Friction

Friction is the force that stops a box sliding when you push it gently, and slows it down once it does slide. It is the one force in Paper 4 whose size is not fixed: it adjusts itself, up to a limit, to whatever is needed to prevent motion. Understanding exactly when friction takes its maximum value $\mu R$, and when it does not, is the key to a large share of the marks on the paper, in equilibrium, Newton's law and energy questions alike.

Friction as part of the contact force

When two rough surfaces touch, the contact force between them has two components: the normal reaction $R$ perpendicular to the surfaces and the friction $F$ along them.

Friction acts to oppose relative motion of the two surfaces. If the surfaces are sliding, friction opposes the sliding. If they are not sliding, friction opposes the motion that would happen without it (the "tendency to move").

How big is friction?

Push gently on a heavy box resting on a rough floor. It does not move, so it is in equilibrium: friction exactly balances your push. Push a little harder and friction increases to match. This continues until friction reaches a maximum value; push harder than that and the box slides.

So friction behaves in two different ways:

• Before slipping, friction is whatever is needed to keep the particle in equilibrium, which can be anything from $0$ up to the maximum.
• At the point of slipping, and while sliding, friction takes its maximum value.

Experiment shows that the maximum friction is proportional to the normal reaction: press the surfaces together twice as hard and the maximum friction doubles. The constant of proportionality is the coefficient of friction.

Some facts about $\mu$:

• $\mu \ge 0$ and has no units (it is a ratio of two forces).
• Typical values: about $0.05$ for ice, $0.3$ to $0.6$ for wood on wood, around $0.8$ for rubber on dry road. There is no rule that $\mu < 1$, though most exam values are.
• The model assumes $\mu$ depends only on the two surfaces, not on the area of contact or the speed. This is a good approximation, but only an approximation.

The normal reaction is not always mg

Because $F_{\max} = \mu R$, anything that changes $R$ changes the maximum friction. On a horizontal floor $R = mg$ only if no other force has a vertical component.

• A force pulling upwards at an angle (a rope pulling a sledge) lifts part of the weight, so $R$ is less than $mg$ and friction is reduced.
• A force pushing downwards at an angle (pushing a lawnmower) presses the object into the floor, so $R$ is more than $mg$ and friction is increased.

This is why it is easier to pull a heavy box with a rope angled upwards than to push it with a force angled downwards. Example 3 below puts numbers on it.

The total contact force

Sometimes a question asks for the total contact force (or "the magnitude of the contact force") rather than its components. Combine $R$ and $F$ like any two perpendicular forces:

Friction that could act either way

When two applied forces act in opposite directions, the particle could be about to move either way, and friction opposes whichever way that is. A question asking for the range of values of a force for equilibrium needs both limiting cases.

Limitations of the friction model

The law $F \le \mu R$ is a model. Its main simplifications:

• In reality the friction while sliding (kinetic friction) is usually a little less than the maximum static friction. Paper 4 uses a single $\mu$ for both.
• $\mu$ can vary with speed, temperature, wear and contamination of the surfaces.
• The model ignores the area of contact. That is reasonably accurate for rigid surfaces but not for, say, tyres.

Common mistakes

Exam technique

Summary

Practice

Answers

1. $R = 60$, $\mu R = 21\ \text{N}$. Only $18\ \text{N}$ of friction is needed, and $18 < 21$, so the box stays at rest with friction $18\ \text{N}$.
2. Constant speed means equilibrium, and the sledge is sliding, so $F = \mu R$: $30 = \mu \times 100$, giving $\mu = 0.3$.
3. $R = 500 - T\sin 20^\circ$. Limiting: $T\cos 20^\circ = 0.4(500 - T\sin 20^\circ)$, so $T(\cos 20^\circ + 0.4\sin 20^\circ) = 200$ and $T = \dfrac{200}{1.07650} = 186\ \text{N}$.
4. $R = 120 + 60\sin 30^\circ = 150$, so $\mu R = 0.45 \times 150 = 67.5\ \text{N}$. The horizontal component of the push is $60\cos 30^\circ = 52.0\ \text{N}$. Since $52.0 < 67.5$, friction can balance it: the box does not move, and the frictional force is $52.0\ \text{N}$.
5. $R = 60$. $F = \sqrt{65^2 - 60^2} = 25$, so $P = 25$. At rest: $25 \le 60\mu$, so $\mu \ge \tfrac{5}{12} = 0.417$.
6. $R = 10m - 100\sin 30^\circ = 10m - 50$. Limiting: $100\cos 30^\circ = 0.25(10m - 50)$, so $86.60 = 2.5m - 12.5$ and $m = 39.6$.
7. (a) $12 = 30\mu$, so $\mu = 0.4$. (b) $P\cos 30^\circ = 0.4(30 - P\sin 30^\circ)$, so $P(\cos 30^\circ + 0.2) = 12$ and $P = 11.3\ \text{N}$. (c) $P\cos 30^\circ = 0.4(30 + P\sin 30^\circ)$, so $P(\cos 30^\circ - 0.2) = 12$ and $P = 18.0\ \text{N}$.
8. $R = 50 - P\sin 30^\circ = 50 - 0.5P$. About to move in the direction of $P$: $P\cos 30^\circ = 15 + 0.5(50 - 0.5P)$, so $P(0.8660 + 0.25) = 40$ and $P = 35.8$. About to move the other way: $P\cos 30^\circ + 0.5(50 - 0.5P) = 15$, so $P(0.8660 - 0.25) = -10$, which is negative. This means that even with $P = 0$ friction alone ($\mu R = 25 \ge 15$) holds the particle, so there is no lower limit beyond $P \ge 0$. The particle is in equilibrium for $0 \le P \le 35.8$.
9. (a) $R = 20 - P\sin\theta$; limiting: $P\cos\theta = 0.6(20 - P\sin\theta)$, so $P(\cos\theta + 0.6\sin\theta) = 12$. (b) $\theta = 0^\circ$: $P = 12.0\ \text{N}$. $\theta = 31^\circ$: $P = \dfrac{12}{0.85717 + 0.30902} = 10.3\ \text{N}$. $\theta = 60^\circ$: $P = \dfrac{12}{0.5 + 0.51962} = 11.8\ \text{N}$. Pulling at a moderate upward angle reduces the force needed, because it reduces $R$ and hence friction; but too steep an angle wastes force in the vertical direction. (The minimum occurs when $\tan\theta = \mu$, here $\theta = 31.0^\circ$, giving $P = 10.3\ \text{N}$.)