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Lesson 4 of 14

Laws of Motion · Lesson 4 of 14

Newton’s Third Law and System Boundaries

“Identify mutual force pairs and explain why they cancel inside a system without cancelling the force on each part.”

Learning Objectives

• Identify a third-law pair by naming the two interacting bodies. • Distinguish mutual pairs from balanced forces on one body. • Explain how internal forces cancel in a combined system. • Apply the ideas to walking, vehicles and a horse-cart system.

An interaction acts on both participants

Press a spring with your hand. The hand pushes the spring, while the spring pushes the hand. Earth pulls a falling stone, and the stone pulls Earth. In Newtonian mechanics, these are two sides of the same interaction. The third law states that the force on A by B equals the negative of the force on B by A. They are simultaneous and act on different bodies.

Third-law pairLaTeX
The arrows in the subscripts identify the body receiving the force and the body exerting it. Either member may be called action or reaction.

The names “action” and “reaction” do not imply that one force happens first or causes a delayed response. The equality does not require equal masses or equal accelerations. If two interacting bodies have very different masses, equal force magnitudes produce very different acceleration magnitudes. Earth’s response to a small stone is extremely small because Earth’s mass is enormous.

Balanced forces are a different relationship

For a book resting on a table, weight is the force on the book by Earth and normal reaction is the force on the book by the table. These can balance on the same book, but they are not a third-law pair. The partner to weight is the gravitational force on Earth by the book. The partner to the normal force is the downward contact force on the table by the book.

N: table on bookmg: Earth on bookArrows show forces on the selected body only.
Forces on a book at rest— These two forces act on the same book and balance. Their separate third-law partners act on Earth and on the table, so those partners are absent from this diagram. Arrow lengths are schematic unless stated otherwise.

If the table accelerates vertically with the book, normal force and weight need not be equal. The mutual contact forces between book and table still remain equal and opposite. This is a useful diagnostic: a third-law equality persists regardless of acceleration, whereas a balancing relation depends on the selected body’s motion.

The boundary determines what counts as internal

Choose the book alone as the system: the table’s contact force is external. Choose book plus table as the system: their contact forces are internal and cancel in the system total. Forces from Earth and the floor remain external to that larger system. Changing the boundary changes the force inventory, so write down the system before summing forces.

A pair cancels in the combined sumLaTeX
For ordinary Newtonian interactions with paired forces, the internal contributions cancel. Individual parts may still accelerate relative to one another.
Why a horse can pull a cart on a road

Problem
A horse pulls a cart forward, while the cart pulls the horse backward. Why can the pair accelerate?

  1. 1.For the cart alone, the horse’s pull is an external forward force; any rolling resistance acts backward.
  2. 2.For the horse-cart system, the mutual pull cancels internally. The horse pushes backward on the ground, and the ground exerts a forward friction force on the horse.
  3. 3.That ground force is external to the system and can exceed resistance, giving forward acceleration. Equal horse-cart forces do not cancel on the cart alone.

Why the same manoeuvre fails in empty space

An isolated horse-cart assembly in empty space cannot accelerate its centre of mass by pulling internally. The horse and cart could move relative to each other, but their total momentum would not change without an external impulse. The road supplies the missing external interaction in the ordinary case. A person can propel a spacecraft by expelling material, but then the expelled material carries momentum; it is a different system arrangement.

Walking and driving need external traction

A foot pushes backward against the ground while the ground pushes forward on the foot. This static friction can accelerate the person forward even though it prevents slipping at the contact. A driven tyre similarly pushes on the road, and the road’s force accelerates the car. Muscles or an engine supply energy and internal forces, but acceleration of the whole person or car requires a net external force.

Equal forces, unequal accelerations

Problem
A 60 kg skater and a 40 kg skater push on each other on ideal frictionless ice. At one instant the interaction force magnitude is 120 N. Find their accelerations.

  1. 1.The forces are 120 N in opposite directions by the third law.
  2. 2.The 60 kg skater has acceleration magnitude 120/60 = 2 m/s²; the 40 kg skater has 120/40 = 3 m/s².
  3. 3.The combined external horizontal force is zero. Their centre of mass has zero horizontal acceleration even though both skaters accelerate individually.

A reliable pair-finding question

For every arrow, say “force on A by B.” To find its partner, swap A and B and reverse the direction. If both proposed arrows act on the same object, they cannot be a third-law pair. This rule works for contact and gravitational interactions and prevents the mistaken argument that all forces must cancel on every body because action equals reaction.

Quiz

Quick check

Which is the third-law partner of Earth’s gravitational force on a book?

Key Takeaways

Key Takeaways

• A mutual pair acts simultaneously on two different bodies. • Equal and opposite forces on one body can balance but are not a third-law pair. • Internal pairs cancel only when both interacting bodies are included in the chosen system. • External traction allows walking and vehicle acceleration. • Name the receiving and exerting bodies to identify force pairs.