PhysicsLesson 5 of 18About 40 minutes

Weight, the normal force and friction

Three forces show up in almost every mechanics problem: the weight of an object, the push of the surface it rests on, and friction. This lesson shows how to find each one.

By the end of this lesson you can

explain the difference between mass and weight, and calculate weight;
find the direction and size of the normal force, on flat ground and on a ramp;
tell static friction from kinetic friction;
use coefficients of friction to calculate friction forces.

Weight

Think first

An astronaut travels from Earth to the Moon. Does her mass change? Does her weight change?

Compare your answer

Her mass, the amount of matter in her body, stays the same. Her weight is the gravitational pull on her, and the Moon pulls much more weakly than Earth, so her weight drops to about one sixth.

Definition

The weight of an object is the size of the gravitational force acting on it. Unlike mass, weight depends on where the object is.

Weight

$$F_g = mg$$

$g$ is the free-fall acceleration where the object is. At Earth's surface $g = 9.81\ \text{m/s}^2$; on the Moon it is about $1.62\ \text{m/s}^2$.

Even on Earth, $g$ varies slightly: it is a little smaller on a high mountain than at sea level. For problems, use 9.81 m/s² unless told otherwise.

Weight on Earth, the Moon and Jupiter

Interactive

$\vec F_g$

Mass

Scroll sideways to see the whole demo, or turn your phone sideways.

Try this

Set the mass to your own mass in kilograms and read your weight on Earth.
Switch to the Moon and to Jupiter. What changes, and what stays the same?

Quick check

What is the weight of a 65.0 kg student on Earth?

Show solution

$F_g = (65.0\ \text{kg})(9.81\ \text{m/s}^2) = 638\ \text{N}$. On the Moon it would be $(65.0)(1.62) = 105\ \text{N}$.

The normal force

A laptop sits on a desk. Gravity pulls it down, yet it does not sink into the desk. The desk pushes up on it. This push is called the normal force, because in geometry "normal" means perpendicular.

Definition

The normal force $\vec F_n$ is the push a surface exerts on an object in contact with it. It always points perpendicular to the surface, away from it.

The normal force adjusts itself to whatever is needed to stop the object from moving into the surface. On a flat surface, with only gravity and the normal force acting vertically, $F_n = mg$. But that is a special case:

On a ramp tilted at angle $\theta$, only part of the weight presses into the surface, so $F_n = mg\cos\theta$.
If you push down on a box, the floor must push up harder: $F_n$ is larger than $mg$.
If you pull up on a box with a rope, the floor pushes up less: $F_n$ is smaller than $mg$.

The normal force on a ramp

Interactive

$\vec F_g$

$\vec F_n$

$F_{g,\parallel}$

$F_{g,\perp}$

$\theta$

AngleMass

Scroll sideways to see the whole demo, or turn your phone sideways.

Try this

Start at 0°. How does the normal force compare with the weight?
Raise the angle slowly. Which component of the weight grows, and which shrinks?
At what angle are the two components equal?

Worked example 1

A 12.0 kg box rests on a ramp tilted at 30.0°. Find the normal force on the box.

Find the weight. $F_g = mg = (12.0\ \text{kg})(9.81\ \text{m/s}^2) = 118\ \text{N}$
Take the component of the weight into the ramp. Only gravity and the normal force act perpendicular to the ramp, and the box does not move in that direction, so they balance: $$F_n = mg\cos\theta = (118\ \text{N})\cos 30.0^\circ$$
Calculate.
$F_n = 102\ \text{N}$, perpendicular to the ramp surface

Static friction

Push gently on a heavy bookcase and it does not budge. Some force must be cancelling your push: static friction, the force between two surfaces that are touching but not sliding over each other.

Static friction is a responsive force. Push with 20 N and static friction pushes back with 20 N. Push with 50 N and it pushes back with 50 N. But it has a limit. Once your push goes past the maximum static friction, $F_{s,\text{max}}$, the bookcase breaks free and starts to slide.

Kinetic friction

Once something is sliding, the friction on it is kinetic friction. For most pairs of surfaces it is smaller than the maximum static friction, and it stays roughly the same whatever the speed. That is why it is harder to get a heavy box started than to keep it going.

Friction comes from the roughness of surfaces at a microscopic level: tiny bumps catch on each other, and atoms in the two surfaces attract each other where they touch.

Push harder and harder

Interactive

$\vec F_{\text{applied}}$

$\vec F_{\text{friction}}$

applied force (N)

friction (N)

static

kinetic

Push

Scroll sideways to see the whole demo, or turn your phone sideways.

Try this

Move the push slider up slowly from zero. Watch the friction arrow and the graph.
Find the push at which the box starts to slide. What happens to the friction force at that moment?
Describe the shape of the graph in one sentence.

The coefficient of friction

Friction depends on two things: how hard the surfaces press together (the normal force) and what the surfaces are made of. The second factor is captured by a number called the coefficient of friction, $\mu$ (the Greek letter mu). It has no units.

Friction forces

$$F_{s,\text{max}} = \mu_s F_n \qquad\qquad F_k = \mu_k F_n$$

$\mu_s$ is the coefficient of static friction and $\mu_k$ the coefficient of kinetic friction. Static friction can be anything from zero up to $F_{s,\text{max}}$.

