Newton’s third law: whenever two objects interact, they exert forces on each other that are equal in magnitude and opposite in direction—F_A on B = −F_B on A. Those are action–reaction pairs: they act on different objects, are the same type of force (gravity on each other, contact forces like normal or friction, or tension), and don’t cancel when you draw the free-body diagram of a single object. Internal forces between parts of a system won’t change the system’s center-of-mass motion. How to remember it: always look for pairs. If object A pushes B, ask “what pushes A?” and draw two arrows on different objects, equal length, opposite direction. Use concrete examples: when you push a wall you feel the wall push back (normal force); Earth pulls you and you pull Earth (gravity). Practice drawing paired forces on FBDs—that’s exactly what AP will test (CED Topic 2.3: represent paired forces and link to momentum). For a quick study review, see the topic guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc) and extra practice (https://library.fiveable.me/practice/ap-physics-1-revised).

Newton’s third law says forces come in action–reaction pairs: FAF on B = −FB on A (equal magnitude, opposite direction)—that’s about the forces, not the motions (CED 2.3.A.1). Why motions differ: each object has its own net force and mass, and motion follows Fnet = ma (CED 2.5). The pair forces act on different objects, so other forces (gravity, normal, friction, tension) and each object’s mass change the net force and therefore the acceleration. Example: when you push a wall the wall pushes back equally, but the wall’s huge mass and attachments make its acceleration essentially zero while you move. Also, internal force pairs within a system don’t change the system’s center-of-mass motion (CED 2.3.A.2). For more examples and AP-style practice on action–reaction pairs, see the Topic 2.3 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc), the Unit 2 overview (https://library.fiveable.me/ap-physics-1-revised/unit-2), and extra practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Think of Newton’s Third Law as always pairing forces between two different objects: FA on B = −FB on A. Key points: the two forces are equal in magnitude, opposite in direction, and act on different objects—so they never cancel on the same free-body diagram. Examples: - Book on a table: gravity from Earth pulls the book down (Fearth on book). The book pulls Earth up with an equal opposite gravitational force (Fbook on earth). On the book’s FBD you also draw the table’s normal force upward; the table feels an equal downward force from the book (Fbook on table = −Ftable on book). - Person pushing a wall: you push wall to the right; wall pushes you left with equal magnitude. - Rocket: hot gas pushed backward; gas pushes rocket forward (this pair changes the rocket’s momentum). Remember internal forces (forces between parts of a system, like tension segments) don’t change the system’s center-of-mass motion. For more examples and AP-aligned practice, check the Topic 2.3 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc) and try practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Newton’s second law relates the net force on a single object to its motion: ΣF = ma. It tells you how an object’s center-of-mass acceleration depends on the vector sum of all external forces on that object (useful for free-body diagrams and solving for acceleration). Newton’s third law describes interactions between two objects: for every force A exerts on B, B exerts an equal and opposite force on A (F_A on B = −F_B on A). Key differences: second-law forces sum on one object and produce acceleration; third-law forces come in pairs acting on two different objects and never cancel when applied to the same object. Also note internal forces (third-law pairs inside a system) don’t change the motion of the system’s center of mass. These distinctions show up a lot on the AP exam in FBD and momentum problems (Topic 2.3; see the topic study guide for examples) (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc). For unit review and lots of practice questions, check the Unit 2 overview and practice set (https://library.fiveable.me/ap-physics-1-revised/unit-2) (https://library.fiveable.me/practice/ap-physics-1-revised).

