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GRADE 10 · DYNAMICS

Newton's laws

How forces relate to acceleration and interactions between bodies.

20 minutes8 theory cards
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Read the full lesson theory here without an account. In the interactive lesson, the same theory is presented as cards, followed by questions with feedback and practice problems. Your progress is saved.

Understanding step by step

A trolley stops quickly on a rough track but rolls much longer on a smooth one. Must a force keep motion going, or is force needed for something else: changing velocity?

Rough roadNearly smooth icev₀v₀stops quicklykeeps moving a long timeInitial instant: the push has endedWhat changes velocity?
Two experiments with one trolley start at the same velocity v₀ immediately after the push ends. Labels under the tracks describe what follows: a rough road stops it quickly, while nearly smooth ice lets it travel a long time. What changes its velocity?

When velocity stays unchanged

Reducing friction lets the trolley keep moving longer after a push. Imagine removing resistance entirely: nothing remains to change its velocity. The persistence of rest or uniform straight-line motion without external influence is called inertia. Frames in which a free body behaves this way are inertial frames. An Earth-based laboratory is approximately inertial in many school experiments; a sharply braking carriage is not a suitable example of this model.

Nmgv = constΣF = 0, a = 0
In an inertial frame, the trolley moves uniformly in a straight line with no horizontal forces. The total normal reaction N and gravitational force mg balance; they are drawn at one trolley point for translational-motion calculations. With zero initial velocity the trolley would stay at rest.

Who acts on whom?

When a hand pushes a trolley, they interact. The effect on the chosen body is described by force \vec F, in newtons. First isolate the trolley mentally: which objects act on it—the hand, Earth, the floor? Add these forces as vectors to obtain the net force \sum\vec F. Opposing effects can balance partly or fully. The trolley's force on the hand is not on this list: it acts on a different recipient.

Choose one trolleyNmgpullFсопрneighbourNet force: sum of forces on it
The dashed line isolates the trolley whose equation we are writing. Only forces acting on it are shown: pull, resistance, gravity mg and total normal reaction N. For translational motion, arrows are moved to one trolley point; forces on the grey neighbouring object are excluded.

From force to acceleration

The same trolley can be accelerated more or less strongly. Experiments show that doubling the net force doubles acceleration. Doubling mass at the same force makes velocity change half as fast. Newton's second law combines these relations: \sum\vec F=m\vec a. Mass m, in kg, measures inertia; acceleration \vec a, in m/s², points along the net force, not necessarily along the existing velocity.

Net force determines acceleration6 Н2 НΣF = 4 Нa = ΣF / m, to the right
In this example the trolley experiences 6 N right and 2 N left; vertical forces balance and are omitted. For translational motion, the arrows are drawn at one point. Their sum, 4 N right, is shown separately below on the same scale. It is the net force, not an extra force.

Limits of the law

Why not simply take one prominent force and write F=ma? Acceleration comes from all the forces acting together. Here F is the magnitude of their vector sum. We use \sum\vec F=m\vec a for a constant-mass body in an inertial frame, at speeds much smaller than light speed. When fuel leaves a rocket, the chosen body's composition changes; this equation cannot be transferred to it without accounting for mass flow.

Where the law appliesinertialframemassconstantspeedswell below cF in F = ma is the net force,not just any selected forceotherwise the answer is wrong
This form of the second law applies in an inertial frame, with constant mass and speeds far below light speed. In the magnitude formula, the force is the net force.

What a newton means

The force unit is consistent with mass and acceleration units: 1\,\text{Н}=1\,\text{кг}\cdot\text{м/с}^2. A net force of 1 N gives a 1 kg body acceleration 1 m/s²: its velocity changes by 1 m/s each second. The same force gives a 2 kg body acceleration 0.5 m/s². If the net force falls to zero, acceleration vanishes for finite positive mass, but the velocity already gained remains.

1 Н = 1 кг · м/с²1 кгΣF = 1 Нa = 1 м/с²2 кгΣF = 1 Нa = 0,5 м/с²Same net force, mass ×2 → a / 2
The net force is 1 N in both experiments: red arrows have equal lengths. The trolley masses are 1 and 2 kg; accelerations are 1 and 0.5 m/s². Green acceleration arrows are in the ratio 2:1; force and acceleration have different scales and units.

Balanced forces

A book rests on a table: gravity acts, but the support balances it. A trolley can also move at constant velocity with zero net force. In both cases the law gives \vec a=0, so the velocity vector stays constant; rest is the special case of zero initial velocity. Motion alone therefore does not show that the net force points forwards. We must observe the change in velocity.

Forces act, but acceleration is zero10 Н10 Нv = constΣF = 0, a = 0Motion continues
A trolley moves in a straight line in an inertial frame. Two equal opposite horizontal forces of 10 N act on it; vertical forces balance and are omitted. Force arrows are drawn at one point for translational-motion calculations. Their zero sum gives a=0 with nonzero constant velocity.

An interaction pair

Push a wall with your palm: your palm feels a force too. For a hand and trolley, call the hand's force on the trolley \vec F_1 and the trolley's force on the hand \vec F_2. Newton's third law relates them: \vec F_1=-\vec F_2. Their magnitudes are equal and their directions opposite at the same instant. They are two sides of one interaction, not a sequence of action followed by response. Different masses mean equal forces need not give equal accelerations.

Equal, but acting on different bodiesF₁F₂handtrolley|F₁| = |F₂|, opposite directions
The hand pushes the trolley right with F₁. The trolley pushes the hand left with F₂. Equal opposite arrows start at the contact point and refer to different bodies. Only this interaction pair is shown: F₂ is not in the equation for the trolley alone.

