R1–R5MechanicsCore

Newton’s Laws of Motion

Q: Does a larger mass always accelerate less?

For the same resultant force, yes — a = F/m. But if the force also scales with mass, as gravity does, the mass cancels and the acceleration is unchanged.

Kinematics described motion without asking what caused it. Newton’s three laws supply the cause: forces change motion, and the relationship between them is $F = ma$ . Everything in the rest of the Mechanics strand is an application of those three statements.

35 min Forces and Newton’s Laws

The big picture

The second law does the calculating, but the other two are what let you set the calculation up. The first converts "moving at constant velocity" into "resultant force is zero", turning a description into an equation. The third tells you which forces belong on your diagram: a third-law pair acts on $\textbf{different objects}$ , so the two halves never appear together in one free-body diagram. Nearly every wrong answer in this topic traces back to a diagram with a force missing or one included that should not be — which is why the diagram, not the algebra, is what repays practice.

What you'll be able to do

State the three laws and apply them to described situations
Draw free-body diagrams and resolve to find a resultant force
Apply $F = ma$ in one and two dimensions
Identify third-law pairs and distinguish them from balanced forces

The first law, and what it lets you write down

A body remains at rest or moves with constant velocity unless acted on by a resultant force. The everyday intuition that motion requires a continuous push is wrong — what requires a force is a $\textit{change}$ in motion.

The examinable use is a translation. Whenever a question says an object is stationary, moving at constant speed, or in equilibrium, it is telling you the resultant force is zero — and that gives you an equation in each direction.

The property that makes an object resist changes in motion is its $\textbf{inertia}$ , measured by its mass.

\Sigma \mathbf{F} = \mathbf{0} \iff \text{equilibrium}

Constant velocity or at rest.

Example 1A crate of mass 40 kg is pulled along rough horizontal ground at constant speed by a horizontal force of 90 N. Find the frictional force.

1Constant speed means zero resultant force horizontally.

2So friction exactly balances the applied force.

Answer

90\ \mathrm{N}

, opposing the motion.

Tip — The phrase "constant velocity" is worth a mark on its own — write "resultant force $= 0$ by Newton’s first law" before doing anything else.

The second law and free-body diagrams

A resultant force produces acceleration in the same direction, with $\mathbf{F} = m\mathbf{a}$ where $\mathbf{F}$ is the $\textbf{resultant}$ — never a single force unless it acts alone.

The method is a free-body diagram: draw the object by itself and mark every force acting $\textbf{on}$ it. Weight always acts vertically downwards; normal reaction acts perpendicular to the contact surface; tension acts along a string away from the object; friction opposes relative motion.

Choose a positive direction, sum the forces along it with signs, and apply $F = ma$ . In two dimensions, do this along two perpendicular directions independently.

\mathbf{F}_{\text{resultant}} = m\mathbf{a}

The second law.

W = mg

Weight, downwards.

Example 1A 5 kg block on a smooth horizontal surface is pulled by a horizontal force of 20 N. Find its acceleration.

1Smooth, so no friction. Vertically the weight and normal reaction balance.

2Horizontally the resultant is 20 N.

a = \dfrac{20}{5}

Answer

4\ \mathrm{m\,s^{-2}}

Example 2A lift of mass 500 kg accelerates upwards at

1.2\ \mathrm{m\,s^{-2}}

. Find the tension in the cable. Take

g = 9.8\ \mathrm{m\,s^{-2}}

1Forces on the lift: tension

T

up, weight

500 \times 9.8 = 4900

N down.

2Taking upwards positive:

T - 4900 = 500 \times 1.2

T = 4900 + 600

Answer

T = 5500\ \mathrm{N}

Why this works

The tension exceeds the weight because the lift is accelerating upwards — an unbalanced upward force is required. If it moved at constant speed the tension would be exactly 4900 N, and if it accelerated downwards it would be less.

The third law, and the pair that is not one

If A exerts a force on B, then B exerts an equal and opposite force on A. The two forces are the same type, act along the same line, and — the condition that matters — act on $\textbf{different objects}$ .

The classic trap is a book resting on a table. Its weight (Earth pulls book) and the normal reaction (table pushes book) are equal and opposite, but they are $\textit{not}$ a third-law pair: both act on the book, and they are different types of force. They balance because of the $\textbf{first}$ law, since the book is not accelerating.

The genuine partners are elsewhere: the book pulls the Earth upwards with an equal gravitational force, and the book pushes down on the table with an equal contact force.

\mathbf{F}_{A \to B} = -\mathbf{F}_{B \to A}

Equal, opposite, and on different bodies.

Example 1A person stands on the floor. State the third-law partner of the normal reaction from the floor on the person.

