## Friction

Friction is a contact force that opposes motion. It acts whenever two surfaces move over one another. The amount of friction depends on four things:

surface roughness
Rougher surfaces experience more friction. With very smooth surfaces, there is still friction because of van der Waals forces.
contact
More surface area of contact results in more friction. For example, sled runners reduce surface area of contact and thus reduce friction.
materials
There is a coefficient of friction (denoted by mu) for every combination of two materials. The friction associated with steel & wood is different from the friction associated with steel & ice, for example.
normal force
A greater normal force results in more friction. This means that heavier objects will experience more friction, and it also means that pressing down on an object from above will increase friction.

We calculate friction using

|vec F_"f"| = mu|vec F_"N"|.

If the only forces are gravity and the normal force, this can be rewritten as

|vec F_"f"| = mum|vec g|.

There are two types of friction. Static friction must be overcome to move an object that is initially at rest with respect to the other surface. Kinetic friction acts when an object is in motion. There are separate coefficients for these two cases: mu_"s" for static and mu_"k" for kinetic, where mu_"s" > mu_"k". The values of both are determined experimentally for every pair of materials.

Keep in mind that vec F_"f" will be negative if you define the direction of motion to be positive (which is what you should usually be doing).

### Example

You are standing on a roof inclined at an angle of 31º with the coefficient of static friction being 0.21. Your mass is 68.5 kg. You are hanging on to a cable that runs parallel to the roof. What must the force of tension in the cable be to stop you from sliding down the roof?

First, we draw a sketch:

Now we can write our net force equation:

vec F_"net" = vec F_"g" + vec F_"N" + vec F_"f" + vec F_"T" = mvec a.

The gravitation force and the normal force combine, giving us

vec F_"net" = F_"g,ramp" + vec F_"f" + vec F_"T" = mvec a.

Since we are solving for the case where the mass does not slide down the roof, acceleration is zero:

F_"g,ramp" + vec F_"f" + vec F_"T" = 0.

Now we can solve for the tension force, giving us

vec F_"T" = -vec F_"f" - vec F_"g,ramp",

and, keeping in mind that [fwd] as indicated in the sketch will be positive, we can substitute to get

vec F_"T" = -(-mu_"s"|vec F_"N"|) - m|vec g|sin theta = mu_"s"m|vec g|cos theta - m|vec g|sin theta.

This simplifies to

vec F_"T" = m|vec g|(mu_"s"cos theta - sin theta),

and now we can finally substitute our known values:

vec F_"T" = (68.5\ "kg")(9.80\ "N/kg")(0.21cos31º - sin31º) = -224.9\ "N".

Therefore, the cable must provide a tension force of 225 N [bwd], otherwise it will snap and you will slide off the roof.