🔄 Action-Reaction Pairs

Part 1 of 7 — Newton's Third Law

Newton's Third Law is one of the most frequently misunderstood ideas in physics. It's simple to state but tricky to apply — and it's a favorite on the AP exam.

For every action, there is an equal and opposite reaction.

But what does this really mean? Let's break it down carefully.

Newton's Third Law — Precise Statement

If object A exerts a force on object B, then object B exerts a force on object A that is equal in magnitude and opposite in direction.

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

Three Critical Properties of Action-Reaction Pairs

Additional Rules

• Action-reaction pairs are the same type of force (both gravitational, both normal, both contact, etc.)
• They exist simultaneously — one cannot exist without the other
• They never cancel each other because they act on different objects

Action-Reaction Examples

Example 1: You Push a Wall

• Action: You push the wall with 50 N to the right
• Reaction: The wall pushes you with 50 N to the left
• These act on different objects (you and the wall)

Example 2: Earth Pulls an Apple

• Action: Earth pulls the apple down with gravitational force $mg$
• Reaction: The apple pulls Earth up with gravitational force $mg$
• Earth accelerates negligibly because $a = F/M_{\text{Earth}} \approx 0$

Example 3: Foot on Ground (Walking)

• Action: Your foot pushes backward on the ground
• Reaction: The ground pushes your foot forward (friction)
• This is what propels you forward!

Common Misconception Alert ⚠️

The normal force on a book sitting on a table is NOT the reaction to the book's weight!

• Weight = Earth pulls book down (gravitational)
• Normal = Table pushes book up (contact)
• These are different types of forces acting on the same object

Actual pairs:

• Earth pulls book ↔ Book pulls Earth (gravitational pair)
• Book pushes table ↔ Table pushes book (contact pair)

How to Identify Action-Reaction Pairs

The "Flip Test"

To find the reaction to any force:

1. Identify the two objects — the one exerting the force and the one receiving it
2. Swap the objects — the reaction force has the same two objects but swapped

Template:

• Action: "Object A pushes/pulls Object B"
• Reaction: "Object B pushes/pulls Object A"

Example

Every force in the universe has a reaction partner. You can never have an isolated single force.

Action-Reaction Concept Check 🎯

Force Pair Reasoning 🧮

1. A person pushes a 30 kg box with 150 N. How much force does the box exert on the person (in N)?

2. Earth exerts a gravitational force of 9.8 N on a 1 kg apple. What gravitational force does the apple exert on Earth (in N)?

3. A 1000 kg car pushes backward on the road with 3000 N (via its tires). The road pushes the car forward with how many N?

Round all answers to 3 significant figures.

Classify the Force Pairs 🔍

Exit Quiz — Action-Reaction Pairs ✅

🤝 Identifying Force Pairs

Part 2 of 7 — Newton's Third Law

Identifying action-reaction pairs correctly is one of the most tested skills on the AP Physics 1 exam. In this lesson, we'll practice recognizing force pairs across different types of interactions: contact forces and gravitational forces.

Contact Force Pairs

Contact forces arise when two objects touch. The Third Law pair involves both objects.

Normal Force Pair

When a block sits on a table:

• Block pushes down on table (contact force)
• Table pushes up on block (normal force)

$\vec{F}_{\text{block on table}} = -\vec{F}_{\text{table on block}}$

Friction Force Pair

When you slide a box across the floor:

• Box pushes backward on floor (friction)
• Floor pushes forward on box (friction)

Tension Force Pair

When a rope connects to a block:

• Rope pulls block (tension on block)
• Block pulls rope (tension on rope)

Note: Every contact point between two objects creates a Third Law pair.

Gravitational Force Pairs

Gravity is a mutual force between any two masses:

$F = G\frac{m_1 m_2}{r^2}$

Earth-Object Pair

• Earth pulls object down with force $mg$
• Object pulls Earth up with force $mg$

Earth barely accelerates because $a_{\text{Earth}} = mg/M_{\text{Earth}} \approx 0$.

