UK · A-Levels

Forces and Newton's laws

This topic applies vectors to forces and motion. It covers forces as vectors, resultant and equilibrium, resolving forces, Newton's three laws of motion and the equation F = ma, the modelling of weight, normal reaction, tension, thrust and friction, and the analysis of connected particles over pulleys and on strings.

4 sections~11 min reading time3 competencies

T·191919 / 20

Exam profile

AO1 · Apply F = ma and resolve forces to solve motion problemsAO2 · Justify force diagrams, equilibrium and modelling assumptionsAO3 · Model connected systems and motion on inclines

Operators:findshow thatdetermineresolvecalculatemodel

basic level

AS-Level covers forces as vectors, equilibrium, Newton's laws and F = ma in one dimension.

higher level

The full A-Level adds friction, motion on inclined planes, resolving forces at angles and connected particles.

Depth

Text

Contents · 4 sections

Forces and Newton's laws

4 sections8 key takeaways5 formulas8 mistake warnings

Forces, resultants and equilibrium#

~3 min reading timeStandardDfE GCE Mathematics — forcesAQA 7357 (Mechanics)

A particle in equilibrium under three forces

Fig. 1For equilibrium the two tensions must balance the weight, so the horizontal and vertical components each sum to zero.

A force is a vector, with magnitude and direction, and when several forces act on a particle their combined effect is their vector sum, the resultant. Forces are added by the triangle or parallelogram law, or, more usually in calculation, by resolving each into perpendicular components and adding the components. The resultant then has components equal to the sums of the horizontal and vertical components of the individual forces.

A particle is in equilibrium when the resultant force on it is zero, so it remains at rest or moves with constant velocity. In component form this means the horizontal components balance and the vertical components balance separately:

\sum F_x = 0

and

\sum F_y = 0

. These two equations are the standard tool for finding unknown forces in a static situation.

Resolving a force

F

acting at an angle

\theta

to a chosen direction gives components

F\cos\theta

along that direction and

F\sin\theta

perpendicular to it. Choosing sensible directions to resolve along — often horizontal and vertical, or along and perpendicular to a slope — is what makes a problem tractable, and is a decision worth making deliberately.

A clear, labelled force (free-body) diagram is the foundation of every forces problem: draw the particle, mark every force acting on it with its direction, and only then resolve and form equations. Omitting a force, or including one that does not act on the chosen body, is the commonest source of error, so the diagram must be complete and accurate.

Conditions for equilibrium

\text{Equilibrium: } \sum F_x = 0 \ \text{and} \ \sum F_y = 0

The horizontal and vertical components of the forces each balance to zero.

Worked example

Equilibrium of a particle

A box of weight $60\ \text{N}$ rests on a smooth horizontal floor while a horizontal force $P$ and a rope at $30^\circ$ above the horizontal with tension $20\ \text{N}$ act on it, keeping it in equilibrium. Find the normal reaction $N$ .

Resolve vertically
Upward forces: $N$ and the vertical component of the tension $20\sin 30^\circ = 10\ \text{N}$ ; downward: weight $60\ \text{N}$ .
Apply equilibrium
$N + 10 = 60$ .
$N + 20\sin 30^\circ = 60$
Solve
$N = 60 - 10 = 50\ \text{N}$ .

Result: The normal reaction is $N = 50\ \text{N}$ .

Typical mistakes

Omitting a force (such as the normal reaction or weight) from the diagram.
Interchanging $\cos\theta$ and $\sin\theta$ when resolving a force at an angle.

Active revision

A particle of weight $50\ \text{N}$ hangs in equilibrium from two strings making $40^\circ$ and $40^\circ$ with the vertical. Find the tension in each string.

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Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Newton's laws and F = ma#

~3 min reading timeStandardDfE GCE Mathematics — forcesAQA 7357 (Mechanics)

Unbalanced forces on a lift

Fig. 2When the tension exceeds the weight the resultant is upward, so by

F = ma

the body accelerates upward.

Newton's first law states that a body continues at rest or moving with constant velocity unless a resultant force acts on it. So constant velocity (including rest) is equivalent to zero resultant force — this is why the equilibrium conditions of the previous section describe both static and uniformly-moving bodies. A change in motion always signals a resultant force.

