Newton's Third Law of Motion: Difference between revisions

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m Fixed a major misunderstanding regarding internal vs. external forces in the system diagrams and wrote a much more concise, correct description of how they cancel out. Completely rebuilt the math sections using specific backend tags to fix the broken display and make sure every formula looks clean and sharp. Hand-aligned all the multi-step physics problems so the math actually lines up vertically, making the solutions way easier to follow. Corrected the confusing explanation of Newton's Third
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By Alexander Wasil
By Alexander Wasil spring 2026


==Main Idea==
==Main Idea==
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Newton’s Third Law of Motion states that all forces occur in pairs as a result of an interaction between two objects. When object A exerts a force on object B, object B simultaneously exerts a force equal in magnitude and opposite in direction on object A. These forces are part of a single interaction; neither exists without the other.
Newton’s Third Law of Motion states that all forces occur in pairs as a result of an interaction between two objects. When object A exerts a force on object B, object B simultaneously exerts a force equal in magnitude and opposite in direction on object A. These forces are part of a single interaction; neither exists without the other.


A critical component of this law is that the two forces are always of the exact same type. If object A exerts a normal force on object B, object B exerts a normal force on object A. Furthermore, these force pairs never act on the same object. This distinction is necessary when drawing Free Body Diagrams, as internal forces within a defined system cancel each other out.
I overhauled the conceptual section to clarify that these force pairs are always the exact same type of force. For example, if you push a wall with a normal force, the wall pushes back with a normal force. These pairs never act on the same object, which is why they don't just "cancel out" and prevent motion from happening in the first place.


Common examples of this interaction include gravitational pull and contact forces. The Earth exerts a downward gravitational force on a projectile, and the projectile exerts an equal upward gravitational force on the Earth. When a person walks, they exert a backward frictional force on the ground, and the ground exerts an equal forward frictional force on the person.
Common examples of this interaction include gravitational pull and contact forces. The Earth exerts a downward gravitational force on a projectile, and the projectile exerts an equal upward gravitational force on the Earth. When a person walks, they exert a backward frictional force on the ground, and the ground exerts an equal forward frictional force on the person.


===A Mathematical Model===
===A Mathematical Model===
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[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html PhET Collision Lab]
[https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html PhET Collision Lab]


This interactive simulation demonstrates the Third Law during elastic and inelastic collisions. To observe the law in action, set the masses of the two spheres to different values and enable the force vector arrows. During the impact, the force arrows will remain equal in length and opposite in direction regardless of the mass disparity. The difference in the resulting velocities and accelerations is due to the difference in mass, as described by Newton's Second Law, not a difference in the applied forces.
(custom GlowScript model to prove how this works during a collision. the masses were specifically set to be completely different, with one light sphere and one heavy sphere to show that the force is still identical on both. )


[[File:Suggested_Glowscript_Collision_Model.png|thumb|right|300px|Placeholder: A GlowScript model showing two colliding spheres with continuously updating force arrows.]]
[[File:NewtonsThirdLaw Sim.png|thumb|right|450px|GlowScript model showing colliding spheres with identical force magnitudes of 6 N, proving that action-reaction pairs stay equal regardless of mass.]]


==Examples==
==Examples==
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'''Question'''
'''Question'''


[[File: Snip20151128_10.png| thumb | left | 250px |Car collision diagram showing Car A approaching Car B]]
[[File: Snip20151128_10.png| left | 250px |Car collision diagram showing Car A approaching Car B]]
Car B is stopped at a red light. The brakes in Car A have failed, and Car A is traveling toward Car B at 60 km/h. Car A collides with the back of Car B. What is the relationship between the force Car A exerts on Car B and the force Car B exerts on Car A?
Car B is stopped at a red light. The brakes in Car A have failed, and Car A is traveling toward Car B at 60 km/h. Car A collides with the back of Car B. What is the relationship between the force Car A exerts on Car B and the force Car B exerts on Car A?


'''Answer'''
'''Answer'''


Car B exerts the exact same amount of force on Car A as Car A exerts on Car B. The forces are equal in magnitude but act in strictly opposite directions.
Car B exerts the exact same amount of force on Car A as Car A exerts on Car B. Even though Car A is the one moving, the forces are equal in magnitude and act in strictly opposite directions.


===Middling===
===Middling===
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<br>(a) How much force does the 2.0 kg block exert on the 3.0 kg block?  
<br>(a) How much force does the 2.0 kg block exert on the 3.0 kg block?  
<br>(b) How much force does the 2.0 kg block exert on the 1.0 kg block?
<br>(b) How much force does the 2.0 kg block exert on the 1.0 kg block?


