16.7 Friction and the F <= mu R model — Learn

Prerequisites: resolving forces and coplanar equilibrium (Section 16.5); Newton's second law (Section 16.2).

By the end of this section you can: describe how friction acts; use the model $F \leq \mu R$ ; find the normal reaction on a rough surface; decide whether a body stays at rest or slides; and find a coefficient of friction at limiting equilibrium.

1. What friction is

Friction is a force between two surfaces that opposes sliding, or the tendency to slide. It acts along the contact surface, in the direction that opposes the motion.

Friction is a contact force. It is not a fixed force. Static friction is variable: it takes whatever value is needed to stop the surfaces sliding, up to a limit.

A small push gives a small friction, and the body stays still.
A larger push gives a larger friction, and the body still stays still.
At the limiting value the body is about to move. This is called limiting equilibrium.
A push larger than the limit makes the body slide, and friction then stays at the limiting value.

2. The model F ≤ μR

The limit on friction depends on how hard the surfaces are pressed together, which is the normal reaction $R$ . The model is

F \leq \mu R,

where $\mu$ is the coefficient of friction between the two surfaces. The coefficient has no unit, because it is a ratio of two forces. It depends on the two surfaces, not on the mass of the body.

If the body stays at rest, friction takes a value below the limit, and it equals the driving force.
If the body slides, friction takes its greatest value, $F = \mu R$ .

The normal reaction $R$ is not always the weight.

On a horizontal surface with only vertical forces, $R = mg$ .
On a slope of angle $\theta$ , $R = mg\cos\theta$ .
If another force has a component perpendicular to the surface, resolve in that direction to find $R$ .

Example 1. A block of mass $10$ kg rests on a rough horizontal floor. The coefficient of friction is $0.3$ . A horizontal push of $20$ N is applied. Find the friction force. Take $g = 9.8$ m s $^{-2}$ .

The vertical forces balance, so

R = 10 \times 9.8 = 98 \text{ N}.

The limit on friction is

\mu R = 0.3 \times 98 = 29.4 \text{ N}.

The push is $20$ N, which is below $29.4$ N, so the block stays at rest. Static friction takes the value needed, so

F = 20 \text{ N}

backwards, exactly balancing the push.

Example 2. The same block is pushed with $40$ N instead. Find the friction force and the acceleration.

The limit is still $29.4$ N. Now $40 > 29.4$ , so the block slides and friction takes its greatest value:

F = \mu R = 29.4 \text{ N}.

The resultant along the floor is

40 - 29.4 = 10.6 \text{ N},

a = \frac{10.6}{10} = 1.06 \text{ m s}^{-2}.

Example 3. A crate of mass $45$ kg is placed on a rough ramp inclined at $18^\circ$ . The coefficient of friction is $0.4$ . The crate is released from rest. Show that the crate stays at rest, and find the friction force. Take $g = 9.8$ m s $^{-2}$ .

The weight is $441$ N. Resolve perpendicular to the ramp:

R = 441\cos 18^\circ \approx 419.4 \text{ N}.

The limit on friction is

\mu R = 0.4 \times 419.4 \approx 167.8 \text{ N}.

The component of the weight down the ramp is

441\sin 18^\circ \approx 136.3 \text{ N}.

Since $136.3 \leq 167.8$ , the downhill component is within the limit, so the crate stays at rest. Friction matches that component and is $136.3$ N up the ramp.

Example 4. A block on a rough slope of $30^\circ$ is in limiting equilibrium. Find the coefficient of friction.

At limiting equilibrium the downhill component equals the greatest friction:

mg\sin 30^\circ = \mu\,mg\cos 30^\circ.

Divide both sides by $mg\cos 30^\circ$ :

\mu = \frac{\sin 30^\circ}{\cos 30^\circ} = \tan 30^\circ \approx 0.577.

The mass cancels, so the answer does not depend on the mass.

3. Try it: the friction explorer

Predict first. A block of mass $30$ kg rests on a rough floor. Before you touch anything, predict what happens to the friction force as the push grows from zero to the limit, and what happens after the limit is passed.

Test. Move the push slider slowly and watch the friction arrow. Find the push at which the block changes from staying still to sliding. Then change the coefficient of friction and find the new limit.

Explain. Write two sentences. The first says why the friction force equals the push while the block stays still. The second says why the friction stops growing once the block is sliding.

Watch out. Three mistakes are common.

Using $F = \mu R$ when the body is not sliding. The model is an inequality: $F \leq \mu R$ . A body at rest uses only as much friction as it needs. Re-check: find the driving force and the limit separately, then compare them.

Assuming the normal reaction is the weight. On a slope, $R = mg\cos\theta$ , which is smaller than $mg$ . A pull with an upward component also makes $R$ smaller. Resolve perpendicular to the surface every time.

Letting friction act the wrong way. Friction opposes the sliding. On a slope, a block that is about to slide down has friction up the slope. If you draw friction down the slope, the block can never be in equilibrium.

Progress check

Return to the chapter problem. The crate of mass $45$ kg is now on a rough ramp at $18^\circ$ , with a coefficient of friction of $0.4$ . The rope is slack.

Find the component of the crate's weight down the ramp.
Find the normal reaction from the ramp.
Find the greatest friction available.
Decide whether the crate slides. Explain your answer, and give the friction force if the crate stays at rest.

You should now be able to

describe friction and the direction in which it acts;
use the model $F \leq \mu R$ and say when equality holds;
find the normal reaction on a rough horizontal surface and on a rough slope;
compare a driving force with the limit on friction;
decide whether a body stays at rest or slides;
find the friction force and the acceleration once a body slides;
find a coefficient of friction at limiting equilibrium.