Newton's Third Law — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: definition of Newton’s Third Law, identification of action-reaction force pairs, third-law analysis of contact, normal, tension, and gravitational forces, system selection, and correction of common free-body diagram errors.

You should already know: How to draw free-body diagrams for isolated objects. Newton’s First and Second Laws of motion. The definition of contact and fundamental forces.

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 C: Mechanics 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?

Newton's Third Law (often called the action-reaction law) is a foundational rule of classical mechanics, accounting for approximately 10-15% of the Unit 2 exam weight per the official AP Physics C: Mechanics CED, and appearing in both MCQ and FRQ sections. It almost never is tested in isolation; instead, it is a required intermediate step for multi-object force problems, connected motion, and system analysis.

The formal definition uses standard notation where ${\vec{F}}_{AB}$ means the force exerted by body A on body B (so this force acts on body B). The law states: If body A exerts a force ${\vec{F}}_{AB}$ on body B, then body B exerts a force ${\vec{F}}_{BA}$ on body A that satisfies: $${\vec{F}}_{BA}=-{\vec{F}}_{AB}$$ This means the two forces are always equal in magnitude, opposite in direction, and act on different objects. Synonyms for the two forces are third-law pairs or action-reaction pairs, where the labels "action" and "reaction" are arbitrary—neither force causes the other; they exist simultaneously as an interaction between two bodies. Unlike Newton's First or Second Law, which describe the motion of a single object, Newton's Third Law describes the mutual interaction between two separate objects.

2. Identifying Correct Action-Reaction Force Pairs

The most frequently tested skill for Newton's Third Law on the AP exam is correctly identifying which two forces form a valid third-law pair, and distinguishing third-law pairs from balanced forces. There are three non-negotiable criteria for a valid third-law pair:

1. Both forces are the same type of force (e.g., both gravitational, both contact/normal).
2. Each force acts on a different object.
3. Each force is exerted by the opposite object (the first force is A on B, the second is B on A).

A common example of a common mistake: A book rests on a table. The gravitational force on the book from Earth ($F_{g,\text{Earth on book}}$) is often incorrectly paired with the normal force on the book from the table. This is wrong: both forces act on the same object (the book), and they are different force types (gravitational vs. contact). The correct third-law pair for the gravitational force on the book is the gravitational force exerted by the book on Earth, upward. The correct third-law pair for the normal force on the book from the table is the normal force exerted by the book on the table, downward. Balanced forces (two forces on the same object that sum to zero for zero acceleration) are never third-law pairs.

Worked Example

A student pushes a 10 kg box to the right along a frictionless horizontal floor, accelerating the box. Which of the following is a valid third-law pair? A) The student’s weight downward, and the normal force from the floor upward on the student B) The push force the student exerts on the box rightward, and the friction force the box exerts on the floor leftward C) The push force the student exerts on the box rightward, and the push force the box exerts on the student leftward D) The weight of the box downward, and the push force the student exerts on the box rightward

1. Apply the three criteria to each option, starting with the "different objects" check.
2. Option A has both forces acting on the student (same object), so eliminate A. Option B has forces from two different interactions (student-box and box-floor), so eliminate B. Option D has two different force types (gravitational vs. contact), so eliminate D.
3. Option C: First force is exerted by student on box (rightward contact), second force is exerted by box on student (leftward contact). All three criteria are satisfied.
4. The correct answer is Option C.

Exam tip: When asked to identify a third-law pair on an MCQ, first eliminate any option where both forces act on the same object—this will eliminate ~50% of wrong answers immediately.

3. Third Law and System Selection

Newton's Third Law is the foundation of system-based problem solving for multi-object connected systems, a key AP C FRQ topic. When solving for the acceleration of multiple connected objects, you can choose any combination of objects as your system. Forces that act between objects inside the system are called internal forces. By Newton's Third Law, every internal force has an equal and opposite partner that is also inside the system. When you sum all forces on the entire system, the sum of internal forces is always zero: $$\sum {\vec{F}}_{\text{int}}=\sum \left({\vec{F}}_{AB}+{\vec{F}}_{BA}\right)=0$$ This means only external forces (forces exerted by objects outside the system on objects inside) contribute to the acceleration of the system's center of mass. This is a massive simplification: you can find the acceleration of the entire connected system in one step, without solving for internal forces first. If you need to find the magnitude of an internal force (e.g., tension between two blocks, contact force between two blocks), you split the system after finding acceleration and isolate one object to solve for the internal force.

