Resultant forces (incl. resolving HT) — AQA GCSE Physics

SPEC 4.5.1.4 • FORCES — RESULTANT FORCES

Resultant Forces

How to combine multiple forces into one. The foundation for everything in Newton's laws — and the lesson with the most examined HT content in Module 5.

Recap — Forces are Vectors

From lesson 5.01: a force is a vector. That means every force has both:

A size (magnitude), in newtons (N)

A direction (which way it acts)

Why this matters here

Most objects have multiple forces on them at once: a car has thrust and friction, a parachutist has weight and air resistance, a box on a slope has weight and reaction force. To work out what happens, we need to combine all those vectors into a single force with the same effect — the resultant force.

Resultant Force — Definition

Resultant Force

The single force that has the same effect as all the original forces acting on an object combined.

It's the "net force" — what's left after all the forces are added together as vectors.

Vector addition rule

To combine forces:

Same direction → add the magnitudes

Opposite direction → subtract the magnitudes (resultant points the direction of the larger force)

At an angle → use a vector diagram (scale drawing) or Pythagoras for right-angled cases

Adding Forces in a Straight Line

The Foundation core skill. Two forces on the same line — work out the resultant by inspection.

The two rules

Same direction: add → e.g. 5 N forward + 3 N forward = 8 N forward

Opposite directions: subtract → e.g. 5 N forward − 3 N back = 2 N forward

Worked Example — Resultant in a Straight Line

A box is pushed across a floor. The pushing force is 30 N to the right. The friction force is 12 N to the left.

State the magnitude and direction of the resultant force. 2 marks

1

Identify directions: the two forces act in opposite directions → subtract.

2

Subtract: 30 − 12 = 18

3

Direction: The pushing force (30 N) is larger than friction (12 N), so the resultant acts in the direction of the push: to the right.

A

Answer: Resultant force = 18 N to the right.

Always state direction

Resultant force is a vector. The mark scheme requires both magnitude AND direction in the answer. "18 N" alone may not score full marks — write "18 N to the right" or "18 N in the direction of the push".

Equilibrium — Resultant Force = Zero

When the forces on an object are balanced, the resultant force is zero. The object is then in equilibrium.

Equilibrium

A state where the resultant force on an object is zero.

Two cases — and most students forget one

When resultant force = 0, the object can be either:

Stationary (at rest, not moving), OR

Moving at constant velocity (constant speed in a straight line)

In both cases, the object's velocity is not changing — so the acceleration is zero. This is Newton's First Law (formally covered in lesson 5.10) — a foundational forward-link.

Real-world example

A car cruising at constant speed on a motorway has thrust = drag (balanced forces, resultant = 0). It's not stationary, but it IS in equilibrium. A parachutist at terminal velocity is the same — falling at constant velocity because weight = air resistance.

Examiner Alert — Equilibrium = No Acceleration

Jun 2024 P2 Q02.1 — the most recent test of this content

When asked "the child is standing still — what is the resultant vertical force? Give a reason", the answer is 0 N, and the reason must link the zero resultant to no acceleration or balanced forces.

The Jun 2024 mark scheme accepts any one of these three reasons (the second mark depends on stating "0 N" first):

"The object isn't accelerating" — credited

"The upwards forces are equal to the downwards forces" — credited

"The forces are balanced" — credited

The mark-scheme chain

Equilibrium questions reward the full chain: balanced forces → resultant = 0 → no acceleration → constant velocity (or stationary). State "resultant = 0" plus any one of "balanced" / "not accelerating" / "constant velocity" and you score every mark available.

Stationary vs constant velocity — both count

Don't forget: when resultant = 0, the object can be moving at constant velocity, not just stationary. Use "either stationary or moving at constant velocity" to cover both cases.

Higher Tier Only — From Here On

The next 6 slides cover content that only appears on Higher Tier papers: free body diagrams, adding forces at right angles, and resolving forces into components.

Foundation students can skip ahead to the practice question (slide 15), but understanding free body diagrams will help with all of Module 5.

Free Body Diagrams

A free body diagram shows all the forces acting on a single object as labelled arrows. The object is often drawn as a simple shape (a dot, a box) with forces emerging from its centre.

Rules for free body diagrams

Draw only the object being analysed (often as a dot or box)

Draw all forces as labelled arrows from the object, with length proportional to magnitude

Show the direction of each force clearly with an arrowhead

Adding Forces at Right Angles

When two forces act at 90° to each other, they can't simply be added or subtracted. Use one of these methods:

Method 1 — Scale Drawing

Draw both arrows tip-to-tail to scale. The resultant is the arrow from the start of the first to the tip of the second. Measure length (with ruler → scale → newtons) and angle (with protractor).

The spec method.

Method 2 — Pythagoras (right angles only)

For perpendicular forces:

resultant² = F₁² + F₂²

Then measure the angle from the scale drawing.

Faster for clean numbers.

Mark scheme requires the diagram

Real mark schemes (e.g. Jun 2021 P2 Q07.5) state: " to gain full marks a vector diagram must have been drawn ". Pythagoras can give a calculated value that the mark scheme will accept as an alternative for the magnitude, but the scale drawing itself must exist on the page. Use Pythagoras as a check or quick calculation, but always draw the diagram. The spec's "scale drawings only" requirement is enforced literally.

