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āš™ļøAP Physics C: Mechanics Unit 2 Review

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2.3 Newton's Third Law

2.3 Newton's Third Law

Written by the Fiveable Content Team • Last updated June 2026
Verified for the 2027 exam
Verified for the 2027 exam•Written by the Fiveable Content Team • Last updated June 2026
āš™ļøAP Physics C: Mechanics
Unit & Topic Study Guides
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Newton's Third Law says that when two objects interact, they push or pull on each other with forces that are equal in size and opposite in direction. The key catch is that these paired forces act on different objects, so they never cancel each other out.

Why This Matters for the AP Physics C: Mechanics Exam

Force and translational dynamics is one of the heaviest parts of the course, and Newton's Third Law shows up constantly. You use it to set up free-body diagrams, connect forces between objects in a system, and reason about tension in strings and pulleys. It also supports the kind of translation between words and math that the free-response section rewards, where you make a claim, back it with physics principles, and then derive supporting equations. Getting comfortable with paired forces now makes second-law problems, momentum, and collisions easier later.

Key Takeaways

  • Paired forces are equal in magnitude and opposite in direction: Fāƒ—AĀ onĀ B=āˆ’Fāƒ—BĀ onĀ A\vec{F}_{\mathrm{A}\text{ on }\mathrm{B}}=-\vec{F}_{\mathrm{B}\text{ on }\mathrm{A}}, and they always act on two different objects.
  • Third-law pairs never appear on the same free-body diagram because they act on different objects, so they do not cancel.
  • Internal forces inside a system come in equal and opposite pairs and cannot change the motion of the system's center of mass.
  • Only external forces can accelerate a system's center of mass.
  • Tension is the net result of tiny segments of a string pulling on each other in response to an external force.
  • An ideal string has negligible mass and does not stretch, so its tension is the same at every point; an ideal pulley has negligible mass and friction and only redirects the force.

Paired Forces Between Objects

When two objects interact, they apply forces to each other that are equal in magnitude and opposite in direction. This relationship is the heart of Newton's Third Law.

Fāƒ—AĀ onĀ B=āˆ’Fāƒ—BĀ onĀ A\vec{F}_{\mathrm{A} \text { on } \mathrm{B}}=-\vec{F}_{\mathrm{B} \text { on } \mathrm{A}}

The negative sign means the forces point in opposite directions. These paired forces always act on different objects, which is the most important detail to keep straight.

  • When you push on a wall, the wall pushes back on you with an equal force.
  • When a hammer strikes a nail, the nail exerts an equal force back on the hammer.
  • As a swimmer pushes water backward, the water pushes the swimmer forward.

These paired forces do not cancel each other out because they act on different objects. That is why you cannot lift yourself by pulling on your own shoelaces. Both of those forces act on you, so they sum to zero on you and nothing happens.

Representing Third-Law Pairs

To represent a third-law pair, draw the two interacting objects separately and label the forces as Fāƒ—AĀ onĀ B\vec{F}_{A\text{ on }B} on object B and Fāƒ—BĀ onĀ A\vec{F}_{B\text{ on }A} on object A. These two forces are equal in magnitude and opposite in direction, but they appear on different free-body diagrams because they act on different objects.

For example, if a person pushes a wall, you draw two separate diagrams: one of the wall showing Fāƒ—personĀ onĀ wall\vec{F}_{\text{person on wall}} pointing into the wall, and one of the person showing Fāƒ—wallĀ onĀ person\vec{F}_{\text{wall on person}} pointing back into the person. Each force in the pair acts on a different object, so they never appear together on the same free-body diagram. Always identifying "who pushes whom" keeps your diagrams correct.

Internal Forces and Center of Mass

You can analyze the motion of a system by sorting forces into internal and external. This split tells you which forces affect the overall motion and which do not.

