Forces and Equilibrium

Syllabus coverage

identify the forces acting in a given situation, for example by drawing a force diagram
understand the vector nature of force, and find and use components and resultants, by calculation rather than by scale drawing
use the principle that, when a particle is in equilibrium, the vector sum of the forces acting is zero, or equivalently that the sum of the components in any direction is zero
understand that the contact force between two surfaces can be represented by two components, the normal component and the frictional component
use the model of a smooth contact, and understand the limitations of this model
understand the concepts of limiting friction and limiting equilibrium, recall the definition of the coefficient of friction, and use the relationship $F = μ R$ or $F \leq μ R$ , as appropriate
use Newton's third law, for example that the force exerted by a particle on the ground is equal and opposite to the force exerted by the ground on the particle

Part of the AS & A Level Maths syllabus 9709.

A block pulled along rough horizontal ground: the tension acts at 30° above the horizontal, the weight and the normal reaction are vertical, and friction opposes the intended motion.

F = μ R

holds only at limiting equilibrium.

Forces are vectors, and a particle is a single point

A force is a push or a pull, measured in newtons (N), and it has both a magnitude and a direction. That is what it means to say that force is a vector. Two forces of 3 N and 4 N do not always add to 7 N: at right angles they add to 5 N, and in opposite directions they add to 1 N.

The syllabus works with a particle - a body small enough that every force acting on it can be taken to act at a single point. A question may describe a box, a car or a sledge, but you treat it as a particle: one dot, with every arrow drawn from it. Extended bodies are treated as particles too, so there are no turning effects to consider in this topic.

Vector notation is not used in the mechanics papers, so you will not see $\vec{F}$ . Direction is described in words - 'due east', 'at 30° to the horizontal', 'along the line of greatest slope' - and the working is done by resolving, which is what the next few blocks are about.

The weight of a particle of mass $m$ kilograms is $m g$ newtons acting vertically downwards. Cambridge uses $g = 9.8 m s^{- 2}$ unless the question says otherwise; questions sometimes say to take $g = 10 m s^{- 2}$ , which keeps the arithmetic tidy.

Quantity	Symbol	Unit and direction
mass	$m$	kg - not a force
weight	$m g$	N, vertically downwards
normal reaction	$R$	N, perpendicular to the surface
friction	$F$	N, along the surface
tension	$T$	N, along the string, pulling
resultant	$\sum F$	N, in the direction of the combined effect

A 5 kg box has a weight of

5 \times 9.8 = 49

N, not 5 N. Confusing mass with weight is the commonest way to lose the first mark of a mechanics question.

The forces you have to identify

Every question starts by naming the forces acting on the particle, and the list is short. Weight $m g$ acts vertically downwards. The normal reaction $R$ comes from a surface, perpendicular to that surface and away from it - on a slope it is perpendicular to the slope, not vertical. Friction $F$ acts along the surface, opposing the tendency of the body to slide. Tension $T$ acts in a string or rope, along it, and always pulls the particle towards the other end. Thrust acts in a rod, along it, and pushes. An applied force is whatever the question pushes or pulls with, such as a horizontal force of 20 N.

The normal reaction and friction are the two components of the single contact force from the surface. That is why a block on rough ground has two arrows from the ground: one perpendicular to it and one along it.

Strings pull and rods push. A string cannot push a particle away from the other end, so a string that would need to push simply goes slack. Rods and struts can push as well as pull, which is why they can hold something up.

Force	Acts	Direction	Note
weight	at the particle	vertically downwards	$m g$ , always present
normal reaction	from the surface	perpendicular to the surface, away from it	$R$
friction	from the surface	along the surface, opposing sliding	$F \leq μ R$
tension	in a string	along the string, towards the other end	pulls only
thrust	in a rod	along the rod, pushing	rods can push
applied force	as the question states	as the question states	e.g. 20 N at 40° above the horizontal

Never put on the diagram a force that the particle exerts on something else. If a block pushes the ground, that arrow belongs to the ground, not to the block, and including it makes the equilibrium equations wrong.

Drawing the force diagram

Draw the particle as a dot or a small box, then draw one arrow for each force acting on it, starting at the particle. Label each arrow with its magnitude if the question gives one, or with its symbol if it does not. The diagram is not decoration: it is where the marks for identifying the forces are earned, and it is how you avoid forgetting a component.

For a body on a slope, mark the angle of the slope at one end and then use the two directions that keep the algebra short: parallel to the plane and perpendicular to the plane. Resolving in those directions puts the normal reaction in one equation only and the friction in the other.

