Newton’s Third Law (AP Physics C: Mechanics 2.3)

A horse pulls a cart forward. The cart pulls back on the horse with a force of exactly the same size. So how does anything ever move? The answer depends on two facts this topic makes precise: the two forces in such a pair act on different objects, and only forces from outside a system can change how the system as a whole moves.

Forces come in pairs

Whenever object A exerts a force on object B, object B exerts a force on object A that is equal in magnitude and opposite in direction:

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

This is Newton's third law. It follows from the idea in Topic 2.2 that a force is an interaction: one interaction between two objects shows up as two forces, one on each. The two forces in a third-law pair always:

have equal magnitudes and opposite directions;
act on different objects, one on each;
are the same kind of force (both gravitational, both normal, both friction);
exist at the same time. Neither one is a delayed "reaction"; they start and stop together.

To find a force's partner, swap the two objects in its name. The partner of "the gravitational force on the apple by Earth" is "the gravitational force on Earth by the apple". If an apple of mass $0.15\ \text{kg}$ falls with $g = 9.8\ \text{m/s}^2$ , Earth pulls it down with $1.47\ \text{N}$ , and the apple pulls Earth up with $1.47\ \text{N}$ . Equal forces don't mean equal effects: Earth's mass is $5.97 \times 10^{24}\ \text{kg}$ , so its acceleration toward the apple is

a = \frac{1.47\ \text{N}}{5.97\times 10^{24}\ \text{kg}} = 2.5\times 10^{-25}\ \text{m/s}^2,

far too small to detect.

A book resting on a table feels two forces of equal size, gravity down and the normal force up. These are not a third-law pair: they act on the same object (the book), and they are different kinds of force. They are equal only because the book isn't accelerating (Topic 2.4); in an elevator that speeds up, they aren't. The partner of the normal force on the book by the table is the normal force on the table by the book, pushing down on the tabletop.

Internal pairs cancel, which is why systems need outside forces

If both objects in a pair are inside the system you have chosen, both forces of the pair act on the system. They are equal and opposite, so they add to zero. Every internal force has its partner inside the system too, so all the internal forces together add to zero. Since the motion of the center of mass depends on the total force on the system (Topic 2.5), internal forces cannot change the velocity of the system's center of mass. They can only move the parts relative to each other.

Now the horse and cart make sense. For the horse-plus-cart system, the horse's pull on the cart and the cart's pull on the horse are internal, and they cancel. The horizontal forces that matter are external: the ground pushes forward on the horse's hooves (the partner of the hooves pushing back on the ground), and the ground and axles resist the cart's wheels. When the ground's forward push on the horse is bigger than the resistance on the cart, the system speeds up. If you choose the cart alone as the system, the horse's pull is external, and it is the force that accelerates the cart.

The choice of system decides which forces are internal, so it is your choice, not the problem's. Worked example 2 uses it both ways.

Tension: what a rope does

A rope looks like one object, but it is a chain of tiny segments, each pulling on its neighbors. Those pulls are third-law pairs: segment 1 pulls segment 2 toward itself, and segment 2 pulls segment 1 back with the same force. The tension at a point in a rope is the magnitude of the pull that the parts on either side of that point exert on each other. It is what any object tied to the rope at that point feels.

Here is why tension can change along a rope. Take a horizontal rope accelerating with acceleration $a$ , and look at a short segment of length $dx$ and mass $dm = \lambda\,dx$ . The rope ahead pulls it forward with tension $T + dT$ , and the rope behind pulls it back with $T$ . Newton's second law for the segment gives

(T + dT) - T = dm\,a \quad\Rightarrow\quad \frac{dT}{dx} = \lambda a.

If the rope's mass is negligible ( $\lambda \to 0$ ), $\frac{dT}{dx} = 0$ : the tension is the same everywhere along it, even while it accelerates. A rope with real mass needs a net force on each segment to accelerate it, so the tension grows toward the end being pulled. Gravity does the same thing to a rope hanging straight down: the rope at height $y$ above the bottom end holds up the rope below it, so $T(y) = \lambda g y$ , largest at the top.

