Forces and Force Diagrams

AS · M1 · 18 min

Every mechanics question starts the same way: work out which forces act on the object and draw them. Get the force diagram right and the rest of the question is usually routine algebra; get it wrong (a missing friction force, a normal reaction drawn vertically on a slope) and every later mark is lost. This note covers the forces that appear in Paper 4, the modelling words Cambridge uses to describe them, Newton's third law, and a reliable method for drawing force diagrams.

What a force is

A force is a push or a pull. It is measured in newtons (N). One newton is the force that gives a mass of $1\ \text{kg}$ an acceleration of $1\ \text{m s}^{-2}$ ; you will meet this properly in Newton's laws of motion (note not yet published).

A force has both a size and a direction, so it is a vector. Saying "a force of 20 N" is incomplete: 20 N pulling up a slope and 20 N pulling down it have opposite effects. On a diagram a force is an arrow: the arrow points the way the force acts, and the label gives its magnitude.

Because forces are vectors, two forces only cancel if they are equal in size and opposite in direction and act along the same line. Much of this course is about splitting forces into perpendicular parts so that you can add them; that is the subject of Resolving forces.

Modelling: the particle and its friends

Real objects are complicated. A car has wheels, an engine and air flowing over it; a crate has corners and a rough base. Mechanics replaces real objects with models: simplified versions that keep what matters and throw away what does not. Cambridge uses a fixed vocabulary for these simplifications, and every word in a question carries information.

Definition

A particle is an object whose size can be ignored, so that all the forces on it act at a single point. In Paper 4 every object, including cars, crates, people and lifts, is modelled as a particle.

The particle model means you never worry about objects turning or toppling. All forces are drawn as acting at one point, and the only question is whether they balance.

Word in the question	What it means	What it tells you to do
particle	size ignored; forces act at one point	no rotation; draw all forces from one point
light (string, rod, pulley)	mass is zero	the string or rod has no weight; tension is the same along it
inextensible string	does not stretch	connected particles move together with the same speed and acceleration
smooth surface	no friction	the contact force is just the normal reaction
rough surface	friction may act	include a friction force along the surface
smooth pulley or smooth peg	no friction at the pulley	tension is the same on both sides
rod	rigid, can push or pull	can carry a tension or a thrust
at rest, in equilibrium	no acceleration	the forces balance
about to slip, on the point of sliding	limiting equilibrium	friction has its maximum value $\mu R$

Tip

Read a mechanics question once for the story and once for the modelling words. Underline "smooth", "rough", "light", "inextensible" and "about to". Each one either removes a force or tells you its exact value.

The forces you need

There are only a handful of forces in Paper 4. Learn what each one is, which way it acts, and what produces it.

Weight

The weight of an object is the force of gravity on it. It always acts vertically downwards, whatever the object is resting on.

Key result

W = mg, \qquad g = 10\ \text{m s}^{-2} \text{ in Paper 4}

Mass $m$ is in kilograms; weight $W$ is a force in newtons. A $5\ \text{kg}$ box has weight $50\ \text{N}$ .

Mass and weight are different things. Mass is the amount of matter and is the same on the Moon; weight is a force and depends on gravity. Examiners penalise "the weight is 5 kg".

Watch out

The syllabus specifies $g = 10\ \text{m s}^{-2}$ . Using $9.8$ or $9.81$ gives answers the mark scheme does not accept. Use $g = 10$ unless a question explicitly says otherwise.

Normal reaction

When two surfaces touch, each pushes on the other. The part of that push which is perpendicular to the surfaces is the normal reaction (or normal contact force), usually labelled $R$ or $N$ . "Normal" is the mathematical word for "at right angles".

On horizontal ground the normal reaction is vertical.
On a slope the normal reaction is perpendicular to the slope, not vertical.
The normal reaction is whatever size it needs to be to stop the object sinking into the surface. It is not automatically equal to the weight: pull up on a box and the floor pushes less; push down and it pushes more.
If $R$ becomes zero the surfaces are about to separate. A question asking when an object "is about to lose contact" or "leaves the floor" is telling you to set $R = 0$ .

