**Pre reqs:** [Free body diagrams](http://scienceblogs.com/dotphysics/2008/09/basics-free-body-diagrams.php)

Friction is an interaction between two objects in contact that opposes relative motion of those two objects. It is not something fundamental (like gravity, or electromagnetic force), but it comes up enough that it will be worthwhile to talk about it. Let me start with a simple example. Suppose I have a book on a table. Here is the free body diagram for the book:

![Screenshot 27](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-27.jpg)

Simple enough – right? There are two forces on the book. A contact force (the table pushing up) and a long range force (the gravitational force of the Earth pulling down on the book). These two forces have the same magnitude, so when added together, they give a total of zero vector. This means the book is in equilibrium.

Now, what if I push on the book from the side? Suppose I push with 1 Newton. If the book is still in equilibrium, what does that mean? It means the free body diagram must look like this:

![Screenshot 28](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-28.jpg)

If the book is still in equilibrium, then the force of the table on the book (due to friction) would have to have the same magnitude as me pushing on the book. Note: Even though my push and friction are equal and opposite, these are not Newton’s third law force pairs – I talked about that in the free-body-diagram post.

Now suppose I had no nose? Or suppose I pushed on it a little harder, say with 2 Newtons. If the book still does not move, then the free body diagram could look something like this:

![Screenshot 29](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-29.jpg)

If the book does not move, then the force of friction MUST be the same magnitude as the force I am pushing with. What if friction were greater in magnitude than the force I push on it. If this were the case, the book would increase in speed in the direction opposite I push on it. That would freak me out. Anyway, perhaps you see an important aspect of the friction force. The force depends on the circumstances. Compare this with the gravitational force. Near the surface of the Earth, I can say the gravitational force is *F = m(9.8 N/kg)*. This is a force that can be determined independent of other forces acting on it. For friction (and for the normal force) the force depends on other forces acting on it. In a sense, friction COULD be calculated like the gravitational force. What is friction after all? It is an interaction between the two surfaces. This is due to the electrical nature of matter. However, I could not easily determine a value for friction this way unless I calculated the position and electrical force on all the particles in both objects. (I don’t even think 10,000 8th graders could do that).

Never fear, there is still a model for the frictional force. It does not depend on the positions of the two materials, but rather the force pushing these two materials together (the normal force). If there is no relative motion between the two surfaces, then friction is:

![Screenshot 30](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-30.jpg)

Some important notes:

- This is not written in vector form. The normal force is a force perpendicular to the surface. The friction force is parallel to the surface in the opposite direction that the two surfaces want to slide.
- ? is the coefficient of friction (actually, static friction). It is a value that depends on the two surfaces interacting. The lower this number, the more “slippery” they are. Typically, the value is between 0 and 1. (unitless)
- Look at the less than or equal to sign. This goes along with the two examples above. Be very very careful. Many times I see students calculating the MAX friction force instead of the actual friction force.
- In this model, the frictional force DOES NOT depend on the surface area.
- This is just a model. This model doesn’t always work. Take for instance the Mythbusters episode where they tried to pull two phone books apart that were only held together by friction. In this case, the friction force was so high because the surface area was so high.
- This is for the case when the two surfaces do not move relative to each other. A good example is the tires on a car while it is driving and the tires are not slipping.

**What happens when it slides?**

Once the two surfaces move relative to each other, the model for friction changes to this:

![Screenshot 31](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-31.jpg)

Notice that it is now “equal to” not less than or equal to. Once the two surfaces are moving relative to each other, the frictional force is constant. There is a different coefficient friction for static and kinetic friction. Typically the value for the coefficient of kinetic friction is less than that for static friction.

**Enough talk, how about an example**

Suppose I have a wooden ramp with a wooden block on it. Further suppose I raise the incline of the ramp until it slides. What angle will it slide if the coefficient of static friction is 0.3?

Here is a diagram and free body diagram for the case JUST when the block is about to slide:

![Screenshot 33](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-33.jpg)

First – notice that the normal force between plane and the block is PERPENDICULAR to the plane. This means that the normal force is NOT in the opposite direction as the gravitational force.

Next – notice that since the x-y axis is not real, I can put it where ever and however I want. I have chosen to make the x-axis parallel to the plane. The result is that there is only one force (the gravitational force) that is not on either the x- or y-axis. Finally, notice that the angle between the gravitational force and the y-axis is the same as the angle the plane is inclined.

Ok, what is next? Well, if this is in equilibrium, all the forces still add up to the zero vector. So, for the x-direction this is:

![Screenshot 34](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-34.jpg)

Note that I called the force between the block and the Earth as “F_{grav}” and then I called it “mg” where m is the mass of the block and g is the gravitational field near the Earth (*g = 9.8 N/kg*). Also, these are scalar (not vector) equations since they are components in the x-direction. And for the y-direction:

![Screenshot 35](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-35.jpg)

Now there is one other expression – the relationship between friction and the normal force. Since this is at the angle that it JUST starts to slide, I can write:

![Screenshot 36](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-36.jpg)

Now I can use this to substitute in for the friction force. I then get the following two equations:

![Screenshot 37](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-37.jpg)

Solving the first equation for N and substituting into the second equation:

![Screenshot 38](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-38.jpg)

So, if the coefficient of friction is 0.3, then:

![Screenshot 39](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/10/screenshot-39.jpg)

That is it.