Newton's laws of motion

Newton’s laws connect forces to motion. Their central calculation is

$$\boxed{\sum \mathbf F=m\mathbf a},$$

where $\sum \mathbf F$ is the vector sum of all external forces acting on one chosen body. A force does not cause velocity. A resultant force causes acceleration, meaning a change in velocity.

Prerequisites

You should be able to:

• distinguish mass, weight, velocity and acceleration;
• choose a positive direction and use signed quantities;
• identify forces and draw a free body diagram;
• use SI units from quantities and units in mechanics;
• rearrange linear equations.

Unless a question states otherwise, use $g=9.8\ \mathrm{m\,s^{-2}}$.

Newton’s first law

An object remains at rest, or continues moving with constant velocity in a straight line, unless a non-zero resultant external force acts on it.

In symbols,

$$\sum \mathbf F=\mathbf 0 \quad\Longrightarrow\quad \mathbf a=\mathbf 0.$$

Zero acceleration means constant velocity, not necessarily zero velocity. A car travelling east at a steady $20\ \mathrm{m\,s^{-1}}$ has zero acceleration. Its driving force balances resistance, so the resultant force is zero.

The law also describes equilibrium:

• static equilibrium: the object is at rest;
• dynamic equilibrium: the object moves with constant velocity.

Balanced forces do not disappear

If weight and a normal reaction balance, both forces still act. Their vector sum is zero. Do not omit real forces merely because the object is in equilibrium.

Worked example 1: steady motion

A car travels along a straight horizontal road at a constant speed. Its engine provides a driving force of $1.6\ \mathrm{kN}$. Find the total resistance.

Constant velocity means $a=0$, so the horizontal resultant is zero. Converting units,

$$1.6\ \mathrm{kN}=1600\ \mathrm N.$$

Taking forwards as positive and writing the resistance as $R$,

$$1600-R=0.$$

Therefore

$$\boxed{R=1600\ \mathrm N}.$$

The car does not need a resultant forward force to keep moving. It needs a driving force to balance resistance.

Self-check 1

A lift moves vertically upwards at a constant speed. Its mass is $750\ \mathrm{kg}$. Find the tension in its cable.

Answer

Constant speed in a fixed direction means constant velocity, so $a=0$. Taking upwards as positive,

$$T-750g=0.$$

Hence

$$T=750(9.8)=\boxed{7350\ \mathrm N}.$$

Newton’s second law

For constant mass, the resultant external force on a body equals its mass multiplied by its acceleration:

$$\boxed{\sum \mathbf F=m\mathbf a}.$$

The equation is vectorial. In one dimension, choose a positive direction and use

$$\boxed{\text{forces in the positive direction}-\text{forces in the negative direction}=ma}.$$

In two dimensions, apply the law separately to perpendicular components:

$$\sum F_x=ma_x, \qquad \sum F_y=ma_y.$$

One newton is the force that gives a mass of $1\ \mathrm{kg}$ an acceleration of $1\ \mathrm{m\,s^{-2}}$:

$$1\ \mathrm N=1\ \mathrm{kg\,m\,s^{-2}}.$$

More generally, Newton’s second law is $\sum \mathbf F=\dfrac{\mathrm d\mathbf p}{\mathrm dt}$, where momentum is $\mathbf p=m\mathbf v$. For the constant mass models used here, this becomes $\sum \mathbf F=m\mathbf a$.

A reliable method

1. Isolate one body and draw every external force acting on it.
2. Choose and state a positive direction.
3. Find the weight as $mg$, not merely $m$.
4. Write one equation of the form $\sum F=ma$ in each required direction.
5. Solve before rounding.
6. Interpret the sign and check units.

If the calculated acceleration is negative, the body accelerates opposite to your chosen positive direction. The equation has not failed.

Worked example 2: thrust and resistance

A boat of mass $1200\ \mathrm{kg}$ has a forward thrust of $3.8\ \mathrm{kN}$ and experiences resistance of $1.4\ \mathrm{kN}$. Find its acceleration.

Take forwards as positive. The resultant force is

$$3800-1400=2400\ \mathrm N.$$

Apply Newton’s second law:

$$2400=1200a.$$

Thus

$$\boxed{a=2.0\ \mathrm{m\,s^{-2}}}.$$

It would be wrong to use $3800=1200a$, because resistance is also an external force on the boat.

Worked example 3: finding an unknown force

A $6\ \mathrm{kg}$ box accelerates to the right at $2.5\ \mathrm{m\,s^{-2}}$. A horizontal pull of $28\ \mathrm N$ acts to the right and resistance $R$ acts to the left. Find $R$.

Take right as positive:

$$28-R=6(2.5).$$

Therefore

$$28-R=15 \quad\Longrightarrow\quad \boxed{R=13\ \mathrm N}.$$

Notice that resistance is smaller than the pull because the resultant must point right.

Self-check 2

A cyclist and bicycle have total mass $85\ \mathrm{kg}$. The driving force is $210\ \mathrm N$ and resistance is $74\ \mathrm N$. Find the acceleration.

Answer

Taking forwards as positive,

$$210-74=85a.$$

Hence

$$a=\frac{136}{85}=\boxed{1.6\ \mathrm{m\,s^{-2}}}.$$

Vertical motion and apparent weight

For a body moving vertically, weight $mg$ acts downwards whether the body is moving up, moving down or instantaneously at rest. The direction of acceleration is determined by the resultant force, not automatically by the direction of motion.

Worked example 4: an accelerating lift

A passenger of mass $60\ \mathrm{kg}$ stands on scales in a lift accelerating upwards at $1.2\ \mathrm{m\,s^{-2}}$. Find the reading of the scales.

