AP Physics 1 · Topic 2.3

Topic 2.3: Newton's Third Law

Unit 2: Force and Translational Dynamics18-23% of the multiple-choice section

Newton's third law says that when object A exerts a force on object B, B exerts a force on A that is equal in magnitude and opposite in direction. The two forces act on different objects, which is why they never cancel and never belong on the same free-body diagram.

AP Physics: Unit 2 (topics 2.3 Newton's Third Law). AP Physics 1 Unit 2, Topic 2.3, covering learning objective 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 2.3.A.1 states the law as the force on B by A equals the negative of the force on A by B, 2.3.A.2 rules internal forces out of the motion of a system's center of mass, and 2.3.A.3 defines tension along with ideal strings and ideal pulleys. Boundary statements keep tension in a massive string qualitative and limit interaction at a distance to gravitational forces. The CED lists suggested skills 1.A, 2.D, 3.B, and 3.C here, and weights Unit 2 at 18 to 23 percent of the multiple-choice section.

What the third law states

Newton's third law describes the interaction of two objects in terms of the paired forces that each exerts on the other. The CED writes it as one equation:

\vec{F}_{A \text{ on } B} = -\vec{F}_{B \text{ on } A}

Read the subscripts, because they carry the whole content. On the left is a force exerted on B; on the right is a force exerted on A. The minus sign says the two point in opposite directions, and the equality says their magnitudes match. Nothing in the statement mentions mass, speed, or which object did the pushing. A mosquito hitting a windshield and the windshield hitting the mosquito exchange forces of exactly equal magnitude.

Learning objective 2.3.A asks you to describe the interaction of two objects using Newton's third law and a representation of paired forces exerted on each object. Note representation and each object. The expected answer is usually two free-body diagrams, one per object, with the two paired arrows appearing on different pictures. Topic 2.3 sits in Unit 2, weighted at 18 to 23 percent of the multiple-choice section.

Paired forces never cancel

Cancellation requires two forces on the same object. A third law pair is, by construction, two forces on different objects, so the question of cancellation never arises. If pairs did cancel, nothing could ever accelerate, since every force in the universe has a partner.

A book resting on a table has two different pairings in play, and mixing them is the classic error.

The table pushes up on the book, and the book pushes down on the table. That is a third law pair: one force on the book, one force on the table.
The book also feels a gravitational force from Earth pulling it down. That force acts on the book, and it happens to equal the table's upward push in magnitude, so the two balance and the book stays put.

The second pairing is not a third law pair at all. It is two separate interactions that happen to balance on one object, which is Newton's first law. The partner of the gravitational force on the book is the book's gravitational pull on Earth, which acts on Earth. The quick test: if two forces appear on the same free-body diagram, they are not a third law pair.

Naming the partner in one step

Every force has exactly one third law partner, and finding it is mechanical. Write the force with both nouns, the force on A by B, then swap the two names to get the force on B by A. That swapped force is the partner. It has the same magnitude, the same type of interaction, and the opposite direction, and it exists for exactly as long as the interaction does.

Force on your diagram	Its third law partner
Force on the box by the floor, up	Force on the floor by the box, down
Force on the crate by the rope, forward	Force on the rope by the crate, backward
Force on the skater by the wall, away from the wall	Force on the wall by the skater, into the wall
Force on the ball by Earth, down	Force on Earth by the ball, up

Two checks keep the swap honest. Same type of interaction: the partner of a contact push is a contact push, and the partner of a gravitational pull is a gravitational pull. And different objects: if your candidate partner acts on the same object as the original force, you have found a balanced force, not a partner. Practicing the swap on every arrow you draw in Topic 2.2 makes it automatic.

Equal forces, unequal accelerations

The third law fixes the forces, not the motion. Newton's second law then separates the two objects, because acceleration is force divided by mass. Same force magnitude, different masses, different accelerations, in inverse proportion. That is suggested skill 2.D for this topic, predicting new values using functional dependence between variables: double one object's mass and its acceleration from the same paired force is halved.

This settles the objection students raise first. If a car and a truck collide and the two forces are equal, why is the car the one wrecked? Because the same force magnitude divided by a much smaller mass is a much larger acceleration, and it is the acceleration, not the force, that the occupants and the bodywork experience.

