AP Physics C: Mechanics · Topic 2.3

Topic 2.3: Newton's Third Law

Unit 2: Force and Translational Dynamics20-25% of the multiple-choice section

If object A exerts a force on object B, B exerts a force on A that is equal in magnitude and opposite in direction. AP Physics C states the law as AP Physics 1 does but drops the boundary limiting tension in a massive rope to qualitative description, so a C question can ask you to calculate it.

AP Physics: Unit 2 (topics 2.3 Newton's Third Law). AP Physics C: Mechanics Unit 2, Topic 2.3. One learning objective, 2.3.A, describe the interaction of two objects or systems using Newton's third law and a representation of paired forces exerted on each object or system. Supported by 2.3.A.1 (the paired-force statement, with F_A on B = -F_B on A printed), 2.3.A.2 (interactions between objects within a system, that is internal forces, do not influence the motion of a system's center of mass), 2.3.A.3 (tension is the macroscopic net result of forces that INFINITESIMAL segments of a string, cable, chain or similar system exert on each other in response to an external force; the AP Physics 1 wording of the same statement omits the word infinitesimal), 2.3.A.3.i (an ideal string has negligible mass and does not stretch under tension), 2.3.A.3.ii (the tension in an ideal string is the same at all points), 2.3.A.3.iii (in a string with nonnegligible mass, tension may not be the same at all points) and 2.3.A.3.iv (an ideal pulley has negligible mass and rotates about an axle through its center of mass with negligible friction). AP Physics C: Mechanics prints NO boundary statement on this topic. AP Physics 1 prints TWO that the C course does not: one limiting students to describing tension qualitatively in a string, cable, chain or similar system with mass, and one limiting interaction at a distance to gravitational forces. Removing the first is what allows a quantitative massive-rope question in AP Physics C. Suggested skills are 1.A, 2.C, 3.B and 3.C; AP Physics 1 lists 2.D where the C course lists 2.C. The exam-conventions box on the C: Mechanics Table of Information states that springs and strings are assumed to be ideal unless otherwise stated. One of the unit's six optional sample instructional activities is on Topic 2.3, the tug-of-war discussion.

What changes here, and it is not the law

The law itself is identical in both courses. Statement 2.3.A.1 reads the same in the AP Physics 1 and AP Physics C: Mechanics frameworks, and both print

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

What differs is the fencing around it. AP Physics 1 prints two boundary statements on Topic 2.3 and AP Physics C: Mechanics prints none.

The first AP Physics 1 boundary statement reads: "AP Physics 1 only expects students to describe tension qualitatively in a string, cable, chain, or similar system with mass. For example, students might note that the tension in a hanging chain is greater toward the top of the chain." The second reads: "The interaction between objects or systems at a distance is limited to gravitational forces in AP Physics 1. In AP Physics 2, gravitational, electric, and magnetic forces may be considered."

Remove the first fence and a whole class of question opens up. A rope with mass has a tension that varies continuously along its length, and finding that variation means writing the mass of the part of the rope ahead of a point as a function of position and applying the second law to it. Statement 2.3.A.3 in the AP Physics C framework even changes one word to point at this: it defines tension as the macroscopic net result of forces that infinitesimal segments of a string, cable, chain, or similar system exert on each other in response to an external force. AP Physics 1's version of the same statement says "segments" without the qualifier.

That is the differentiator on this page. Everything else about the third law is shared, and the second worked example below is the one an algebra-based course cannot set.

What the CED requires of Topic 2.3

One learning objective, and the shortest topic in Unit 2 by statement count.

Objective 2.3.A: describe the interaction of two objects or systems using Newton's third law and a representation of paired forces exerted on each object or system. Suggested skills 1.A, 2.C, 3.B and 3.C.

2.3.A.1: Newton's third law describes the interaction of two objects or systems in terms of the paired forces that each exerts on the other, with the equation above.
2.3.A.2: interactions between objects within a system (internal forces) do not influence the motion of a system's center of mass.
2.3.A.3: tension is the macroscopic net result of forces that infinitesimal segments of a string, cable, chain, or similar system exert on each other in response to an external force.
2.3.A.3.i: an ideal string has negligible mass and does not stretch when under tension.
2.3.A.3.ii: the tension in an ideal string is the same at all points within the string.
2.3.A.3.iii: in a string with nonnegligible mass, tension may not be the same at all points within the string.
2.3.A.3.iv: an ideal pulley is a pulley that has negligible mass and rotates about an axle through its center of mass with negligible friction.

