Friction

A LevelAQAEdexcelOCR

Friction

In the Forces section, we mentioned that friction was calculated by the equation $F \leq \textcolor{red}{\mu} R$, where $\textcolor{red}{\mu}$ is known as the coefficient of friction, which has no units. As a general rule of thumb, the rougher a surface is, the higher its coefficient of friction.

We also have the term limiting friction, when friction is at its maximum. This is shown when $F = \textcolor{red}{\mu} R$.

Make sure you are happy with the following topics before continuing.

A LevelAQAEdexcelOCR

Fundamentals

Example: Here’s a box on a flat surface with two people pushing and pulling in the same direction.

Let’s now say that the weight of the box, $W$, is $100\text{ N}$, and that the friction is limiting (i.e. applying any more force in $T$ or Thrust will cause the box to accelerate). Say, also, that the friction $F$ is $50\text{ N}$.

Then we have the normal reaction force $R = 100\text{ N}$.

Since $F = \textcolor{red}{\mu} R$, the coefficient of friction, $\textcolor{red}{\mu} = \dfrac{50}{100} = 0.5$.

A LevelAQAEdexcelOCR

Note:

By definition, the coefficient of friction $\mu$ is such that $\textcolor{red}{\mu} \geq 0$. It is usually, but not always, less than $1$.

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Friction on a Tilted Plane

Example: Here’s a $1500 \text{ kg}$ car being towed up an incline at an angle of $30\degree$. Say the slope has a coefficient of friction $\textcolor{red}{\mu} = 0.4$. What would be the required combined force of the tension in the rope, $T$, and the thrust from the car? Assume the car is travelling at constant velocity up the hill.

First, adapt this system by rotating it, to visualise the perpendicular forces a little more easily.

Perpendicular to $F$, we have $R = W\cos 30°$.

$W = 1500 \times 9.8 = 14700\text{ N}$, so $R = 12730.57\text{ N}$

So, parallel to $F$, we have $T + \text{Thrust} = F + W\sin 30°$.

By $F_{max} = \textcolor{red}{\mu} R$, the limit of friction is $F_{max} = 5092.23\text{ N}$

Since we have $W$ and $F_{max}$, we can see that the minimum thrust and tension required to move the car up the slope is $5092.23 + W\sin 30° = 12442.23\text{ N}$.

A LevelAQAEdexcelOCR

Friction Example Questions

We have $F \leq \mu R$.

Since no other vertical forces are applied, we have $W = R$.

We also have $W = mg$.

Therefore, we have the expression

$F \leq \mu mg$

Resolving perpendicular to $F$:

$R = W\cos 20° = mg\cos 20°$

Resolving parallel to $F$:

Resultant Force = $F_{res} = ma = 0.2m$

We also have

$F_{res} = W\sin 20° - \mu R$

$= mg\sin 20° - \mu mg\cos 20°$

So,

$g\sin 20° - \mu g\cos 20° = 0.2$

Giving

$\mu = \dfrac{9.8\sin 20° - 0.2}{9.8\cos 20°} = 0.342\text{ (to }3\text{ dp)}$

For $v = u + at$, we have

$0 = 80 + 20a\\$

$a = \dfrac{-80}{20} = -4\text{ ms}^{-2}$ horizontally.

From Newton’s Second Law, $F = ma = 0.01 \times -(-4) = 0.04\text{ N}$.

Vertically, we have $R = 0.01 \times 9.8 = 0.098\text{ N}$

With no other horizontal forces acting on the system, we assume the friction is limiting, so

$\mu = \dfrac{0.04}{0.098} = 0.408\text{ (to }3\text{ dp)}$

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