This video gives a brief introduction of dry friction. As the term indicates, 
dry friction is friction between two dry surfaces without liquid or lubricant. 
I’m sure we all have experience with friction. Imagine a crate with certain weight sitting on the floor 
and imagine yourself trying to pull it and move it. 
From experience, we know that if the force P that we exert is not big enough, the crate will not move.
Only when the force P exceeds a critical, limiting amount the crate will start moving. 
So what happens before the applied force P reaches this limiting amount? 
According to Newton’s second law, unbalanced force along this horizontal direction will sure cause motion, therefore, 
apparently there must be another force acting in the opposite direction that balances the 
applied force P, preventing the crate from moving. 
And from experience, we know this force is the frictional force between the crate and the floor. 
And also from experience, we know if the surfaces are rough, frictional force is large, versus 
if the contacting two surfaces are both smooth, the frictional force is small. 
Realistically, there’s no perfectly smooth surface, therefore friction always exits, although sometimes it can be negligible. 
Now let’s analyze this further. 
If we zoom in on a small portion of the contacting surfaces of this system, 
then the seemingly smooth surfaces are actually rough, with ups and downs on both sides.
And if we zoom in even further.
And imagine you are trying to pull the top surface to move to the right. 
At these contacting points the top surface will be stopped by the bottom surface, and the bottom surface will exert forces, R_1, R_2 etc.  
to resist the motion of the top surface. 
How many of those forces are there? It’s impossible for us to tell. 
What are the directions and magnitudes of these forces?
Also impossible to tell.
However, we do know that for each of these forces, we can surely resolve it into two components, 
one vertical component force N, and one horizontal component force F.
Let’s first focus on the vertical component forces N_1, N_2, N_3 etc. 
Although now we know the directions of these forces, we still don’t know the magnitudes. 
But this can be considered as a distributed load that we are already familiar with,
with a load intensity function still unknown. 
But we can still find a concentrated load to replace this distributed load, and let’s call it N,
which in magnitude equals to the integration of the load intensity function.
And also it needs to be placed at a certain location.
Let’s call it point O for now.
And we can do the same thing and summarize all the horizontal component forces and replace them with a concentrated force F.
We don’t know for sure exactly where we should position this F force. But that’s less important. We know this force must act between 
the two contacting surfaces, tangential to both surfaces, which is good enough for now. 
And lastly let’s add the weight force, acting at the weight center of the body, and now we have completed the free body diagram of this crate. 
And from your past experience probably from studying physics we know this force is known
as the normal force, because it is always normal to the contacting surfaces,
and this force is the frictional force.
When the rigid body is static,
this force is the static frictional force, sometimes with a subscript s standing for static,
and we can apply the rigid body equilibrium analysis that we have learned to solve for the 
unknown forces N and F, as well as the location of point O. 
So we put the free body diagram in an x-y coordinate system, 
and start writing the three equilibrium equations for 2-D rigid body analysis. 
First,
resultant force along the x direction equals to the applied force P minus F, that equals to zero, therefore F equals to P. 
Secondly,
resultant force along the y direction equals to N minus W, and it equals to zero, therefore N equals to W.
And lastly, let’s write the moment equilibrium equation about point O,
therefore, resultant moment M_O equals to W times x, minus P times h, and it equals to zero, and from here we can determine the position 
for point O, x, which is P times h over W. 
Sometimes we encounter this type of situation when we want to move a relatively tall object. 
From experience, we know that if we place the force too high, the object will tip over.
From analysis, we realize that if the calculated x 
equals to half of a, 
and that places the point O
at the edge of the box,
then the box is about to tip over.
Let’s go back to our original scenario, again, imagine yourself trying to move this crate on the floor by applying a force P on it. 
We know that any force P bigger than zero will cause a tendency to move the crate to the right,
and as a response, frictional force develops 
in order to resist this tendency of motion.
The frictional force is always tangent to the contacting surfaces and always in the opposite direction of the impending motion. 
Imagine you are gradually increasing this force P you apply, and we want to know how the frictional force changes as a response to that. 
Let's use this axis to represent the increase in the applied force P,
and use this axis to represent the increase in the frictional force F.
Initially, the applied force P is small, not enough to surpass the maximum static frictional force, 
therefore, not enough to move the crate. So the crate is static, 
or we say it's in equilibrium, and the frictional force during this stage is static frictional force, 
F_s, and according to the equilibrium analysis we did earlier, F_s simply equals to P.
It stays this way until the applied force P reaches the limiting value, 
and at this point, the frictional force reaches a maximum static frictional force ,
F_s equals to miu_s times normal force N. miu_s is the coefficient for static friction, and it is determined through experiments. 
Up to this point the crate is still motionless 
but it is on the edge of slipping. If you continue to increase the applied force P, the crate will start sliding on the surface. 
Motion starts, and the friction between the two contacting surfaces is now kinetic friction, F_k, which equals to 
miu_k times the normal force N. miu_k is the coefficient for kinetic friction. It is normally smaller than the coefficient for static friction, 
and it is also determined through experiments. 
As a summary,
when the object is static and is not on the edge of motion,
the static frictional force should be expressed through equilibrium conditions. Please do not readily assume F_s equals to miu_s times N,
because that only applies to maximum static frictional force.
When the object is static but it's on the edge of sliding, this is when the static frictional force 
reaches the maximum value miu_s times N, with miu_s being the coefficient of static friction.
Or when the object is static but is on the edge of tipping over, this is when x calculated from 
moment equilibrium places the location of the normal force at the edge of this object.
When the object is sliding,
now the frictional force is kinetic frictional force, and it equals to niu_k times N, with miu_k being the coefficient of kinetic friction.
Both coefficient of static friction and coefficient of kinetic friction are determined experimentally and
you can find these coefficients for common surfaces on the Internet. For example here is a screen shot of
a page from Wikipedia that lists the coefficients of static friction for several commonly seen contacting surfaces.
