Forces and Friction

What you'll learn

How to draw and use a force diagram for a particle in equilibrium.
How to resolve forces horizontally/vertically or parallel/perpendicular to a plane.
How friction works, including limiting friction and the coefficient of friction.
How to decide whether an object accelerates, stays still, or is about to slip.

The core idea: forces and modelling

A force is a push or pull, measured in newtons, N. In this topic, you usually model objects as particles, meaning all forces act through one point and you do not worry about rotation.

Definition

Particle model

A particle is a model of an object where its size and shape are ignored. This lets you draw all forces acting at a single point.

Common forces you will meet:

Weight: the gravitational force acting vertically downwards. For mass $m$ , weight is $m g$ , where $g = 9.8$ .
Tension: the pulling force in a light string, acting along the string away from the particle.
Normal reaction: the contact force from a surface, acting perpendicular to the surface.
Friction: a contact force that opposes actual or possible sliding.

Key Idea

Start with a force diagram

Almost every question in this topic becomes easier once you draw all forces and choose sensible directions to resolve them.

Equilibrium and resolving forces

An object is in equilibrium when its acceleration is zero. So the resultant force is zero in every direction.

Definition

Equilibrium

A particle is in equilibrium if the resultant force on it is zero. At A-Level, this usually means resolving forces in two perpendicular directions and setting both totals equal.

When a force acts at an angle, split it into components. If a force $F$ makes angle $θ\theta$ with the horizontal:

A force at angle θ to the horizontal resolved into horizontal and vertical components.

horizontal component: $Fcos⁡θF\cos\theta$
vertical component: $Fsin⁡θF\sin\theta$

If the angle is measured from the vertical, swap the roles.

Common Mistake

Using the wrong component

If the angle is with the vertical, the vertical component is usually the cosine part, not the sine part. Sketch a right-angled triangle if unsure.

Example

A particle held by a sloping string

A particle is attached to a wall by a light string. A horizontal force of 18 N pulls it away from the wall. The string is taut and makes an angle of $30∘30^\circ$ with the vertical. Find the tension and the weight of the particle.

Force diagram for a particle held by a sloping string with a horizontal pull.

Draw the forces on the particle: tension $T$ along the string, weight $W$ vertically downwards, and the 18 N force horizontally away from the wall.
Resolve horizontally. The horizontal component of tension balances the 18 N force:
$Tsin⁡30∘=18T\sin 30^\circ = 18$
Solve for the tension:
$\frac{18}{\sin 30^\circ} = 36$
Resolve vertically. The vertical component of tension balances the weight:
$T\cos 30^\circ$
Substitute $T = 36$ :
$36\cos 30^\circ \approx 31.2$
The tension is 36 N and the weight is approximately 31.2 N.

Two strings supporting a particle

When two strings support a mass, each string has its own tension. You normally get two simultaneous equations by resolving horizontally and vertically.

If a particle of mass $m$ is at rest, its weight is $m g$ acting downwards.

Example

Two angled strings in equilibrium

A particle of mass 8 kg is held by two light strings. String A makes an angle of $55∘55^\circ$ to the horizontal on the left, and string B makes an angle of $35∘35^\circ$ to the horizontal on the right. Find the two tensions.

Two-string equilibrium diagram showing both tensions and the weight of the particle.

Let the tension in string A be $T_A$ and the tension in string B be $T_B$ .
Resolve horizontally. The horizontal components balance:
$TAcos⁡55∘=TBcos⁡35∘T_A\cos 55^\circ = T_B\cos 35^\circ$
Resolve vertically. The upward components balance the weight:
$TAsin⁡55∘+TBsin⁡35∘=8gT_A\sin 55^\circ + T_B\sin 35^\circ = 8g$
From the horizontal equation, write $T_B$ in terms of $T_A$ :
$TB=TAcos⁡55∘cos⁡35∘T_B = \frac{T_A\cos 55^\circ}{\cos 35^\circ}$
Substitute this into the vertical equation:
$TAsin⁡55∘+TAcos⁡55∘cos⁡35∘sin⁡35∘=78.4T_A\sin 55^\circ + \frac{T_A\cos 55^\circ}{\cos 35^\circ}\sin 35^\circ = 78.4$
Solve:
$TA≈63.9T_A \approx 63.9$
Use the horizontal equation to find $T_B$ :
$TB=63.9cos⁡55∘cos⁡35∘≈44.7T_B = \frac{63.9\cos 55^\circ}{\cos 35^\circ} \approx 44.7$
So the tensions are approximately 63.9 N and 44.7 N.

