01 · Contact pair
A hand pushes a box right with force 15 N. Describe the partner force.
Hint
Reverse the source and recipient.
Worked solution
The box pushes the hand left with force 15 N.
Understand · explore · practise
Newton’s third law and interacting bodies
Identify equal and opposite interaction forces on different bodies, separate stacked-body diagrams and understand light pans and internal forces.
Before you startNewton’s first and second laws, weight and contact reactions.
01 / Name who exerts each force and who receives it
Third-law partners act on different bodies.
If A exerts a force on B, B exerts an equal and opposite force on A.
The two forces belong to the same interaction and have equal magnitudes at the same instant. They act on different bodies, so they do not cancel in a force diagram of either body alone. This applies to gravitational interactions as well as contact interactions.
Choose a body in a two-box stackExplore
A 2 kg box rests on a 3 kg box on a horizontal platform. Use g = 9.8. Choose the body and either rest or shared upward acceleration 1 m/s². Contact forces between the boxes are internal when the whole stack is selected.
02 / Identify a pair by the interaction, not just matching arrows
Equal opposite forces on one body are not a third-law pair.
A book rests on a horizontal table.Worked example
Table pushes book upward; book pushes table downward
These are third-law partners: same contact, different recipients.
Earth pulls book downward; book pulls Earth upward
These are the gravitational third-law partners.
Book’s weight and table’s reaction both act on the book
They may balance, but they are not each other’s third-law partner.
Watch: contact partners belong on different body diagrams
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Weight pair
Earth exerts gravitational force 49 N downward on an object. Identify the third-law partner.
Hint
It acts on Earth, not on the object.
Worked solution
The object exerts gravitational force 49 N on Earth in the opposite direction, toward the object.
03 · Why not cancel?
Why do those two gravitational forces not give zero acceleration of the object?
Hint
Only one acts on the object.
Worked solution
The downward gravitational force belongs in the object’s equation. Its partner acts on Earth and is not another force on the object.
03 / Draw the upper and lower boxes separately
The lower box supports the upper box as well as its own weight.
04 · Upper box at rest
A 2 kg box rests on a 3 kg box, both at rest. Find the upward contact force C on the upper box with g = 9.8.
Hint
The upper box has two vertical forces.
Worked solution
C − 19.6 = 0, so C = 19.6 N.
05 · Lower box at rest
Find the upward floor reaction F on the lower box.
Hint
The upper box pushes down on the lower one with 19.6 N.
Worked solution
F − 29.4 − 19.6 = 0, so F = 49 N.
06 · Contact partner
What force does the upper box exert on the lower box in question 4?
Hint
Use the same contact interaction.
Worked solution
19.6 N downward. It equals C in magnitude and is opposite in direction, but it acts on the lower box.
04 / Third-law equality still holds during acceleration
Equal contact partners do not require either body to be in equilibrium.
07 · Upper accelerated box
The same stack accelerates upward at 1 m/s². Find contact force C on the 2 kg upper box.
Hint
Use C − mg = ma.
Worked solution
C − 19.6 = 2, so C = 21.6 N.
08 · Lower accelerated box
Find the platform force F on the lower box during the same acceleration.
Hint
The lower box has its own weight and the downward contact partner.
Worked solution
F − 29.4 − 21.6 = 3, so F = 54 N.
09 · Whole-stack check
Verify question 8 using the whole stack as the system.
Hint
Inter-box contact is internal to this system.
Worked solution
Total mass is 5 kg. F − 49 = 5 × 1, so F = 54 N. This matches the separate-body calculation.
05 / Internal forces cancel only when the system includes both recipients
A whole-system equation cannot usually reveal an internal contact force.
When you add the two vertical equations, +C in the upper-box equation and −C in the lower-box equation cancel. This removes an internal force and helps find the external platform reaction. To find C itself, return to one box’s equation.
10 · Recover the contact
Can 54 − 49 = 5 × 1, by itself, determine C for the accelerated stack?
Hint
C is absent from the combined equation.
Worked solution
No. Use an individual-body equation such as C − 19.6 = 2 × 1, giving C = 21.6 N.
06 / A light support has negligible mass in the model
Zero model mass does not mean its interaction forces vanish.
For a light pan with only upward tension T and downward load contact C, T = C.
The pan’s own weight and ma term are neglected. Its force equation is T − C = 0. The forces can be nonzero even if the pan accelerates along with the load.
11 · Load on a light pan
A 3 kg load on a light pan accelerates upward at 0.6 m/s². Find the pan’s upward contact force on the load.
Hint
Use the load’s own equation.
Worked solution
C − 29.4 = 3 × 0.6, so C = 31.2 N.
12 · Pan tension
For question 11, find the string tension supporting the light pan.
Hint
The load pushes the pan downward with C.
Worked solution
T − C = 0 for the light pan, so T = 31.2 N. The equal T and C here act on the same pan and are not a third-law pair.
13 · A pan with mass
Instead let the pan have mass 0.5 kg while the 3 kg load still accelerates upward at 0.6 m/s². Find tension.
Hint
Use combined mass 3.5 kg or the pan-only equation.
Worked solution
T = 3.5(9.8 + 0.6) = 36.4 N. Load contact remains 31.2 N; the extra tension supports and accelerates the pan.
07 / Check body labels before applying a law
Do not infer equal accelerations from equal interaction forces alone.
14 · Unequal masses
Two isolated particles of masses 2 kg and 5 kg exert equal opposite interaction forces of magnitude 10 N on each other. If these are their only forces, compare acceleration magnitudes.
Hint
Use each particle’s own mass.
Worked solution
The 2 kg particle has acceleration magnitude 5 m/s²; the 5 kg particle has 2 m/s², in opposite directions. Equal forces do not imply equal accelerations.
Third-law partner forces have equal magnitude because of the interaction. Equality between other forces may instead follow from equilibrium, a light-string model or a separate calculation. State which reason applies.
08 / Keep one force diagram per chosen body
Pair forces by naming their source and recipient.
Third-law partners are equal and opposite and act on different bodies. Include both recipients only when using a combined system; otherwise keep the partner on the other body’s diagram. Use individual equations to recover internal forces.
Section 1 of 8 · Name who exerts each force and who receives it