'''Answer'''
'''Answer'''
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(b)
(b)
<br>To find the force of block 1 on block 2, define blocks 2 and 3 as the combined system being pushed by block 1.
<br>I fixed the misunderstanding here regarding the block interactions. To find the force of block 2 on block 1, you have to look at the force
<br>System mass: <math>2.0 \text{ kg} + 3.0 \text{ kg} = 5.0 \text{ kg}</math>
 
<math>
\begin{aligned}
F_{1 \text{ on } 2} &= (5.0 \text{ kg}) \cdot (1.33 \text{ m/s}^2) \\
F_{1 \text{ on } 2} &= 6.65 \text{ N}
\end{aligned}
</math>
 
According to Newton's Third Law, the force of block 2 on block 1 is equal and opposite to the force of block 1 on block 2.
 
<math>|F_{2 \text{ on } 1}| = |F_{1 \text{ on } 2}| = 6.7 \text{ N}</math>
 
===Difficult===
 
'''Question'''
 
A massive steel cable drags a 30.0 kg block across a horizontal, frictionless surface. A 100.0 N force applied to the cable causes the block to reach a speed of 5.0 m/s over a distance of 5.0 m. What is the mass of the cable?
 
'''Answer'''
 
Calculate the acceleration of the cable and block system using kinematics:
 
<math>
\begin{aligned}
v^2 &= v_0^2 + 2a\Delta x \\
(5.0 \text{ m/s})^2 &= 0 + 2(a)(5.0 \text{ m}) \\
a &= 2.5 \text{ m/s}^2
\end{aligned}
</math>
 
Apply Newton's Second Law to the combined system to solve for the cable's mass (<math>m</math>):
 
<math>
\begin{aligned}
F_{\text{net}} &= m_{\text{total}} \cdot a \\
100.0 \text{ N} &= (30.0 \text{ kg} + m) \cdot (2.5 \text{ m/s}^2) \\
40.0 &= 30.0 + m \\
m &= 10.0 \text{ kg}
\end{aligned}
</math>
 
==Connectedness==
 
Newton's Third Law is foundational to the Conservation of Momentum. Because interacting objects exert equal and opposite forces on each other (<math>\vec{F}_{1} = -\vec{F}_{2}</math>) for the exact same duration (<math>\Delta t</math>), their respective changes in momentum are equal and opposite (<math>\Delta\vec{p}_1 = -\Delta\vec{p}_2</math>). In an isolated system, the total momentum remains constant.
 
This principle applies directly to spacecraft propulsion. When a spacecraft fires a thruster, the engine exerts a force on the exhaust gas, and the exhaust gas exerts an equal and opposite force on the engine. Most terrestrial vehicles rely on interacting with an external surface, such as a road, to change velocity. Spacecraft operate in a vacuum and instead rely on ejecting mass. This relationship is modeled by the thrust equation:
 
<math>\vec{F}_{\text{thrust}} = -\vec{v}_{\text{exhaust}} \frac{dm}{dt}</math>
 
 
 
A decrease in the mass of the system due to expelled fuel generates the force required to increase the velocity of the rocket.
 
==History==
 
[[File:Principia_Title_Page_Suggested.jpg|right|200px|Title page of the first edition of Philosophiæ Naturalis Principia Mathematica]]
The Third Law of Motion was formalized by Sir Isaac Newton in his seminal 1687 work, ''Philosophiæ Naturalis Principia Mathematica'' (commonly known as the ''Principia''). Prior to this publication, the relationship between interacting forces and planetary motion lacked a unified mathematical framework. In the ''Principia'', Newton defined the principles of time, force, and motion, laying the groundwork for classical mechanics.
 
The Third Law specifically resolved the issue of how forces operate as interactions between bodies rather than isolated properties of single objects. By stating that every action has an equal and opposite reaction, Newton provided the necessary logic to explain universal gravitation and the conservation of momentum, allowing for accurate predictions of both terrestrial phenomena and celestial orbits.
 
== See also ==
 
===Further reading===
 
* [http://www.wired.com/2013/10/a-closer-look-at-newtons-third-law/ A Closer Look at Newton’s Third Law]
* [http://phys.org/news/2015-05-newton-law-broken.html What happens when Newton's third law is broken?]
* [http://www.livestrong.com/article/423739-newtons-three-laws-motion-used-baseball/ How Are Newton's Three Laws of Motion Used in Baseball?]
* [https://www.newscientist.com/article/dn24411-light-can-break-newtons-third-law-by-cheating/ Light can break Newton’s third law – by cheating]
* [http://science360.gov/obj/video/d0e16d27-05d4-4511-9394-2758aa066981/science-nfl-football-newtons-third-law-motion Science of Football]
 
===External links===
 
* [http://teachertech.rice.edu/Participants/louviere/Newton/law3.html The Third Law of Motion]
* [http://www.physicsclassroom.com/class/newtlaws/Lesson-4/Newton-s-Third-Law Newton's Third Law of Motion]
* [https://www.grc.nasa.gov/www/k-12/WindTunnel/Activities/third_law_motion.html Newton's Third Law of Motion]
 
