# 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

$\left|{\overrightarrow{F}}_{\text{f}}\right|=\mu \left|{\overrightarrow{F}}_{\text{N}}\right|\text{.}$

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

$\left|{\overrightarrow{F}}_{\text{f}}\right|=\mu m\left|\overrightarrow{g}\right|\text{.}$

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: ${\mu}_{\text{s}}$ for static and ${\mu}_{\text{k}}$ for kinetic, where ${\mu}_{\text{s}}>{\mu}_{\text{k}}\text{.}$ The values of both are determined experimentally for every pair of materials.

## 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:

${\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_{\text{g}}+{\overrightarrow{F}}_{\text{N}}+{\overrightarrow{F}}_{\text{f}}+{\overrightarrow{F}}_{\text{T}}=m\overrightarrow{a}\text{.}$

The gravitation force and the normal force combine, giving us

${\overrightarrow{F}}_{\text{net}}={F}_{\text{g,ramp}}+{\overrightarrow{F}}_{\text{f}}+{\overrightarrow{F}}_{\text{T}}=m\overrightarrow{a}\text{.}$

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

${F}_{\text{g,ramp}}+{\overrightarrow{F}}_{\text{f}}+{\overrightarrow{F}}_{\text{T}}=0\text{.}$

Now we can solve for the tension force, giving us

${\overrightarrow{F}}_{\text{T}}=-{\overrightarrow{F}}_{\text{f}}-{\overrightarrow{F}}_{\text{g,ramp}}\text{,}$

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

${\overrightarrow{F}}_{\text{T}}=-(-{\mu}_{\text{s}}\left|{\overrightarrow{F}}_{\text{N}}\right|)-m\left|\overrightarrow{g}\right|\phantom{\rule{0.1667em}{0ex}}\mathrm{sin}\phantom{\rule{0.08335em}{0ex}}\theta ={\mu}_{\text{s}}m\left|\overrightarrow{g}\right|\phantom{\rule{0.1667em}{0ex}}\mathrm{cos}\phantom{\rule{0.08335em}{0ex}}\theta -m\left|\overrightarrow{g}\right|\phantom{\rule{0.1667em}{0ex}}\mathrm{sin}\phantom{\rule{0.08335em}{0ex}}\theta \text{.}$

This simplifies to

${\overrightarrow{F}}_{\text{T}}=m\left|\overrightarrow{g}\right|({\mu}_{\text{s}}\mathrm{cos}\phantom{\rule{0.08335em}{0ex}}\theta -\mathrm{sin}\phantom{\rule{0.08335em}{0ex}}\theta )\text{,}$

and now we can finally substitute our known values:

${\overrightarrow{F}}_{\text{T}}=(68.5\text{}\text{kg})(9.80\text{}\text{N/kg})(0.21\phantom{\rule{0.1667em}{0ex}}\mathrm{cos}\phantom{\rule{0.08335em}{0ex}}31\text{\xba}-\mathrm{sin}\phantom{\rule{0.08335em}{0ex}}31\text{\xba})=-224.9\text{}\text{N}\text{.}$

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