What is the difference between static and kinetic friction?

Static friction acts when surfaces are not sliding and grows to match the applied force, up to a maximum of μsFN. Kinetic friction acts while they slide and is roughly constant at μkFN. Usually μk is smaller than μs, so it's easier to keep something sliding than to start it.

Are the weight and the normal force a third law pair?

No. They often have the same size and opposite directions, but they act on the same object, and third-law pairs always act on two different objects. They balance because of the first law, not the third.

Why is the normal force on a slope less than the weight?

The surface only pushes at right angles to itself, so it only has to balance the part of the weight that is into the slope, mg cos θ. The other part, mg sin θ, acts down the slope.

Does a moving object always have a net force on it?

No. By Newton's first law, an object moving at constant velocity has zero net force on it. A net force changes the velocity: it speeds the object up, slows it down or changes its direction.

A book rests on a table. Earth pulls down on the book with its weight. Which force is the Newton's third law partner of that weight?

The table pushing up on the book

The book pushing down on the table

A car travels at a constant 100 km h^−1 along a straight, flat highway. The net force on the car is:

forward, equal to the driving force

backward, equal to air resistance

Year 11 Physics · Fundamentals of mechanics · Forces · 14 min read

Newton's laws of motion

Newton's three laws connect forces to motion. Almost every Year 11 forces question follows the same pattern: draw a free-body diagram, find the net force, then use the first law (net force zero) or the second law (F_net = ma).

This guide covers free-body diagrams, the three laws, static and kinetic friction, connected bodies and inclined planes, including the derivation of a = g sin θ − μg cos θ. Then the mistakes that cost marks, a quick quiz and a full worked example with a box, a ramp and a pulley.

What you need to know

Determine the net force on an object with vector analysis and free-body diagrams.
Apply Newton's first law to objects at rest or moving at constant velocity, and analyse the forces on them.
Analyse the forces on an accelerating object, and use F_net = ma (Newton's second law) to solve problems.
Investigate how acceleration depends on force and on mass.
Describe everyday situations that show the outcomes of Newton's third law.
Use Newton's laws and vector diagrams to analyse the forces on connected bodies, whether touching or roped together.
Compare kinetic and static friction, and use f = μF_N to solve problems.
Apply Newton's laws to a range of real-world problems, static and moving.
On an incline, find the normal force, draw free-body diagrams showing the net, frictional, normal and gravitational forces, and resolve the weight into F_⊥ = mg cos θ and F_∥ = mg sin θ.
Apply Newton's second law along an incline with friction to derive a = g sin θ − μg cos θ, and solve incline problems for objects accelerating or moving at constant velocity.
Explain the physics of why ramps were used by Aboriginal and/or Torres Strait Islander Peoples for heavy loads.

The equations you will use:

F_net = maThe net force (the vector sum of all the forces) and the acceleration point the same way.

F_g = mgWeight, with g = 9.8 m s⁻² near Earth's surface.

f = μF_NKinetic friction while sliding; for static friction, μ_sF_N is the largest value it can reach.

F_⊥ = mg cos θ F_∥ = mg sin θThe weight's parts across and along an incline at angle θ.

a = g sin θ − μg cos θSliding down an incline with friction (no other forces along it).

Dot points: PY11-FM-18, PY11-FM-19, PY11-FM-20, PY11-FM-21, PY11-FM-22, PY11-FM-23, PY11-FM-24, PY11-FM-25, PY11-FM-26, PY11-FM-27, PY11-FM-28, PY11-FM-29, PY11-FM-30, PY11-FM-31, PY11-FM-32, PY11-FM-33, PY11-FM-34 (2025 Physics syllabus, Year 11), in our own words.

Mini lesson

Free-body diagrams and net force

A free-body diagram shows one object on its own, with an arrow for every force acting on it, drawn from the object and labelled. The usual forces are the weight (mg, down), the normal force (F_N, at right angles to the surface), tension (T, along a rope), friction (f, along the surface) and any applied push or pull.

The net force is the vector sum of all of them. Add the forces in each direction separately (resolving any that are at an angle), then combine. Forces exerted by the object on other things never go on its free-body diagram.

A box pulled along a floor. Vertically F_N = mg, so those cancel. Horizontally F_net = F − f, to the right, so the box accelerates to the right.

Newton's first law: balanced forces

If the net force on an object is zero, it stays at rest or keeps moving at a constant velocity (same speed, same direction). That is the first law.

Constant velocity does not mean no forces. A car cruising at 100 km h⁻¹ on a straight road has a forward driving force from the road on its tyres, and backward air resistance and rolling friction. They balance, so the net force is zero and the velocity doesn't change. If the driving force stopped, the net force would point backwards and the car would slow down.

Newton's second law: F_net = ma

A non-zero net force makes an object accelerate in the direction of the net force: a = F_net ÷ m. Double the net force and the acceleration doubles; double the mass and it halves.

In the lab you can show this with a trolley pulled along a track by a string over a pulley to a hanging mass. To change the force while keeping the total mass constant, move slotted masses from the trolley to the hanger. To change the mass, add masses to the trolley and keep the hanging mass the same. Measure the acceleration with light gates, a ticker timer or video.