Friction

Friction is the part of the contact force that acts along the surfaces. It opposes the relative motion of the surfaces, or the motion that would happen if friction were absent. On a rough horizontal floor a box being dragged to the right feels friction to the left; a box at rest on a rough slope feels friction up the slope, because without it the box would slide down.

Friction is not a fixed force. It can be anything from zero up to a maximum value $\mu R$ , where $\mu$ is the coefficient of friction. The full story is in Friction.

The contact force as a whole

Normal reaction and friction are not really two separate forces: they are two components of a single contact force between two surfaces.

Definition

The contact force between two surfaces can be represented by two components: the normal component $R$ , perpendicular to the surfaces, and the frictional component $F$ , along the surfaces. If the contact is smooth, the frictional component is zero.

The total contact force has magnitude $\sqrt{R^2 + F^2}$ and acts at an angle to the normal. Questions occasionally ask for it directly; most of the time you work with $R$ and $F$ separately.

The smooth model and its limitations

A smooth surface is one that exerts no friction. It is a model, not reality: every real surface has some friction. The smooth model is reasonable for ice, a polished table, a well-oiled pulley or a wheel on a bearing, where friction is small compared with the other forces.

Its limitations are worth knowing, because "state a modelling assumption" or "explain why this answer may be unrealistic" occasionally appears:

On a smooth slope nothing can stop a particle sliding down, so a particle can never rest on a smooth inclined plane without some other force.
Without friction, a car's wheels could not grip the road, so the car could not accelerate or brake.
Predictions made with the smooth model (speeds, distances) are overestimates when the real surface has friction.

Tension and thrust

A tension is the pulling force in a string, rope, cable or rod. It always acts along the string, away from the object it is attached to: a string can only pull.

A thrust (or compression) is the pushing force in a rod that is being squashed. A rod can push or pull, so the force in a rod is either a tension or a thrust. In a tow-bar question, the tow-bar is in tension when the car is accelerating and pulls the trailer, but in thrust when the car brakes and the trailer pushes against the car.

Key result

For a light inextensible string passing over a smooth pulley or peg, the tension is the same throughout the string.

Driving force and resistance

A car's engine produces a driving force (sometimes "tractive force") forwards along the road. Resistance to motion (air resistance, rolling resistance) acts backwards, opposite to the velocity. The syllabus states that resistances other than friction are only included when the question says so; if a question says nothing about air resistance, leave it out.

Newton's third law

Forces always come in pairs. When you push on a wall, the wall pushes back on you.

Definition

Newton's third law: if body A exerts a force on body B, then body B exerts a force on body A that is equal in magnitude and opposite in direction.

The syllabus example: the force exerted by a particle on the ground is equal and opposite to the force exerted by the ground on the particle.

The two forces in a third-law pair:

act on different bodies (one on A, one on B), so they never appear on the same force diagram;
are the same type of force (both contact forces, or both gravitational);
are equal in size and opposite in direction, always, whether or not anything is accelerating.

The classic misconception is that the weight of a book on a table and the normal reaction from the table form a third-law pair. They do not. They act on the same body (the book) and are different types of force. They happen to be equal when the book is at rest, but that is because the book is in equilibrium (a first-law fact), and they stop being equal if the table is in an accelerating lift. The true partners are:

weight of the book (Earth pulls book down) pairs with the book pulling the Earth up;
table pushes book up pairs with book pushing table down.

The man pushes down on the box with 700 N and the box pushes up on the man with 700 N: a Newton’s third law pair, drawn on different diagrams. The ground pushes up on the box with 900 N.