Friction and the coefficient of friction

Friction acts parallel to the surface and opposes motion or the tendency to move.

Definition

Coefficient of friction

The coefficient of friction, usually $μ\mu$ , measures how rough two surfaces are. The maximum possible friction is $μR\mu R$ , where $R$ is the normal reaction.

If an object is just about to slip, it is in limiting equilibrium and the friction has its maximum value:

Ffriction=μRF_{\text{friction}} = \mu R

If the object is not on the point of slipping, then friction may be less than $μR\mu R$ .

Common Mistake

Only use $F = μ R$ at the limit

You may only set friction equal to $μR\mu R$ when the object is moving, or is on the point of moving. If it is just resting normally, friction adjusts as needed up to its maximum.

Example

Block pulled along a rough horizontal floor

A 20 kg block rests on a rough horizontal floor. The coefficient of friction is 0.3. A force $P$ pulls the block at $25∘25^\circ$ above the horizontal. The block is on the point of sliding. Find $P$ .

Free-body diagram for a block pulled at an angle on a rough horizontal floor.

Draw the forces: weight $20 g$ downwards, normal reaction $R$ upwards, pull $P$ at $25∘25^\circ$ , and friction acting backwards.
Since the block is on the point of sliding, friction is limiting:
$\mu R = 0.3R$
Resolve vertically. The upward forces balance the downward forces:
$P\sin 25^\circ = 20g$
Rearrange for $R$ :
$P\sin 25^\circ$
Resolve horizontally. The pulling component balances friction:
$Pcos⁡25∘=0.3RP\cos 25^\circ = 0.3R$
Substitute for $R$ :
$Pcos⁡25∘=0.3(196−Psin⁡25∘)P\cos 25^\circ = 0.3(196 - P\sin 25^\circ)$
Solve:
$P(cos⁡25∘+0.3sin⁡25∘)=58.8P(\cos 25^\circ + 0.3\sin 25^\circ) = 58.8$
Hence:
$\approx 56.9$
The required pulling force is approximately 56.9 N.

Tip

Pulling versus pushing

A pull angled upwards reduces the normal reaction, so friction decreases. A push angled downwards increases the normal reaction, so friction increases.

Inclined planes

For a block on a slope, choose axes:

parallel to the plane
perpendicular to the plane

If the plane is inclined at angle $θ\theta$ to the horizontal, the weight $m g$ has components:

Weight on an inclined plane resolved into components parallel and perpendicular to the plane.

down the plane: $mgsin⁡θmg\sin\theta$
into the plane: $mgcos⁡θmg\cos\theta$

Key Idea

Resolve along the slope

On an inclined plane, do not resolve horizontally and vertically unless you have a special reason. Parallel and perpendicular to the plane are usually much cleaner.

Example

Finding the coefficient of friction on a slope

A block of weight 12 N rests on a rough plane inclined at $28∘28^\circ$ to the horizontal. It is on the point of sliding down the plane. Find $μ\mu$ .

Limiting equilibrium on a rough inclined plane with friction acting up the slope.

The component of weight down the plane is $12sin⁡28∘12\sin 28^\circ$ .
The normal reaction is found perpendicular to the plane:
$12\cos 28^\circ$
Since the block is about to slide down, friction acts up the plane and is limiting:
$\mu R$
Resolve parallel to the plane. Friction balances the down-slope component of weight:
$μR=12sin⁡28∘\mu R = 12\sin 28^\circ$
Substitute $12\cos 28^\circ$ :
$μ(12cos⁡28∘)=12sin⁡28∘\mu(12\cos 28^\circ) = 12\sin 28^\circ$
Cancel 12 and solve:
$μ=tan⁡28∘≈0.532\mu = \tan 28^\circ \approx 0.532$
The coefficient of friction is approximately 0.532.

