Newton's Third Law — Action and Reaction

Spec mapping: OCR H556 Module 3.2 — Forces in action (Newton's three laws of motion; identification of action-reaction pairs; common misconceptions). Refer to the official OCR H556 specification document for exact wording.

Newton's Third Law is the most frequently misquoted and misunderstood of the three. Generations of students recite the mantra "every action has an equal and opposite reaction" without ever being taught the crucial qualifiers — that the two forces in a pair act on different bodies and are of the same physical type. Getting this right is a standard A-Level exam question, and OCR examiners will penalise any student who claims that the weight of a book and the normal force on it form a Newton-3 pair (they do not).

This lesson fixes the wording, works through canonical examples (walking, propulsion, gravitational attraction, contact forces), and hunts down the misconceptions that cost marks. By the end you should be able to look at any force and write down its Newton-3 partner with the confidence of a graduate.

Statement of the 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, opposite in direction, and of the same physical type.

In compact symbolic form:

$\vec{F}_{A \to B} = -\vec{F}_{B \to A}$

Three essential features that mark schemes test directly:

Different bodies. The two paired forces act on different objects — never on the same object. This is the single biggest discriminator between Newton-3 (pair on two bodies) and Newton-1 (balanced forces on one body).
Same physical type. If A pulls B gravitationally, B pulls A gravitationally. If A pushes B by contact, B pushes A by contact. A gravitational force is never paired with a contact (normal) force, even if their magnitudes happen to be equal.
Simultaneous and equal-magnitude. Both forces exist at the same instant. There is no delay, no "action then reaction in sequence". Both pop into existence together and disappear together when the interaction ends.

A fourth, often-implicit, feature is that both partners lie along the line joining the two bodies (for radial interactions like gravity, electrostatics, normal contact). For non-radial forces (friction between sliding surfaces, lift on an aerofoil), the Newton-3 partner is still equal-magnitude and opposite-direction, but the geometry can be less obvious.

Diagram — A Correct Newton-3 Pair

flowchart LR
  A["Body A"] -->|"F (A on B)"| B["Body B"]
  B -->|"F' = −F (B on A), same type"| A

Every valid Newton-3 pair has exactly this structure: two forces, one on each body, equal magnitude, opposite direction, same physical mechanism. If any one of those conditions fails, the two forces are not a Newton-3 pair — even if they happen to be equal and opposite for some other reason.

Example 1 — Walking

When you walk you push your foot backwards against the ground (via friction). By Newton 3 the ground pushes your foot forwards by an equal amount. It is that forward force on you, from the ground that propels you forward — Newton 2 then converts this resultant force into your forward acceleration.

$\vec{F}_{\text{foot} \to \text{ground}}$ : friction, directed backward, on the ground.
$\vec{F}_{\text{ground} \to \text{foot}}$ : friction, directed forward, on you.

Same magnitude (Newton 3), opposite direction, both contact friction, acting on different bodies — a textbook Newton-3 pair.

Without friction (a smooth icy lake), the pair shrinks toward zero magnitude. You cannot walk forward not because Newton 3 fails — it doesn't — but because there is no friction to push backward against the ground in the first place, so the reaction force is correspondingly weak. This is why ice-skaters dig the edges of their blades into the ice: it generates a horizontal friction component that, by Newton 3, produces the forward push.

Example 2 — Jet and Rocket Propulsion

A jet engine forces hot exhaust gas backwards at very high speed. By Newton 3 the gas pushes the engine forwards — that forward push on the rocket, from the exhaust gas is what we call thrust.

$\vec{F}_{\text{rocket} \to \text{exhaust}}$ : contact/pressure, backward, on the gas.
$\vec{F}_{\text{exhaust} \to \text{rocket}}$ : contact/pressure, forward, on the rocket.

Crucially, rockets work in vacuum. They do not push against the atmosphere; they push against their own exhaust. A persistent myth — that a rocket "needs air to push against" — is a Newton-3 misunderstanding. The interaction is between the rocket and the gas it ejects, not the rocket and the surrounding medium. This is also why the Space Shuttle's main engines worked just as well above the atmosphere as within it (and indeed better, because no atmospheric back-pressure opposes the exhaust).

Lesson 10 returns to rockets quantitatively using momentum conservation, which is a direct consequence of Newton 3 applied to the rocket-plus-exhaust system.

Example 3 — Gravitational Attraction

Earth pulls an apple of mass $0.10$ kg downward with $W = mg \approx 1$ N. By Newton 3 the apple pulls Earth upward with $1$ N.

$\vec{F}_{\text{Earth} \to \text{apple}}$ : gravitational, $1$ N downward.
$\vec{F}_{\text{apple} \to \text{Earth}}$ : gravitational, $1$ N upward.

Both forces are gravitational (same type), they have equal magnitude, opposite directions, and act on different bodies. A textbook Newton-3 pair.