No boundary statement. Four of the seven statements are about strings and pulleys, which tells you where the questions come from.

One more thing to know before you assume every rope is massless. The exam-conventions box printed with the AP Physics C: Mechanics Table of Information has three bullets, and the third is: "Springs and strings are assumed to be ideal unless otherwise stated." So the default is 2.3.A.3.i and 2.3.A.3.ii, and 2.3.A.3.iii applies only when a question says the rope has mass. When it does say so, it is saying so for a reason.

Paired forces never cancel, and why

The third-law pair acts on two different objects. Forces only add up, or cancel, when they act on the same object, so a third-law pair can never cancel anything.

This is the reason the classic objection fails. If the horse pulls the cart forward with the same force the cart pulls the horse backward, why does anything move? Because those two forces are on different bodies. Draw the cart's free-body diagram and the horse's pull appears there, along with friction from the ground; draw the horse's diagram and the cart's pull appears there, along with the ground's forward push on the horse's hooves. Each diagram has its own net force.

A reliable way to name the partner of any force is to write the force as "X on Y" and swap the names. The partner is "Y on X", it has the same magnitude, it points the opposite way, it is the same kind of force, and it acts at the same instant.

Force in your diagram	Its third-law partner	Acts on
Earth pulls block down	block pulls Earth up	Earth
Table pushes book up	book pushes table down	table
Rope pulls crate right	crate pulls rope left	rope
Road pushes tyre forward	tyre pushes road backward	road

Notice the first row. The partner of the gravitational force on a block is not the normal force from the table. It is the block's gravitational pull on the Earth, and it acts on the Earth. Confusing the normal force with the third-law partner of weight is a standing trap here, and the giveaway is that the two are only equal in magnitude in the special case of a level surface with no vertical acceleration. Put the book in a lift and they immediately differ, while the true pair stays equal forever.

Equal forces, unequal accelerations. Statement 2.3.A.1 fixes the forces, not the accelerations. Since $\vec{a} = \vec{F}_{\text{net}}/m$ for each body separately, the lighter object gets the larger acceleration. A ball hits the floor with the same force the floor uses on the ball; the ball bounces and the Earth does not visibly move, because the Earth's mass is about $10^{23}$ times larger.

Internal forces and the center of mass

Statement 2.3.A.2 is the one that reaches furthest into the rest of the course: interactions between objects within a system (internal forces) do not influence the motion of a system's center of mass.

The reason is the third law. Every internal force comes in a pair, both members of the pair are inside the system, and they are equal and opposite, so they cancel in the sum over all forces on the system. Only external forces survive that sum. Statement 2.5.A.3 states the consequence in Topic 2.5: the velocity of a system's center of mass will only change if a nonzero net external force is exerted on that system.

Two practical uses.

Choosing the system to make a force disappear. Two blocks in contact pushed across a table exert contact forces on each other. Treat the pair as one system and those forces vanish from the analysis, so a single equation gives the acceleration. Then, if you need the contact force itself, draw one block on its own and apply the second law to it with the acceleration you already have. The first worked example below is exactly this, done twice over.

Predicting what cannot happen. An astronaut who throws a wrench, two carts released by a compressed spring, an exploding firework: the center of mass of the whole system carries on exactly as it was, because the forces involved are internal. This is the fact that Unit 4 later turns into conservation of momentum.

A warning that follows from the same statement: the internal forces are still real and can still break things. They do not move the center of mass, which is not the same as doing nothing.

Ideal strings, real strings, and ideal pulleys

Ideal string. Statement 2.3.A.3.i says an ideal string has negligible mass and does not stretch under tension. Statement 2.3.A.3.ii gives the consequence you actually use: the tension in an ideal string is the same at all points within the string. That is why you can write one symbol $T$ for the whole rope in an Atwood machine.

Why negligible mass gives constant tension is worth one line, because the C exam can ask you to justify it. Take a tiny segment of the string of mass $\Delta m$ . The forces on it are the tensions pulling from either side, and Newton's second law gives $T_{\text{ahead}} - T_{\text{behind}} = \Delta m\, a$ . If $\Delta m \to 0$ , the difference goes to zero and the tension is uniform. If $\Delta m$ is not negligible, the difference is not zero, and that is statement 2.3.A.3.iii.