Lifts are a classic second-law problem. For a person standing on scales in a lift accelerating upward, F_N − mg = ma, so the scales read F_N = m(g + a), more than their weight.

Newton's third law: force pairs

When object A pushes or pulls on object B, B pushes or pulls on A with a force of the same size in the opposite direction. The two forces are the same type, and they act on different objects, which is why they never cancel each other out.

Everyday examples: when you walk, your foot pushes the ground backwards and the ground pushes you forwards. A swimmer pushes water backwards and the water pushes the swimmer forwards. A rocket pushes exhaust gas out the back and the gas pushes the rocket forwards. A rifle recoils because the bullet pushes back on it.

A book resting on a table has two forces on it, its weight and the normal force, and they balance. They are not a third-law pair, because they act on the same object. The partner of the book's weight is the book's gravitational pull on Earth.

Friction: static and kinetic

Static friction acts when surfaces are not sliding. It is only as big as it needs to be to stop sliding, up to a maximum of μ_sF_N. Push a heavy crate gently and static friction matches your push exactly, so it doesn't move.

Kinetic friction acts once the surfaces slide, with a fairly constant size f = μ_kF_N. For most pairs of surfaces μ_k is less than μ_s, which is why something is harder to start moving than to keep moving.

Use the actual normal force. It equals mg only on a level surface with no other vertical forces; on an incline, or with a pull at an angle, find it from the forces at right angles to the surface.

Connected bodies

For objects joined by a string, or pushing against each other, use two views. First treat them as one system: only external forces count, and the total mass accelerates, so a = F_net,external ÷ m_total. Then look at one body on its own to find the force between them (the tension or the contact force).

Example: trolleys A (2.0 kg) and B (3.0 kg) are joined by a light string, and B is pulled with 10 N on a smooth track. As a system, a = 10 ÷ 5.0 = 2.0 m s⁻². A alone has only the tension pulling it, so T = m_Aa = 2.0 × 2.0 = 4.0 N.

System: a = 10 ÷ (2.0 + 3.0) = 2.0 m s⁻². A alone: T = 2.0 × 2.0 = 4.0 N. The tension pulls each trolley towards the other, and it is less than the 10 N pull.

Inclined planes, and why ramps help

On an incline at angle θ, tilt your axes: along the slope and at right angles to it. The weight splits into mg sin θ down the slope and mg cos θ into the slope. Nothing moves at right angles to the slope, so the normal force is F_N = mg cos θ, less than mg.

For an object sliding down with kinetic friction (acting up the slope): F_net = mg sin θ − μmg cos θ = ma, and the mass cancels to give a = g sin θ − μg cos θ. If it slides at constant velocity, a = 0, so μ = tan θ. If something is pulled up the slope, friction acts down the slope instead.

The 2025 syllabus asks you to explain the physics of why ramps were used by Aboriginal and/or Torres Strait Islander Peoples to move heavy loads. The physics of any ramp is the same: on a smooth ramp, the force needed to push a load up steadily is mg sin θ, a fraction of the mg needed to lift it straight up. The trade-off is distance: the load moves further along the ramp, so the work done is still at least mgh. A gentler ramp needs a smaller force over a longer distance.

Sliding down: F_N balances mg cos θ, and the net force along the slope is mg sin θ − f. With f = μmg cos θ this gives a = g sin θ − μg cos θ.

Common mistakes

Calling the weight and the normal force on a book a Newton's third law pair.They act on the same object, so they can't be a pair. The partner of the book's weight is the book's pull on Earth; the partner of the normal force is the book's push on the table.
Thinking an object moving at constant velocity needs a forward net force to keep it going.Constant velocity means zero net force (first law). A forward force is needed only to balance friction and drag.
Using F_N = mg on an incline, in an accelerating lift, or with a pull at an angle.Find F_N from the forces at right angles to the surface: mg cos θ on an incline, m(g + a) in a lift accelerating upward.
Always setting static friction equal to μ_sF_N.μ_sF_N is only the maximum. Below that, static friction equals whatever force is trying to slide the object.
Assuming the string tension equals the hanging mass's weight while the system accelerates.If the hanging mass accelerates downward, its net force is down, so T = m(g − a), less than mg. Find a from the whole system first.
Drawing friction down the slope for an object sliding down.Friction opposes the sliding: up the slope when sliding down, down the slope when moving up.

Worked example

A 3.0 kg box sits on a ramp inclined at 25° to the horizontal. A light string tied to the box runs parallel to the ramp, over a light, frictionless pulley at the top, to a 2.0 kg mass hanging freely. When released, the box slides up the ramp. The coefficient of kinetic friction between the box and the ramp is 0.20.

(a) Calculate the acceleration of the box. (b) Calculate the tension in the string.

(6 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. The hanging weight is about 2 × 10 = 20 N. Against it: the box's weight along the ramp, about 3 × 10 × sin 25° ≈ 13 N, and friction, about 0.2 × 27 ≈ 5 N. That leaves only 2 or 3 N to accelerate 5 kg, so I predict a small acceleration, under 0.5 m s⁻², and a tension a little below 20 N, because the hanging mass is accelerating down.