Look for two rules: (1) action and reaction act on two different objects, and (2) they are equal in magnitude, opposite in direction, and part of the same interaction (F_A on B = −F_B on A). In free-body diagrams, pair forces across objects, not on the same FBD. Quick checklist: - Identify the two objects interacting (e.g., book and table). - Ask “what force does object 1 exert on object 2?” and then draw the opposite force that object 2 exerts on object 1 (same magnitude, opposite direction). - Pair types must match the interaction: contact forces pair with contact forces (normal ↔ normal, friction ↔ friction); gravitational forces pair between masses (Earth ↔ object). - Don’t pair internal forces inside a single system when analyzing center-of-mass motion (internal forces cancel). Example: book on table—book feels normal up from table; table feels equal normal down from book. Also weight: Earth pulls book down (mg); book pulls Earth up with equal magnitude. For AP practice and examples see the Topic 2.3 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc) and extra problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Because the action and reaction forces in Newton’s third law act on different objects, they don’t cancel on one object. When you walk you push your foot backward on the ground (action). The ground pushes you forward with an equal and opposite force (reaction). The ground’s push acts on you, the push you made acts on the Earth—so they’re a pair but on different bodies (CED EK 2.3.A.1). Your body doesn’t get pushed backward because the net force on your center of mass is the forward ground force minus any other forces (like air resistance). If that forward ground force is larger than opposing forces, your net force is forward and you accelerate forward. If there were no friction (e.g., on ice) your foot couldn’t push the ground effectively and you’d slip instead—then you wouldn’t move forward. For the AP framing, emphasize action–reaction pairs always act on different objects (2.3.A) and that internal forces don’t change a system’s center-of-mass motion (2.3.A.2). For a quick review, see the Topic 2.3 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc) and try related practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Saying “internal forces don’t affect the center of mass motion” means that forces objects in a system exert on each other (internal forces) cancel in pairs and cannot change the motion of the system’s center of mass—only external forces can. By Newton’s 3rd law each internal force F_A on B has an equal and opposite F_B on A, so their vector sum is zero for the whole system. Therefore the center-of-mass acceleration is determined only by the net external force: a_cm = (∑F_ext)/M_total. That’s why collisions between parts of a closed system can change internal motion but leave the COM motion tied to external pushes (or none at all—so momentum of the whole system is conserved). This is exactly the point in the AP CED (2.3.A.2) and is useful on FRQs about momentum/COM or systems with internal interactions. For a quick review see the Topic 2.3 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc) and try related practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Tension is the net result of many tiny internal contact forces between adjacent bits of a string. In an ideal string (negligible mass, inextensible) each infinitesimal segment must have zero net force, so the pull from the left on a segment equals the pull from the right but opposite in direction. By Newton’s third law those internal forces come in equal-and-opposite pairs, so the scalar tension magnitude is the same everywhere in the string (CED 2.3.A.3.i–ii). That’s why, with an ideal pulley (negligible mass, negligible friction), the tension on both sides of the pulley is the same. If the string has nonnegligible mass, segments need net force to accelerate the string’s mass, so tension can vary along its length (CED 2.3.A.3.iii). For AP exam problems, draw free-body diagrams that treat the string’s tension as a single force on each connected object when the string is ideal (see the Topic 2.3 study guide: https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc and unit overview: https://library.fiveable.me/ap-physics-1-revised/unit-2). For extra practice, try problems at (https://library.fiveable.me/practice/ap-physics-1-revised).

An ideal string (CED 2.3.A.3.i–ii) is a model: it has negligible mass, doesn’t stretch, and therefore the tension is the same at every point. That lets you treat the string as transmitting a single force T to connected objects and apply Newton’s laws simply (useful on the AP free-response and multiple-choice problems). A real string can have mass and stretch. If the string has nonnegligible mass, different pieces need net force to accelerate, so tension can vary along its length (CED 2.3.A.3.iii). Stretching (elasticity) means the relationship between forces and extension matters (Hooke’s law). Also, nonideal pulleys (mass/friction) change tensions on either side. Quick tip: if a problem says “ideal string/pulley,” assume uniform tension and no stretch. If not stated or a string’s mass is given, expect a tension gradient and account for internal forces and distributed mass. For review and examples, see the Topic 2.3 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc) and more unit practice (https://library.fiveable.me/ap-physics-1-revised/unit-2, practice problems: https://library.fiveable.me/practice/ap-physics-1-revised).

If the string is massless and ideal, its tension is the same at every point: a single T transmits the force, and each tiny segment’s forces are equal and opposite (CED 2.3.A.3.i–ii). That’s why you can treat T as constant in many pulley/blocked problems on the AP exam. If the string has non-negligible mass, tension can change along its length (CED 2.3.A.3.iii). Each segment must accelerate (or support the weight of segments below it), so adjacent tensions differ by the net force needed for that segment. For a hanging chain, T is smallest at the bottom and grows toward the top because each higher point supports more weight. For an accelerating heavy rope, ΔT over a piece = (mass of piece) × (acceleration). Use Newton’s third-law pairs locally (each segment pulls its neighbor equally and oppositely) but remember the net within the string isn’t zero when segments have mass/acceleration. Review this in the Topic 2.3 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc) and practice similar problems (https://library.fiveable.me/practice/ap-physics-1-revised).