Real string. Statement 2.3.A.3.iii: in a string with nonnegligible mass, tension may not be the same at all points within the string. The word "may" is doing work. A massive rope lying slack on a table has zero tension everywhere; a massive rope being accelerated, or hanging vertically, does not. The general rule is that the tension at any point equals the mass of everything on the far side of that point times the acceleration, plus whatever external forces act on that far side. A hanging chain has more tension at the top because the top has to hold more chain.

Ideal pulley. Statement 2.3.A.3.iv defines it as a pulley with negligible mass that rotates about an axle through its center of mass with negligible friction. That is what lets a pulley redirect a string without changing the tension in it. A pulley with mass has rotational inertia and takes a net torque to spin up, so the tensions on the two sides of it differ. That is a Unit 5 problem, and Topic 5.6 is where it belongs; in Unit 2 the pulley is ideal unless the question says otherwise.

If you want a step-by-step routine for the algebra of a rope problem, the site's how to find tension guide covers it, and the tension glossary entry is a one-paragraph definition.

If you want the algebra-based version of this topic

If you are taking AP Physics 1, the page for you is AP Physics 1 Topic 2.3: Newton's Third Law. It covers the same law, the same seven-statement structure, and the same pulley and string vocabulary, and it stays inside the two AP Physics 1 boundary statements. If you are taking AP Physics C: Mechanics, this page is the one that goes past them.

	AP Physics 1 Topic 2.3	AP Physics C Topic 2.3
The law	$\vec{F}_{\text{A on B}} = -\vec{F}_{\text{B on A}}$	identical
Statements	2.3.A.1 to 2.3.A.3.iv	2.3.A.1 to 2.3.A.3.iv
Wording of 2.3.A.3	"segments" of a string	"infinitesimal segments"
Massive rope	qualitative description only	no such limit
Forces at a distance	limited to gravitational	no such limit
Suggested skills	1.A, 2.D, 3.B, 3.C	1.A, 2.C, 3.B, 3.C
Boundary statements	2	0

The suggested skills differ by one entry: AP Physics 1 lists 2.D, predict new values or factors of change using functional dependence between variables, where AP Physics C lists 2.C, compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario. That fits the rope problem, where you are comparing tension at two locations within one scenario.

Related reading on this site: the Newton's third law glossary entry for the definition, and Topic 2.1 on systems and center of mass, which statement 2.3.A.2 depends on.

How Topic 2.3 is tested

Unit 2 carries 20 to 25 percent of the multiple-choice section across about 15 to 25 class periods, the highest minimum weighting of any unit in AP Physics C: Mechanics (only Unit 3, at 15 to 25 percent, reaches the same 25 percent ceiling). Topic 2.3 has one learning objective out of the unit's nineteen.

The suggested skills are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.C, compare physical quantities between two or more scenarios or at different times and/or locations within a single scenario; 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.

Two of those four are Science Practice 3, which is 30 to 35 percent of the free-response section and 20 to 35 percent of the multiple-choice section between skills 3.B and 3.C. That is the signature of a topic tested through claims and justifications rather than through calculation, which matches the objective's own wording: describe the interaction using a representation of paired forces.

The CED lists an optional sample instructional activity on Topic 2.3: students in groups of two or three explain why a strong adult will win against a small child in tug of war, even though the rope always has the same tension at both ends, then support the reasoning with free-body diagrams. That question is a complete third-law exam item in itself. The answer lives outside the rope: the pair of forces at the rope's ends are equal, and the contest is decided by the friction each competitor can get from the ground.

The unit opener also lists as an essential question: if you apply the same amount of push to a car as you would a shopping cart, why doesn't the car move? Same structure, same answer shape.

For practice on the surrounding material, the forces and Newton's laws practice set is the closest fit on this site.

Three blocks in a line, and the pairs between them

Three blocks rest in contact on a frictionless horizontal surface: $m_1 = 1.0$ kg, $m_2 = 2.0$ kg and $m_3 = 3.0$ kg, in that order left to right. A horizontal force $F = 12$ N pushes on the left face of $m_1$ , toward the right. Find the acceleration, the force $m_1$ exerts on $m_2$ , and the force $m_2$ exerts on $m_3$ . Then name the third-law partner of each.