Motion on a rough inclined plane

If the block is moving, use Newton’s second law:

Fresultant=maF_{\text{resultant}} = ma

The friction force still acts against the motion. If the block slides down the plane, friction acts up the plane.

Example

Acceleration down a rough slope

A 3 kg block is released from rest on a rough plane. The plane is inclined at angle $θ\theta$ where $tan⁡θ=0.75\tan\theta = 0.75$ . The coefficient of friction is 0.2. Find the acceleration.

Moving block on a rough slope with acceleration down the plane and friction up the plane.

Use $tan⁡θ=0.75=34\tan\theta = 0.75 = \frac{3}{4}$ . This suggests a 3-4-5 triangle, so:
$sin⁡θ=35,cos⁡θ=45\sin\theta = \frac{3}{5}, \qquad \cos\theta = \frac{4}{5}$
Resolve perpendicular to the plane:
$mg\cos\theta$
Friction acts up the plane because the block moves down the plane:
$\mu R = 0.2mg\cos\theta$
Apply Newton’s second law down the plane:
$mgsin⁡θ−0.2mgcos⁡θ=mamg\sin\theta - 0.2mg\cos\theta = ma$
Cancel $m$ :
$g(\sin\theta - 0.2\cos\theta)$
Substitute the exact trig values:
$9.8\left(\frac{3}{5} - 0.2 \times \frac{4}{5}\right)$
Calculate:
$a = 4.312$
The acceleration is approximately 4.31 m s−2 down the plane.

Common Mistake

Keeping the mass unnecessarily

In many sliding-block problems, the mass cancels. If two blocks have the same $μ\mu$ on the same slope, their accelerations are the same, even if their masses differ.

Deciding whether the block moves

Sometimes a force is removed, and you must decide whether the object starts moving. Compare the force trying to make it move with the maximum possible friction.

For a block on a slope with no applied force:

force down the plane is $mgsin⁡θmg\sin\theta$
maximum friction is $μmgcos⁡θ\mu mg\cos\theta$

If $mgsin⁡θ>μmgcos⁡θmg\sin\theta > \mu mg\cos\theta$ , it slides down.

If $mgsin⁡θ≤μmgcos⁡θmg\sin\theta \le \mu mg\cos\theta$ , it stays at rest.

Example

After an applied force is removed

A brick of weight 10 N is on a rough plane at $30∘30^\circ$ . The coefficient of friction is 0.7. A force that had been holding it is removed. Decide whether the brick moves.

Force comparison for deciding whether a brick remains at rest on a rough slope.

The component of weight down the plane is:
$10sin⁡30∘=510\sin 30^\circ = 5$
The normal reaction is:
$10\cos 30^\circ$
The maximum friction available is:
$μR=0.7×10cos⁡30∘≈6.06\mu R = 0.7 \times 10\cos 30^\circ \approx 6.06$
Compare the two forces: the down-slope component is 5 N, while the maximum friction is about 6.06 N.
Since friction can be large enough to prevent sliding, the brick remains at rest.

Exam technique

In the exam

Draw a force diagram first, including directions for friction and the normal reaction.
Choose axes that match the situation: horizontal/vertical for hanging particles, parallel/perpendicular for inclined planes.
Use $F=μRF=\mu R$ only when the object is moving or in limiting equilibrium.
If asked to “determine whether it moves”, compare the driving force with the maximum possible friction.

Self review

Check yourself

If a force pulls a block upwards at an angle, what happens to the normal reaction?
On a slope, what are the two components of weight parallel and perpendicular to the plane?
When is it valid to write friction as exactly $μR\mu R$ ?

Forces and Friction

What you'll learn

How to draw and use a force diagram for a particle in equilibrium.
How to resolve forces horizontally/vertically or parallel/perpendicular to a plane.
How friction works, including limiting friction and the coefficient of friction.
How to decide whether an object accelerates, stays still, or is about to slip.