A pulley is just another object in an interaction pair described by Newton’s third law: the string pulls on the pulley and the pulley pulls back on the string with an equal and opposite force. If a mass hangs from a string over a pulley, each contact (mass↔string, string↔pulley, pulley↔axle) has its own action–reaction pair. We assume ideal pulleys/strings on the AP because the CED defines an ideal string as massless and unstretchable and an ideal pulley as having negligible mass and friction. That means (1) tension is the same everywhere in the string (so you can treat the string’s force on each object as equal in magnitude), and (2) the pulley doesn’t store rotational energy or introduce extra torques/energy losses. Those assumptions let you apply Newton’s 3rd law and Newton’s 2nd law cleanly in free-body diagrams and center-of-mass reasoning (CED 2.3.A.1, 2.3.A.3.i–iv). For AP practice and worked examples, check the Topic 2.3 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc) and unit review (https://library.fiveable.me/ap-physics-1-revised/unit-2).

You’re right that Newton’s third law says forces come in equal-and-opposite pairs, but the key is they act on different objects. For the book on the table: - Gravity (weight) mg acts downward on the book from Earth. - The table exerts a normal force N upward on the book. Those two can be equal (N = mg), so the net force on the book is zero and it doesn’t accelerate. The action–reaction pair for gravity is actually Earth on book (mg down on the book) and book on Earth (an equal upward force on Earth). Those two don’t cancel because one acts on the book and the other on Earth. Similarly, the table’s upward normal on the book pairs with the book’s downward force on the table. Internal forces within a system don’t change its center-of-mass motion (CED 2.3.A.2). If you want a short CED-aligned refresher on action–reaction, check the Topic 2.3 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc). For more practice, see the unit resources (https://library.fiveable.me/ap-physics-1-revised/unit-2) and thousands of problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Short answer: treat each object as its own system, draw separate free-body diagrams, use Newton’s 2nd law for each, and remember Newton’s 3rd law gives equal-and-opposite tensions at contact points—not a cancellation inside one FBD. How to do it step-by-step: 1. Draw separate FBDs for every mass (label T on each string segment, gravity, normal, friction as needed). AP wants clear diagrams and labels (CED 2.3.A, 2.2). 2. Write ΣF = m a for each object in the appropriate direction. For a pulley/string system, each mass may feel a different T only if the string/pulley aren’t ideal; for an ideal string T is the same everywhere (CED 2.3.A.3.i–ii). 3. Use Newton’s 3rd law to relate interaction forces: the string pulls the mass with T and the mass pulls the string with −T (these act on different objects). 4. If objects share acceleration (connected string), impose kinematic constraints (same magnitude, maybe opposite direction). 5. Solve the simultaneous equations for T and a. If pulleys/strings have mass or friction, note tensions can differ (CED 2.3.A.3.iii–iv). On the exam: draw clear FBDs, state assumptions (ideal string/pulley), show algebra—FRQs reward diagrams + equations (skills 1.A, 3.B). For a focused review, see the Topic 2.3 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc). For extra practice, use the unit practice sets (https://library.fiveable.me/practice/ap-physics-1-revised).

Newton’s third law (F_A on B = −F_B on A) tells you every interaction between two objects comes with equal and opposite forces. Those paired forces produce equal and opposite impulses (∫F dt), so the change in momentum of one object is exactly the negative change in momentum of the other: Δp_A = −Δp_B. If you treat the two as a single isolated system (no external net force), those internal, equal-and-opposite momentum changes cancel, so the total momentum of the system stays constant—that’s conservation of momentum. Key CED points: paired forces are internal to a system and don’t affect the system’s center-of-mass motion (2.3.A.1–2.3.A.2). This is exactly why momentum is conserved in collisions when external forces are negligible (Unit 2 ties into Unit 4 on momentum). For more practice and a focused review of Newton’s third law, see the Topic 2.3 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc) and extra problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Rockets work in space because of Newton’s third law: for every force there’s an equal and opposite force. The rocket pushes gas out the back (action); the gas pushes the rocket forward (reaction). You don’t need air to “push against”—the interaction is between the rocket and the exhaust. Think of the rocket+exhaust as one system: when the rocket expels mass backward at high speed, the system’s center-of-mass momentum is conserved, so the rocket’s center of mass moves forward. For the rocket alone, once the gas is expelled it’s no longer part of the rocket, so the expelled gas exerts a thrust force on the rocket (Frocket on gas = −Fgas on rocket). That thrust changes the rocket’s momentum and produces acceleration. This is exactly the Newton’s third-law / momentum idea in Topic 2.3 (see F_A on B = −F_B on A). For more review, check the Topic 2.3 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/3-newtons-third-law/study-guide/WPQ8Q8B1IdMQ6Uoc) and practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).