Friction in Engineering

This is my solution to a question I asked in the forums I will now answer on frictional force, I do it in detail it was,

Coefficient of Friction

There is a rule of thumb in mechanics where the force of resistance is F =  mu.N where N is the normal force of an object on a surface and mu is called the coefficient of friction.

My question is, how can one explain it to only depend on N? And why is 0 <  mu < 1 ?

Formulation of the Problem

We will assume there is a block on a level smooth plane under the following forces,

The applied force F needed to move the block, F countering and equal to the drag force due to friction and N the vertical force of the surface plane opposing the fixed weight W of the block, while R is the vector resultant of N and F, as the hypotenuse of the formed 90deg triangle. The Greek letter for alpha would be the angle between R and F, as the x-axis.

F is then parallel to the x-axis and N is perpendicular, i.e. parallel to the y-axis. The block is at rest  or moving at constant speed which means the forces all balance out, the total resultant must be zero.

As customary "mu" will be the Greek letter used for the coefficient of friction and mu is defined by
N = mu.F. We want to show that N depends only on F, as given in the formula.

Calculations

The reasoning might not be so easy to follow, if interested one should make your own sketches and write the arguments out.

There are in effect two questions,

The Formula

As a rule commonly used in Engineering Mechanics we want to explain why F depends only on N. We would think that as contact area is doubled and all else kept the same, friction will double, and in the same way F doubles as pressure does, everything else staying exactly the same.

As follows, it stands to reason that all else unchanged, the friction force will be proportional to the contact area of our block, say F = c.A with c constant. In the same way keeping all else the same, (including the contact area now) we see that the friction force should be proportional to the normal pressure N = d.P 

Of course pressure is given as force per area,

From the frictional force it then follows that F = d.P / c.A = (d/c) P.A = ( (d/c) (N/A) x A) = (d/c) N leaving only the normal force N = (c/d).F or N = mu.F, with mu = c/d

In our model thus, frictional force F depends only on the normal force, in this case thus on the weight of our block. This then is the "rule of thumb" in mechanical engineering and it works quite well. In effect the areas cancel leaving only the normal force N.

The Coefficient

We want to explain why 0 < mu < 1 The reasoning might be not so easy to follow, one should make your own sketches and write the arguments down.

There are illustrations I made on https://www.instagram.com/full.force.gale/ 

The definition of mu is as N = mu.F ; Also mu < 1  is the same as,  N = mu.F and N < F ;
This means mu < 1 is equivalent to N < F ; Since mu > 0 is trivial, therefore 0 < mu < 1

Further Remarks

The situation of an object on a level plane surface at standstill is in essence the same as at a constant speed, because there are no unbalanced forces and therefore no acceleration.

All the reasoning is sound, except for restrictions on typesetting and with only some minor technicalities which are easily accommodated.