The ideal pulley's job is redirection. A rope that runs horizontally to a pulley and then hangs down pulls on objects at both ends with the same tension, one horizontally and one vertically. A pulley with mass or axle friction needs different tensions on its two sides to start it spinning; Unit 5 handles that case.

Because an ideal string doesn't stretch, two objects tied together by it have accelerations of equal magnitude, which is the constraint that lets you solve for connected objects in Topic 2.5.

Worked examples

Worked exampleThe pairs on a fridge magnet

A magnet is stuck to the vertical side of a steel refrigerator and doesn't slide. List every force on the magnet, then name the third-law partner of each one.

Show solution

Search for forces on the magnet, naming the source of each (Topic 2.2): Earth pulls it, and it touches only the refrigerator, which can exert a normal force and friction. The magnet also attracts the steel, and the steel attracts it. So the forces on the magnet and their partners are:

Force on the magnet	Direction	Partner
gravity, by Earth	down	gravity on Earth by the magnet, up
magnetic attraction, by the fridge	toward the fridge	magnetic attraction on the fridge by the magnet, toward the magnet
normal force, by the fridge	away from the fridge	normal force on the fridge by the magnet, into the fridge
static friction, by the fridge	up	friction on the fridge by the magnet, down

Notice that the magnetic pull and the normal force on the magnet balance each other, and so do gravity and friction, but none of those balancing pairs is a third-law pair. Every partner acts on a different object (Earth or the fridge), which is why none of them appears on the magnet's free-body diagram.

Worked exampleA push on the ice

A $60\ \text{kg}$ student on skates stands at rest on frictionless ice next to a $15\ \text{kg}$ crate. The student pushes the crate horizontally with a force of $30\ \text{N}$ .

(a) What force does the crate exert on the student?

(b) Find the acceleration of each while they are in contact.

Show solution

(a) By Newton's third law, the crate pushes on the student with $30\ \text{N}$ , directed opposite to the push, toward the student's back.

(b) Choose each object as its own system, so the push is external to it. Vertically, gravity and the normal force balance for each. Horizontally, each has one force:

a_{\text{crate}} = \frac{30\ \text{N}}{15\ \text{kg}} = 2.0\ \text{m/s}^2 \ \text{(forward)}, \qquad a_{\text{student}} = \frac{30\ \text{N}}{60\ \text{kg}} = 0.50\ \text{m/s}^2 \ \text{(backward)}.

The forces are equal; the accelerations are not, because the masses differ by a factor of 4.

(c) Now choose the student and crate together. The push and its partner are internal and add to zero, and the ice is frictionless, so the net external horizontal force is zero. The center of mass doesn't accelerate. Check it from (b), with forward positive:

a_{cm} = \frac{(15)(2.0) + (60)(-0.50)}{75} = 0\ \text{m/s}^2.

The center of mass stays at rest while the two slide apart.

Worked exampleA heavy rope pulls a block

A block of mass $M = 8.0\ \text{kg}$ sits on a frictionless horizontal floor. A uniform rope of mass $m = 2.0\ \text{kg}$ and length $L = 3.0\ \text{m}$ is tied to it, and a student pulls the free end of the rope horizontally with a force $F = 40\ \text{N}$ . The rope stays horizontal.

(a) Calculate the acceleration of the block.

(b) Calculate the tension in the rope where it attaches to the block.

(c) Derive an expression for the tension $T(x)$ in the rope at a distance $x$ from the student's hand, in terms of $F$ , $M$ , $m$ , $L$ and $x$ .

(d) A classmate says the tension should be $40\ \text{N}$ everywhere in the rope, as for the ropes in most problems. Explain why the classmate is wrong here, and state when the classmate's claim would be correct.

Show solution

(a) Choose the block and rope together as the system. The tension forces inside the rope and the pull between rope and block are internal. The only external horizontal force is the student's pull (gravity and the floor's normal force are vertical and balance):

a = \frac{F}{M + m} = \frac{40\ \text{N}}{10.0\ \text{kg}} = 4.0\ \text{m/s}^2.

