# Friction

Frictional force opposes relative motion between two objects in contact. Its magnitude depends on the surface properties of the objects and the normal force between them.

**Static friction** is the force that must be overcome in order to start moving an object at rest. It has a limiting value given by
\begin{align}
f_\text{max}=\mu_s N,
\end{align}
where $\mu_s$ is the coefficient of static friction and $N$ is the normal reaction.

**Dynamic (or kineatic or sliding) friction** is the force that acts on an object that is already in motion. Its value is given by
\begin{align}
f_k=\mu_k N,
\end{align}
where $\mu_k$ is the coefficient of static friction.

**Rolling friction** is the force that opposes the motion of a rolling object, such as a wheel or a ball. Rolling friction is generally weaker than sliding friction.

## Problems from IIT JEE

**Problem (IIT JEE 1980):**
A block of mass 2 kg rests on a rough inclined plane making an angle of ${30}^\mathrm{o}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$. The frictional force on the block is,

- 9.8 N
- $0.7\times 9.8\times\sqrt{3}\;\mathrm{N}$
- $9.8\times \sqrt{3}\;\mathrm{N}$
- $0.7\times 9.8\;\mathrm{N}$

**Solution:**
The forces on the block of mass $m={2}\;\mathrm{kg}$ are its weight $mg$, normal reaction $N$, and the frictional force $f$ (see figure).

The net force on the block is zero because it is at rest. Resolve $mg$ in the directions parallel and perpendicular to the inclined plane. Apply Newton's second law in these directions to get,
\begin{align}
N&=mg\cos30 \\
&=(2)(9.8)(0.866)={16.97}\;\mathrm{N},\\
f&=mg\sin30\\
&=(2)(9.8)(0.5)={9.8}\;\mathrm{N}.
\end{align}
Note that $f$ is less than $f_\text{max}=\mu N=(0.7)(16.97)={11.88}\;\mathrm{N}$.

## Related

- Newton's Laws of Motion
- Rolling Friction (Demo)
- Direction of the Frictional Forces on the Bicycle Wheels during Pedaling