Why the trolley accelerates

Why does the trolley accelerate anyway? Only the hand's force from the interaction pair acts on it; the other force acts on the hand. Forces can balance on the same chosen body, so this pair cannot cancel in the trolley's equation. For a system containing person and trolley, the pair becomes internal and cancels in the total; external forces, such as those from the floor, remain. The chosen system changes the force list, not the motion itself.

Choose a body or a systemF₁F₂Trolley: only F₁ from this pairF₁F₂Hand + trolley: internal pair
Both rows show the same hand pushing the same trolley. F₁ acts right on the trolley, F₂ left on the hand. Above, the dashed line isolates the trolley: F₂ is not in its equation. Below, hand and trolley form the chosen system: the pair is internal and cancels in the total, while external forces remain.

Work through an example

Horizontal forces of 6 N right and 2 N left act on a 2 kg trolley. Find its acceleration.

Two horizontal forcesm = 2 кг6 Н2 НVertical forces balanceFind acceleration
One 2 kg trolley experiences 6 N right and 2 N left. Arrows use the same scale and lengths in the ratio 3:1. For translational motion, forces are drawn at one trolley point; vertical forces balance and are omitted. Find acceleration.

Let's look at an example · 1

Choose the trolley and take right as positive.

Choose the trolley and axism = 2 кг6 Н2 НxRight is positive
The trolley, mass and original forces are unchanged. The dashed line selects the body and the x axis points right. The 6 N force component is therefore positive and the 2 N component negative.

Let's look at an example · 2

Add force components: ΣFx=6−2=4 Н.

Add componentsm = 2 кг6 Н2 НxΣFₓ = 4 НΣFₓ = 6 − 2 = 4 Н
On the same diagram, components sum to 6 − 2 = 4 Н. The net force is shown separately below; on the same scale its arrow is twice the 2 N arrow and two thirds of the 6 N arrow. This is the result of addition, not a third extra force.

Let's look at an example · 3

By Newton's second law, ax=ΣFx/m=4/2=2 м/с².

Divide by massm = 2 кг6 Н2 НxΣFₓ = 4 Нaₓ = ΣFₓ / m = 4 / 2 = 2 м/с²
The 2 kg trolley, both forces and their resultant remain. Substitute ΣFₓ = 4 Н into the second law: aₓ = 4 Н / 2 кг = 2 м/с². The positive sign refers to the rightward axis.

Worked example · 4

Check: the net force points right, so acceleration also points right.

Check directionm = 2 кг6 Н2 НxΣFₓ = 4 Нaₓ = 2 м/с²
The same trolley and forces. Below the separate net-force arrow is rightward acceleration: their directions agree. Do not compare force and acceleration arrow lengths with each other, since they represent different quantities and units.

Let's check the result

Acceleration is 2 m/s² to the right: each second the rightward velocity component increases by 2 m/s. If the trolley initially moved left, it would first slow down.

Key idea · 1

Force changes velocity. With zero net force, both rest and existing uniform straight-line motion persist.

Key idea · 2

Acceleration increases with net force and decreases with mass; write the equation for a specific body in a chosen reference frame.

Key idea · 3

Interaction forces are equal but act on different bodies. They cannot cancel in the equation for one body.

Where do we go next?

Next we study models for particular forces and conditions for equilibrium.

Basic Concepts

2 concepts

Inertia

The persistence of rest or uniform straight-line motion in an inertial reference frame when the net force is zero.

No net force is needed to maintain uniform motion. A net force is needed to change a body's velocity.

Newton's laws

Basic laws of dynamics that connect the movement of a body with the forces acting on it.

Choose the body, show all forces acting on it, and add them as vectors: the net force determines acceleration.

Physical quantities

2 quantities

Force

F

A measure of mechanical interaction between bodies.

Unit: newton · \mathrm{Н}

Net force

\sum F

The vector sum of all forces acting on a body.

Unit: newton · \mathrm{Н}

Lesson formulas

4 formulas

Newton's second law

F=ma

The resultant force is equal to the product of mass and acceleration.

Definition of the newton

1\,\mathrm{Н}=1\,\frac{\mathrm{кг}\cdot\mathrm{м}}{\mathrm{с}^2}

One newton gives a one-kilogram mass an acceleration of one metre per second squared.

Second law with several forces

\sum\vec{F}=m\vec{a}

The vector sum of forces is equal to the product of mass and acceleration.

Newton's third law

\vec{F_1}=-\vec{F_2}

The forces of interaction between two bodies are equal in magnitude and opposite in direction.

Test the relationship

Make a prediction first. The button changes the stated quantity; then compare the result and explore other values.

Force, mass and friction

ΣF = F + Fтр; a = ΣF/m; |Fтр| ≤ μmg

The object starts at rest on a horizontal surface; g = 10 m/s². Limiting static and kinetic friction coefficients are taken equal. N = mg.

At F = 6 N and μ = 0, mass increases from 2 to 4 kg. How will a change?

Your prediction

Initial values

Show vectors
F6 НFтр0 НΣF6 Н

Calculated result

a
3 м/с²
ΣF
6 Н
Fтр
0 Н
Explaining the result

Static friction matches the applied force up to μmg, so a small F produces zero acceleration. After slipping starts, acceleration is set by the net force.

Related formulas

The same formulas in the reference, with symbols and units.

Problems on this topic

Statements, answer options and the walkthrough.