The scales exert an upward normal reaction $R$ on the passenger. Their weight $60g$ acts downwards. Taking upwards as positive,

$$R-60g=60(1.2).$$

Therefore

$$\begin{aligned} R&=60(9.8)+60(1.2)\\ &=660\ \mathrm N. \end{aligned}$$

The scale reading is the contact force, so it is $\boxed{660\ \mathrm N}$. It exceeds the passenger’s weight because an upward resultant is required.

If the lift accelerated downwards at $1.2\ \mathrm{m\,s^{-2}}$, the same upward-positive convention would give $a=-1.2$ and

$$R-60g=60(-1.2),$$

so $R=516\ \mathrm N$. Direction of acceleration, not direction of travel, controls the reading.

Self-check 3

A $50\ \mathrm{kg}$ person stands on scales in a lift. The scales read $440\ \mathrm N$. Find the lift’s acceleration, including its direction.

Answer

Take upwards as positive:

$$440-50g=50a.$$

Thus

$$a=\frac{440-490}{50}=-1.0\ \mathrm{m\,s^{-2}}.$$

The negative sign means the acceleration is $\boxed{1.0\ \mathrm{m\,s^{-2}}\text{ downwards}}$. The lift could be moving upwards and slowing down, or moving downwards and speeding up.

Newton’s third law

If body $A$ exerts a force on body $B$, then body $B$ simultaneously exerts a force of equal magnitude and opposite direction on body $A$:

$$\boxed{\mathbf F_{A\text{ on }B}=-\mathbf F_{B\text{ on }A}}.$$

Third law pairs:

• are the same type of interaction;
• have equal magnitude and opposite direction;
• act at the same time;
• act on different bodies.

Because they act on different bodies, they do not cancel in a free body diagram for either body.

Weight and reaction are not a third law pair

For a book resting on a table:

• the Earth’s force on the book is the book’s weight;
• the table’s force on the book is the normal reaction.

Both act on the book, so they cannot be a third law pair. They happen to balance while the book has no vertical acceleration.

The third law partner of the table’s force on the book is the book’s force on the table. The partner of the Earth’s force on the book is the book’s gravitational force on the Earth.

Worked example 5: two blocks in contact

Blocks $A$ and $B$, of masses $3\ \mathrm{kg}$ and $5\ \mathrm{kg}$, touch on a smooth horizontal surface. A force of $24\ \mathrm N$ pushes $A$ towards $B$. Find their acceleration and the contact force between them.

Treat both blocks as one system. The contact forces are internal and cancel when the system equations are added:

$$24=(3+5)a.$$

Hence

$$a=3\ \mathrm{m\,s^{-2}}.$$

Now isolate block $B$. Its only horizontal force is the contact force $C$ exerted by $A$:

$$C=5a=5(3)=\boxed{15\ \mathrm N}.$$

By Newton’s third law, $B$ exerts a $15\ \mathrm N$ force on $A$ in the opposite direction. Checking block $A$:

$$24-15=3(3),$$

as required.

Internal forces and systems

A force can be internal for one chosen system but external for another. The contact force is internal to the system consisting of both blocks, but external when block $A$ or block $B$ is considered alone.

Self-check 4

A hand pushes a wall with a horizontal force of $40\ \mathrm N$. State the third law partner of this force. Do the two forces cancel?

Answer

The wall pushes the hand with a horizontal force of $40\ \mathrm N$ in the opposite direction. The forces do not cancel because one acts on the wall and the other acts on the hand. They would cancel only when considering the combined hand and wall as one system, where both are internal forces.

Common misconceptions

Misconception	Correction
A moving object must have a force in the direction of motion	A resultant force is needed to change velocity, not to maintain constant velocity
$F=ma$ uses one selected force	$F$ means the resultant of all external forces on the chosen body
If $a=0$, no forces act	Forces may act and balance
Acceleration points in the direction of motion	Acceleration points in the direction of the resultant force
Action and reaction cancel on one object	A third law pair acts on two different objects
A heavier object must accelerate faster	For the same resultant force, $a=F/m$, so greater mass gives smaller acceleration

Mixed exam style problem

A van of mass $1500\ \mathrm{kg}$ travels along a straight horizontal road. Its engine supplies a constant driving force of $2.4\ \mathrm{kN}$.

1. While the van travels at constant speed, find the resistance.
2. The resistance then decreases to $1650\ \mathrm N$. Find the van’s acceleration.
3. Starting at $12\ \mathrm{m\,s^{-1}}$, find its speed after $8$ seconds if the forces remain constant.

Solution

At constant speed, $a=0$. Therefore the forces balance and

$$\boxed{R=2400\ \mathrm N}.$$

After resistance decreases, the forward resultant is

$$2400-1650=750\ \mathrm N.$$

Newton’s second law gives

$$750=1500a,$$

so

$$\boxed{a=0.5\ \mathrm{m\,s^{-2}}}.$$

The forces and mass are constant, so the acceleration is constant. Using $v=u+at$,

$$v=12+0.5(8)=\boxed{16\ \mathrm{m\,s^{-1}}}.$$

Final checklist

Before accepting a solution, ask:

• Did I isolate the correct body or system?
• Did I include every external force and exclude forces acting on other bodies?
• Did I choose a positive direction and keep signs consistent?
• Did I use the resultant force in $\sum F=ma$?
• If $a=0$, did I recognise equilibrium or constant velocity?
• If I named a third law pair, do its forces act on different bodies?
• Are all quantities in compatible SI units?

Next steps

Use Newton’s laws with resolving forces to handle inclined and two dimensional problems. Then apply them to friction and connected particles and pulleys. Once an acceleration is known and constant, connect dynamics to the SUVAT equations.