1The force is the floor pushing up on the person — a contact force.

2Swap the objects and keep the type.

AnswerThe person pushes down on the floor with an equal contact force.

Tip — Test any claimed pair with the sentence "A on B" and "B on A". If both forces act on the same object, or they are different types, it is not a third-law pair.

Forces at an angle

When a force does not act along the direction of motion, resolve it into components before applying $F = ma$ . The component along the motion drives the acceleration; the perpendicular component affects the normal reaction.

For a force at angle $\theta$ above the horizontal on a body on level ground, the vertical component reduces the normal reaction: $R = mg - F\sin\theta$ . Since friction depends on $R$ , pulling at an angle reduces friction as well as reducing the forward component.

Resolve in two perpendicular directions and treat each independently — the same technique used throughout the strand.

F\cos\theta = ma \text{ (horizontal)}

Component along the motion.

R = mg - F\sin\theta

Vertical balance with an upward pull.

Example 1A 10 kg box on smooth level ground is pulled by a 50 N force at

30^{\circ}

above the horizontal. Find the acceleration and the normal reaction. Take

g = 9.8

1Horizontal:

50\cos30^{\circ} = 43.30

N, so

a = \dfrac{43.30}{10}

2Vertical:

R + 50\sin30^{\circ} = 10(9.8)

, so

R = 98 - 25

Answer

a = 4.33\ \mathrm{m\,s^{-2}}

;

R = 73\ \mathrm{N}

Why this works

The normal reaction has dropped from 98 N to 73 N because the rope is partly supporting the box. On a rough surface that reduction would cut the friction too, which is why an angled pull can be easier despite delivering less forward force.

Think like an examiner

Draw and label a free-body diagram for every force question — it is often credited directly.
Use the resultant in $F = ma$ , and state which direction you are taking as positive.
For third-law pairs, name both objects and the type of force.
Resolve before applying $F = ma$ whenever a force acts at an angle.

Common misconceptions

Calling the weight and normal reaction on a resting book a third-law pair.

They act on the same object and are different types. They balance by the first law.

Substituting a single force into

F = ma

when others act.

F

must be the resultant of all forces in that direction.

Putting forces the object exerts on other things on its free-body diagram.

Only forces acting

\textit{on}

the object belong there.

Taking the normal reaction as

mg

when a force acts at an angle.

Resolve vertically: an upward component reduces

R

below

mg

Newton’s laws

\Sigma F = 0 \Rightarrow \text{equilibrium}

First law.

F_{\text{resultant}} = ma

Second law.

F_{A\to B} = -F_{B\to A}

Third law — different objects.

W = mg

Weight.

R = mg - F\sin\theta

Normal reaction with an angled pull.

Stretch yourself

A person of mass 70 kg stands on a set of scales in a lift. The scales read 630 N. Determine the magnitude and direction of the lift’s acceleration. Take $g = 9.8\ \mathrm{m\,s^{-2}}$ .

Hint — The scale reading is the normal reaction on the person. Apply $F = ma$ to the person alone.

Reveal solution

SolutionThe scales read the normal reaction

R = 630

N acting upwards on the person, whose weight is

70 \times 9.8 = 686

N downwards. Taking upwards positive and applying the second law to the person:

R - W = ma

, so

630 - 686 = 70a

, giving

-56 = 70a

and

a = -0.8\ \mathrm{m\,s^{-2}}

. The negative sign means the acceleration is

0.8\ \mathrm{m\,s^{-2}}

\textbf{downwards}

. Note this does not tell you which way the lift is moving — it could be moving downwards and speeding up, or moving upwards and slowing down. Acceleration and velocity are independent, and the scale reading alone cannot distinguish the two cases.

\blacksquare

Questions students ask

Why are the weight and normal force on a book not a third-law pair?

Because both act on the book, and third-law pairs act on different objects. They are also different types of force. Their equality here follows from the first law, since the book is not accelerating.

What forces go on a free-body diagram?

Everything touching the object (normal reaction, friction, tension, applied forces) plus anything acting at a distance (weight). Nothing the object exerts on other bodies.

Does a larger mass always accelerate less?

For the same resultant force, yes —

a = F/m

. But if the force also scales with mass, as gravity does, the mass cancels and the acceleration is unchanged.

Can an object be moving with no resultant force?

Yes. The first law says zero resultant force means constant velocity, which includes any non-zero constant speed in a straight line.

Key takeaways

First law: zero resultant force means at rest or constant velocity, and vice versa.
Second law: resultant force equals mass times acceleration, in the same direction.
Third law: equal and opposite forces acting on two different objects.
A free-body diagram shows only forces acting on the chosen object.
Resolve angled forces before applying $F = ma$ , and check the effect on the normal reaction.