Internal forces occur between objects within a defined system. These forces:

  • Are interactions between parts of the system
  • Always come in equal and opposite pairs by Newton's Third Law
  • Do not affect the motion of the system's center of mass

External forces involve interactions between system objects and the outside environment:

  • Act on system objects from sources outside the defined system
  • Can change the motion of the center of mass
  • Determine the acceleration of the center of mass through Newton's Second Law

The center of mass of a system moves as if all the system's mass were concentrated at that point and all external forces were applied there. Because internal forces always come in equal and opposite pairs, they cancel within the system and cannot change the motion of the center of mass.

Tension in Strings and Cables

Tension is the pulling force transmitted through a string, cable, or similar object when it is pulled tight by forces acting at each end. It shows up in many mechanical systems.

At the system level, tension is the macroscopic net result of adjacent infinitesimal segments of a string, cable, or chain pulling on each other in response to an external force. Each tiny segment exerts a force on the neighboring segment, and that distributed interaction is what you observe as tension.

  • The tension force always points away from a point on the string, pulling inward at both ends.
  • In an ideal string with negligible mass, the tension is uniform throughout the entire length.

That uniform-tension simplification makes many systems much easier to analyze.

Properties of Ideal Strings

Ideal strings are simplified models. Real strings have mass and can stretch, but the ideal string model is a useful approximation for many situations.

An ideal string has these key features:

  • Negligible mass compared to the objects it connects
  • Does not stretch when force is applied
  • Transmits tension undiminished from one end to the other

Because the tension force has the same magnitude at all points along an ideal string, you can analyze systems like connected masses and pulleys more easily.

Tension Variations in Strings

Real strings with non-negligible mass have tension that changes along their length. Each segment of the string must support not only the attached load but also the weight of the string below it.

The tension in a hanging heavy string:

  • Is largest at the top, where it supports both the load and the entire string's weight
  • Decreases toward the bottom as less string weight needs to be supported
  • Is smallest at the bottom, where it only supports the attached load

How much the tension varies depends on the string's mass and length. For short, lightweight strings, treating tension as constant is usually a reasonable approximation.

Ideal Pulleys

An ideal pulley changes the direction of a tension force without changing its magnitude.

Properties of an ideal pulley include:

  • Negligible mass, so its rotational inertia can be neglected
  • Rotates about an axle through its center of mass
  • Negligible friction at the axle

When combined with an ideal string, an ideal pulley keeps the tension the same throughout the string while redirecting the force. This is why pulleys are useful for lifting loads and applying forces in different directions.

How to Use This on the AP Physics C: Mechanics Exam

Free Response

When a question asks you to explain an interaction, name the two objects and write the pair clearly, like Fāƒ—personĀ onĀ boat=āˆ’Fāƒ—boatĀ onĀ person\vec{F}_{\text{person on boat}}=-\vec{F}_{\text{boat on person}}. Justify claims by pointing to the equal-and-opposite relationship and to which object each force acts on. If the problem then asks for an equation, derive it from Newton's Second Law applied to one object at a time.

Problem Solving

For connected masses over an ideal pulley, write a separate Newton's Second Law equation for each mass, treat the tension as the same throughout the ideal string, and pick a consistent positive direction for the system. Add the equations to solve for acceleration, then plug back in to find tension.

Common Trap

If your free-body diagram shows two equal and opposite forces on the same object and you call them a third-law pair, stop and recheck. A real pair always splits across two diagrams. Forces that balance on one object (like normal force and gravity on a book at rest) are not a third-law pair.

Common Misconceptions

  • Thinking action-reaction forces cancel. They are equal and opposite but act on different objects, so they never cancel on a single object.
  • Confusing balanced forces with a third-law pair. Gravity and the normal force on a resting book act on the same object and are not a pair.
  • Believing the larger object always feels the larger force. Both objects feel the same size force; the lighter one just accelerates more because of its smaller mass.
  • Assuming tension is always uniform. That holds only for an ideal (massless) string. A heavy string can have different tension at different points.
  • Thinking internal forces can move a system. Internal forces cancel in pairs and cannot accelerate the center of mass; only external forces can.