Why We Don't Notice Earth's Acceleration

For a 1 kg ball:

• Ball: $a = 9.8$ m/s² (very noticeable!)
• Earth: $a = 9.8/(6 \times 10^{24}) \approx 1.6 \times 10^{-24}$ m/s² (completely undetectable)

Important Distinction

These are two separate pairs — the normal force is NOT the pair of the weight!

Force Pairs with Multiple Objects

Block A on Block B on Table

Consider block A (2 kg) stacked on block B (5 kg) on a table.

Forces on Block A:

• Weight: Earth pulls A down ($W_A = 19.6$ N)
• Normal: B pushes A up ($N_{BA}$)

Third Law pairs involving A:

1. Earth pulls A ↔ A pulls Earth
2. B pushes A up ↔ A pushes B down ($N_{AB} = N_{BA}$)

Forces on Block B:

• Weight: Earth pulls B down ($W_B = 49$ N)
• Normal from A: A pushes B down ($N_{AB} = 19.6$ N)
• Normal from table: Table pushes B up ($N_{\text{table}})$

Equilibrium of B: $N_{\text{table}} = W_B + N_{AB} = 49 + 19.6 = 68.6 \text{ N}$

The table supports the weight of both blocks.

Force Pair Identification 🎯

Force Pair Calculations 🧮

Block A (3 kg) sits on top of Block B (7 kg), which sits on a table. Use $g = 9.8$ m/s².

1. What is the magnitude of the force that Block A exerts on Block B (in N)?

2. What is the normal force the table exerts on Block B (in N)?

3. What is the magnitude of the gravitational force that Block A exerts on Earth (in N)?

Round all answers to 3 significant figures.

Classify Each Force Pair 🔍

Exit Quiz — Identifying Force Pairs ✅

💡 Why Action-Reaction Doesn't Mean Nothing Moves

Part 3 of 7 — Newton's Third Law

If the forces are always equal and opposite, why doesn't everything just cancel out? Why can anything move at all? This is the #1 conceptual trap with Newton's Third Law, and this lesson will clear it up once and for all.

The "Paradox"

Student question: "If the horse pulls the cart with force $F$, and the cart pulls the horse back with force $-F$, the total force is zero, so nothing should move. Right?"

Wrong! Here's why:

The Critical Error

The student is adding forces that act on different objects. You can only add forces that act on the same object to find the net force.

Correct Analysis

Forces on the cart:

• Horse pulls cart forward: $+F$
• Friction/ground on cart: $-f_{\text{cart}}$
• Net force on cart: $F - f_{\text{cart}}$

Forces on the horse:

• Cart pulls horse backward: $-F$
• Ground pushes horse forward: $+f_{\text{ground}}$
• Net force on horse: $f_{\text{ground}} - F$

If $f_{\text{ground}} > F > f_{\text{cart}}$, both the horse and the cart accelerate forward!

Key rule: To determine if an object accelerates, look at the forces ON THAT OBJECT ONLY.

System vs. Individual Object Analysis

Individual Object Analysis

To find the acceleration of a single object, draw its FBD and apply $F_{\text{net}} = ma$ to that object alone.

Action-reaction pairs always involve two different objects, so they appear on two different FBDs — never on the same one.

System Analysis

When analyzing multiple objects as a system:

• Internal forces (Third Law pairs between objects in the system) cancel
• Only external forces determine the system's acceleration

Example: Person Pushing a Box

Person (60 kg) pushes box (20 kg) with 100 N on a frictionless floor.

