Newton's Third Law

Lesson ~12 min read 8 MCQs

In simple terms: In simple terms, Newton's Third Law is about how forces always come in equal and opposite pairs. When you push on something, it pushes back on you with the exact same amount of force.

Why this matters

But here's a question that puzzles a lot of physics students: if you push on the car, and the car pushes back on you with an equal force, why does the car move at all? Shouldn't the forces cancel out, resulting in a stalemate?

This is the classic riddle of Newton's Third Law. In this lesson, we'll unravel this puzzle. We'll see how to identify these force pairs, understand why they don't cancel each other out, and learn how this simple rule governs everything from a rocket launch to the tension in a rope.

Diagram

This diagram illustrates Newton's Third Law using the example of a person pushing a crate. The diagram is split into three parts. On the left is a pictorial representation of Person A pushing Crate B. In the middle is a free-body diagram for the crate, showing the person's push force to the right, and the normal and weight forces. On the right is a free-body diagram for the person, showing the crate's push-back force to the left, along with friction, normal, and weight forces. The action-reaction forces between the person and crate are highlighted in red and shown to be equal in length but opposite in direction, reinforcing the principle that they act on different objects.

Concept map

flowchart TD
    A[Start: Two objects, A and B, are interacting.] --> B{Are you analyzing the motion of Object B?};
    B -->|Yes| C[Draw Free-Body Diagram for B ONLY];
    C --> D[Identify all forces acting ON B];
    D --> E{Is F_A on B one of these forces?};
    E -->|Yes| F[Include F_A on B in ΣF_B = m_B * a_B];
    F --> G[Solve for the motion of B];
    B -->|No, I'm analyzing A| H[Draw Free-Body Diagram for A ONLY];
    H --> I[Identify all forces acting ON A];
    I --> J{Is F_B on A one of these forces?};
    J -->|Yes| K[Include F_B on A in ΣF_A = m_A * a_A];
    K --> L[Solve for the motion of A];
    subgraph Legend
        M[Note: F_A on B and F_B on A are a Newton's 3rd Law pair. They are equal and opposite, but they NEVER appear on the same FBD.]
    end

Core explanation

Alright, let's dive into one of the most famous—and most misunderstood—laws in all of physics. You've probably heard the sound bite: "For every action, there is an equal and opposite reaction." That's Newton's Third Law in a nutshell. But what does it really mean?

Let's be more precise. A better way to say it is: If object A exerts a force on object B, then object B simultaneously exerts a force on object A that is equal in magnitude and opposite in direction.

Mathematically, we write this as: F⃗_A on B = -F⃗_B on A

The little arrows mean these are vectors, which have both magnitude (size) and direction. The minus sign tells us the direction is perfectly opposite. The subscripts are your best friends here—they tell you who is doing the pushing/pulling and who is being pushed/pulled. F⃗_A on B is the force exerted by A on B.

The "Why Does Anything Move?" Problem

Here is the secret, the single most important takeaway for this topic: Action-reaction pairs act on different objects.

They can never cancel each other out because you only sum up the forces acting on a single object to find its net force and acceleration (that's Newton's Second Law, ΣF = ma).

Let's look at the example of you (Person A) pushing a crate (Object B) across a floor.

You push the crate to the right. That's F⃗_A on B.
The crate pushes you to the left. That's F⃗_B on A.

These two forces are an action-reaction pair. They are equal in magnitude and opposite in direction.

To see if the crate moves, we only look at the forces on the crate. This includes your push, F⃗_A on B, and maybe a friction force from the floor. To see if you move, we only look at the forces on you. This includes the crate's push back, F⃗_B on A, and the crucial force of static friction between your shoes and the floor.

The crate moves because the force you apply to it is greater than the friction force on it. You don't go flying backward because the friction force from the floor on your shoes pushes you forward, counteracting the push from the crate. The system moves because of an external force—the friction from the floor on you!

Systems and Internal Forces

When we group multiple interacting objects together, we call it a "system." In our example, the "you + crate" system is interacting. The forces F⃗_A on B and F⃗_B on A are internal forces within this system.

Here's a key idea from the College Board (EK 2.3.A.2): Internal forces do not change the motion of the system's center of mass. Because they come in equal and opposite pairs, they all cancel out when you look at the system as a whole. To make the whole system accelerate, you need an external force, like the friction from the floor pushing on your feet.

Think of trying to move a car by sitting inside it and pushing on the dashboard. It's impossible! Your push on the dash is an internal force, and the dash pushes back on you. Nothing happens. You need to get out and push on the ground (an external interaction) to move the car.