The same argument explains why Earth does not visibly respond when you jump. You pull up on Earth with precisely the force Earth pulls down on you. Earth's mass is so much greater than yours that the acceleration this produces is far too small to notice. Nothing is being violated. The mass ratio simply runs the other way, and the worked examples below put numbers on a version you can actually measure.

Internal forces do not move the center of mass

Essential knowledge 2.3.A.2 is the third law's biggest structural payoff. Interactions between objects within a system, meaning internal forces, do not influence the motion of a system's center of mass.

The pairing is the reason. Draw a boundary around both objects in an interaction and both halves of the pair land inside it. They are equal in magnitude and opposite in direction, so they add to zero, and a sum of zero cannot accelerate the system's center of mass. Only forces from outside the boundary survive the sum.

That is why the system shortcut from Topic 2.1 is safe rather than sloppy. Treat two blocks and their connecting string as one object and the tension vanishes from the equation, not because you neglected it but because it cancels itself. It is also why no amount of internal effort relocates a system's center of mass. Two skaters shoving apart on frictionless ice, starting at rest, fly off in opposite directions while the center of mass of the pair stays exactly where it began. The second worked example below runs that calculation on a track.

Tension, ropes, and ideal pulleys

The CED files tension under Newton's third law, and the definition explains why. 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. Imagine cutting the rope anywhere: the two cut faces pull on each other with a third law pair, and that mutual pull is the tension at that point.

Four essential knowledge statements set the rules.

An ideal string has negligible mass and does not stretch when under tension.
The tension in an ideal string is the same at all points within the string.
In a string with nonnegligible mass, tension may not be the same at all points within the string.
An ideal pulley is a pulley that has negligible mass and rotates about an axle through its center of mass with negligible friction.

The ideal case is the exam's default: the Table of Information lists among its conventions that springs and strings are assumed to be ideal unless otherwise stated. For a string with mass, the topic's boundary statement keeps things qualitative. AP Physics 1 only expects students to describe tension qualitatively in a string, cable, chain, or similar system with mass, and the CED offers the example that the tension in a hanging chain is greater toward the top. Numerical setups live in how to find tension.

How Topic 2.3 is tested

The CED lists four suggested skills for this topic: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Skill 1.A is the two-diagram representation the learning objective names, and skill 2.D is the mass-ratio argument.

One boundary is worth carrying. In AP Physics 1 the interaction between objects or systems at a distance is limited to gravitational forces, with AP Physics 2 adding electric and magnetic ones. Every non-gravitational pair in this course is therefore a contact pair, and it lasts only while the surfaces touch.

Practice by drawing both diagrams for a two-object setup, then naming the partner of every arrow and saying which object that partner acts on. Gravity's partner will always land on Earth, which is a useful reminder that partners live off the page. The free-body diagram builder is built for that check, and Topic 2.2 covers the drawing rules the exam applies.

Contact forces between two crates

Two crates sit in contact on a frictionless level floor: crate A has mass 2.0 kg and crate B has mass 5.0 kg. You push horizontally on the outer face of crate A with a 21 N force, so A shoves B along. Find the acceleration and the magnitude of the contact force pair between the crates. Then find the contact force if you push crate B with the same 21 N instead.

Take the direction of the push as positive and hold it to the end.
Treat both crates as one system, since they move together. The contact forces between them are internal and cancel, so the only external horizontal force is the push: $a = F_{net}/m_{sys} = (21 \, \mathrm{N})/(2.0 \, \mathrm{kg} + 5.0 \, \mathrm{kg}) = (21 \, \mathrm{N})/(7.0 \, \mathrm{kg}) = 3.0 \, \mathrm{m/s^2}$ .
Isolate crate B. The only horizontal force on it is the push from A, so $F_{A \text{ on } B} = m_B a = (5.0 \, \mathrm{kg})(3.0 \, \mathrm{m/s^2}) = 15 \, \mathrm{N}$ , forward.
Newton's third law fixes the partner without any new calculation: $F_{B \text{ on } A} = 15 \, \mathrm{N}$ backward, and that force acts on crate A, on a different diagram.
Check on crate A: the two horizontal forces give $21 \, \mathrm{N} - 15 \, \mathrm{N} = 6.0 \, \mathrm{N}$ , and $m_A a = (2.0 \, \mathrm{kg})(3.0 \, \mathrm{m/s^2}) = 6.0 \, \mathrm{N}$ . The two agree, so the pair is consistent.
Now push crate B instead with the same 21 N. The system acceleration is unchanged at $3.0 \, \mathrm{m/s^2}$ , because the total mass and the external force are the same. Isolate crate A: its only horizontal force is now the push from B, so $F_{B \text{ on } A} = m_A a = (2.0 \, \mathrm{kg})(3.0 \, \mathrm{m/s^2}) = 6.0 \, \mathrm{N}$ .

