Newton's Third Law and Free-Body Diagrams — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Newton's third law action-reaction pair identification, free-body diagram (FBD) drawing conventions, common force classification, internal/external force separation, and system selection for dynamics problems consistent with the AP Physics 1 CED.

You should already know: Classification of common contact and non-contact forces, Newton's first and second laws of motion, how to resolve force vectors into components.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Newton's Third Law and Free-Body Diagrams?

This topic is the core foundational skill for Unit 2: Dynamics, which accounts for 12–18% of the total AP Physics 1 exam score per the official College Board CED. Concepts from this topic appear in both multiple-choice (MCQ) and free-response questions (FRQ), as it is the required first step for nearly all dynamics problems — mistakes here propagate to wrong answers even if you apply Newton's second law correctly.

Newton's third law is the fundamental rule describing how two objects interact via force: whenever object A exerts a force on object B, object B exerts an equal-magnitude, opposite-direction force of the same type back on object A. Free-body diagrams (FBDs, also called force diagrams) are the standardized graphical tool to isolate all forces acting on a single object or system, so you can correctly apply Newton's second law. Unlike other interaction diagrams, FBDs never include forces acting on other objects, only the target body. This topic is required for all subsequent dynamics work, from incline problems to circular motion to collisions.

2. Newton's Third Law and Action-Reaction Pairs

The formal statement of Newton's third law (N3L) for two interacting objects A and B is: $${\vec{F}}_{A\text{ on }B}=-{\vec{F}}_{B\text{ on }A}$$ Three non-negotiable properties of N3L force pairs are tested repeatedly on the AP exam: 1) They always act on two different objects, never the same object; 2) They are always the same type of force (e.g., gravitational, normal, frictional, tension); 3) They are equal in magnitude at all times, even if the objects are accelerating, have different masses, or move at different speeds.

The best way to avoid confusion is to always use "X on Y" notation for every force, which makes it easy to confirm which object each force acts on. A common confusion is mixing up balanced forces (two forces acting on the same object that sum to zero, per Newton's first law) with N3L pairs. For example, a book resting on a table: the weight of the book (Earth pulling on the book) and the normal force (table pushing on the book) are balanced forces on the same object, so they are not an N3L pair. The actual N3L pairs are: (1) Earth pulls on book / book pulls on Earth (both gravitational), and (2) table pushes on book / book pushes on table (both normal).

Worked Example

Problem: A 60 kg astronaut pushes off a 120 kg stationary space capsule in deep space (no gravity or air resistance). Immediately after they lose contact, what is the relationship between the magnitude of the force the astronaut exerts on the capsule ($F_{AonC}$) and the magnitude of the force the capsule exerts on the astronaut ($F_{ConA}$)? A. $F_{AonC}=F_{ConA}$ B. $F_{AonC}>F_{ConA}$ because the capsule has more mass C. $F_{AonC}<F_{ConA}$ because the astronaut will accelerate more D. There is not enough information to determine the relationship

Solution:

1. The interaction between the astronaut and capsule is a contact force interaction, so Newton's third law applies directly to the force pair between them.
2. Newton's third law requires that interaction force pairs are always equal in magnitude, regardless of the masses or accelerations of the two objects.
3. The different accelerations of the astronaut and capsule are explained by Newton's second law ($a=F_{net}/m$): same magnitude force, different masses give different accelerations, but this does not change the equal magnitude of the interaction force pair.
4. The only correct option is A.

Exam tip: On any question asking you to identify an action-reaction pair, first eliminate any pair that acts on the same object — that is never a Newton's third law pair, no matter what the direction or magnitude.

3. Drawing Correct Free-Body Diagrams

A free-body diagram (FBD) is a simplified diagram that shows only the forces acting on the single object (or system) of interest. The AP Physics 1 exam regularly requires you to draw or interpret FBDs, with points awarded for correct force direction, correct labeling, and not including extra invalid forces.

The AP-accepted drawing conventions are: 1) Represent the object of interest as a single point (all forces act on the center of mass for rigid objects, so this is acceptable); 2) Draw each force vector starting at the point, pointing outward in the direction the force acts; 3) Label every force with a standard label that clearly identifies its type and source: $F_g$ for gravitational force (weight), $F_N$ for normal force, $T$ for tension, $f$ for friction, $F_{push}$ for an applied contact push; 4) Never include the net force in the FBD, and never include forces that the object exerts on other objects. The most important pre-step for any FBD is explicitly defining your system: which object (or group of objects) are you analyzing? Changing your system changes the FBD, because forces internal to the system are excluded.