You do not notice the Earth accelerating toward the apple because Earth's mass is $\sim 6 \times 10^{24}$ kg, giving an acceleration:

$a_{\text{Earth}} = \frac{F}{M_{\text{Earth}}} = \frac{1}{6 \times 10^{24}} \approx 1.7 \times 10^{-25} \text{ m s}^{-2}$

— around $10^{25}$ times smaller than the apple's acceleration. But it is non-zero; in principle the Earth jiggles a tiny bit every time an apple falls. The Moon's effect (much larger mass, much larger gravitational pull) is observable: Earth and Moon both orbit their common centre of mass (the barycentre), which lies about $4700$ km from Earth's centre — inside Earth, but offset from the geometric centre. Earth's wobble around this barycentre is the gravitational Newton-3 partner of the Moon's orbital motion.

Example 4 — Book on a Table: The Classic Trap

A $2.0$ kg book lies at rest on a table. Forces on the book:

Weight $W = mg = 19.6$ N downward (gravitational, from Earth).
Normal contact force $N = 19.6$ N upward (contact, from the table).

Students constantly claim $W$ and $N$ form a Newton-3 pair. They do not. Here is why:

$W$ and $N$ both act on the same body (the book). A Newton-3 pair must act on different bodies.
$W$ is gravitational; $N$ is contact. Different physical types.

$W$ and $N$ happen to be equal-and-opposite — but only because the book is in equilibrium (Newton 1). If the table were on a lift accelerating upward, $N$ would exceed $W$ — and yet $W$ and $N$ would still not be a Newton-3 pair. They are simply two unrelated forces that happen to act on the same body, and whose net effect depends on Newton 1 or Newton 2 according to whether the book is accelerating.

The true Newton-3 partners are:

Force on book	Type	Newton-3 partner	Acts on
Weight $W = 19.6$ N down	Gravitational	Book pulls Earth up with $19.6$ N	Earth
Normal force $N = 19.6$ N up	Contact	Book pushes table down with $19.6$ N	Table

Notice that each partner acts on a different body and is the same type as its partner. This is the canonical exam trap, and OCR mark schemes will penalise the " $W$ and $N$ are a Newton-3 pair" mistake explicitly.

Example 5 — Horse and Cart Paradox

An old paradox: "If the cart pulls the horse back with the same force the horse pulls the cart forward (Newton 3), how can the system ever accelerate? Shouldn't they exactly cancel?"

Resolution: the Newton-3 pair (cart-on-horse, horse-on-cart) acts on different bodies, so the forces never cancel on any single body. They live on separate free-body diagrams:

Free-body diagram of the horse: the road pushes the horse forward (friction on hooves, $F_{\text{road}}$ ), the cart pulls the horse backward via the harness ( $F_{\text{cart on horse}}$ , the Newton-3 partner of the horse-on-cart pull). For the horse to accelerate, $F_{\text{road}} > F_{\text{cart on horse}}$ , and the resultant on the horse is forward.
Free-body diagram of the cart: the horse pulls the cart forward via the harness ( $F_{\text{horse on cart}}$ ), and the cart's wheels feel some friction ( $F_{\text{cart resistance}}$ , much smaller). The resultant on the cart is forward, accelerating it.

Both bodies have forward resultants. Newton 3 is fully consistent with this. The mistake is to apply Newton-1 reasoning (forces cancel ⇒ no motion) to forces on different bodies, which is conceptually wrong.

Example 6 — Two Magnets and Two Charges

Two bar magnets face each other, N-pole to S-pole. The first attracts the second with force $\vec{F}$ ; the second attracts the first with $-\vec{F}$ . Both electromagnetic, equal magnitude, opposite direction — textbook Newton-3 pair.

Replace the magnets with two charged balloons rubbed against wool. They repel each other electrostatically with equal-and-opposite forces — another textbook Newton-3 pair, this time repulsive.

The "type" condition still holds: both forces are electromagnetic. What changes is the sign — attractive vs repulsive — but the Newton-3 structure is identical.

Worked Example 1 — Contact Forces Between Two Blocks

Two blocks A ( $3.0$ kg) and B ( $2.0$ kg) sit on a frictionless surface, touching. You push block A from the left with a force $F = 15$ N so both accelerate together.

Whole system: $a = F/(m_A + m_B) = 15/5.0 = 3.0$ m s $^{-2}$ .

Contact force between the blocks ( $C$ ):

On B (Newton 2): $C = m_B a = 2.0 \times 3.0 = 6.0$ N forward.

By Newton 3, B pushes A backward with $6.0$ N.

Cross-check with Newton 2 on A: resultant on A = $F - C = 15 - 6 = 9.0$ N forward; $a_A = 9.0/3.0 = 3.0$ m s $^{-2}$ . ✓ Consistent with the system result.

The Newton-3 pair here is:

$\vec{F}_{A \to B}$ : contact, $6.0$ N forward, on block B.
$\vec{F}_{B \to A}$ : contact, $6.0$ N backward, on block A.

Same magnitude, opposite direction, both contact forces, different bodies. Note that the $15$ N push on A and the $6$ N back-reaction on A are not a Newton-3 pair: they act on the same body (A), and (probably) come from different sources.

