Newton's Third Law of Motion — Action & Reaction (NCERT 4.6)
1. What Does the Third Law Claim?
Newton's third law says forces are never solo acts: if body A exerts a force on body B, then B simultaneously exerts an equal and opposite force on A along the same line — FAB = −FBA. Where the second law told one body how to respond to a net force, the third law is a statement about interactions: every force in nature is one member of a pair, and the pair is born and dies together. The law sounds trivial and is the most failed topic in the chapter, because its only hard consequence is bookkeeping discipline: each force acts on a specific body, and a body's motion is decided solely by the forces acting on it.
Examine the pair rules precisely — NTA tests them one at a time. A genuine action–reaction pair is (i) equal in magnitude, (ii) opposite in direction, (iii) collinear (same line), (iv) simultaneous (born and dying together), (v) acting on two different bodies, and (vi) the same kind of force (both contact, both gravitational, both electrostatic). Fail any one test and the two forces merely coexist — they are not partners. The book on the table is the canonical exam trap, and the horse-cart paradox is the canonical discussion question; both are resolved below with numbers.
2. Complete Theory: Pairs, Non-Pairs, and Why Nothing Cancels
Why the pair never cancels. Cancellation is an operation on one body: the book stays at rest because N and mg — both acting on the book — sum to zero. An action–reaction pair never meets this condition, since its members live on different bodies: your push acts on the wall, the wall's reaction acts on you. Each body feels exactly one member and responds to it via the second law. When you "feel the wall push back", that is not the wall helping you cancel your own push — it is the wall opposing you, exactly as hard as you pushed. The student error NTA banks on is adding FAB and FBA to claim "net force zero, so nothing can ever move"; the correction is one sentence — forces on different bodies never share an equation.
The non-pair everyone mistakes for a pair. Book on table, four forces, two true pairs: (1) Earth pulls book down (mg, on book) ⇔ book pulls Earth up (on Earth, absurdly tiny because Earth's mass is huge); (2) book presses table down (on table) ⇔ table pushes book up (N, on book). Meanwhile N and mg — the two forces on the book — balance each other in equilibrium but are not partners: they are different kinds of force (contact vs gravitational), and they would part ways the moment the table accelerates upward (N grows) or the book and table fall freely (N vanishes, mg survives). "Equal and opposite at this instant" is not the definition of a third-law pair — different bodies and same type are the tests that N and mg fail.
Everyday demonstrations, exam-ready. Walking: you push the ground backward (action on ground); the ground's frictional reaction pushes you forward — the force that accelerates you is exerted by the ground, which is why you cannot walk on frictionless ice. Swimming/rowing: water pushed backward → swimmer/boat pushed forward. Gun firing: gun pushes bullet forward → bullet pushes gun backward (recoil — quantified on the next page). Rocket: rocket pushes exhaust gases down → gases push rocket up; a rocket needs no "air to push against", a favourite NEET assertion–reason. Book pressing hand: you feel the book's weight as the book pressing your palm — the reaction to your palm's upward push.
The horse-cart paradox — resolved. The paradox: horse pulls cart forward with F; cart pulls horse backward with the same F (third law — verified in the figure). Since the forces are equal, "the system can never accelerate". The resolution: the cart's acceleration is decided by forces on the cart — the horse's pull minus ground friction on the cart's wheels. The horse's acceleration is decided by forces on the horse — ground friction on its hooves minus the cart's backward pull. The third-law pair cancels only in the combined system's books, where it becomes internal and drops out; what remains external is the ground's friction on the horse's hooves, and that is what moves the whole outfit. Example 2 runs the numbers.
3. Visualising Two Bodies and Their Pair
Figure 4.3 — One pair, two bodies: forces never share a body with their partner
Top: A pushes B (red, on B); B pushes A (green, on A) — equal, opposite, and on different boxes. Bottom: the book-on-table non-pair autopsy.
Exam read-out: a pair test is a checklist: equal? opposite? collinear? simultaneous? different bodies? same type? Six yeses make partners. N and mg fail on "different bodies" and "same type" — that failure is the exam answer.