Tension: A Chain of Third-Law Pairs

Now let's talk about ropes, strings, and cables. When you pull on a rope, you create tension. What is tension? It's the force transmitted through the rope.

Imagine a rope is made of a billion tiny segments, all linked together. When you pull the first segment, it pulls the second, which pulls the third, and so on. Each of these tiny pulls is an action-reaction pair.

Segment 1 pulls Segment 2.
Segment 2 pulls Segment 1 back.

Tension is the macroscopic effect of all these tiny internal forces.

For most AP Physics 1 problems, we use a few simplifying assumptions:

Ideal String
It has negligible mass and doesn't stretch. This is a huge help because it means the tension is the same at every point in the string. If it's 10 N at one end, it's 10 N in the middle and 10 N at the other end.
Ideal Pulley
It has negligible mass and no friction in its axle. This means a pulley's only job is to redirect the force. It doesn't "use up" any of the tension. The tension on one side of an ideal pulley is the same as the tension on the other side.

What if a string isn't ideal and has mass? Think about a heavy anchor chain being lifted by a crane. The top link of the chain has to support the weight of all the links below it. The bottom link only has to support its own weight. So, in a massive chain, the tension is greatest at the top and least at the bottom. You won't see this as often, but it's important to know the "ideal" model has its limits.

By mastering how to identify force pairs and remembering they act on different objects, you'll have the key to solving a huge range of dynamics problems.

Worked examples

Example 1

Pushing a Crate

Priya, who has a mass of 60 kg, pushes a 20 kg crate on a frictionless surface. She pushes the crate with a horizontal force of 40 N.

A) What is the magnitude of the force the crate exerts on Priya? B) What is the acceleration of the crate?

Solution:

Part A: Find the force from the crate on Priya. This is a direct application of Newton's Third Law. The law states that if Priya exerts a force on the crate (F⃗_P on C), the crate exerts an equal and opposite force on Priya (F⃗_C on P).

F⃗_C on P = -F⃗_P on C

We are given that the magnitude of Priya's push is 40 N. Therefore, the magnitude of the force the crate exerts back on Priya is also 40 N. The direction is opposite, but the magnitude is the same.

This is a classic "gotcha" question. Don't overthink it! The Third Law guarantees the forces are equal in magnitude.

Part B: Find the acceleration of the crate. To find the crate's acceleration, we use Newton's Second Law (ΣF = ma) and we only consider the forces acting on the crate.

1
Identify forces on the crate
- F⃗_P on C: Priya's push (40 N to the right).
- w⃗_C: Weight of the crate (down).
- N⃗_C: Normal force from the floor (up).
- The surface is frictionless, so f = 0.
2
Apply Newton's Second Law
We only care about the horizontal motion. The only horizontal force on the crate is Priya's push. ΣF_x = m_crate * a_x F_P on C = m_crate * a_crate
3
Solve for acceleration
40 N = (20 kg) * a_crate a_crate = 40 N / 20 kg a_crate = 2 m/s²

The acceleration of the crate is 2 m/s² to the right. Notice we used the crate's mass (20 kg), not Priya's mass or the combined mass. We isolated the object we cared about.

Example 2

A Book on a Table

A 1.5 kg textbook is resting on a table.

A) Identify the action-reaction pair for the gravitational force on the book. B) Identify the action-reaction pair for the normal force the table exerts on the book.

Solution:

Part A: Gravitational Force Pair

1
Action
The Earth exerts a downward gravitational force on the book (F⃗_Earth on Book). This is the book's weight.
2
Reaction
The book must exert an equal and opposite force on the Earth. So, the reaction is the upward gravitational force the book exerts on the planet Earth (F⃗_Book on Earth).

Yes, you read that right. The book is pulling the entire Earth up with a force equal to its own weight! This force is tiny compared to the Earth's mass, so the Earth's resulting acceleration is immeasurably small, but the force is there.

Part B: Normal Force Pair

1
Action
The table exerts an upward contact force on the book to support it. This is the normal force, F⃗_Table on Book.
2
Reaction
The book must exert an equal and opposite force on the table. So, the reaction is the downward contact force the book exerts on the table, F⃗_Book on Table.

Try it yourself

Problem 1

Carlos (80 kg) and Maya (60 kg) are on a frozen, frictionless pond. They stand face-to-face and push off each other. Carlos experiences an acceleration of 1.2 m/s². What is the magnitude of Maya's acceleration?

