The coefficient of friction

In a nutshell

Friction is a force which exists when there is a rough surface and acts in the opposite direction to motion. The coefficient of friction is represented by the letter $\mu$ (pronounced "mew").

Equations

description	equation
Maximum value of friction between two surfaces	$F=\mu R$

Variable definitions

quantity name	symbol	unit name	unit
$Friction$	$F$	$Newton$	$N$
$Coefficient\ of\ friction$	$\mu$	$Unitless$	$None$
$Normal\ reaction$	$R$	$Newton$	$N$

Frictional force

Frictional forces only exist when there is an object in motion on a rough surface, or if the object has a tendency to move, such as a ball at rest on a rough slope. When there is a force acting on an object, it will not move if the frictional force is greater than or equal to the force applied. When the force applied is greater than the maximum frictional force, the object will move.

The maximum value of friction is based on two things: the normal reaction, $R$ , and the roughness of the surface. Roughness is measured using the coefficient of friction, $\mu$ . Rougher surfaces have a higher value of $\mu$ . The maximum value of friction between two surfaces is given by $F_{Max}=\mu R$ .

Example 1

A block with a mass of $10kg$ is being pushed along a rough surface with a force of $50N$ . The coefficient of friction between the block and the floor is $0.3$ . Calculate:

a: The magnitude of the frictional force.

b: The acceleration of the block.

Maths; Forces and friction; KS5 Year 13; The coefficient of friction

a: The block is moving, therefore the frictional force is the maximum value of friction:

$\begin{aligned}Friction&= \mu R\\ F&= 0.3\times 10g\\ F&= 29.4N\end{aligned}$

The frictional force is $\underline{29.4N}$ .

b: Resolve horizontally:

$\begin{aligned}50-\mu R&=10a\\ 50-29.4&=10a\\ 10a&=20.6\\ a&=2.06ms^{-2}\end{aligned}$

The acceleration of the block is $\underline{2.06ms^{-2}}$ .

Example 2

A ball of mass $15kg$ is on a rough slope at an angle of $20\degree$ to the horizontal. There are no external forces acting on the ball. If the ball is at rest, what is the coefficient of friction between the ball and the slope?

Maths; Forces and friction; KS5 Year 13; The coefficient of friction

If the ball is at rest, then the parallel component of the weight of the ball is equal to the frictional force. Resolve horizontally:

$\begin{aligned}F&= 15g\sin(20)\\ F&= 147\sin(20) = 50.2769611...N\end{aligned}$

Find the reaction force of the slope on the ball. resolve perpendicularly:

$\begin{aligned}R&= 15g\cos(20)\\ R&= 147\cos(20) = 138.13482...N\end{aligned}$

Find $\mu$ :

$\begin{aligned}F&= \mu R\\ \mu &=\dfrac{F}{R}\\ \mu &= \dfrac{147\sin(20)}{147\cos(20)}\\ \mu &= \tan(20)\\ \mu&= 0.36397...\end{aligned}$

The coefficient of friction is $\underline{0.364} \ (3\ d.p.)$