Newton's second law states that the resultant force on a body equals its mass times its acceleration,

F = ma

, with the acceleration in the direction of the resultant force. This is the central equation of dynamics: given the forces you find the acceleration, or given the motion you find an unknown force. Applied in each direction separately, it gives

\sum F_x = ma_x

and

\sum F_y = ma_y

Newton's third law states that if body A exerts a force on body B, then B exerts an equal and opposite force on A. These action-reaction pairs act on different bodies, which is why they never cancel on a single body; the law is essential for connected particles, where the tension one particle feels is matched by the pull it exerts back through the string.

The standard forces to model are weight

W = mg

(always vertically down), the normal reaction

N

(perpendicular to a surface), tension in a taut string (along the string, pulling), and thrust in a rod (a push). Applying

F = ma

then means drawing the free-body diagram, resolving, and equating the resultant in each direction to

ma

Newton's second law

F = ma

The resultant force equals mass times acceleration, in the direction of that resultant.

Worked example

Applying F = ma

A block of mass $5\ \text{kg}$ is pulled along a smooth horizontal floor by a horizontal force of $20\ \text{N}$ . Find its acceleration.

Identify the resultant
The floor is smooth, so horizontally the only force is the $20\ \text{N}$ pull; vertically weight and normal balance.
Apply F = ma horizontally
$20 = 5a$ .
$20 = 5a$
Solve
$a = \dfrac{20}{5} = 4\ \text{m s}^{-2}$ .

Result: The acceleration is $4\ \text{m s}^{-2}$ in the direction of the pull.

Typical mistakes

Setting the resultant force to zero when the body is accelerating (it equals $ma$ , not zero).
Treating an action-reaction pair as acting on the same body, so they wrongly cancel.

Active revision

A lift of mass $800\ \text{kg}$ accelerates upwards at $1.5\ \text{m s}^{-2}$ . Taking $g = 9.8\ \text{m s}^{-2}$ , find the tension in the supporting cable.

Practise matching questions8 questions on this topic

Active recall

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Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Friction and the inclined plane#

~3 min reading timeAdvancedDfE GCE Mathematics — forcesAQA 7357 (Mechanics)

Forces on a block on a rough incline

Fig. 3On the incline the weight has a component down the slope; friction acts up the slope, and

R

is perpendicular to it.

Friction is a force that opposes relative sliding between surfaces in contact, acting along the surface in the direction that resists motion. It is modelled by the inequality

F \leq \mu R

, where

R

is the normal reaction and

\mu

the coefficient of friction. While a body is static, friction takes exactly the value needed to maintain equilibrium, up to its maximum; at the point of slipping (limiting equilibrium) it reaches its maximum value

F = \mu R

The distinction between

F \leq \mu R

(not yet slipping) and

F = \mu R

(on the point of slipping or actually sliding) is examined carefully. A body remains still as long as the friction required is within its capacity

\mu R

; if more is demanded than

\mu R

can supply, the body slides and friction stays at

\mu R

while opposing the motion.

On an inclined plane it is almost always best to resolve along and perpendicular to the slope rather than horizontally and vertically. For a plane at angle

\alpha

, the weight

mg

resolves into

mg\sin\alpha

down the slope and

mg\cos\alpha

into the slope, so the normal reaction is

R = mg\cos\alpha

and the component tending to slide the body is

mg\sin\alpha

. This choice of directions keeps the equations simple.

A body on a rough slope stays still if the component down the slope does not exceed the maximum friction, i.e.

mg\sin\alpha \leq \mu\,mg\cos\alpha

, which simplifies to

\tan\alpha \leq \mu

. If it slides, applying

F = ma

along the slope with friction

\mu R

acting up the slope gives the acceleration. Recognising which case applies is the crux of inclined-plane problems.

The friction model

F \leq \mu R, \qquad F = \mu R \text{ (limiting)}

Friction can be anything up to $\mu R$ ; it reaches $\mu R$ at the point of slipping.

On an inclined plane

R = mg\cos\alpha, \qquad \text{slides if } \tan\alpha > \mu

Resolve along and perpendicular to the slope; the body slips once $\tan\alpha$ exceeds $\mu$ .

Worked example

Sliding on an incline

A $2\ \text{kg}$ block is released on a rough plane inclined at $30^\circ$ with $\mu = 0.2$ ( $g = 9.8\ \text{m s}^{-2}$ ). Find its acceleration down the slope.