The weight is the force that needs care. On a slope at angle $θ$ , the weight $m g$ has component $m g \sin θ$ down the slope and $m g \cos θ$ perpendicular to the slope. The angle between the weight and the perpendicular to the slope is $θ$ , which is why the perpendicular component is the cosine one.

a block on horizontal ground: weight down, normal reaction up, friction along the ground, plus any applied force
a block on a slope with nothing else acting: weight down, normal reaction perpendicular to the slope, friction along the slope
a particle hanging from two strings: weight down, and a tension along each string, both pulling
a particle pulled by a rope at an angle: the tension splits into a horizontal component and a vertical component

Every arrow in a force diagram starts at the particle and points away from it. An arrow drawn pointing into the dot, or drawn somewhere else on the page, is the first thing a mark scheme ignores.

Resolving a force into components

To resolve a force is to replace it by two perpendicular forces that together have the same effect. A force $F$ acting at an angle $θ$ to a chosen direction has component $F \cos θ$ along that direction and $F \sin θ$ perpendicular to it. The component along the direction the force makes the angle with is the cosine one.

The two components are the sides of a right-angled triangle whose hypotenuse is $F$ . The side next to the angle is the cosine one and the side opposite the angle is the sine one. Getting them the wrong way round is the most expensive mistake in this topic, and one check catches it: a component can never be larger than the force itself.

The syllabus assumes these trigonometric results, worth a glance before you resolve anything: $\sin (90^{\circ} - θ) = \cos θ$ , $\cos (90^{\circ} - θ) = \sin θ$ , $\tan θ = \frac{\sin θ}{\cos θ}$ and $\sin^{2} θ + \cos^{2} θ = 1$ . The first two are why the component perpendicular to the direction you measured from is the sine one: the angle between the force and the perpendicular is $90^{\circ} - θ$ .

Angle $θ$	0°	30°	45°	60°	90°
$\sin θ$	0	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	1
$\cos θ$	1	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	0
$\tan θ$	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$	undefined

Resolving is a replacement, not an addition. Once a force has been resolved into two components, the original force is gone - never count the force and its components together.

Resultants

The resultant of two or more forces is the single force that has the same effect on the particle. For two forces at right angles the magnitude is $R = \sqrt{F_{1}^{2} + F_{2}^{2}}$ , and the direction satisfies $\tan θ = \frac{F_{2}}{F_{1}}$ where $θ$ is measured from the $F_{1}$ direction. So 9 N and 12 N at right angles have a resultant of 15 N at $\tan^{- 1} (\frac{9}{12}) = {36.9}^{\circ}$ to the 12 N force.

For three or more forces, or forces that are not at right angles to each other, resolve every force in two perpendicular directions, add the components in each direction, and then combine the two totals with Pythagoras. That method always works, which is why resolving is the method the syllabus expects.

A resultant of zero is not a small force: it means the forces cancel exactly, which is equilibrium. The two ideas are the same statement seen from opposite sides.

The syllabus is explicit that calculations are always required. An answer obtained by drawing the arrows to scale and measuring the result with a ruler does not score, however careful the drawing.

Equilibrium

A particle is in equilibrium when it stays at rest, or continues to move at constant velocity. The syllabus states the condition as: the vector sum of the forces acting is zero, or equivalently, the sum of the components in any direction is zero. In symbols, $\sum F = 0$ in each of two perpendicular directions.

The method is always the same. Choose two perpendicular directions, state which way is positive, resolve every force in those two directions, and write one equation for each. Two equations for two unknowns is the normal shape of a 9709 equilibrium question.

On horizontal ground the two directions are horizontal and vertical. On a slope they are parallel to the plane and perpendicular to the plane. Either choice is correct, but the slope directions make the algebra shorter because the reaction and the friction then each appear in only one equation.

When exactly three forces act and the particle is in equilibrium, the three arrows can be drawn head to tail to form a closed triangle. That is the triangle of forces. It is a legitimate alternative to resolving and is sometimes quicker, but the syllabus says such methods are not required knowledge and questions will not point you towards them.

state the positive direction: 'taking the direction up the plane as positive'
resolve every force in that direction, including the ones you expect to cancel
write $\sum F = 0$ for each of the two directions and solve the equations together
if a force of unknown magnitude acts at an unknown angle, resolve it as $F \cos θ$ and $F \sin θ$

Equilibrium is a statement about components, not about forces. A force of 50 N can be part of an equilibrium if another force cancels its components, and two forces of 50 N at 90° to each other can never be in equilibrium on their own.