Worked Example

Problem: A 5 kg block is sliding upward along a rough inclined plane (no external push acts on the block after it was set in motion). Draw a correct free-body diagram for the block, and list all forces with their sources.

Solution:

1. Define the system as the 5 kg block, our object of interest.
2. Identify all non-contact forces acting on the block: the gravitational force from Earth, pointing vertically downward, labeled $F_g$.
3. Identify all contact forces: the block is only in contact with the incline, which exerts two forces: (a) a normal force perpendicular to the incline surface, pointing outward toward the block, labeled $F_N$; (b) kinetic friction parallel to the incline, opposing the block's upward motion, so it points down the incline, labeled $f$.
4. No other forces act on the block: there is no "force of motion" pushing it upward — that is a fictitious force, so it is not included. The final FBD has three correctly directed and labeled vectors, no extra forces.

Exam tip: If you are ever tempted to draw a "force of motion" or "force of inertia" on an FBD, stop: all forces must come from an interaction with another object, so these fictitious forces are never allowed, and including them will lose points.

4. Internal vs. External Forces and System Selection

When solving problems involving multiple interacting objects, you can choose to treat each object as a separate system or treat multiple objects together as a single combined system. This choice is valid because of Newton's third law: any force internal to the system (acting between two objects inside the system) forms an action-reaction pair. When you sum all forces on the whole system, the two equal and opposite forces in the pair add to zero, so internal forces can be ignored entirely when calculating the net acceleration of the whole system. Only external forces (forces exerted by an object outside the system on an object inside the system) contribute to the net force of the system.

This technique drastically simplifies problems with connected objects, such as two blocks pulled by a string or a car pulling a trailer, by letting you find the acceleration of the whole system first without solving for internal tension or normal forces first. You can then go back and solve for internal forces after you know the acceleration.

Worked Example

Problem: Two blocks of mass $m_1=2\text{kg}$ and $m_2=3\text{kg}$ are connected by a massless string on a frictionless horizontal table. A horizontal applied force $F=10\text{N}$ pulls on $m_1$ to the right. What is the acceleration of the two-block system?

Solution:

1. Choose the system to be both blocks together, so total mass $M=m_1+m_2=2\text{kg}+3\text{kg}=5\text{kg}$.
2. Classify forces: the tension between the blocks is internal to the system, so per Newton's third law, the tension force on $m_1$ and the tension force on $m_2$ cancel out, so we ignore them.
3. Vertically, weight and normal force from the table balance to zero. The only net external horizontal force on the system is the applied force $F=10\text{N}$.
4. Apply Newton's second law: $F_{net}=Ma$, so $a=\frac{F_{net}}{M}=\frac{10\text{N}}{5\text{kg}}=2{\text{m/s}}^2$ to the right. This matches the result from solving for each block separately, but is much faster.

Exam tip: If an FRQ asks for the acceleration of a system of multiple connected objects, always use the combined system method first to get acceleration quickly, then solve for internal forces (like tension) separately.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Labeling the normal force on a resting book and the weight of the book as an action-reaction pair. Why: Students confuse balanced forces (two forces on the same object that sum to zero) with Newton's third law pairs, which always act on two different objects. Correct move: For any candidate N3L pair, check what each force acts on: if both act on the same object, it cannot be an N3L pair, and confirm both are the same type of force.
• Wrong move: Drawing the normal force on an incline pointing straight upward instead of perpendicular to the incline surface. Why: Students confuse normal force direction with the direction of weight, or assume normal force always opposes gravity. Correct move: Normal force is always perpendicular to the contact surface between two objects, so point it perpendicular to the surface, regardless of gravity's direction.
• Wrong move: Including a "force of motion" on an FBD for a sliding object. Why: Students intuitively think moving objects need a force to keep them moving, from everyday experience with friction, leading to this misconception. Correct move: Every force on an FBD must come from an interaction with an identifiable object outside the system; if you can't name the source object, the force doesn't exist.
• Wrong move: Including internal forces in the net force calculation for a combined system. Why: Students forget internal forces cancel per Newton's third law, leading to under or over-counting net force and wrong acceleration. Correct move: When drawing an FBD for a combined system, cross out any force that acts between two objects inside the system before calculating net force.
• Wrong move: Claiming action-reaction forces cancel out when calculating acceleration of a single object. Why: Students remember "equal and opposite" so they assume the forces sum to zero, leading to a wrong conclusion of zero acceleration. Correct move: Action-reaction forces act on different objects, so you only include one of the two forces in the FBD of your target object, so they never cancel in the $F_{net}$ for that object.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A student places a full coffee mug on a horizontal wooden desk, which sits on the classroom floor. Which of the following options correctly identifies a Newton's third law action-reaction pair in this scenario? A. The gravitational force of Earth on the mug, and the normal force of the desk on the mug B. The gravitational force of Earth on the mug, and the gravitational force of the mug on Earth C. The normal force of the desk on the mug, and the weight of the mug D. The gravitational force of the mug on the desk, and the normal force of the desk on the mug