K · Knowns

I sketch the set-up with every value on it. The box moves up the ramp, so kinetic friction on it acts down the ramp. I take the direction of motion as positive for both bodies: up the ramp for the box, down for the hanging mass.

m₁ = 3.0 kg (box), θ = 25°, μ = 0.20
m₂ = 2.0 kg (hanging), g = 9.8 m s⁻²
Light string, frictionless pulley: same a and same T for both

Knowns on the sketch: 3.0 kg box, 25° ramp, μ = 0.20, 2.0 kg hanging mass. Unknowns: the acceleration a (box up the ramp) and the tension T.

N · Not known

Two answers: the acceleration, with a direction, and the tension.

(a) a = ? m s⁻² (b) T = ? N

O · Organise

I treat the box and the hanging mass as one system along the string, so the tension is internal and drops out. Driving the system is the hanging weight; against it are the box's weight along the ramp and friction. The normal force comes from the across-ramp balance. Then I find T from the hanging mass on its own.

F_N = m₁g cos θ, f = μF_N
System: m₂g − m₁g sin θ − f = (m₁ + m₂)a
Hanging mass alone: m₂g − T = m₂a, so T = m₂(g − a)

W · Work it out

One force per line, full calculator values kept, then the net force and the acceleration. I'll check T on the box as well, because it has to give the same acceleration.

m₁g sin θ = 3.0 × 9.8 × sin 25° = 12.42 N
F_N = 3.0 × 9.8 × cos 25° = 26.65 N, f = 0.20 × 26.65 = 5.329 N
m₂g = 2.0 × 9.8 = 19.6 N
F_net = 19.6 − 12.42 − 5.329 = 1.846 N
(a) a = 1.846 ÷ (3.0 + 2.0) = 0.3692 m s⁻²
(b) T = 2.0 × (9.8 − 0.3692) = 18.86 N
Check on the box: T − m₁g sin θ − f = 18.86 − 12.42 − 5.329 = 1.11 N = 3.0 × 0.369 ✓

U · Units

Newtons divided by kilograms gives m s⁻², and the tension is a force in newtons. The data are given to two significant figures.

a ≈ 0.37 m s⁻², T ≈ 19 N

D · Direction

Both answers are vectors. The box accelerates up the ramp (and the hanging mass down). The string pulls the box up the ramp, parallel to its surface. As a sense check, T is less than the 19.6 N hanging weight, as it must be for the hanging mass to accelerate down.

a ≈ 0.37 m s⁻² up the ramp; T ≈ 19 N on the box, up the ramp along the string

(a) 0.37 m s⁻² up the ramp. (b) The tension is about 19 N (18.9 N).

Check against the prediction. I predicted under 0.5 m s⁻² and a tension a little below 20 N. 0.37 m s⁻² and 18.9 N both fit: only 1.8 N of the forces is left over to accelerate 5.0 kg.

Exam advice

What markers look for

Draw a free-body diagram with every force labelled, and deal with both directions: what happens along the motion and what happens at right angles to it. Answers that handled only one force, or one direction, lost marks.NESA marking feedback, 2024, summarised
When you use Newton's third law, name both objects: the force A exerts on B, and the equal and opposite force B exerts on A. The stronger answers said exactly which object the partner force acts on.NESA marking feedback, 2025, summarised
In explanations, link each step with a cause and effect: the forces are unbalanced, so there is a net force down the slope, so the object accelerates down the slope. A list of facts with no links scores less.NESA marking feedback, 2023, summarised
Show the equation, the substitution and the answer with a unit, and keep a sign convention for direction. Correct working earns marks even when a number slips.General advice from NESA marking feedback, 2021 to 2025, summarised

Summaries in our own words. Read NESA's own feedback with the exam packs: 2023 · 2024 · 2025.

Where this leads in Year 12

Year 12 mechanics is Newton's second law with new net forces. In uniform circular motion the net force (friction on a flat bend, tension in a string, or gravity) points to the centre and equals mv² ÷ r. For satellites, gravity is the net force. In electromagnetism, the force on a charged particle in an electric or magnetic field goes into the same F_net = ma. In every case a labelled free-body diagram is still the first step.

FAQ

What is the difference between static and kinetic friction?: Static friction acts when surfaces are not sliding and grows to match the applied force, up to a maximum of μ_sF_N. Kinetic friction acts while they slide and is roughly constant at μ_kF_N. Usually μ_k is smaller than μ_s, so it's easier to keep something sliding than to start it.
Are the weight and the normal force a third law pair?: No. They often have the same size and opposite directions, but they act on the same object, and third-law pairs always act on two different objects. They balance because of the first law, not the third.
Why is the normal force on a slope less than the weight?: The surface only pushes at right angles to itself, so it only has to balance the part of the weight that is into the slope, mg cos θ. The other part, mg sin θ, acts down the slope.
Does a moving object always have a net force on it?: No. By Newton's first law, an object moving at constant velocity has zero net force on it. A net force changes the velocity: it speeds the object up, slows it down or changes its direction.

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