Worked Example

Two blocks $m_1=2$ kg and $m_2=3$ kg are connected by a massless string, with $m_1$ on a frictionless horizontal table and $m_2$ hanging vertically off the edge of the table via a massless pulley. Find the acceleration of the system and the tension in the string, using $g=9.8{\text{ m/s}}^2$.

1. Treat both blocks as a single combined system. All tension forces between the string and blocks are internal, so they cancel by Newton's Third Law.
2. Sum external forces: The weight of $m_1$ cancels with the normal force from the table, so the only net external force is the weight of $m_2$. Apply Newton's Second Law: $$m_{2}g=(m_1+m_2)a$$
3. Solve for acceleration: $$a=\frac{m_{2}g}{m_1+m_2}=\frac{(3)(9.8)}{5}=5.88{\text{ m/s}}^2$$
4. Isolate $m_1$ to find tension. The only horizontal force on $m_1$ is tension $T$, so $T=m_{1}a=(2)(5.88)=11.76\text{ N}$. The equal magnitude of tension pulling $m_1$ and $m_2$ for a massless string is a direct result of Newton's Third Law.

Exam tip: When solving for acceleration of a connected system, always use the combined system first to avoid solving a system of two equations—this saves 2-3 minutes on FRQ and reduces arithmetic error.

4. Third Law in Accelerating Contact Problems

A common AP context for Newton's Third Law is calculating contact forces between accelerating objects, most famously apparent weight in an accelerating elevator. Apparent weight is defined as the force a person exerts on a weighing scale (or the floor) in an accelerating system. By Newton's Third Law, this is equal in magnitude to the normal force the scale exerts on the person, so you first solve for the normal force on the person, then use the third law to get the apparent weight.

Another common context is collision problems: a frequent misconception is that a larger, heavier object exerts a larger force on a smaller lighter object during a collision. This is wrong: by Newton's Third Law, the forces are equal in magnitude. The larger acceleration of the smaller object comes from its smaller mass ($a=F/m$), not a larger force.

Worked Example

A 60 kg person stands on the floor of an elevator that accelerates upward at $2.0{\text{ m/s}}^2$. What is the magnitude of the force the person exerts on the elevator floor? Use $g=9.8{\text{ m/s}}^2$.

1. Draw a free-body diagram for the person: two forces act on the person: weight $mg$ downward (from Earth) and normal force $N$ upward (from the elevator floor).
2. Set upward as positive, and apply Newton's Second Law: $$\sum F=N-mg=ma$$
3. Solve for $N$, the force of the floor on the person: $$N=m(g+a)=60(9.8+2.0)=708\text{ N}$$
4. By Newton's Third Law, the force of the person on the floor is equal in magnitude to $N$, so the final answer is $708\text{ N}$.

Exam tip: Always circle what the question asks for at the start: if it asks for the force the object exerts on the surface, not the other way around, explicitly confirm the magnitude is the same via Newton's Third Law to earn full points on FRQ.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming that the gravitational force on a book from Earth and the normal force on the book from the table form a third-law pair. Why: Students confuse balanced forces (same object, sum to zero for zero acceleration) with third-law pairs (different objects, same interaction). Correct move: Always check if both forces act on the same object—if yes, they cannot be a third-law pair.
• Wrong move: Arguing that a large truck pushing a small car must exert a larger force on the car than the car exerts on the truck, because the truck is bigger. Why: Students confuse the force of interaction with the resulting acceleration ($a=F/m$). Correct move: Always remember third-law pairs are always equal in magnitude, regardless of mass or acceleration; differing accelerations come from differing masses, not differing forces.
• Wrong move: Including internal third-law pair forces when summing forces for a combined system. Why: Students forget internal forces cancel and double-count them, leading to an incorrect net force. Correct move: When you define your system, cross out all forces that act between objects inside the system—only keep forces from objects outside the system.
• Wrong move: Reporting the wrong magnitude for the force exerted by object B on A, after solving for the force exerted by A on B. Why: Students forget the question asks for the force on the second object, especially in apparent weight problems. Correct move: Explicitly apply ${\vec{F}}_{BA}=-{\vec{F}}_{AB}$ before writing your final answer.
• Wrong move: Treating tension at the two ends of a massive accelerating rope as equal in magnitude. Why: Students generalize the massless rope result to massive ropes incorrectly. Correct move: For a massive rope, calculate the tension difference using Newton's Second Law ($T_1-T_2=m_{\text{rope}}a$).

