Newton's Third Law

AP Physics C: Mechanics· 45 min read

1. Core Definition of Newton's Third Law★★☆☆☆⏱ 10 min

All forces in classical mechanics arise from interactions between two objects. Newton's Third Law formalizes this relationship, often misstated as "for every action, there is an equal and opposite reaction" — this phrasing is ambiguous without clear context about what the forces act on.

📘 Definition

Newton's Third Law of Motion

\vec{F}_{AB} = -\vec{F}_{BA}

If object A exerts a force of a given type on object B, then object B exerts a force of the same type, equal magnitude, and opposite direction on object A. The two forces always act on different objects.

Example:

A book resting on a table: the book exerts a downward normal force on the table, and the table exerts an equal-magnitude upward normal force on the book.

📐 Worked Example

A 0.5 kg apple falls from a tree toward Earth. What is the magnitude of the gravitational force Earth exerts on the apple, and what is the magnitude of the gravitational force the apple exerts on Earth?

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1. Calculate the gravitational force Earth exerts on the apple using the definition of weight:
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$F_{g,Earth\ on\ apple} = mg = (0.5\ \text{kg})(9.8\ \text{m/s}^2) = 4.9\ \text{N}$
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1. By Newton's Third Law, the force the apple exerts on Earth is equal in magnitude:
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$|F_{g,apple\ on\ Earth}| = |F_{g,Earth\ on\ apple}| = 4.9\ \text{N}$
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1. Even with equal force magnitude, Earth's acceleration is negligible due to its large mass:
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$a_{Earth} = \frac{F}{M_{Earth}} \approx 8 \times 10^{-25}\ \text{m/s}^2$

2. Action-Reaction Pairs vs Balanced Forces★★★☆☆⏱ 15 min

The most frequently tested distinction on the AP exam is between action-reaction pairs (Newton's Third Law) and balanced forces (Newton's First Law for equilibrium). The key difference is what object each force acts on.

Action-reaction pairs: Always act on two different objects, and are always the same type of force.
Balanced forces: Always act on the same object, and can be different types of forces.

📐 Worked Example

A box is pulled at constant velocity along a rough horizontal floor by a rope. Identify the Newton's Third Law pair for the tension force exerted by the rope on the box.

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1. Name the original force properly: $T_{r o p e o n b o x}$ = tension exerted by the rope on the box.
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1. By Newton's Third Law, the pair must be the same force type, exerted by the box on the rope.
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1. Eliminate the common wrong answer: friction on the box acts on the same object (the box) as tension, so it balances tension but is not a Third Law pair.
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1. The correct action-reaction pair is the tension force exerted by the box on the rope, equal magnitude and opposite direction to $T_{r o p e o n b o x}$ .

3. Application to Multi-Object Dynamics★★★☆☆⏱ 20 min

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When solving dynamics problems for systems of multiple connected objects, Newton's Third Law lets you relate interaction forces between objects. You can calculate acceleration for the entire system first, then isolate individual objects to find internal interaction forces.

📐 Worked Example

A 2 kg block A sits on a frictionless horizontal surface, connected by a massless rope to a 1 kg block B hanging vertically over a massless, frictionless pulley. Find the tension force that block A exerts on the rope.

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1. Treat both blocks as a single system to find overall acceleration. Net force equals the weight of hanging block B:
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$a = \frac{F_{net}}{m_{total}} = \frac{m_B g}{m_A + m_B} = \frac{(1\ \text{kg})(9.8\ \text{m/s}^2)}{3\ \text{kg}} = 3.27\ \text{m/s}^2$
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1. Isolate block B and apply Newton's Second Law to find tension the rope exerts on B:
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$m_B g - T = m_B a \implies T = m_B(g-a) = 1(9.8 - 3.27) = 6.53\ \text{N}$
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1. By Newton's Third Law, the tension the rope exerts on B equals the tension B exerts on the rope, which equals the tension A exerts on the rope for a massless rope.
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Final answer: The tension block A exerts on the rope is $6.53 N$ .

4. Common Pitfalls

Wrong move:

Claiming gravitational force pulling a book down and normal force pushing the book up are an action-reaction pair.

Why:

Both forces act on the same object (the book) and are different types of force.

Correct move:

The pairs are (1) Earth gravity on book / book gravity on Earth, and (2) book normal on table / table normal on book.

Wrong move:

Assuming action-reaction forces always cancel out when calculating net force.

Why:

The two forces act on different objects, so you only ever include one force when calculating net force for a single object.

Correct move:

Only internal forces between objects inside a system cancel out for the whole system's net force; you never include both forces in a single object's net force calculation.

Wrong move:

Claiming a larger truck exerts a larger force on a small car during a collision.

Why:

Newton's Third Law requires equal magnitude force regardless of mass or speed.

Correct move:

The force each exerts on the other is equal magnitude; the smaller car has a larger acceleration due to its smaller mass, per Newton's Second Law.

Wrong move:

Forgetting that your gravitational force on Earth equals the gravitational force Earth exerts on you.

Why:

Confuses force magnitude with acceleration; people incorrectly assume Earth's larger mass gives it a larger force.

Correct move:

The magnitude of the gravitational force you exert on Earth is exactly equal to your weight, matching the force Earth exerts on you.

5. Quick Reference Cheatsheet

Property	Action-Reaction Pairs	Balanced Forces
Acts on	Two different objects	Same object
Force type	Always the same type	Can be different types
Follows	Newton's Third Law	Newton's First Law (equilibrium)
Cancel for system net force	Yes (if both in system)	Always cancel for the object

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Force pair identification
2022 · FRQ
Atwood machine tension pairs
2021 · MCQ
Balanced force vs pair distinction

What's Next

Newton's Third Law is the foundation for all dynamics problems in AP Physics C Mechanics, from multi-object pulley systems to collisions, circular motion, and systems of particles. Correctly identifying force pairs is required to draw accurate free-body diagrams, which are the starting point for almost every FRQ and MCQ on the exam. Mastering the distinction between pairs and balanced forces will eliminate many common points of lost points on test day. Build on this knowledge with the following topics to continue your unit progress.