Declare the convention: rightward positive, the direction of $F$ .
Use 2.3.A.2 first. The contact forces between the blocks are internal to the three-block system, so they do not affect the motion of that system's center of mass. Treat all three as one object: $a = \dfrac{F}{m_1 + m_2 + m_3} = \dfrac{12}{6.0} = 2.0\ \mathrm{m/s^2}$ .
Now isolate to get an internal force. Take blocks 2 and 3 together as a system. The only horizontal force on that system is the push from block 1, so $F_{1\text{ on }2} = (m_2 + m_3)a = (5.0)(2.0) = 10$ N.
Take block 3 alone. Its only horizontal force is the push from block 2, so $F_{2\text{ on }3} = m_3 a = (3.0)(2.0) = 6.0$ N.
Check every block against the second law independently. Block 1: $12 - 10 = 2.0$ N and $m_1 a = 2.0$ N. Block 2: $10 - 6.0 = 4.0$ N and $m_2 a = 4.0$ N. Block 3: $6.0$ N and $m_3 a = 6.0$ N. All three agree.
Name the partners. The partner of $F_{1\text{ on }2} = 10$ N rightward is $F_{2\text{ on }1} = 10$ N leftward, acting on block 1. The partner of $F_{2\text{ on }3} = 6.0$ N rightward is $F_{3\text{ on }2} = 6.0$ N leftward, acting on block 2.
Read the pattern. The contact force at each interface equals the mass ahead of that interface times the acceleration, so it decreases from 12 N to 10 N to 6.0 N as you move along the chain. Each block has to accelerate less mass than the one behind it.
The trap to avoid: $F_{1\text{ on }2} = 10$ N and $F_{2\text{ on }1} = 10$ N are a third-law pair, so they are equal and opposite, but they do not cancel, because one acts on block 2 and the other on block 1. If they cancelled, nothing would accelerate.

$a = 2.0\ \mathrm{m/s^2}$ ; $F_{1\text{ on }2} = 10$ N and $F_{2\text{ on }3} = 6.0$ N, both rightward. Their partners are 10 N leftward on block 1 and 6.0 N leftward on block 2. Each contact force equals the mass ahead of it times the shared acceleration.

Tension along a rope that has mass

A uniform rope of mass $m_r = 1.2$ kg and length $L = 2.5$ m lies on a frictionless horizontal surface with one end attached to a block of mass $M = 4.0$ kg. A horizontal force $F = 22$ N is applied to the free end of the rope. Derive an expression for the tension a distance $x$ from the pulled end, then evaluate it at $x = 0$ , at the midpoint, and at the rope-block junction.

Note first why this problem exists in AP Physics C and not in AP Physics 1: the AP Physics 1 boundary statement on Topic 2.3 limits students to describing tension qualitatively in a string with mass. There is no such boundary statement in AP Physics C: Mechanics, so the calculation is fair game.
Declare the convention: positive in the direction of $F$ , along the rope toward the block.
Whole system first. The rope and block accelerate together: $a = \dfrac{F}{M + m_r} = \dfrac{22}{5.2} = 4.2308\ \mathrm{m/s^2}$ .
Set up the mass distribution. The rope is uniform, so its linear mass density is $\lambda = \dfrac{m_r}{L} = \dfrac{1.2}{2.5} = 0.48$ kg/m, using the definition of statement 2.1.B.3.i.
Cut the rope at position $x$ and look at everything ahead of the cut. That is the block plus a length $L - x$ of rope, so its mass is $M + \lambda(L - x)$ . The only horizontal force on that piece is the tension $T(x)$ pulled from behind, because the surface is frictionless.
Apply the second law to the piece ahead of the cut: $T(x) = \left[M + \lambda (L - x)\right] a$ . That is the derivation, and the numbers follow.
At $x = 0$ , the pulled end: $T = (4.0 + 0.48 \times 2.5)(4.2308) = (5.2)(4.2308) = 22.0$ N. This must equal $F$ , and it does, which is the check that the setup is right.
At the midpoint $x = 1.25$ m: $T = (4.0 + 0.48 \times 1.25)(4.2308) = (4.6)(4.2308) = 19.5$ N.
At $x = L = 2.5$ m, the junction with the block: $T = (4.0)(4.2308) = 16.9$ N. That is the force the rope actually delivers to the block, and it is 5.1 N less than the force you applied at the other end.
Differentiate to see the shape: $\dfrac{dT}{dx} = -\lambda a = -(0.48)(4.2308) = -2.03$ N/m. The tension falls linearly, at a constant 2.03 newtons per metre, because the rope is uniform. A nonuniform rope would give a curved graph, and the slope at any point would still be $-\lambda(x) a$ .
Check the limit that recovers the ideal case. Let $m_r \to 0$ and $\lambda \to 0$ , and $T(x) \to Ma = F$ everywhere: the tension becomes uniform, which is statement 2.3.A.3.ii.