The core idea: forces and modelling

A force is a push or pull, measured in newtons, N. In this topic, you usually model objects as particles, meaning all forces act through one point and you do not worry about rotation.

Definition

Particle model

A particle is a model of an object where its size and shape are ignored. This lets you draw all forces acting at a single point.

Common forces you will meet:

Weight: the gravitational force acting vertically downwards. For mass $m$ , weight is $m g$ , where $g = 9.8$ .
Tension: the pulling force in a light string, acting along the string away from the particle.
Normal reaction: the contact force from a surface, acting perpendicular to the surface.
Friction: a contact force that opposes actual or possible sliding.

Key Idea

Start with a force diagram

Almost every question in this topic becomes easier once you draw all forces and choose sensible directions to resolve them.

Equilibrium and resolving forces

An object is in equilibrium when its acceleration is zero. So the resultant force is zero in every direction.

Definition

Equilibrium

A particle is in equilibrium if the resultant force on it is zero. At A-Level, this usually means resolving forces in two perpendicular directions and setting both totals equal.

When a force acts at an angle, split it into components. If a force $F$ makes angle $θ\theta$ with the horizontal:

A force at angle θ to the horizontal resolved into horizontal and vertical components.

horizontal component: $Fcos⁡θF\cos\theta$
vertical component: $Fsin⁡θF\sin\theta$

If the angle is measured from the vertical, swap the roles.

Common Mistake

Using the wrong component

If the angle is with the vertical, the vertical component is usually the cosine part, not the sine part. Sketch a right-angled triangle if unsure.

Example

A particle held by a sloping string

Force diagram for a particle held by a sloping string with a horizontal pull.

Draw the forces on the particle: tension $T$ along the string, weight $W$ vertically downwards, and the 18 N force horizontally away from the wall.
Resolve horizontally. The horizontal component of tension balances the 18 N force:
$Tsin⁡30∘=18T\sin 30^\circ = 18$
Solve for the tension:
$\frac{18}{\sin 30^\circ} = 36$
Resolve vertically. The vertical component of tension balances the weight:
$T\cos 30^\circ$
Substitute $T = 36$ :
$36\cos 30^\circ \approx 31.2$
The tension is 36 N and the weight is approximately 31.2 N.

Two strings supporting a particle

When two strings support a mass, each string has its own tension. You normally get two simultaneous equations by resolving horizontally and vertically.

If a particle of mass $m$ is at rest, its weight is $m g$ acting downwards.

Example

Two angled strings in equilibrium

Two-string equilibrium diagram showing both tensions and the weight of the particle.

Let the tension in string A be $T_A$ and the tension in string B be $T_B$ .
Resolve horizontally. The horizontal components balance:
$TAcos⁡55∘=TBcos⁡35∘T_A\cos 55^\circ = T_B\cos 35^\circ$
Resolve vertically. The upward components balance the weight:
$TAsin⁡55∘+TBsin⁡35∘=8gT_A\sin 55^\circ + T_B\sin 35^\circ = 8g$
From the horizontal equation, write $T_B$ in terms of $T_A$ :
$TB=TAcos⁡55∘cos⁡35∘T_B = \frac{T_A\cos 55^\circ}{\cos 35^\circ}$
Substitute this into the vertical equation:
$TAsin⁡55∘+TAcos⁡55∘cos⁡35∘sin⁡35∘=78.4T_A\sin 55^\circ + \frac{T_A\cos 55^\circ}{\cos 35^\circ}\sin 35^\circ = 78.4$
Solve:
$TA≈63.9T_A \approx 63.9$
Use the horizontal equation to find $T_B$ :
$TB=63.9cos⁡55∘cos⁡35∘≈44.7T_B = \frac{63.9\cos 55^\circ}{\cos 35^\circ} \approx 44.7$
So the tensions are approximately 63.9 N and 44.7 N.

Friction and the coefficient of friction

Friction acts parallel to the surface and opposes motion or the tendency to move.