System approach (person + box):

• External forces: only friction from ground on person's feet = 100 N forward
• Wait — on a frictionless floor, the person pushes the box and the floor provides no traction. Let's say the person can push off a wall.
• The person pushes the wall with force $F$; the wall pushes back on the person with $F$
• System mass: $60 + 20 = 80$ kg
• If the person exerts 100 N: $a = 100/80 = 1.25$ m/s² for the whole system

Individual analysis (box only):

• Person pushes box: 100 N... but wait, we need to find the actual contact force
• If $a = 1.25$ m/s²: $F_{\text{on box}} = m_{\text{box}} \times a = 20 \times 1.25 = 25$ N

Real-World Examples

Rocket Propulsion

• Rocket pushes exhaust gases backward
• Exhaust pushes rocket forward (Third Law)
• Net force on rocket alone: thrust forward > weight → acceleration up!

Swimming

• Swimmer pushes water backward with hands
• Water pushes swimmer forward
• Net force on swimmer alone: forward push from water > drag → swimmer moves forward

Walking

• Foot pushes ground backward
• Ground pushes foot forward (static friction)
• Net force on person alone: ground's push > air resistance → you walk forward

Key Pattern

In every case, the object accelerates because the net external force on that object is nonzero. The reaction force acts on a different object and doesn't affect the first object's motion.

Resolving the "Paradox" 🎯

System and Object Analysis 🧮

A person (80 kg) pushes a cart (40 kg) on a frictionless surface. The system accelerates at 2 m/s².

1. What is the net external force on the system (in N)?

2. What force does the person exert on the cart (in N)?

3. What force does the cart exert on the person (in N)?

Conceptual Reasoning 🔍

Exit Quiz — Why Things Move ✅

🔗 Connected Objects — Strings, Pulleys, and the Atwood Machine

Part 4 of 7 — Newton's Third Law

When objects are connected by strings or ropes over pulleys, they form systems where the motion of one affects the motion of all others. The Atwood machine is the classic AP Physics 1 problem of this type.

Ideal Strings and Pulleys

Ideal (Massless) String

In AP Physics 1, strings are assumed to be:

• Massless — the string has negligible mass
• Inextensible — the string doesn't stretch

Key consequence: Tension is the same throughout an ideal string, even if it bends around a pulley.

Ideal (Massless, Frictionless) Pulley

An ideal pulley:

• Has negligible mass
• Has no friction at the axle
• Only redirects the tension — it doesn't change the magnitude

Constraint: Same Magnitude of Acceleration

If two objects are connected by an inextensible string:

• They have the same magnitude of acceleration
• If one speeds up, the other speeds up at the same rate
• If one moves 1 m, the other moves 1 m

The Atwood Machine

Two masses ($m_1$ and $m_2$, with $m_2 > m_1$) connected by a string over a frictionless, massless pulley.

Setting Up the Problem

FBD of $m_1$ (lighter, accelerates up): $T - m_1 g = m_1 a$

FBD of $m_2$ (heavier, accelerates down): $m_2 g - T = m_2 a$

Solving for Acceleration

Add the two equations: $m_2 g - m_1 g = (m_1 + m_2)a$

$\boxed{a = \frac{(m_2 - m_1)}{(m_1 + m_2)} g}$

Solving for Tension

Substitute $a$ back: $T = m_1(g + a) = m_1 g + m_1 \frac{(m_2 - m_1)g}{m_1 + m_2}$

$\boxed{T = \frac{2m_1 m_2}{m_1 + m_2} g}$

Special Cases

Table-Pulley System

A block on a frictionless table ($m_1$) connected by a string over a pulley to a hanging block ($m_2$).

FBD of $m_1$ (on table, accelerates horizontally): $T = m_1 a$

FBD of $m_2$ (hanging, accelerates down): $m_2 g - T = m_2 a$

Solving

Add the equations: $m_2 g = (m_1 + m_2)a$

$\boxed{a = \frac{m_2}{m_1 + m_2} g}$

$\boxed{T = \frac{m_1 m_2}{m_1 + m_2} g}$

Key Observations

• Only the hanging mass provides the driving force ($m_2 g$)
• Both masses resist acceleration (total inertia = $m_1 + m_2$)
• The tension is always less than $m_2 g$ (otherwise $m_2$ wouldn't accelerate)

Connected Object Concepts 🎯

Atwood Machine Calculations 🧮

An Atwood machine has $m_1 = 4$ kg and $m_2 = 6$ kg. Use $g = 10$ m/s².