(b) Now choose the block alone. The only horizontal force on it is the rope's pull, so

T_{\text{block}} = Ma = (8.0\ \text{kg})(4.0\ \text{m/s}^2) = 32\ \text{N}.

(c) Choose as the system everything beyond the point $x$ : the rope from $x$ to $L$ plus the block. Its mass is $M + \frac{m}{L}(L - x)$ , and the only horizontal force on it is the tension $T(x)$ from the rope on the student's side:

T(x) = \left[M + \frac{m}{L}(L - x)\right]\frac{F}{M + m}.

Equivalently, the piece of rope between the hand and $x$ , of mass $\frac{m x}{L}$ , has $F$ forward and $T(x)$ backward, so $T(x) = F - \frac{m x}{L}\cdot\frac{F}{M+m}$ , which is the same expression. Checks: at $x = 0$ it gives $T = F = 40\ \text{N}$ ; at $x = L$ it gives $T = \frac{MF}{M+m} = 32\ \text{N}$ , matching (b). At the middle of the rope, $x = 1.5\ \text{m}$ , it gives $36\ \text{N}$ .

(d) Each segment of a rope with mass is accelerating, so it needs a net forward force: the rope ahead of it must pull harder than the rope behind it holds it back. The tension therefore drops from $40\ \text{N}$ at the hand to $32\ \text{N}$ at the block. The classmate's claim holds for an ideal rope, whose mass is negligible compared with the block's: with $m \to 0$ , the expression in (c) becomes $T = F$ at every $x$ .

Where students lose points

Check yourself

Question 1
A mosquito splatters on the windshield of a moving truck. During the collision, how does the force on the mosquito by the truck compare with the force on the truck by the mosquito?
1. A
  The force on the mosquito is much larger.
2. B
  The force on the truck is much larger.
3. C
  They have equal magnitudes.
4. D
  The force on the truck is zero, because the truck doesn't slow down noticeably.
Show answer
C
The two forces are a third-law pair, so their magnitudes are equal. The mosquito's tiny mass gives it an enormous acceleration, and the truck's large mass gives it an unnoticeable one, which is what makes choice A tempting.
Question 2
A uniform rope with nonnegligible mass hangs straight down from a ceiling hook, with nothing attached to its bottom end. Where is the tension in the rope greatest?
1. A
  At the bottom end
2. B
  At the middle
3. C
  At the top, where it meets the hook
4. D
  It is the same everywhere along the rope.
Show answer
C
The rope at any point holds up all of the rope below it, and the top point holds up the whole rope. Choice D is true only for a massless rope, whose tension doesn't change along its length.
Question 3
A uniform rope of mass $1.5\ \text{kg}$ and length $3.0\ \text{m}$ hangs straight down from a ceiling hook, with nothing attached to its bottom end. Taking $g = 9.8\ \text{m/s}^2$ , what is the tension at the point $1.0\ \text{m}$ below the hook?
Show answer
$9.8\ \text{N}$
Below that point hangs $2.0\ \text{m}$ of rope, two thirds of the rope, of mass $1.0\ \text{kg}$ . That piece is at rest, so the tension above it balances its weight: $T = (1.0)(9.8) = 9.8\ \text{N}$ .
Question 4
Two students pull horizontally on opposite ends of a light rope, each with a force of $50\ \text{N}$ , and the rope stays at rest. What is the tension in the rope?
1. A
  $0\ \text{N}$
2. B
  $25\ \text{N}$
3. C
  $50\ \text{N}$
4. D
  $100\ \text{N}$
Show answer
C
Any piece of the rope is pulled with $50\ \text{N}$ each way, so the two sides of any cut pull on each other with $50\ \text{N}$ . Choice D adds the two pulls, but they act on opposite ends; choice A adds them as vectors, which gives the net force on the rope, not its tension.

Newton’s Third Law

Forces come in pairs

Internal pairs cancel, which is why systems need outside forces

Tension: what a rope does

Worked examples

Where students lose points

Check yourself

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