Definition

Coefficient of friction

The coefficient of friction, usually $μ\mu$ , measures how rough two surfaces are. The maximum possible friction is $μR\mu R$ , where $R$ is the normal reaction.

If an object is just about to slip, it is in limiting equilibrium and the friction has its maximum value:

Ffriction=μRF_{\text{friction}} = \mu R

If the object is not on the point of slipping, then friction may be less than $μR\mu R$ .

Common Mistake

Only use $F = μ R$ at the limit

You may only set friction equal to $μR\mu R$ when the object is moving, or is on the point of moving. If it is just resting normally, friction adjusts as needed up to its maximum.

Example

Block pulled along a rough horizontal floor

Free-body diagram for a block pulled at an angle on a rough horizontal floor.

Draw the forces: weight $20 g$ downwards, normal reaction $R$ upwards, pull $P$ at $25∘25^\circ$ , and friction acting backwards.
Since the block is on the point of sliding, friction is limiting:
$\mu R = 0.3R$
Resolve vertically. The upward forces balance the downward forces:
$P\sin 25^\circ = 20g$
Rearrange for $R$ :
$P\sin 25^\circ$
Resolve horizontally. The pulling component balances friction:
$Pcos⁡25∘=0.3RP\cos 25^\circ = 0.3R$
Substitute for $R$ :
$Pcos⁡25∘=0.3(196−Psin⁡25∘)P\cos 25^\circ = 0.3(196 - P\sin 25^\circ)$
Solve:
$P(cos⁡25∘+0.3sin⁡25∘)=58.8P(\cos 25^\circ + 0.3\sin 25^\circ) = 58.8$
Hence:
$\approx 56.9$
The required pulling force is approximately 56.9 N.

Tip

Pulling versus pushing

A pull angled upwards reduces the normal reaction, so friction decreases. A push angled downwards increases the normal reaction, so friction increases.

Inclined planes

For a block on a slope, choose axes:

parallel to the plane
perpendicular to the plane

If the plane is inclined at angle $θ\theta$ to the horizontal, the weight $m g$ has components:

Weight on an inclined plane resolved into components parallel and perpendicular to the plane.

down the plane: $mgsin⁡θmg\sin\theta$
into the plane: $mgcos⁡θmg\cos\theta$

Key Idea

Resolve along the slope

On an inclined plane, do not resolve horizontally and vertically unless you have a special reason. Parallel and perpendicular to the plane are usually much cleaner.

Example

Finding the coefficient of friction on a slope

A block of weight 12 N rests on a rough plane inclined at $28∘28^\circ$ to the horizontal. It is on the point of sliding down the plane. Find $μ\mu$ .

Limiting equilibrium on a rough inclined plane with friction acting up the slope.

The component of weight down the plane is $12sin⁡28∘12\sin 28^\circ$ .
The normal reaction is found perpendicular to the plane:
$12\cos 28^\circ$
Since the block is about to slide down, friction acts up the plane and is limiting:
$\mu R$
Resolve parallel to the plane. Friction balances the down-slope component of weight:
$μR=12sin⁡28∘\mu R = 12\sin 28^\circ$
Substitute $12\cos 28^\circ$ :
$μ(12cos⁡28∘)=12sin⁡28∘\mu(12\cos 28^\circ) = 12\sin 28^\circ$
Cancel 12 and solve:
$μ=tan⁡28∘≈0.532\mu = \tan 28^\circ \approx 0.532$
The coefficient of friction is approximately 0.532.

Motion on a rough inclined plane

If the block is moving, use Newton’s second law:

Fresultant=maF_{\text{resultant}} = ma

The friction force still acts against the motion. If the block slides down the plane, friction acts up the plane.

Example

Acceleration down a rough slope

A 3 kg block is released from rest on a rough plane. The plane is inclined at angle $θ\theta$ where $tan⁡θ=0.75\tan\theta = 0.75$ . The coefficient of friction is 0.2. Find the acceleration.

Moving block on a rough slope with acceleration down the plane and friction up the plane.