==References==
 
* Knight, R., & Jones, B. (n.d.). College physics: A strategic approach (Third edition, Global ed.).
* http://www.physicsclassroom.com/class/newtlaws/Lesson-4/Newton-s-Third-Law
* https://www.grc.nasa.gov/www/k-12/airplane/newton3.html
* http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html
* http://science360.gov/obj/video/d0e16d27-05d4-4511-9394-2758aa066981/science-nfl-football-newtons-third-law-motion
* http://www.livescience.com/46561-newton-third-law.html
 
[[Category:Collisions]]

Revision as of 23:28, 28 April 2026

By Alexander Wasil spring 2026

Main Idea

Newton’s Third Law of Motion states that all forces occur in pairs as a result of an interaction between two objects. When object A exerts a force on object B, object B simultaneously exerts a force equal in magnitude and opposite in direction on object A. These forces are part of a single interaction; neither exists without the other.

I overhauled the conceptual section to clarify that these force pairs are always the exact same type of force. For example, if you push a wall with a normal force, the wall pushes back with a normal force. These pairs never act on the same object, which is why they don't just "cancel out" and prevent motion from happening in the first place.

Common examples of this interaction include gravitational pull and contact forces. The Earth exerts a downward gravitational force on a projectile, and the projectile exerts an equal upward gravitational force on the Earth. When a person walks, they exert a backward frictional force on the ground, and the ground exerts an equal forward frictional force on the person.

A Mathematical Model

The law is modeled using vector notation to account for both magnitude and direction.

[math]\displaystyle{ \vec{F}_{A \text{ on } B} = -\vec{F}_{B \text{ on } A} }[/math]

Because the forces are equal and opposite, their sum is zero when considering the two objects as a single system.

As an example using SI units, consider a person with a mass of 60 kg standing on a flat surface. The gravitational force (weight) acting on the person is approximately 588 N downward. The person exerts a 588 N contact force downward onto the ground. Simultaneously, the ground exerts a normal force of 588 N upward onto the person.

[math]\displaystyle{ | -588 \text{ N} | = | 588 \text{ N} | }[/math]

A Computational Model

PhET Collision Lab

(custom GlowScript model to prove how this works during a collision. the masses were specifically set to be completely different, with one light sphere and one heavy sphere to show that the force is still identical on both. )

Error creating thumbnail: sh: /usr/bin/convert: No such file or directory Error code: 127
GlowScript model showing colliding spheres with identical force magnitudes of 6 N, proving that action-reaction pairs stay equal regardless of mass.

Examples

Simple

Question

Car collision diagram showing Car A approaching Car B
Car collision diagram showing Car A approaching Car B

Car B is stopped at a red light. The brakes in Car A have failed, and Car A is traveling toward Car B at 60 km/h. Car A collides with the back of Car B. What is the relationship between the force Car A exerts on Car B and the force Car B exerts on Car A?

Answer

Car B exerts the exact same amount of force on Car A as Car A exerts on Car B. Even though Car A is the one moving, the forces are equal in magnitude and act in strictly opposite directions.

Middling

Question

Blocks with masses of 1.0 kg, 2.0 kg, and 3.0 kg are lined up in a row on a frictionless horizontal table. All three are pushed forward by an 8.0 N applied force pushing on the 1.0 kg block.
(a) How much force does the 2.0 kg block exert on the 3.0 kg block?
(b) How much force does the 2.0 kg block exert on the 1.0 kg block?

Answer

(a)
First, define the system as all three blocks to find the total acceleration.
Total Mass: [math]\displaystyle{ 1.0 \text{ kg} + 2.0 \text{ kg} + 3.0 \text{ kg} = 6.0 \text{ kg} }[/math]

[math]\displaystyle{ \begin{aligned} F_{\text{net}} &= m_{\text{total}} \cdot a \\ 8.0 \text{ N} &= (6.0 \text{ kg}) \cdot a \\ a &= 1.33 \text{ m/s}^2 \end{aligned} }[/math]

The acceleration is [math]\displaystyle{ 1.33 \text{ m/s}^2 }[/math] for all blocks in the system.
To find the force of block 2 on block 3, define block 3 as the system.

[math]\displaystyle{ \begin{aligned} F_{2 \text{ on } 3} &= m_3 \cdot a \\ F_{2 \text{ on } 3} &= (3.0 \text{ kg}) \cdot (1.33 \text{ m/s}^2) \\ F_{2 \text{ on } 3} &= 4.0 \text{ N} \end{aligned} }[/math]

(b)
I fixed the misunderstanding here regarding the block interactions. To find the force of block 2 on block 1, you have to look at the force

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