Worked Example 2 — Identifying Pairs in a Traction Problem

A $1000$ kg car tows a $500$ kg trailer along a horizontal road. The engine provides a driving force of $3000$ N via the tyres. Total resistance is $900$ N on the car and $300$ N on the trailer.

(a) Find the acceleration of the system. (b) Find the tension in the tow-bar. (c) Identify the Newton-3 partner of the tow-bar tension on the trailer.

(a) Treat car+trailer as one system: net force = $3000 - 900 - 300 = 1800$ N; total mass = $1500$ kg; $a = 1800/1500 = 1.2$ m s $^{-2}$ forward.

(b) Trailer alone (Newton 2 forward): $T - 300 = 500 \times 1.2 = 600$ ; $T = 900$ N forward (the tow-bar pulls the trailer forward).

(c) The Newton-3 partner is the tow-bar pulling on the car backward with $900$ N. Same magnitude, opposite direction, contact (tension) force, different body. Cross-check with car alone: net forward = $3000 - 900 - 900 = 1200$ N; $a_{\text{car}} = 1200/1000 = 1.2$ m s $^{-2}$ ✓.

Worked Example 3 — Swimmer and Water

A swimmer of mass $60$ kg accelerates from rest by pushing water backward. In one stroke she pushes $4.0$ kg of water backward at $1.2$ m s $^{-1}$ relative to the ground (initially the water was at rest). Find the magnitude of the average force she exerts on the water if the stroke lasts $0.50$ s, and her resulting forward velocity.

Force on the water (Newton 2 applied to the water): $F = \Delta p_{\text{water}}/\Delta t = (4.0 \times 1.2)/0.50 = 9.6$ N backward.

Newton-3 partner — force on the swimmer: $9.6$ N forward (same magnitude, opposite direction, both contact, different bodies).

Newton 2 applied to the swimmer over $0.5$ s (ignoring drag): $\Delta v_{\text{swimmer}} = F\,\Delta t/m = 9.6 \times 0.5/60 = 0.08$ m s $^{-1}$ . So she gains $8$ cm s $^{-1}$ per stroke. Many strokes needed for proper swimming pace — explaining why competitive swimming demands a high stroke rate combined with maximum momentum transferred per stroke.

Note the system (swimmer + water) starts at rest and ends with the swimmer moving forward and the water moving backward — total momentum stays zero (conservation), but each body's momentum has changed by equal and opposite amounts. This is a preview of lesson 6.

Always Has a Partner?

Yes — always. For every real force you can identify on a body, there is a Newton-3 partner force on whichever body is producing it. If you cannot identify the partner, you have mis-identified the original force, or the original "force" is a pseudo-force (centrifugal, Coriolis) introduced because you are working in an accelerating frame.

Quick check procedure

Given a force $\vec{F}$ on a body $Y$ that you suspect has a Newton-3 partner:

Write the original as " $X$ exerts force $\vec{F}$ on $Y$ " — identify $X$ .
Write the partner as " $Y$ exerts force $-\vec{F}$ on $X$ ".
Check: do the two forces act on different bodies ( $X$ and $Y$ , not $Y$ twice)?
Check: are they the same type (both gravitational, or both contact, or both electromagnetic, etc.)?
Check: are they equal in magnitude, opposite in direction?

If all four checks pass, you have identified a genuine Newton-3 pair. If any fails — typically check 3 (different bodies) — then the two forces in question are not a Newton-3 pair, even if they happen to be equal and opposite for some other reason.

Newton 3 and Conservation of Momentum

Newton 3 is the reason that linear momentum is conserved in any isolated (closed) system. Sketch of the argument:

Two bodies A and B interact internally over time $\Delta t$ with average forces $\vec{F}_{A \to B}$ and $\vec{F}_{B \to A} = -\vec{F}_{A \to B}$ .

Impulse on A: $\vec{J}_A = \vec{F}_{B \to A} \Delta t = \Delta \vec{p}_A$ .
Impulse on B: $\vec{J}_B = \vec{F}_{A \to B} \Delta t = \Delta \vec{p}_B$ .

By Newton 3, $\vec{J}_A = -\vec{J}_B$ , so:

$\Delta \vec{p}_A + \Delta \vec{p}_B = 0 \quad \Rightarrow \quad \Delta \vec{p}_{\text{total}} = 0$

— total momentum is unchanged by the interaction. Newton 3 gives conservation of momentum directly, and this is the conceptual foundation of lessons 6-10. We will revisit the derivation in lesson 6 with explicit collision examples.

Newton's Third Law — Action and Reaction

Newton's Third Law — Action and Reaction

Statement of the Third Law

Diagram — A Correct Newton-3 Pair

Example 1 — Walking

Example 2 — Jet and Rocket Propulsion

Example 3 — Gravitational Attraction

Example 4 — Book on a Table: The Classic Trap

Example 5 — Horse and Cart Paradox

Example 6 — Two Magnets and Two Charges

Worked Example 1 — Contact Forces Between Two Blocks

Worked Example 2 — Identifying Pairs in a Traction Problem

Worked Example 3 — Swimmer and Water

Always Has a Partner?

Quick check procedure

Newton 3 and Conservation of Momentum

Synoptic Links

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