Use $tan⁡θ=0.75=34\tan\theta = 0.75 = \frac{3}{4}$ . This suggests a 3-4-5 triangle, so:
$sin⁡θ=35,cos⁡θ=45\sin\theta = \frac{3}{5}, \qquad \cos\theta = \frac{4}{5}$
Resolve perpendicular to the plane:
$mg\cos\theta$
Friction acts up the plane because the block moves down the plane:
$\mu R = 0.2mg\cos\theta$
Apply Newton’s second law down the plane:
$mgsin⁡θ−0.2mgcos⁡θ=mamg\sin\theta - 0.2mg\cos\theta = ma$
Cancel $m$ :
$g(\sin\theta - 0.2\cos\theta)$
Substitute the exact trig values:
$9.8\left(\frac{3}{5} - 0.2 \times \frac{4}{5}\right)$
Calculate:
$a = 4.312$
The acceleration is approximately 4.31 m s−2 down the plane.

Common Mistake

Keeping the mass unnecessarily

In many sliding-block problems, the mass cancels. If two blocks have the same $μ\mu$ on the same slope, their accelerations are the same, even if their masses differ.

Deciding whether the block moves

Sometimes a force is removed, and you must decide whether the object starts moving. Compare the force trying to make it move with the maximum possible friction.

For a block on a slope with no applied force:

force down the plane is $mgsin⁡θmg\sin\theta$
maximum friction is $μmgcos⁡θ\mu mg\cos\theta$

If $mgsin⁡θ>μmgcos⁡θmg\sin\theta > \mu mg\cos\theta$ , it slides down.

If $mgsin⁡θ≤μmgcos⁡θmg\sin\theta \le \mu mg\cos\theta$ , it stays at rest.

Example

After an applied force is removed

A brick of weight 10 N is on a rough plane at $30∘30^\circ$ . The coefficient of friction is 0.7. A force that had been holding it is removed. Decide whether the brick moves.

Force comparison for deciding whether a brick remains at rest on a rough slope.

The component of weight down the plane is:
$10sin⁡30∘=510\sin 30^\circ = 5$
The normal reaction is:
$10\cos 30^\circ$
The maximum friction available is:
$μR=0.7×10cos⁡30∘≈6.06\mu R = 0.7 \times 10\cos 30^\circ \approx 6.06$
Compare the two forces: the down-slope component is 5 N, while the maximum friction is about 6.06 N.
Since friction can be large enough to prevent sliding, the brick remains at rest.

Exam technique

In the exam

Draw a force diagram first, including directions for friction and the normal reaction.
Choose axes that match the situation: horizontal/vertical for hanging particles, parallel/perpendicular for inclined planes.
Use $F=μRF=\mu R$ only when the object is moving or in limiting equilibrium.
If asked to “determine whether it moves”, compare the driving force with the maximum possible friction.

Self review

Check yourself

If a force pulls a block upwards at an angle, what happens to the normal reaction?
On a slope, what are the two components of weight parallel and perpendicular to the plane?
When is it valid to write friction as exactly $μR\mu R$ ?

Forces and Friction

What you'll learn

The core idea: forces and modelling

Equilibrium and resolving forces

Two strings supporting a particle

Friction and the coefficient of friction

Inclined planes

Motion on a rough inclined plane

Deciding whether the block moves

Recap questions

You've reached the end

Forces and Friction

What you'll learn

The core idea: forces and modelling

Equilibrium and resolving forces

Two strings supporting a particle

Friction and the coefficient of friction

Inclined planes

Motion on a rough inclined plane

Deciding whether the block moves

Recap questions

You've reached the end

Forces and Friction

What you'll learn

The core idea: forces and modelling

Equilibrium and resolving forces

Two strings supporting a particle

Friction and the coefficient of friction

Inclined planes

Motion on a rough inclined plane

Deciding whether the block moves

Recap questions

You've reached the end

Forces and Friction Revision Guide

Forces and Friction

What you'll learn

The core idea: forces and modelling

Equilibrium and resolving forces

Two strings supporting a particle

Friction and the coefficient of friction

Inclined planes

Motion on a rough inclined plane

Deciding whether the block moves

Recap questions

You've reached the end

Forces and Friction Revision Guide