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

Three blocks of mass $m_1=1$ kg, $m_2=2$ kg, $m_3=3$ kg are placed in contact with each other on a frictionless horizontal surface. A constant horizontal force $F=12$ N is applied to $m_1$ to the right. What is the magnitude of the force that $m_2$ exerts on $m_3$? A) 2 N B) 4 N C) 6 N D) 12 N

Worked Solution: First, treat all three blocks as a combined system. All contact forces between the blocks are internal, so they cancel by Newton's Third Law. The net external force is $F=12$ N, so Newton's Second Law gives $F=(m_1+m_2+m_3)a⟹12=6a⟹a=2{\text{ m/s}}^2$. Next, isolate $m_3$: the only horizontal force on $m_3$ is the contact force from $m_2$, so $F_{23}=m_{3}a=3\times 2=6$ N. Correct answer: C.

Question 2 (Free Response)

A 5 kg block rests on top of a 10 kg block, which rests on a frictionless horizontal table. The coefficient of static friction between the two blocks is ${\mu }_s=0.3$, and the coefficient of kinetic friction is ${\mu }_k=0.2$. A horizontal force $F$ is applied to the 10 kg lower block. (a) Draw free-body diagrams for the upper block and the lower block, identifying all third-law pairs in your diagram. (b) What is the maximum force $F$ that can be applied such that the two blocks accelerate together without slipping between them? (c) If $F=10$ N is applied, what is the magnitude of the friction force that the upper block exerts on the lower block? Use $g=9.8{\text{ m/s}}^2$.

Worked Solution: (a) Upper block forces: Weight $m_{u}g$ downward (Earth), normal force $N_{l\to u}$ upward (lower block), static friction $f_{l\to u}$ rightward (lower block). Lower block forces: Weight $m_{l}g$ downward (Earth), normal force $N_{table\to l}$ upward (table), applied force $F$ rightward, normal force $N_{u\to l}$ downward (upper block), static friction $f_{u\to l}$ leftward (upper block). Third-law pairs: $(N_{l\to u},N_{u\to l})$ and $(f_{l\to u},f_{u\to l})$. (b) For no slipping, acceleration is the same for both blocks. The upper block is accelerated by static friction, so $f_s=m_{u}a$. Maximum static friction is $f_{s,max}={\mu }_s{N}_{l\to u}={\mu }_s{m}_{u}g$, so maximum acceleration is $a_{max}={\mu }_{s}g$. For the combined system, $F_{max}=(m_u+m_l)a_{max}=(15)(0.3\times 9.8)=44.1\text{ N}$. (c) $F=10$ N < 44.1 N, so blocks accelerate together. System acceleration: $a=F/(m_u+m_l)=10/15=2/3{\text{ m/s}}^2$. Friction on upper block: $f_{l\to u}=m_{u}a=10/3\approx 3.33$ N. By Newton's Third Law, the friction force the upper block exerts on the lower block has the same magnitude, so $f_{u\to l}\approx 3.33\text{ N}$.

Question 3 (Application / Real-World Style)

A 70 kg astronaut is floating at rest outside the International Space Station, and pushes off from a 10000 kg service module to return to the airlock. During the push, the astronaut accelerates at $2{\text{ m/s}}^2$ away from the module. What is the magnitude of the acceleration of the service module away from the astronaut during the push, and what is the magnitude of the interaction force?

Worked Solution: By Newton's Third Law, the force the astronaut exerts on the module is equal in magnitude to the force the module exerts on the astronaut. First calculate the force on the astronaut: $F=m_a{a}_a=(70)(2)=140\text{ N}$. The force on the module is also 140 N, so acceleration of the module is $a_m=F/m_m=140/10000=0.014{\text{ m/s}}^2$. In context, this means the astronaut moves away quickly while the massive service module barely accelerates at all, even though the interaction force is equal for both.

7. Quick Reference Cheatsheet

8. What's Next

Newton's Third Law is a prerequisite for all multi-object force problems coming up next in Unit 2, including friction analysis and uniform circular motion dynamics. It is also the foundational concept for Unit 4's conservation of momentum: the rule that internal forces cancel is required to derive and apply momentum conservation for isolated systems, which is one of the highest-weight topics on the AP exam. Third-law analysis of interaction forces also plays a key role in Unit 3's rotational dynamics, when calculating net torque on systems of rotating objects. Without mastering the skills in this chapter, you will struggle to simplify complex multi-part FRQ problems and earn full points.

• Friction and drag forces
• Uniform circular motion dynamics
• Conservation of momentum
• Rotational dynamics