Drawing force diagrams

Drawing a force diagram

Decide which body you are drawing. Draw it alone, as a simple box or dot. If there are several bodies (a car and trailer, two particles on a string), draw a separate diagram for each, or one for the whole system, but always know which.
Draw the weight $mg$ vertically down.
For every surface the body touches, draw the normal reaction perpendicular to that surface, pushing away from it.
For every rough surface, draw friction along the surface, opposing the motion or the tendency to move. If you cannot tell which way, guess; a negative answer means the other way.
For every string or rod attached, draw a tension along it, away from the body (or a thrust towards the body for a rod in compression).
Add any other forces given in the question: pushes, pulls, driving forces, resistances.
Label every force with a letter or value, and mark every angle that the question gives.
If the body is accelerating, show the acceleration with a separate double-headed or offset arrow, not as a force.

Weight W acts vertically down, the normal reaction R acts at right angles to the surface, friction F acts along the surface against the motion, and the tension T acts along the rope, away from the sledge.

Watch out

Never draw "the force of motion", "the force of the throw" or " $ma$ " as a force on a diagram. Once a ball has left your hand nothing is pushing it forwards; it keeps moving because of its velocity, not because of a force. And $ma$ is the result of the forces, not an extra one.

A block at rest on a rough plane inclined at angle α. The weight W is vertical, the normal reaction R is perpendicular to the plane, and friction F acts up the plane because the block tends to slide down.

Identifying forces on a sledge

A child pulls a sledge across rough horizontal snow using a rope inclined at an angle $\theta$ above the horizontal. List the forces acting on the sledge and state the direction of each.

Solution

There are four forces, shown in the diagram above.

Weight $W = mg$ , vertically downwards.
Normal reaction $R$ from the snow, vertically upwards (perpendicular to the horizontal surface).
Friction $F$ from the snow, horizontally, opposite to the direction of motion.
Tension $T$ in the rope, along the rope, at angle $\theta$ above the horizontal, away from the sledge.

Note that $R$ is not equal to $W$ here: part of the tension pulls upwards, so the snow has to push up less. You will calculate this in Friction.

Normal reaction with a vertical pull

A crate of mass $12\ \text{kg}$ rests on a horizontal floor. A vertical rope attached to the crate pulls upwards with tension $T$ newtons.

(a) Find the normal reaction when $T = 45$ .

(b) Find the least value of $T$ for which the crate leaves the floor.

Solution

The forces on the crate are its weight $12 \times 10 = 120\ \text{N}$ down, the tension $T$ up and the normal reaction $R$ up. The crate is at rest, so the upward forces balance the downward one:

R + T = 120

(a) With $T = 45$ : $R = 120 - 45 = 75\ \text{N}$ .

(b) The crate is about to leave the floor when the floor no longer needs to push, that is when $R = 0$ . Then $T = 120$ , so the least tension is $120\ \text{N}$ .

A third-law pair

A man of mass $70\ \text{kg}$ stands on a box of mass $20\ \text{kg}$ , which rests on horizontal ground. Find

(a) the force exerted by the box on the man,

(b) the force exerted by the man on the box,

Solution

(a) The man is in equilibrium under his weight $700\ \text{N}$ down and the normal reaction $R_1$ from the box up. So $R_1 = 700\ \text{N}$ , upwards.

(b) By Newton's third law, the man pushes on the box with a force equal and opposite to (a): $700\ \text{N}$ , downwards.

(c) The forces on the box are its weight $200\ \text{N}$ down, the man's push $700\ \text{N}$ down and the ground's reaction $R_2$ up:

R_2 = 200 + 700 = 900\ \text{N}, \text{ upwards}

The diagram earlier in this note shows the two separate force diagrams. The $700\ \text{N}$ third-law pair appears once on each, in opposite directions.

Choosing the direction of friction

A particle rests on a rough plane inclined at $20^\circ$ to the horizontal. It is held by a light string parallel to the plane, attached to a point further up the slope. State the forces acting on the particle, and explain why the direction of friction cannot be decided until the tension is known.

Solution

The forces are: the weight $mg$ vertically down; the normal reaction $R$ perpendicular to the plane; the tension $T$ up the plane along the string; and friction $F$ along the plane.

