Friction: static, limiting and kinetic friction: WASSCE Physics

Without friction you could not walk, a car could not brake, a nail would slide out of wood and a knot would come undone. Yet friction also wears out engines and wastes fuel. This lesson explains what friction is, how big it can get, and how to calculate it.

Learning objectives

By the end of this lesson you should be able to:

define friction and distinguish static, limiting and kinetic (dynamic) friction
state the laws of solid friction
define and calculate the coefficient of friction
solve problems involving friction on horizontal surfaces

What is friction?

Definition

Friction is the force that opposes relative motion (or the tendency to move) between two surfaces in contact. It acts along the surfaces, opposite to the direction of motion or attempted motion.

Friction arises because even smooth-looking surfaces are rough on a microscopic scale. Their tiny bumps interlock and form temporary bonds where they touch.

A block on a rough floor pulled to the right by force P; friction F acts to the left, weight W acts down and normal reaction R acts up

Figure 1. The forces on a block pulled along a rough floor. WASSCE Tutor diagram.

Static, limiting and kinetic friction

Graph of friction force against applied force: friction rises equal to the applied force up to the limiting value, then drops slightly to a constant kinetic value once the block slides

Figure 2. How friction responds as the pull increases. WASSCE Tutor diagram.

Type	When	Size
Static friction	the body is at rest and a force tries to move it	adjusts itself to equal the applied force, up to a maximum
Limiting friction	the body is just about to move	the maximum value of static friction, $F = \mu_s R$
Kinetic (dynamic, sliding) friction	the body is sliding	roughly constant, $F = \mu_k R$ , a little less than limiting friction

This explains why it takes more force to start pushing a heavy box than to keep it sliding.

Coefficient of friction

Definition

The coefficient of static friction $\mu_s$ is the ratio of the limiting friction to the normal reaction:

\mu_s = \dfrac{F_{lim}}{R}

The coefficient of kinetic friction $\mu_k$ is the ratio of the kinetic friction to the normal reaction. Both are pure numbers (no unit).

On a horizontal surface with no other vertical forces, $R = mg$ , so $F = \mu mg$ .

Surfaces	Approximate $\mu_s$
Rubber tyre on dry tarmac	0.8–1.0
Rubber tyre on wet road	0.4–0.6
Wood on wood	0.3–0.5
Steel on steel (dry)	0.6
Steel on ice	0.03

The laws of solid friction

Friction opposes the relative motion of the surfaces.
Friction is independent of the area of contact (for the same normal reaction).
Limiting friction is directly proportional to the normal reaction: $F \propto R$ .
Friction depends on the nature of the surfaces in contact.
Kinetic friction is roughly independent of the speed of sliding (for moderate speeds).

Worked examples

Worked example

1. A 20 kg crate rests on a floor. The coefficient of static friction is 0.4 and of kinetic friction 0.3. ( $g = 10$ m s⁻²)

(a) What horizontal force just starts it moving?

$R = mg = 200$ N, so $F_{lim} = \mu_s R = 0.4 \times 200 = 80$ N.

(b) What force keeps it moving at constant speed?

$F = \mu_k R = 0.3 \times 200 = 60$ N.

(c) A force of 100 N is applied. Find the acceleration.

Once sliding, friction is 60 N. Resultant $= 100 - 60 = 40$ N, so $a = \dfrac{40}{20} = 2$ m s⁻².

(d) A force of 50 N is applied to the crate at rest. What is the friction?

50 N is less than 80 N, so the crate does not move: static friction is just 50 N, equal to the push. Friction is not automatically $\mu R$ .

Common mistake

$F = \mu R$ gives the maximum static friction. If the push is smaller than this, friction equals the push and the body stays at rest.

Worked example

2. A car of mass 1000 kg travelling at 20 m s⁻¹ brakes hard and skids to a stop. The coefficient of kinetic friction between tyres and road is 0.8. Find the stopping distance. ( $g = 10$ m s⁻²)

Friction $F = 0.8 \times 1000 \times 10 = 8000$ N; deceleration $a = \dfrac{8000}{1000} = 8$ m s⁻².

0 = 20^2 - 2(8)s \Rightarrow s = \dfrac{400}{16} = 25\ \text{m}

\begin{gathered}0 = 20^2 - 2(8)s \\[6pt] \begin{aligned}\Rightarrow s &= \dfrac{400}{16} \\ &= 25\ \text{m}\end{aligned}\end{gathered}

On a wet road with $\mu = 0.4$ , the deceleration halves and the stopping distance doubles to 50 m.

Pulling at an angle

If a rope pulls at angle θ above the horizontal, part of the pull lifts the body, so the normal reaction is reduced:

R = mg - P\sin\theta \qquad F = \mu(mg - P\sin\theta)

\begin{gathered}R = mg - P\sin\theta \\[6pt] F = \mu(mg - P\sin\theta)\end{gathered}

Worked example

A 10 kg box is pulled by a rope at 30° above the horizontal; $\mu = 0.5$ . Find the tension needed to keep it moving steadily. ( $g = 10$ m s⁻²)

Horizontally: $P\cos 30° = 0.5(100 - P\sin 30°)$

0.866P = 50 - 0.25P \Rightarrow 1.116P = 50 \Rightarrow P = 44.8\ \text{N}

\begin{gathered}\begin{aligned}&0.866P \\ &= 50 - 0.25P\end{aligned} \\[6pt] \Rightarrow 1.116P = 50 \\[6pt] \Rightarrow P = 44.8\ \text{N}\end{gathered}

Pushing down at 30° instead would increase $R$ and need more force.

Key points

Friction opposes relative motion and acts along the surfaces.
Static friction matches the applied force up to the limiting value $\mu_s R$ ; kinetic friction $\mu_k R$ is slightly less.
$\mu = F/R$ has no unit; friction does not depend on contact area.
Use the resultant force (applied minus friction) in $F = ma$ .

WASSCE examination focus

WAEC tests this

You may be asked to state the laws of friction, define the coefficient of static friction, distinguish static and dynamic friction, and calculate the force needed to move a body or its acceleration when friction acts. The friction-against-applied-force graph sometimes appears for interpretation.

Quick check

What is the unit of the coefficient of friction?
A 50 N block needs 20 N to start it sliding on a horizontal bench. Find $\mu_s$ .
Does increasing the area of contact (same weight) increase friction?

Answers: 1. It has no unit. 2. $\mu_s = 20 \div 50 = 0.4$ . 3. No: friction is independent of the area of contact.

Friction: static, limiting and kinetic friction

Learning objectives

What is friction?

Static, limiting and kinetic friction

Coefficient of friction

The laws of solid friction

Worked examples

Pulling at an angle

Key points

WASSCE examination focus

Quick check

Keep going with friction

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