Pushing A, the contact pair has magnitude 15 N. Pushing B, the same two crates under the same 21 N give a contact pair of 6.0 N. The two members of a pair always match each other, but the size of the pair depends on how much mass sits beyond the contact, which is why the contact force is not simply half the push.

Spring plunger between two carts

A 0.50 kg cart carries a compressed spring plunger of negligible mass held against a 2.0 kg cart on a level frictionless track. Both carts start at rest. At one instant during the release, the plunger exerts 6.0 N on the 2.0 kg cart. Find each cart's acceleration at that instant and describe what the system's center of mass does.

Take the direction the 2.0 kg cart is pushed as positive and hold it throughout.
Newton's third law gives the partner force with no calculation. The plunger belongs to the 0.50 kg cart, so if it exerts 6.0 N forward on the 2.0 kg cart, the 2.0 kg cart exerts 6.0 N backward on the 0.50 kg cart.
On each cart the vertical forces, gravity down and the normal force up, cancel, so each cart's horizontal acceleration comes from the plunger pair alone.
Light cart: $a = F/m = (6.0 \, \mathrm{N})/(0.50 \, \mathrm{kg}) = 12 \, \mathrm{m/s^2}$ , in the negative direction.
Heavy cart: $a = F/m = (6.0 \, \mathrm{N})/(2.0 \, \mathrm{kg}) = 3.0 \, \mathrm{m/s^2}$ , in the positive direction.
The magnitudes are in the ratio $12/3.0 = 4.0$ , exactly the inverse of the mass ratio $(2.0 \, \mathrm{kg})/(0.50 \, \mathrm{kg}) = 4.0$ . Equal forces, unequal accelerations.
Now the system. Both plunger forces are internal to the two-cart system, and the external forces cancel vertically, so the net external force is zero. The center of mass started at rest and stays at rest, which is essential knowledge 2.3.A.2 in action.
Confirm it with positions. Holding the center of mass fixed requires $m_1 d_1 = m_2 d_2$ , so the light cart travels $(2.0 \, \mathrm{kg})/(0.50 \, \mathrm{kg}) = 4.0$ times as far as the heavy cart, in the opposite direction.

The 0.50 kg cart accelerates at $12 \, \mathrm{m/s^2}$ and the 2.0 kg cart at $3.0 \, \mathrm{m/s^2}$ , in opposite directions, from a force pair whose magnitude is 6.0 N on both. The system's center of mass does not move, because every horizontal force in play is internal to the system.

Frequently asked questions

Do Newton's third law pairs cancel each other out?

They cannot, because they act on different objects. Cancellation is something that happens between two forces on one free-body diagram, and the two halves of a pair are never on the same diagram. If pairs did cancel, nothing could ever accelerate, because every force has a partner somewhere.

How do you find the third law partner of a force?

Write the force naming both objects, as the force on A by B, then swap the names to get the force on B by A. The partner has equal magnitude, the same type of interaction, and the opposite direction, and it acts on the other object. If your candidate acts on the same object as the original, you have found a balanced force instead.

What is the partner of the weight of a book sitting on a table?

The gravitational pull the book exerts on Earth, directed upward and acting on Earth. It is not the table pushing up on the book. That upward push does balance the book's weight, but it is a contact interaction between book and table, and its own partner is the book pushing down on the table.

If the forces are equal, why does the lighter object accelerate more?

Because acceleration is force divided by mass. The third law fixes the two force magnitudes as equal, and Newton's second law then divides each by a different mass. A 0.50 kg cart and a 2.0 kg cart pushed apart by a 6.0 N pair accelerate at 12 and 3.0 m/s squared, a ratio of 4.0 that matches the inverse mass ratio.

Is the tension the same everywhere in a rope?

In an ideal string, yes. The CED defines an ideal string as having negligible mass and not stretching under tension, and states that its tension is the same at all points. In a string with nonnegligible mass the tension may differ from point to point, and AP Physics 1 keeps that case qualitative, noting for example that a hanging chain has greater tension toward the top.