Approximate coefficients of friction
Surfaces in contact	$\mu_s$	$\mu_k$
Rubber tire on dry concrete	1.0	0.8
Glass on glass	0.9	0.4
Steel on steel (dry)	0.7	0.6
Wood on wood	0.4	0.2
Waxed ski on snow	0.10	0.05
Steel on ice	0.03	0.02
Teflon on Teflon	0.04	0.04

Typical values. Real coefficients vary with how clean, dry and smooth the surfaces are.

Worked example 2

A 30.0 kg desk sits on a level floor. It takes a horizontal push of 120 N to get it moving, and a push of 90.0 N to keep it sliding at constant velocity. Find $\mu_s$ and $\mu_k$.

Find the normal force. The floor is level and the push is horizontal, so $F_n = mg = (30.0\ \text{kg})(9.81\ \text{m/s}^2) = 294\ \text{N}$.
Static: the push that just starts it equals the maximum static friction. $$\mu_s = \frac{F_{s,\text{max}}}{F_n} = \frac{120\ \text{N}}{294\ \text{N}} = 0.41$$
Kinetic: at constant velocity the net force is zero, so kinetic friction equals the push: $$\mu_k = \frac{F_k}{F_n} = \frac{90.0\ \text{N}}{294\ \text{N}} = 0.31$$
$\mu_s = 0.41$, $\mu_k = 0.31$

Worked example 3

A 15.0 kg box of tiles rests on a floor with $\mu_s = 0.50$ and $\mu_k = 0.35$. (a) You push horizontally with 60.0 N. Does it move? (b) You push with 90.0 N instead. What is its acceleration?

Find the normal force and the friction limits. $F_n = (15.0)(9.81) = 147\ \text{N}$. So $F_{s,\text{max}} = 0.50 \times 147 = 73.6\ \text{N}$ and $F_k = 0.35 \times 147 = 51.5\ \text{N}$.
(a) Compare the push with the static limit. 60.0 N is less than 73.6 N, so the box stays put. Static friction is exactly 60.0 N, matching the push.
(b) 90.0 N is more than 73.6 N, so the box slides and kinetic friction (51.5 N) acts. Net force: $90.0 - 51.5 = 38.5\ \text{N}$.
Apply the second law. $$a = \frac{38.5\ \text{N}}{15.0\ \text{kg}} = 2.57\ \text{m/s}^2$$
(a) No, it stays at rest. (b) $a = 2.57\ \text{m/s}^2$ in the direction of the push.

Check yourself

Question 1

Free-fall acceleration on Mars is 3.71 m/s². What would a 40.0 kg rover weigh there?

Show solution

$F_g = (40.0\ \text{kg})(3.71\ \text{m/s}^2) = 148\ \text{N}$. On Earth it weighs 392 N; its mass is 40.0 kg in both places.

Question 2

A suitcase sits on a level airport floor. You slowly lift one end of the floor panel so that it becomes a gentle ramp. What happens to the normal force on the suitcase?

It increases.
It decreases.
It stays equal to the weight.
It becomes zero.

Show solution

$F_n = mg\cos\theta$, and $\cos\theta$ gets smaller as $\theta$ grows from 0°, so the normal force decreases.

Question 3

A 20.0 kg crate slides across a level floor at constant velocity while you push it horizontally with 58.9 N. What is the coefficient of kinetic friction?

Show solution

Constant velocity, so kinetic friction equals the push: $F_k = 58.9\ \text{N}$. $F_n = (20.0)(9.81) = 196\ \text{N}$, so $\mu_k = 58.9 / 196 = 0.30$.

Question 4

A wooden block rests on a board. You tilt the board slowly, and the block starts to slide when the angle reaches $\theta$. Show that $\mu_s = \tan\theta$, then find the angle for $\mu_s = 0.60$.

Show solution

Just before it slides, static friction is at its maximum and balances the downhill part of the weight: $\mu_s mg\cos\theta = mg\sin\theta$. Divide by $mg\cos\theta$: $\mu_s = \tan\theta$. For $\mu_s = 0.60$, $\theta = \tan^{-1}(0.60) = 31^\circ$. The mass cancels, so heavy and light blocks start to slide at the same angle.

Question 5

A box is at rest on the floor. You push it with a force smaller than $F_{s,\text{max}}$. How large is the friction force on the box?

Equal to your push
Equal to $\mu_s F_n$
Equal to $\mu_k F_n$
Zero, because the box does not move

Show solution

The box stays at rest, so the net force is zero and static friction exactly matches your push. $\mu_s F_n$ is only the largest value static friction can reach.

Summary

Weight is the gravitational force on an object: $F_g = mg$. Mass does not change with location; weight does.
The normal force is perpendicular to the surface. It equals $mg$ only on level ground with no other vertical forces; on a ramp $F_n = mg\cos\theta$.
Static friction matches the applied force up to $F_{s,\text{max}} = \mu_s F_n$. Kinetic friction on a sliding object is $F_k = \mu_k F_n$, usually smaller.

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