Worked Solution: We apply the two core rules for N3L pairs: 1) they act on two different objects, 2) they are the same type of force. Option A: both forces act on the mug (same object) and are different types, so it is wrong. Option B: both are gravitational forces, one acts on the mug (Earth on mug) and one acts on Earth (mug on Earth), which fits both rules. Option C: weight is gravitational force on the mug, normal is desk on the mug; both act on the same object, so wrong. Option D: the gravitational force of the mug on the desk is negligible, and it is a different force type than normal, so wrong. The correct answer is B.

Question 2 (Free Response)

A 1000 kg car pulls a 500 kg trailer along a straight horizontal road. The car and trailer accelerate forward at $1.5{\text{m/s}}^2$. Ignore rolling friction and air resistance for all parts. (a) Draw a correct free-body diagram for the trailer, labeling all forces. (b) Find the magnitude of the tension force in the hitch between the car and the trailer. (c) How does the magnitude of the force the trailer exerts on the car compare to the magnitude of the force the car exerts on the trailer? Explain your answer using Newton's third law.

Worked Solution: (a) The system is the trailer. Three forces act on it: 1) Gravitational force $F_g$ from Earth, pointing vertically downward; 2) Normal force $F_N$ from the road, pointing vertically upward (perpendicular to the horizontal road); 3) Tension force $T$ from the hitch, pulling the trailer forward in the direction of acceleration. All vectors are drawn with tails at the center of mass, no extra fictitious forces are included. (b) Apply Newton's second law to the trailer in the horizontal direction: the only horizontal force is tension, so $T=m_{\text{trailer}}a=(500\text{kg})(1.5{\text{m/s}}^2)=\boxed{750\text{N}}$. (c) The magnitudes are equal. The force of the trailer on the car and the force of the car on the trailer form an action-reaction pair per Newton's third law, which requires all interaction force pairs to be equal in magnitude regardless of the acceleration of the objects, so their magnitudes are equal.

Question 3 (Application / Real-World Style)

A 70 kg high jumper pushes off the ground with an upward acceleration of $12{\text{m/s}}^2$ while their feet are still in contact with the ground. What is the magnitude of the force that the high jumper exerts on the ground during this push? Give your answer in newtons, and compare it to the jumper's weight.

Worked Solution: First, define the system as the high jumper. Vertical forces on the jumper are: gravitational force $F_g=mg$ downward, and normal force $F_N$ from the ground upward. Take upward as positive and apply Newton's second law: $$F_{net}=F_N-mg=ma$$ Solve for $F_N$: $F_N=m(g+a)=70\text{kg}(9.8{\text{m/s}}^2+12{\text{m/s}}^2)=1526\text{N}\approx 1530\text{N}$. By Newton's third law, the force the jumper exerts on the ground is equal in magnitude to $F_N$. The jumper's weight is $mg=70\text{kg}\times 9.8{\text{m/s}}^2=686\text{N}$, so the force the jumper exerts on the ground is roughly twice their weight, which is enough to generate the upward acceleration needed to clear a high jump bar.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the non-negotiable foundation for all remaining dynamics topics in Unit 2, and nearly every other topic in AP Physics 1. Next, you will apply the skills you learned here to draw FBDs for objects on inclined planes, solve for acceleration and friction, and analyze connected objects with multiple forces. Without correctly identifying force pairs and drawing accurate FBDs, you cannot correctly set up Newton's second law equations for any dynamics problem, leading to avoidable errors on exam day. Longer term, this topic feeds into circular motion, rotational dynamics, collisions, and simple harmonic motion, all of which require FBDs and Newton's third law as the first step of any problem.

Follow-on topics to study next: Newton's Second Law Applications Circular Motion Dynamics Conservation of Momentum for Systems