Action–reaction pairs
Key idea: When two bodies interact, each exerts an equal and opposite force on the other. These forces act on different bodies, so they do not cancel on one body's free-body diagram.
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Guided lesson
Guided steps
- Name the two bodies that receive the paired forces.
- Swap the bodies and reverse the force direction.
- State the paired force, its body, size and direction.
- Compare balanced forces with each force’s separate interaction pair.
- Calculate the resultant using only forces on the cart.
- Explain which body receives each force and acceleration.
- Compare accelerations using each body’s mass and the equal forces.
- Identify which body receives each lamp or cable force.
- Find the paired force size and both acceleration magnitudes.
- Explain which body receives each force and acceleration.
- Explain why the weight’s pair stays the same as support changes.
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ContinueWhen you push off from the side of a swimming pool, the wall pushes you back just as hard, and that push sends you gliding away. Forces always come in pairs like this, one on each of the two bodies that interact. Telling these action–reaction pairs apart from two forces that simply balance on one body is the key to using them correctly.
Forces come in action–reaction pairs
Two bodies, two forces
A force is always an interaction between two bodies. When body A exerts a force on body B, body B exerts a force on body A at the same time. These two forces are an action–reaction pair. They are:
- equal in size;
- opposite in direction;
- the same kind of force (both gravitational, or both contact forces);
- acting on two different bodies.
To write the pair of a force, swap the two bodies and reverse the direction: the pair of “the force of A on B, to the right” is “the force of B on A, to the left”.
| Force | Its action–reaction pair |
|---|---|
| Earth pulls a student downward (her weight) | the student pulls Earth upward |
| a foot pushes the ground backward | the ground pushes the foot forward |
| a swimmer’s hands push the water backward | the water pushes the swimmer forward |
| a rocket pushes its exhaust gas downward | the exhaust gas pushes the rocket upward |
| a book pushes down on a table | the table pushes up on the book |
Worked example 1
A swimmer and the water
Problem
A swimmer’s hands push the water backward with a force of 80 N. State the other force of the pair and what it does.
Show full solution
Name the interaction
Method
Identify the two bodies and the given force.Reason
An interaction pair always involves the same two bodies.Working
Answer
Swimmer and water; given force: swimmer on water, 80 N backward
Swap the bodies and reverse the direction
Method
Write the force of the water on the swimmer.Reason
The two forces of a pair are equal in size and opposite in direction.Working
Answer
The water pushes the swimmer forward with 80 N
Say what each force does
Working
Answer
Both are contact forces from the same push, acting on different bodies. The forward push on the swimmer is what moves the swimmer forward.
Check your understanding 1
Earth pulls a book downward with a force of 12 N. State the other force of the pair, the body it acts on, and its direction.
Show answer
The book pulls Earth upward with a gravitational force of 12 N. It acts on Earth, not on the book.
A pair is not two balanced forces
A book rests on a table. Two forces act on the book: its weight (Earth pulls it down) and the normal force (the table pushes it up). They are equal and opposite, so the resultant force on the book is zero.
These two forces are not an action–reaction pair: both act on the same body, and they are different kinds of force. Each has its own pair:
- the pair of the book’s weight is the book’s gravitational pull on Earth;
- the pair of the normal force is the book’s push down on the table.
| Two balanced forces | An action–reaction pair |
|---|---|
| act on one body | act on two different bodies |
| can be different kinds of force | are always the same kind of force |
| are equal only when the resultant force is zero | are always equal in size |
| add to zero on that body’s free-body diagram | never appear on the same free-body diagram |
Action–reaction forces do not cancel on either body’s free-body diagram, because they act on different bodies. Each force goes on its own body’s free-body diagram, and whether a body accelerates depends only on the forces acting on that body.
Check your understanding 2 Horse and cart
A horse pulls a cart forwards with 60 N while resistance on the cart is 20 N backwards. A learner also subtracts the cart’s 60 N pull on the horse and obtains −20 N. Find the resultant force on the cart, taking forward as positive.
Show answer
The cart’s 60 N pull acts on the horse, so it does not belong on the cart’s free-body diagram. The forces on the cart give a resultant of 60 - 20 = 40 N forwards. The equal interaction force on the horse does not cancel the horse’s pull on the cart, because they act on different bodies.