The component of the weight down the plane is fixed. If the tension is smaller than this component, the particle tends to slide down, so friction acts up the plane to help the tension. If the tension is larger, the particle tends to be pulled up, so friction acts down the plane. If the two are exactly equal, no friction is needed at all.

In an exam, either work out which case applies, or choose a direction, solve, and interpret a negative value of $F$ as friction acting the other way.

Stacked blocks with a vertical pull

A block $A$ of mass $3\ \text{kg}$ rests on top of a block $B$ of mass $5\ \text{kg}$ , which rests on a horizontal table. A light string attached to $A$ pulls vertically upwards on $A$ with a force of $12\ \text{N}$ . Find the magnitude of the force exerted by $A$ on $B$ and the magnitude of the force exerted by the table on $B$ .

Solution

Block $A$ . Forces: weight $30\ \text{N}$ down, tension $12\ \text{N}$ up, normal reaction $S$ from $B$ up. In equilibrium:

S + 12 = 30 \quad\Rightarrow\quad S = 18\ \text{N}

By Newton's third law, $A$ pushes down on $B$ with $18\ \text{N}$ .

Block $B$ . Forces: weight $50\ \text{N}$ down, the push from $A$ , $18\ \text{N}$ down, normal reaction $R$ from the table up:

R = 50 + 18 = 68\ \text{N}

Check with the whole system ( $A$ and $B$ together, total weight $80\ \text{N}$ ): the external forces are $80$ down, $12$ up and $R$ up, so $R = 80 - 12 = 68\ \text{N}$ . The force between the blocks is internal to the system, so it does not appear in this check.

Tip

Treating several bodies as one system is often quicker: the forces between them are internal and cancel in pairs by Newton's third law. You need separate diagrams only when the question asks about a force between the bodies (a tension, a tow-bar force, a reaction between a person and a lift floor).

Common mistakes

Mistakes examiners see every year

Normal reaction drawn vertically on a slope. It is always perpendicular to the surface.
Weight drawn perpendicular to a slope. Weight is always vertically down.
Assuming $R = mg$ . This is only true on a horizontal surface when no other force has a vertical component. Any pull or push at an angle changes $R$ .
Friction on a smooth surface. Smooth means no friction. Equally, a rough surface does not mean friction must be at its maximum.
Treating weight and normal reaction as a third-law pair. They act on the same body.
Tension pointing into the body. A string can only pull, so tension always points away from the body along the string.
Including air resistance when the question did not mention it.

Exam technique

Exam tip

A clear force diagram is rarely worth marks on its own, but it is where every correct equation comes from. Draw one for every question, even when not asked, and label every force.
When a question says "find the force exerted by $A$ on $B$ ", give the magnitude and direction, and use Newton's third law explicitly if you calculated the force exerted by $B$ on $A$ .
If a question asks you to "state a modelling assumption", good answers name the model and what it implies: "the crate is modelled as a particle, so all forces act at one point", "the string is light, so the tension is the same throughout".
Use $g = 10$ and give non-exact answers to 3 significant figures. Angles are given to 1 decimal place.

Summary

A force is a vector, measured in newtons. Every object in Paper 4 is a particle: all forces act at one point.
Weight $W = mg$ acts vertically down, with $g = 10\ \text{m s}^{-2}$ .
The contact force between surfaces has a normal component $R$ (perpendicular to the surface) and a frictional component $F$ (along it). Smooth means $F = 0$ .
$R$ is not automatically $mg$ ; it adjusts to whatever is needed, and $R = 0$ means contact is about to be lost.
Tension pulls along a string away from the body; a rod can carry tension or thrust. A light string over a smooth pulley has the same tension throughout.
Newton's third law pairs act on different bodies, are the same type of force, and are equal and opposite.
Draw a force diagram for every question: weight, normal reactions, friction, tensions, given forces. Never draw " $ma$ " or "the force of motion".