Worked Example — Right-Angled Resultant

A boat experiences 3 N due east from the engine and 4 N due north from a current. Determine the magnitude and direction of the resultant force. 3 marks

1

Draw tip-to-tail at right angles, to scale (1 cm = 1 N).

2

Pythagoras for magnitude: R² = 3² + 4² = 9 + 16 = 25 → R = 5 N

3

Measure angle on scale drawing with a protractor: θ ≈ 53° from east, toward north (or "N53°E")

A

Answer: Resultant = 5 N at 53° north of east (magnitude AND direction).

Resolving a Force — The Reverse

Just as you can add two forces to get a single resultant, you can do the reverse: take one force and split it into two perpendicular components that have the same combined effect.

Why this matters

Real situations have angled forces — a rope pulling at an angle, a slide tilted on a slope, gravity acting on a ball rolling down a ramp. To analyse the motion, we usually need to split the force into:

A component parallel to the direction of motion (drives the motion)

A component perpendicular to the direction of motion (presses against the surface)

Method (per spec)

Scale drawings only. Draw the original force as an arrow to scale; from the same starting point, draw the two perpendicular axes (e.g. along and across the slope). Project the arrow onto each axis — the projections are the two components. Measure each with a ruler and convert back to newtons.

Example — Block on a Slope

A block of weight W sits on a slope. To analyse why it slides (or doesn't), resolve weight into two components along and perpendicular to the slope.

What each component does

W∥ (parallel to slope): tries to slide the block down the slope

W⊥ (perpendicular to slope): presses the block into the slope surface — friction depends on this

Examiner Alert — Component on a Slope

"Gravity increased causing acceleration" — Jun 2022 + Jun 2024

When a question shows an object on a slope and asks why it accelerates down, students reliably write "gravity increases on the slope" or "the steeper the slope, the bigger the gravity". Wrong.

Gravity (and so weight) is the same regardless of orientation. What changes with slope angle is the component of weight acting along the slope. Steeper slope → larger parallel component → larger accelerating force.

Mark-scheme phrasing

For a slope question, write: "The component of weight along the slope increases as the slope angle increases, producing a larger resultant force and so a larger acceleration." This wording is documented as the credited answer pattern (Jun 2024 P2 Q5.5).

Practice Question — Foundation

A skydiver in free fall has a weight of 700 N acting downward. The air resistance acting upward is 250 N.

Calculate the resultant force on the skydiver. State the direction. 2 marks

1

Directions: Weight acts downward; air resistance acts upward. Opposite directions → subtract.

2

Calculate: 700 − 250 = 450

3

Direction: Weight is larger, so the resultant acts in the direction of the weight: downward.

A

Answer: Resultant force = 450 N downward.

Practice Question — HT HT

A kite is pulled by a wind force of 6 N horizontally and lifted by a vertical force of 8 N from buoyancy.

Determine the magnitude and direction of the resultant force on the kite. 3 marks

1

Draw tip-to-tail: 6 N horizontal arrow, then 8 N vertical arrow at right angles. Choose scale (e.g. 1 cm = 1 N).

2

Pythagoras for magnitude: R² = 6² + 8² = 36 + 64 = 100 → R = 10 N

3

Measure angle on scale drawing with protractor: angle from horizontal ≈ 53° above horizontal.

A

Answer: Resultant = 10 N at 53° above the horizontal.

Mark-scheme requirement

For HT vector questions, the mark scheme requires both magnitude AND direction. An answer of "10 N" alone is incomplete. A direction is one of: an angle relative to a named reference (e.g. "above horizontal"), a compass bearing, or a stated direction (e.g. "north-east").

Weak vs Strong Answers

Q: A car is moving at constant velocity along a straight road. Describe the forces acting on it. (3 marks)

✗ Weak

"There is a thrust force and a friction force. The thrust pushes the car forward."

Names two forces but doesn't link to constant velocity. No mention of balance or resultant. 1/3 marks at best.

✓ Strong

"The forces on the car are balanced — the thrust (forward) is equal in size to the friction and air resistance (backward). The resultant force is zero, so the car is not accelerating — its velocity stays the same, which is why it moves at constant velocity."

Names the forces, states they're balanced, identifies resultant = 0, links to no acceleration and constant velocity. Full marks.

The mark-scheme chain

Equilibrium questions always reward this chain: balanced forces → resultant = 0 → no acceleration → constant velocity (or stationary). Hit all four links and you score every mark available.

Summary & Checkpoint

Foundation core

Resultant force: the single force with the same effect as all the forces combined

In a straight line: add (same direction) or subtract (opposite direction)

Equilibrium = resultant 0 : object is either stationary or moving at constant velocity

Always state direction with magnitude — force is a vector

HT only

Free body diagram: all forces on a single object, as labelled arrows from the object

Right-angled forces: tip-to-tail vector diagram; Pythagoras for magnitude, protractor for angle

Resolving: one force → two perpendicular components (parallel + perpendicular to surface)

Slopes: the component of weight along the slope is what increases with angle — not gravity itself

Spec method: scale drawings only (no trigonometry required)

The two exam-winning sentences

For equilibrium: "The resultant force is zero, so the object is either stationary or moving at a constant velocity."

For slopes: "The component of weight along the slope increases with the slope angle."