Contact forces: the normal reaction and friction

A surface pushes on a body with a single contact force, and it is convenient to split that force into two perpendicular components: the normal reaction $R$ , perpendicular to the surface and away from it, and the frictional component $F$ , along the surface.

The normal reaction is a reaction: it takes whatever value the situation requires. A heavier block gives a bigger $R$ , and pulling upwards on a block reduces $R$ . It is found from the equilibrium equation perpendicular to the surface, never from a formula.

A smooth surface is a model in which there is no friction at all, so $F = 0$ and the contact force is purely the normal reaction perpendicular to the surface. It is a strong idealisation: real surfaces always resist sliding to some extent. On a smooth slope a block always slides, because the component of the weight down the slope is unbalanced and nothing opposes it.

Friction is what lets a block stay put on a rough slope. It adjusts itself to whatever value is needed to prevent sliding, up to a limit, and that limit is the subject of the next block.

A smooth contact means no friction, not no force. The normal reaction is still there, and on a slope it is still perpendicular to the slope rather than vertical.

Limiting friction and the coefficient of friction

Friction is not a fixed force. It is whatever is needed to stop the surfaces sliding, up to a maximum, so $F \leq μ R$ , where $μ$ is the coefficient of friction between the two surfaces. Equality happens only at the point of sliding.

Limiting friction is that maximum value, and a body on the point of sliding is in limiting equilibrium. Then, and only then, $F = μ R$ . The phrases 'about to slip', 'on the point of moving' and 'the block is about to slide' all mean limiting equilibrium.

The coefficient $μ$ is a property of the pair of surfaces, has no units, and for most pairs lies between 0 and 1. It is either given to you or asked for; the working is always the same - find $R$ from the perpendicular equation, find $F$ from the parallel equation, then divide.

Friction acts along the surface in the direction that opposes the tendency to slide. On a slope it points up the slope if the body is about to slide down, and down the slope if the body is about to be pushed up. Decide which way the body is about to move before you draw the arrow.

The standard result: a body in limiting equilibrium on a rough plane inclined at $θ$ , with no other force parallel to the plane, has $R = m g \cos θ$ and $F = m g \sin θ$ , so $μ = \frac{m g \sin θ}{m g \cos θ} = \tan θ$ . The mass and $g$ cancel, which is why the angle at which a body starts to slide depends only on the two surfaces.

not sliding: find $F$ from $\sum F = 0$ and check that $F < μ R$
about to slide: $F = μ R$ , the limiting case
already sliding: friction is at its maximum, $F = μ R$ , and the forces no longer balance

F = μ R

is not a formula for friction in general; it says that friction has reached its maximum. Using it when the body is at rest and not about to move is a reliable way to produce a wrong coefficient.

Newton's third law

Newton's third law: if body A exerts a force on body B, then B exerts a force of the same magnitude and the opposite direction on A. The two forces are of the same type and act on different bodies.

In a mechanics question the useful part is what the law tells you to leave out. A block pushes down on the ground with a force equal in magnitude to the reaction $R$ , but that force acts on the ground, so it does not appear in the block's force diagram. The reaction from the ground on the block does.

The classic trap: the weight of a block and the normal reaction from the table are equal in magnitude and opposite in direction, but they are not a third-law pair. They act on the same body - the block - and they are different types of force. The third-law partner of the weight is the gravitational pull the block exerts on the Earth.

Third-law pairs act on two different bodies. If both forces act on the same body they are not a third-law pair, however equal and opposite they look.

Key terms you must be able to define

Force: A push or a pull, measured in newtons (N), with both a magnitude and a direction.

Newton: The unit of force. A force of 1 N gives a mass of 1 kg an acceleration of $1 m s^{- 2}$ .

Particle: A body whose size is ignored, so that every force acting on it is taken to act at a single point.

Weight: The gravitational force on a particle: $m g$ newtons acting vertically downwards, with $g = 9.8 m s^{- 2}$ unless the question says otherwise.

Normal reaction: The component of the contact force perpendicular to the surface, acting away from the surface. On a slope it is perpendicular to the slope.

Friction: The component of the contact force along the surface, opposing the tendency of the body to slide.

Limiting friction: The maximum value friction can take, $F = μ R$ . It applies only when the body is on the point of sliding.

Limiting equilibrium: The state of a body that is at rest but about to move. Friction is at its maximum, so $F = μ R$ .