Problem 2

A 5 kg box is being pulled upward by a rope. The box is accelerating upward at 2 m/s². What is the tension in the rope? (Use g ≈ 10 m/s²).

TL;DR

In simple terms, Newton's Third Law is about how forces always come in equal and opposite pairs. When you push on something, it pushes back on you with the exact same amount of force.

Key terms

Newton's Third Law AP Physics 1 Action-reaction pairs Force pairs Tension Ideal string Ideal pulley Free-body diagram Internal forces Shrutam US

You can now…

2.3.A: Describe the interaction of two objects using Newton's third law and a representation of paired forces exerted on each object.

Essential knowledge (exam-tested)

2.3.A.1: Newton's third law describes the interaction of two objects in terms of the paired forces that each exerts on the other. F⃗_A on B = -F⃗_B on A
2.3.A.2: Interactions between objects within a system (internal forces) do not influence the motion of a system's center of mass.
2.3.A.3: Tension is the macroscopic net result of forces that segments of a string, cable, chain, or similar system exert on each other in response to an external force.
2.3.A.3.i: An ideal string has negligible mass and does not stretch when under tension.
2.3.A.3.ii: The tension in an ideal string is the same at all points within the string.
2.3.A.3.iii: In a string with nonnegligible mass, tension may not be the same at all points within the string.
2.3.A.3.iv: An ideal pulley is a pulley that has negligible mass and rotates about an axle through its center of mass with negligible friction.

Concept map

flowchart TD
    A[Start: Two objects, A and B, are interacting.] --> B{Are you analyzing the motion of Object B?};
    B -->|Yes| C[Draw Free-Body Diagram for B ONLY];
    C --> D[Identify all forces acting ON B];
    D --> E{Is F_A on B one of these forces?};
    E -->|Yes| F[Include F_A on B in ΣF_B = m_B * a_B];
    F --> G[Solve for the motion of B];
    B -->|No, I'm analyzing A| H[Draw Free-Body Diagram for A ONLY];
    H --> I[Identify all forces acting ON A];
    I --> J{Is F_B on A one of these forces?};
    J -->|Yes| K[Include F_B on A in ΣF_A = m_A * a_A];
    K --> L[Solve for the motion of A];
    subgraph Legend
        M[Note: F_A on B and F_B on A are a Newton's 3rd Law pair. They are equal and opposite, but they NEVER appear on the same FBD.]
    end

Read what Saavi narrates

Have you ever tried to push a friend's stalled car? You plant your feet, lean in, and push with all your might. You can feel the car pushing back against your hands, and you can feel the ground pushing on your shoes.

But here's a question... if you push on the car, and the car pushes back on you with an equal force, why does the car move at all? Shouldn't the forces just cancel out?

This is the classic puzzle of Newton's Third Law. And the solution is surprisingly simple. The law tells us that forces always come in pairs. When you push a car, the car pushes you back. These forces are equal in size and opposite in direction. We call them an action-reaction pair.

The secret is that these two forces act on *different objects*. Your push acts on the car. The car's push acts on you. Since they don't act on the same thing, they can't cancel each other out. The car moves because your push is the main horizontal force on *it*. You don't fly backward because the ground pushes on your feet, keeping you in place.

Let's walk through a quick example. Imagine Priya pushes a 20 kilogram crate with a force of 40 Newtons. What's the force the crate exerts back on Priya? Well, Newton's Third Law says it has to be equal and opposite. So the crate pushes back on Priya with exactly 40 Newtons. Easy, right?

Now, what's the crate's acceleration? For that, we look only at the forces on the crate. The only horizontal force is Priya's 40 Newton push. We use Newton's Second Law, F equals m a. So, 40 Newtons equals 20 kilograms times 'a'. Solving for 'a', we get an acceleration of 2 meters per second squared.

One of the most common mistakes I see is confusing these action-reaction pairs with balanced forces on a single object. For instance, a book on a table has gravity pulling it down and the normal force pushing it up. Those two forces are on the *same object*—the book. They are not a Third Law pair. The Third Law pair to the Earth pulling the book is the book pulling the Earth. Always remember: action-reaction pairs act on two different objects.

Mastering this one idea—that the pairs act on different bodies—is the key. Keep that in mind, and you'll be in great shape for the exam.

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Settings & Accessibility

Newton's Third Law

Why this matters

Diagram

Concept map

Core explanation

The "Why Does Anything Move?" Problem

Systems and Internal Forces

Tension: A Chain of Third-Law Pairs

Worked examples

Pushing a Crate

A Book on a Table

Try it yourself

Problem 1

Problem 2

Practice — 8 questions