Practice

Question

A box of mass $8\ \text{kg}$ rests on a horizontal floor. A boy pushes vertically down on the box with a force of $25\ \text{N}$ . Find the normal reaction between the floor and the box.
A particle is at rest on a rough plane inclined at $30^\circ$ . No other forces act apart from weight and the contact force from the plane. Draw a force diagram and state the direction of the frictional force.
A lamp of mass $1.5\ \text{kg}$ hangs at rest from a light vertical cable. State the tension in the cable and the force the lamp exerts on the cable.
A woman of mass $60\ \text{kg}$ stands on a set of bathroom scales of mass $2\ \text{kg}$ , which rest on the floor. Find the force exerted by the woman on the scales and the force exerted by the floor on the scales.
A crate of mass $40\ \text{kg}$ rests on a horizontal floor. Two vertical ropes are attached to it, with tensions $T$ and $2T$ newtons. Find the value of $T$ for which the crate is just about to leave the floor.
Explain why a particle placed on a smooth inclined plane cannot remain at rest unless another force acts on it.
A car tows a trailer along a straight horizontal road using a light rigid tow-bar. State whether the force in the tow-bar is a tension or a thrust when (a) the car is accelerating forwards, (b) the car is braking and the trailer has no brakes. Explain your answers.
Block $P$ of mass $2\ \text{kg}$ rests on block $Q$ of mass $6\ \text{kg}$ , which rests on a horizontal floor. A vertical force of $X$ newtons pushes down on $P$ . The floor exerts a force of $95\ \text{N}$ on $Q$ . Find $X$ and the force that $Q$ exerts on $P$ .
A particle of mass $m\ \text{kg}$ hangs at rest from a string attached to the ceiling of a stationary lift. A student says "the tension and the weight are a Newton's third law pair, because they are equal and opposite". Explain why the student is wrong, and identify the true third-law partner of the tension acting on the particle.

Answers

Forces on the box: weight $80\ \text{N}$ down, push $25\ \text{N}$ down, $R$ up. $R = 80 + 25 = 105\ \text{N}$ .
Weight $mg$ vertically down; normal reaction $R$ perpendicular to the plane; friction $F$ up the plane, because without friction the particle would slide down.
The lamp is in equilibrium: $T = 1.5 \times 10 = 15\ \text{N}$ . By Newton's third law the lamp pulls down on the cable with $15\ \text{N}$ .
The woman is in equilibrium, so the scales push up on her with $600\ \text{N}$ ; by the third law she pushes down on the scales with $600\ \text{N}$ . Scales: $R = 20 + 600 = 620\ \text{N}$ upwards from the floor.
About to leave the floor means $R = 0$ , so $T + 2T = 400$ , giving $3T = 400$ and $T = 133\ \text{N}$ (3 s.f.).
On a smooth plane the only forces are the weight (vertical) and the normal reaction (perpendicular to the plane). The normal reaction has no component along the plane, but the weight does ( $mg\sin\alpha$ down the plane). Nothing balances that component, so the particle accelerates down the plane.
(a) Tension: the car pulls the trailer forwards, so the bar is stretched and pulls on both. (b) Thrust: the trailer keeps moving and pushes forwards on the car, while the car pushes backwards on the trailer to slow it; the bar is compressed.
Whole system: $R = 20 + 60 + X = 95$ , so $X = 15$ . Block $P$ : weight $20$ down, $X = 15$ down, reaction $S$ from $Q$ up, so $S = 35\ \text{N}$ upwards.
The tension and the weight both act on the same body (the particle) and are different types of force, so they cannot be a third-law pair. They are equal because the particle is in equilibrium. The partner of "string pulls particle up" is "particle pulls string down", a force of the same size acting on the string.

What a force is

Modelling: the particle and its friends

The forces you need

Weight

Normal reaction

Friction

The contact force as a whole

The smooth model and its limitations

Tension and thrust

Driving force and resistance

Newton's third law

Drawing force diagrams

Common mistakes

Exam technique

Summary

Practice

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