Coefficient of friction: The constant $μ$ in $F = μ R$ , a property of the pair of surfaces in contact. It has no units.

Smooth: A model of a contact with no friction at all, so the contact force is just the normal reaction perpendicular to the surface.

Rough: A contact that can exert friction as well as a normal reaction.

Tension: The force in a string or rope, acting along it and pulling the particle towards the other end.

Thrust: The force in a rod or strut, acting along it and pushing. A string cannot exert thrust.

Component: The effect of a force in a chosen direction: $F \cos θ$ along the direction and $F \sin θ$ perpendicular to it.

Resultant: The single force that has the same effect as two or more forces acting together.

Equilibrium: The state in which the vector sum of the forces on a particle is zero, so the particle stays at rest or moves at constant velocity.

Newton's third law: If A exerts a force on B, then B exerts a force of the same magnitude and opposite direction on A. The two forces act on different bodies.

Applied

At AS & A Level the marks sit in the setup: choosing the model, stating the variable and writing the equation before any working. These show the shape of a full-mark answer, then you do the algebra yourself.

A block pulled at an angle

A block of mass $8 k g$ on rough horizontal ground is pulled by a rope at 30° to the horizontal with tension 20 N. Taking $g = 10 m s^{- 2}$ , the vertical equation is $R + 20 \sin 30^{\circ} = 80$ , so $R = 80 - 10 = 70$ N. The horizontal component of the tension is $20 \cos 30^{\circ} = 17.3$ N, so if the block is on the point of moving, $μ = \frac{17.3}{70} = 0.247$ .

Three forces in equilibrium

A particle is in equilibrium under 7 N acting due east, 24 N acting due north and a third force. The resultant of the first two is $\sqrt{7^{2} + 24^{2}} = 25$ N, acting at $\tan^{- 1} (\frac{24}{7}) = {73.7}^{\circ}$ north of east. The third force is equal and opposite to that resultant: 25 N acting 73.7° south of west.

Limiting friction on a slope

A block rests on a rough plane inclined at $θ$ to the horizontal, where $\tan θ = 0.75$ , and is on the point of sliding down the plane. Resolving, $R = m g \cos θ$ and $F = m g \sin θ$ , so $μ = \tan θ = 0.75$ . The mass never enters the calculation.

A smooth slope with a string

A particle of mass $5 k g$ is held on a smooth plane inclined at 30° by a string parallel to the line of greatest slope, with $g = 9.8 m s^{- 2}$ . Parallel to the plane: $T = 5 \times 9.8 \times \sin 30^{\circ} = 24.5$ N. Perpendicular to the plane: $R = 5 \times 9.8 \times \cos 30^{\circ} = 42.4$ N. Because the plane is smooth, no friction term appears in either equation.

Newton's third law in context

A box of mass $4 k g$ rests on the floor of a lift that is not moving. The box pushes down on the floor with 39.2 N and the floor pushes up on the box with 39.2 N - equal and opposite, acting on different bodies, so they are a third-law pair. The weight of the box is also 39.2 N, but its third-law partner is the pull the box exerts on the Earth, not the normal reaction.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What is the weight of a particle of mass

m

m g

newtons, acting vertically downwards, with

g = 9.8 m s^{- 2}

unless the question says otherwise.

Q2State the condition for a particle to be in equilibrium.

The vector sum of the forces acting on it is zero, so the sum of the components in any two perpendicular directions is zero.

Q3In which direction does the normal reaction act?

Perpendicular to the surface and away from it. On a slope that means perpendicular to the slope, not vertical.

Q4In which direction does friction act?

Along the surface, in the direction that opposes the tendency of the body to slide.

Q5What is the difference between tension and thrust?

Tension acts in a string and pulls; thrust acts in a rod and pushes. A string cannot push.

Q6How do you resolve a force

F

acting at an angle

θ

to a direction?

F \cos θ

along that direction and

F \sin θ

perpendicular to it.

Q7What are the components of the weight on a slope at angle

θ

m g \sin θ

down the slope and

m g \cos θ

perpendicular to the slope.

Q8What is the resultant of two perpendicular forces

P

and

Q

\sqrt{P^{2} + Q^{2}}

, acting at

\tan^{- 1} (\frac{Q}{P})

to the direction of

P

Q9What does the model of a smooth contact assume?

That there is no friction, so the contact force is just the normal reaction perpendicular to the surface.