Normal reaction
Perpendicular to the slope: $R = mg\cos 30^\circ = 2 \times 9.8 \times 0.866 = 16.97\ \text{N}$ .
Friction and driving force
Maximum friction $\mu R = 0.2 \times 16.97 = 3.39\ \text{N}$ (up the slope); weight component down the slope $mg\sin 30^\circ = 9.8\ \text{N}$ .
Apply F = ma along the slope
$9.8 - 3.39 = 2a$ , so $a = \dfrac{6.41}{2} = 3.20\ \text{m s}^{-2}$ .
$a = \frac{9.8 - 3.39}{2} = 3.20$

Result: The block accelerates down the slope at about $3.20\ \text{m s}^{-2}$ .

Typical mistakes

Using $F = \mu R$ when the body is not yet on the point of slipping.
Interchanging $\sin\alpha$ and $\cos\alpha$ when resolving the weight on a slope.

Active revision

A block of mass $4\ \text{kg}$ rests on a rough plane inclined at $25^\circ$ . The coefficient of friction is $0.3$ . Determine whether the block slips, and if it does, find its acceleration.

Practise matching questions8 questions on this topic

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Connected particles#

~3 min reading timeAdvancedDfE GCE Mathematics — forcesAQA 7357 (Mechanics)

Free-body diagram of a hanging particle in a system

Fig. 4For a hanging particle in a connected system, tension acts up and weight down; if

W > T

the particle accelerates downward.

Two particles joined by a light inextensible string move together, so they share the same magnitude of acceleration, and the tension is the same throughout the string (assuming a smooth, light pulley where one is involved). The standard method is to draw a separate free-body diagram for each particle and apply

F = ma

to each, giving simultaneous equations that are solved for the common acceleration and the tension.

For two particles connected over a smooth pulley — one hanging, one on a table, or two hanging on either side — the heavier side drives the motion. Writing

F = ma

for each particle (taking the direction of motion as positive for each) gives two equations; adding them eliminates the tension and yields the acceleration, after which either equation gives the tension.

A key modelling point is that 'light and inextensible' is what makes the two accelerations equal in magnitude and the tension uniform: a light string has no mass to accelerate, and an inextensible one keeps the particles a fixed distance apart. A smooth pulley simply redirects the string without changing the tension. Stating these assumptions is expected.

Care with directions is essential: for each particle, take the direction it actually accelerates as positive, so that a hanging mass falling and a mass on a table moving towards the pulley are both treated with positive acceleration. The tension pulls each particle towards the pulley. Keeping the sign conventions consistent across the two free-body diagrams is where accuracy is won.

Connected particles

\text{For each particle: } F_{\text{net}} = ma, \quad \text{same } a, \text{ same } T

Apply $F = ma$ to each particle; the common acceleration and uniform tension solve the pair of equations.

Worked example

Two masses over a pulley

Masses of $2\ \text{kg}$ and $4\ \text{kg}$ hang over a smooth pulley on a light string ( $g = 9.8\ \text{m s}^{-2}$ ). Find the acceleration and the tension.

Equation for the 4 kg mass (falling)
$4g - T = 4a$ , i.e. $39.2 - T = 4a$ .
Equation for the 2 kg mass (rising)
$T - 2g = 2a$ , i.e. $T - 19.6 = 2a$ .
Add to eliminate T
$39.2 - 19.6 = 6a \Rightarrow a = \dfrac{19.6}{6} = 3.27\ \text{m s}^{-2}$ .
$a = \frac{19.6}{6} = 3.27\ \text{m s}^{-2}$
Find T
$T = 19.6 + 2(3.27) = 26.1\ \text{N}$ .

Result: Acceleration $\approx 3.27\ \text{m s}^{-2}$ and tension $\approx 26.1\ \text{N}$ .

Typical mistakes

Assuming the tension differs on the two sides of a smooth pulley (it is the same).
Inconsistent sign conventions between the two particles' equations.

Active revision

Particles of mass $3\ \text{kg}$ and $5\ \text{kg}$ hang from the ends of a light inextensible string over a smooth pulley. Taking $g = 9.8\ \text{m s}^{-2}$ , find the acceleration of the system and the tension in the string.

Practise matching questions8 questions on this topic

Active recall

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Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

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References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Forces and Newton's laws

4 sections~11 min reading time3 competencies

Forces and Newton's laws

Key points

Equilibrium of a particle

Key points

Applying F = ma

Key points

Sliding on an incline

Key points

Two masses over a pulley

Sources

Forces and Newton's laws

Key points

Equilibrium of a particle

Key points

Applying F = ma

Key points

Sliding on an incline

Key points

Two masses over a pulley

Sources