1. What is the acceleration of the system (in m/s²)?

2. What is the tension in the string (in N)?

3. If released from rest, how fast is each mass moving after 3 seconds (in m/s)?

System Analysis Quick Check 🔍

Exit Quiz — Connected Objects ✅

🚗 Two-Body Problems

Part 5 of 7 — Newton's Third Law

Two-body problems involve objects that push or pull each other through direct contact (no string needed). The key: the contact force between them forms a Third Law pair. We analyze each object separately while respecting the constraint that they move together.

Direct-Contact Two-Body Problems

Setup

Two blocks ($m_1$ and $m_2$) are touching on a frictionless surface. An external force $F$ is applied to $m_1$, pushing both blocks together.

The Contact Force

The blocks push on each other with a contact force:

• $m_1$ pushes $m_2$ forward with force $F_{12}$
• $m_2$ pushes $m_1$ backward with force $F_{21} = F_{12}$ (Third Law)

System Analysis (Both Blocks Together)

$F = (m_1 + m_2)a$ $a = \frac{F}{m_1 + m_2}$

Individual Analysis

Block $m_2$ (only force: contact from $m_1$): $F_{12} = m_2 a = \frac{m_2 F}{m_1 + m_2}$

Block $m_1$ (external force minus contact from $m_2$): $F - F_{21} = m_1 a$ $F_{21} = F - m_1 a = F - \frac{m_1 F}{m_1 + m_2} = \frac{m_2 F}{m_1 + m_2}$

Both give the same answer — consistent ✓

Does It Matter Which Block You Push?

Pushing $m_1$ (force applied to the left block)

$a = \frac{F}{m_1 + m_2}$

Contact force between blocks: $F_{\text{contact}} = \frac{m_2}{m_1 + m_2} F$

Pushing $m_2$ (force applied to the right block)

$a = \frac{F}{m_1 + m_2} \quad \text{(same acceleration!)}$

Contact force between blocks: $F_{\text{contact}} = \frac{m_1}{m_1 + m_2} F$

Key Insight

• The acceleration is the same regardless of which block you push
• The contact force is different! It equals the mass of the block NOT being pushed, divided by the total mass, times $F$

Example

$m_1 = 2$ kg, $m_2 = 8$ kg, $F = 50$ N

• Pushing $m_1$: Contact = $\frac{8}{10}(50) = 40$ N
• Pushing $m_2$: Contact = $\frac{2}{10}(50) = 10$ N
• Acceleration either way: $a = 50/10 = 5$ m/s²

Stacked Block Problems

Setup

Block A sits on top of block B on a frictionless surface. A horizontal force $F$ is applied to block B.

If A and B move together (no sliding), they have the same acceleration.

What Accelerates Block A?

The only horizontal force on A is friction from B (static friction).

$f_s = m_A a$

What Does This Mean?

• Friction from B on A is forward (propels A)
• By Third Law, friction from A on B is backward (slows B)

Maximum Acceleration Before Sliding

$f_{s,\max} = \mu_s m_A g$

$m_A a_{\max} = \mu_s m_A g$

$a_{\max} = \mu_s g$

If the applied force produces an acceleration greater than $\mu_s g$, block A slides off block B!

Two-Body Concept Check 🎯

Two-Body Calculations 🧮

Two blocks are in contact on a frictionless surface: $m_1 = 4$ kg and $m_2 = 6$ kg. A force of 30 N is applied horizontally to $m_1$.

1. What is the acceleration of the system (in m/s²)?

2. What is the contact force between the blocks (in N)?

3. If the force were applied to $m_2$ instead, what would the contact force be (in N)?

Two-Body Reasoning 🔍

Exit Quiz — Two-Body Problems ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Newton's Third Law

This workshop focuses on applying Newton's Third Law to solve connected-object and two-body problems systematically. We'll practice the complete workflow: FBD → equations → solve.