Q10When may you use

F = μ R

Only at limiting friction, when the body is on the point of sliding. Otherwise friction is whatever equilibrium requires, and

F \leq μ R

Q11What does 'about to slip' tell you?

The body is in limiting equilibrium, so friction has reached its maximum and

F = μ R

Q12What is the coefficient of friction?

The constant

μ

F = μ R

, a property of the two surfaces in contact. It has no units.

Q13A body is in limiting equilibrium on a rough plane at angle

θ

, with nothing else acting parallel to the plane. What is

μ

μ = \tan θ

, because

F = m g \sin θ

and

R = m g \cos θ

Q14State Newton's third law.

If A exerts a force on B, then B exerts a force of the same magnitude and opposite direction on A. The two forces act on different bodies.

Q15Why are the weight of a book and the normal reaction from the table not a third-law pair?

Both forces act on the book. A third-law pair acts on two different bodies.

Q16Why is a scale drawing not acceptable in a 9709 mechanics question?

The syllabus requires calculations. A resultant measured from a scale diagram does not earn the marks.

0 of 16 marked as known

Exam technique for this topic

Draw the force diagram first and label every arrow. Marks on this topic are earned from the diagram outwards: identify the forces, resolve them, then solve the equations.
State your positive direction before you resolve, and keep the same sign convention in both equations. A sign dropped from one component is the commonest arithmetic error here.
On a slope, resolve parallel and perpendicular to the plane rather than horizontally and vertically. The reaction and the friction then each appear in one equation only, which halves the chance of a slip.
Find $R$ before you use $F = μ R$ . Any force with a component perpendicular to the surface changes $R$ : pulling upwards reduces it, pushing downwards increases it.
Use $F = μ R$ only when the question says 'on the point of moving', 'about to slip' or 'limiting equilibrium'. Otherwise find the friction from the equilibrium equation and check that it is less than $μ R$ .
Give answers to three significant figures unless the question says otherwise, always with the unit, and take $g = 9.8 m s^{- 2}$ unless the question tells you otherwise.

Why students lose marks here

A student wrote

R = m g

always, so the reaction equals the weight.

Why it loses marksA force with a vertical component changes the reaction. Pulling a block upwards at an angle reduces

R

below

m g

; pushing down at an angle increases it.

The answer that scoresResolve perpendicular to the surface. For a block of mass

6 k g

pulled by a rope with a 15 N upward component and

g = 10

R = 60 - 15 = 45

N, not 60 N.

A student wrote9 N and 12 N at right angles give a resultant of 21 N.

Why it loses marksForces are vectors, so they combine as a triangle rather than as numbers. Adding the magnitudes is only correct when the forces act in the same direction.

The answer that scores

R = \sqrt{9^{2} + 12^{2}} = 15

N, at

\tan^{- 1} (\frac{9}{12}) = {36.9}^{\circ}

to the 12 N force.

A student wroteFriction is always equal to

μ R

Why it loses marks

F = μ R

is the maximum friction, reached only at limiting equilibrium. A block at rest on rough ground often has far less friction than the maximum available.

The answer that scoresUse

F = μ R

when the body is about to slip; otherwise find

F

from

\sum F = 0

and note that

F \leq μ R

A student wroteThe weight of a book and the normal reaction from the table are a Newton's third law pair.

Why it loses marksThey act on the same body, and a third-law pair acts on two different bodies. The forces in a pair are also of the same type.

The answer that scoresThe third-law partner of the weight of the book is the gravitational force the book exerts on the Earth.

A student wroteA string can push, so a string below a block can hold it up a slope.

Why it loses marksTension in a string can only pull the particle towards the other end. A string that would have to push simply goes slack.

The answer that scoresThrust requires a rod or strut. A string holding a block on a slope must be on the uphill side, pulling the block up the slope.

A student wroteOn a slope I resolved horizontally and vertically and the equations came out wrong.

Why it loses marksResolving horizontally and vertically is not wrong in principle, but it puts both

R

and

F

into both equations and invites sign errors. Along and perpendicular to the plane is the method the mark scheme expects.

The answer that scoresTake up the slope and perpendicular to the slope as the two directions, giving

F = m g \sin θ

and

R = m g \cos θ

for a block resting in limiting equilibrium.

Syllabus coverage

Forces are vectors, and a particle is a single point

The forces you have to identify

Drawing the force diagram

Resolving a force into components

Resultants

Equilibrium

Contact forces: the normal reaction and friction

Limiting friction and the coefficient of friction

Newton's third law

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here