I sent my answer on your question about friction. But seems it wasn’t properly delivered.

Below I address it once again.

Regards, V.Dimitrov.

Emeritus Professor of Chemical Physics.

22 August 2019

Dear colleague Bhanuday Sharma

To be properly answered you should properly formulate the question. Your statement of this problem is wrong. (…So the question is what is the fundamental microscopic reasoning for this assumption that it the friction is independent of contact area?...)

First of all. The friction (as a phenomenon) is independent on contact area. It depends solely on friction coefficient. Conventional symbol is μ0.

To see the sense of my reply let start with definition.

Friction factor μ0 (drag coefficient, friction coefficient) is the index connecting resistance force FR[N] and uniformly distributed force of normal pressure FN[N]. So, as any ratio of similar indices, it is dimensionless [N/N]. It depends solely on the kind of the frictioning materials - it does not matter is the speech on sliding (gliding) friction, rolling friction, etc.

Leaving aside normal pressure FN (usually it is given data), denotes that (contrary to FN) the resistance force FR indeed depends on many factors (kind of material, state of the surfaces, etc.).

That is why we classify two types of friction indices: the so-called *coefficient of static friction* μst and *coefficient of kinetic (dynamic) friction* μkin. The latter is defined via concept of the *speed of friction* Vμ and can be find via the main equation of uniform motion. What is the difference between these conceptions? If you start to impact on the *body in the rest* you need to overcome initial friction force, (i.e., you deal with the*coefficient of static friction* μst). But then (when the body is in uniform motion), you deal with the *coefficient of kinetic (dynamic) friction μkin * (to provide this motion).

Usually, μst>μkin or even μst>>μkin. (Sometime μst is known as maximum value of friction coefficient, but this is nothing but parlance, which is the source of misunderstanding).

Unfortunately, the very design of this RG-page doesn’t allow to present the proper plot. At this plot *ordinate (friction coefficient) vs absciss (time/distance)* this is the curve with the max at the point of origin, which decreases down some minimal and constant value μkin (*flattening*) as time (distance) increases.

Another case takes place if you bear in mind the *work*. The work provided the very beginning of the motion is indeed higher (sometime, much higher) than the work that is necessary to support the motion, W=F*L. In this sence (i.e. if you provokes the motion) it is *maximal one*, indeed.

If someting of abovestated is unclear, don’t hesitate to be in touch.

Resume. Don’t be misleaded. Follow definitions.

Regards, V.Dimitrov.

Emeritus Professor of Chemical Physics.