$T(x) = \left[M + \lambda(L - x)\right]a$ with $\lambda = m_r/L$ and $a = F/(M + m_r)$ . Numerically $a = 4.23\ \mathrm{m/s^2}$ , and the tension falls linearly from 22.0 N at the pulled end through 19.5 N at the midpoint to 16.9 N where the rope meets the block, at a rate of 2.03 N/m.

Frequently asked questions

Is Newton's third law different in AP Physics C than in AP Physics 1?

The law is the same. Essential knowledge 2.3.A.1 in both course and exam descriptions states that Newton's third law describes the interaction of two objects or systems in terms of the paired forces that each exerts on the other, and both print that the force of A on B equals the negative of the force of B on A. The difference is the boundary statements. AP Physics 1 prints two on this topic, one limiting tension in a massive string to qualitative description and one limiting forces at a distance to gravitational. AP Physics C: Mechanics prints neither, so a C question can ask you to calculate the tension at a point inside a rope that has mass.

How do you find the tension in a rope that has mass?

Cut the rope at the point of interest and apply Newton's second law to everything on the far side of the cut. The tension at that point is the only force the near side exerts on the far side, so it equals the mass beyond the cut times the acceleration, plus any external forces acting beyond the cut. For a uniform rope of linear mass density lambda pulling a block of mass M on a frictionless surface, the tension a distance x from the pulled end is the quantity M plus lambda times the remaining length, all multiplied by the acceleration. It falls off linearly, with slope minus lambda times a. Essential knowledge 2.3.A.3.iii is the statement that allows this: in a string with nonnegligible mass, tension may not be the same at all points.

Why is the tension the same everywhere in an ideal string?

Because an ideal string has negligible mass. Essential knowledge 2.3.A.3.i defines an ideal string as one with negligible mass that does not stretch under tension, and 2.3.A.3.ii states that its tension is the same at all points. The argument is one line of Newton's second law: for a small segment of string of mass delta m, the difference between the tensions pulling on its two ends equals delta m times the acceleration. Send delta m to zero and the difference goes to zero. The AP Physics C: Mechanics exam conventions state that springs and strings are assumed to be ideal unless otherwise stated, so uniform tension is the default.

Is the normal force the third-law partner of weight?

No, and it is an easy one to make. The gravitational force on a book is exerted by the Earth, so its third-law partner is the book's gravitational pull on the Earth, which acts on the Earth. The normal force is exerted by the table, so its partner is the book pushing down on the table. The two happen to be equal in magnitude when the surface is level and there is no vertical acceleration, which is why they get confused, but put the book in an accelerating lift and they differ immediately while the true third-law pairs stay equal.

Why do third-law pairs not cancel each other out?

Because they act on different objects. Forces add or cancel only within a single free-body diagram, and a third-law pair is split across two diagrams by construction: the force of A on B belongs on B's diagram and the force of B on A belongs on A's. That is also why essential knowledge 2.3.A.2 says internal forces do not influence the motion of a system's center of mass. When you draw a boundary around both objects, both members of the pair are inside it, and only then do they cancel in the sum, leaving the external forces to determine how the system as a whole moves.

What is an ideal pulley in AP Physics C?

Essential knowledge 2.3.A.3.iv defines an ideal pulley as a pulley that has negligible mass and rotates about an axle through its center of mass with negligible friction. Those properties are what let a pulley change the direction of a string without changing the tension in it, so the same symbol T can be used on both sides. A pulley with nonnegligible mass has rotational inertia and needs a net torque to change its angular speed, which means the tensions on either side of it must differ. That case belongs to Unit 5, Newton's second law in rotational form, not to Unit 2.

If tension is the same at both ends of a tug-of-war rope, how does anyone win?

The contest is decided at the ground, not at the rope. The rope pulls each competitor toward the other with the same magnitude, which is Newton's third law together with the ideal-string result that tension is uniform. Each competitor's own free-body diagram also contains the friction force from the ground, and it is that force that differs. Whoever can get more friction accelerates the other way. This is one of the AP Physics C: Mechanics course and exam description's own optional sample instructional activities for Unit 2, and the unit opener lists a version of it among its essential questions.