Check your understanding 3 Explain the extra force in the cart calculation
The learner subtracted the cart’s pull on the horse when finding the cart’s resultant. Explain why that force belongs on a different free-body diagram.
Show answer
The cart’s 60 N pull acts on the horse. The horse’s 60 N pull acts on the cart; its interaction pair therefore goes on the horse’s diagram. Only the horse’s forward pull and the backward resistance act horizontally on the cart. Paired forces act on different bodies, so they do not cancel on one diagram.
Worked example 2
Equal forces, different accelerations
Problem
A truck crashes into a small car. The truck exerts a force on the car. Compare the force of the car on the truck, and explain why the car has a larger change in velocity. During the short impact, assume other horizontal forces are negligible.
Show full solution
Compare the pair of forces
Method
The two forces form an action–reaction pair.Reason
Paired forces are always equal in size and opposite in direction.Working
Answer
Force of car on truck = force of truck on car, opposite in direction. The truck does not push harder because it is bigger.
Apply F = ma to each vehicle
Method
Find each acceleration from its own force and mass.Reason
The forces have the same size, but the car has the smaller mass.Working
Answer
a = F/m is larger for the car
Conclude
Working
Answer
The car has the larger acceleration, so its velocity changes more during the crash.
Check your understanding 4
A lamp hangs at rest from a cable. Classify each pair: (i) the cable’s pull up on the lamp and the lamp’s weight; (ii) the cable’s pull up on the lamp and the lamp’s pull down on the cable.
Show answer
(i) Both forces act on the lamp: they are balanced forces, not an action–reaction pair. (ii) One force acts on the lamp and one on the cable, and both come from the same lamp–cable interaction: they are an action–reaction pair.
Check your understanding 5 Two skaters
A 60 kg skater and a 30 kg skater stand still on ice and push each other apart. The 60 kg skater pushes the other with a force of 120 N. Find the size of the force on the 60 kg skater and the magnitude of each skater’s acceleration. Ignore friction.
Show answer
The 60 kg skater is pushed with 120 N in the opposite direction. Accelerations: a = 120/60 = 2.0 m/s² for the 60 kg skater and a = 120/30 = 4.0 m/s² for the 30 kg skater, in opposite directions.
Check your understanding 6 Compare the skaters’ directions
Suppose the 60 kg skater pushes the 30 kg skater to the right. Which way does each accelerate? Explain why equal force sizes give different accelerations. You may draw one force arrow on each skater on paper.
Show answer
The 30 kg skater accelerates right and the 60 kg skater left. The interaction forces are equal and opposite, but each acts on a different skater. Dividing the same 120 N magnitude by 30 kg gives 4 m/s², whereas dividing by 60 kg gives 2 m/s². Compare one force arrow on each body; the lighter skater’s acceleration is twice as large.
Check your understanding 7 The interaction pair survives a change of motion
A 10 N book first rests on a table, then hangs stationary from a string, then is released and falls with negligible air resistance. For each stage, state the force paired with the book’s weight. Does its pair change from the table’s normal force to string tension and then disappear? Explain which body receives the paired force and how that differs from forces that balance the book’s weight.
Show answer
In all three stages, the pair is the book’s gravitational pull on Earth, 10 N in the direction towards the book. It acts on Earth, not on the book. While the book is stationary, the table’s normal force or the string’s tension balances the weight on the book; neither is its gravitational interaction pair. After release the book has no upward support force, yet its gravitational interaction with Earth and the paired force remain.
Challenge yourself
Two identical trolleys are pushed apart by a compressed spring. One trolley is then loaded and the test is repeated. Compare the forces on the trolleys and predict which trolley has the larger acceleration.
Check your thinking
The spring exerts equal-sized forces in opposite directions on the two trolleys. The loaded trolley has the smaller acceleration because a = F/m. Equal force does not mean equal acceleration when the masses differ.
Common mistakes
- Calling weight and the normal force an action–reaction pair. Both act on the same body, so they are balanced forces.
- Saying the heavier or faster body pushes harder. The two forces of a pair are always equal in size.
- Saying action and reaction cancel, so nothing can move. They act on different bodies.
- Saying the reaction comes after the action. Both forces act at the same time, for as long as the bodies interact.
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Syllabus and review details
- SEC G2 Science Physics component 2027 · 2027
Physics section, PDF pages 8–23
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