Practice Problem 1: Newton's Third Law Paired Forces

A 60 kg person stands on a 75 kg rowboat. The person jumps horizontally from the boat onto a dock with a force of 300 N. What is the initial acceleration of the boat as the person jumps?

Solution

Apply Newton's Third Law. When the person pushes forward with 300 N to jump, the boat experiences an equal and opposite force of 300 N backward.

Using Newton's Second Law for the boat: F=maF = ma

Rearranging to solve for acceleration: a=Fm=300Ā N75Ā kg=4Ā m/s2a = \frac{F}{m} = \frac{300 \text{ N}}{75 \text{ kg}} = 4 \text{ m/s}^2

The boat initially accelerates backward at 4 m/s² when the person jumps.

Practice Problem 2: Tension in an Ideal String

A 5 kg mass and a 3 kg mass are connected by an ideal string that passes over an ideal pulley. If the system is released from rest, find the tension in the string and the acceleration of the masses.

Solution

Because the 5 kg mass is heavier, it accelerates downward and the 3 kg mass accelerates upward. Let downward be positive for the 5 kg mass and upward be positive for the 3 kg mass.

For the 5 kg mass: 5gāˆ’T=5a5g - T = 5a

For the 3 kg mass: Tāˆ’3g=3aT - 3g = 3a

Add the two equations: 5gāˆ’T+Tāˆ’3g=5a+3a5g - T + T - 3g = 5a + 3a

2g=8a2g = 8a a=2g8=2Ɨ9.88=2.45Ā m/s2a = \frac{2g}{8} = \frac{2 \times 9.8}{8} = 2.45\ \text{m/s}^2

Now solve for the tension using either mass. Using the 3 kg mass: Tāˆ’3g=3aT - 3g = 3a

T=3g+3a=3(9.8)+3(2.45)=29.4+7.35=36.75Ā NT = 3g + 3a = 3(9.8) + 3(2.45) = 29.4 + 7.35 = 36.75\ \text{N}

The tension in the string is 36.75Ā N36.75\ \text{N}, and the 5 kg mass accelerates downward at 2.45Ā m/s22.45\ \text{m/s}^2 while the 3 kg mass accelerates upward at 2.45Ā m/s22.45\ \text{m/s}^2.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term

Definition

center of mass

The point in a system where the entire mass can be considered to be concentrated for the purposes of analyzing motion and forces.

ideal pulley

A theoretical pulley with negligible mass that rotates about its center of mass with negligible friction.

ideal string

A theoretical string with negligible mass that does not stretch when under tension.

internal forces

Forces that objects within a system exert on each other, which do not affect the motion of the system's center of mass.

Newton's third law

The principle that when one object exerts a force on another object, the second object exerts an equal and opposite force on the first object.

paired forces

Two equal and opposite forces that act on different objects as a result of their interaction, as described by Newton's third law.

tension

The macroscopic net force that segments of a string, cable, chain, or similar system exert on each other in response to an external force.

Frequently Asked Questions

What does Newton's Third Law say?

Newton's Third Law says that when object A exerts a force on object B, object B exerts an equal-magnitude, opposite-direction force on object A: F_A on B = -F_B on A.

Why do third-law force pairs not cancel?

Third-law force pairs act on different objects, so they do not cancel on a single free-body diagram. Forces only cancel when they act on the same object.

How do I identify a third-law pair?

Name the two interacting objects and reverse the wording: force of A on B pairs with force of B on A. Each force has the same magnitude and opposite direction.

What is tension in AP Physics C Mechanics?

Tension is the macroscopic result of tiny segments of a string, cable, or chain pulling on one another in response to an external force.

When is tension the same throughout a string?

Tension is the same throughout an ideal string, which has negligible mass and does not stretch. If the string has nonnegligible mass, tension can vary along the string.

What is an ideal pulley?

An ideal pulley has negligible mass and negligible friction about its axle. With an ideal string, it changes the direction of the tension force without changing its magnitude.

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