Problem-Solving Strategy for Multi-Body Problems

Step 1: System Analysis

• Identify all objects in the system
• Determine the system acceleration: $a = F_{\text{net, external}}/(m_{\text{total}})$

Step 2: Individual Object Analysis

• Draw an FBD for each object
• Apply $\sum F = ma$ to each object
• Use Third Law pairs: $F_{A \text{ on } B} = F_{B \text{ on } A}$
• Use constraint: connected objects have the same $|a|$

Step 3: Solve and Verify

• Solve the system of equations
• Check: Does $T$ fall between the weights? Does the contact force make sense?

Worked Example: Three Blocks

Blocks A (2 kg), B (3 kg), C (5 kg) on a frictionless surface. $F = 40$ N pushes A.

System: $a = 40/10 = 4$ m/s²

Contact force between B and C: $F_{BC} = m_C \times a = 5 \times 4 = 20$ N

Contact force between A and B: $F_{AB} = (m_B + m_C) \times a = 8 \times 4 = 32$ N

Pattern: Each contact force accelerates all the blocks "downstream" from that contact point.

Worked Example: Modified Atwood Machine

A 3 kg block on a frictionless table is connected via a string over a pulley to a 2 kg block hanging off the edge. Find $a$ and $T$.

Block on table ($m_1 = 3$ kg): $T = m_1 a = 3a \quad \text{...(1)}$

Hanging block ($m_2 = 2$ kg, taking down as positive): $m_2 g - T = m_2 a$ $2(9.8) - T = 2a$ $19.6 - T = 2a \quad \text{...(2)}$

Substitute (1) into (2): $19.6 - 3a = 2a$ $19.6 = 5a$ $a = 3.92 \text{ m/s}^2$

Tension: $T = 3(3.92) = 11.76 \text{ N}$

Verification:

• $T = 11.76$ N $< m_2 g = 19.6$ N ✓ (hanging block accelerates down)
• $T > 0$ ✓ (string is taut)

Workshop Multiple Choice 🎯

Workshop Calculations 🧮

An Atwood machine has $m_1 = 5$ kg and $m_2 = 15$ kg. Use $g = 10$ m/s².

1. What is the acceleration (in m/s²)?

2. What is the tension in the string (in N)?

3. How far does $m_2$ fall from rest in 2 seconds (in m)?

Quick Verification Checks 🔍

Exit Quiz — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Newton's Third Law

This final part brings together all Newton's Third Law concepts: action-reaction pairs, force identification, connected objects, Atwood machines, two-body problems, and system analysis. Expect AP-level questions that combine multiple ideas.

Concept Summary

Newton's Third Law

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

• Equal magnitude, opposite direction, different objects
• Same type of force, exist simultaneously, never cancel

Force Pair Identification

• Use the "flip test": swap the two objects
• Weight ↔ gravitational pull (NOT normal force!)
• Contact ↔ contact (normal, friction, tension)

Connected Object Formulas

Two-Body Push Problems

• System: $a = F/(m_1 + m_2)$
• Contact force: $F_{\text{contact}} = m_{\text{other}} \times a$

Key Problem-Solving Checks

• $T$ should be between the two weights (Atwood)
• Contact force $<$ applied force (push problems)
• Internal forces cancel in system analysis

AP-Style Multiple Choice — Set 1 🎯

AP-Style Free Response 🧮

A modified Atwood machine: $m_1 = 8$ kg on a frictionless table, connected to $m_2 = 2$ kg hanging off the edge. Use $g = 10$ m/s².

1. Find the acceleration of the system (in m/s²).

2. Find the tension in the string (in N).

3. If $m_2$ starts 1.5 m above the floor, how long until it hits the floor (in s)? Round to 3 significant figures.

Synthesis Quick Check 🔍

AP-Style Multiple Choice — Set 2 🎯

Final Exit Quiz — Newton's Third Law ✅