4. Solved Examples
Example 1 — Force inventory of a 2 kg book at rest on a table
List every force acting on and by a 2 kg book resting on a horizontal table (g = 10 m s⁻²), marking which are third-law pairs and which merely balance.
Solution (step by step): Forces on the book: weight 20 N down (Earth → book) and normal 20 N up (table → book) — these balance (ΣF = 0, so a = 0) but are not a pair. Forces by the book: 20 N up on Earth (book → Earth) and 20 N down on table (book → table) — these are the reactions completing the two true pairs: {Earth↕book} and {book↕table}. Note the asymmetry of consequence: the 20 N on Earth is unmeasurable because Earth's mass is ~6 × 10²⁴ kg (same force, no visible acceleration), while the 20 N on the table slightly compresses its surface — equal forces, unequal visible effects, because acceleration depends on the receiving mass.
Example 2 — The horse-cart paradox with numbers
A horse (400 kg) pulls a cart (400 kg) forward with 400 N. Ground friction on the horse's hooves is 600 N forward; friction on the cart's wheels is 200 N backward. Find each body's acceleration and explain why the outfit moves at all.
Solution: On the cart: pull 400 N forward − 200 N friction = 200 N net → acart = 200/400 = 0.5 m s⁻². On the horse: ground push 600 N forward − cart's backward pull 400 N = 200 N net → ahorse = 200/400 = 0.5 m s⁻² — consistent, as it must be, since they move together. Whole system: the third-law pair (400 N ↔ 400 N) is internal and cancels; external net = 600 − 200 = 400 N on 800 kg → a = 0.5 m s⁻² ✓. The outfit moves because of the ground's friction on the horse — an external force entirely unrelated to the action–reaction pair. No contradiction, just careful bookkeeping.
5. Practice Questions
Q1. A 60 kg sprinter accelerates from rest to 9 m/s in 3 s. What force accelerates him, which body exerts it, and what is the reaction's role?
ANSWER: a = 3 m s⁻²; F = ma = 180 N. The force is ground friction, exerted by the track on the sprinter — the reaction to his 180 N backward push on the track. His muscles generate the push on the ground; the ground's reaction accelerates him forward. On a frictionless track the reaction would be zero and no horizontal acceleration would be possible.
Q2 (MCQ). A man stands on a weighing scale inside a stationary lift. The scale reads his weight because: (a) his weight and the scale's normal force form an action–reaction pair (b) the scale's normal force equals his weight as a second-law consequence, while the true reaction to his weight is his pull on Earth (c) the Earth pushes him down and the scale pushes him up as a pair (d) none of these
ANSWER: (b) — the reading N = mg is equilibrium bookkeeping on the man, not a third-law statement. The genuine pairs: {Earth pulls man, man pulls Earth} and {man presses scale, scale pushes man}. Options (a) and (c) commit the classic same-body confusion.
Q3. A 20 g bullet is fired from a 4 kg gun at 300 m/s. Both feel the same explosion force for 0.002 s. Find the force on each, the accelerations, and the recoil speed.
ANSWER: Impulse on bullet: Δp = 0.02 × 300 = 6 kg m/s → F = 6/0.002 = 3000 N on each (third law). abullet = 3000/0.02 = 150 000 m s⁻²; agun = 3000/4 = 750 m s⁻² — same force, accelerations inverse to mass. Recoil speed V = Δp_gun/M = 6/4 = 1.5 m/s backward (full treatment on the conservation-of-momentum page).
6. Key Formulas & Takeaways
| Relation / rule | Condition / remark |
|---|---|
| FAB = −FBA | Always, for every interaction, same instant |
| Pair checklist: equal · opposite · collinear · simultaneous · different bodies · same type | All six required; failing one disqualifies the pair |
| N and mg are never partners | Same body, different force types — balance ≠ third law |
| Internal pairs never enter ΣF of a system | Basis of momentum conservation for isolated systems |
| a = F/m on each body separately | Equal forces → unequal accelerations (bullet vs gun) |
The third law is the accounting rule; the next page is the audit it enables: sum every dp/dt in an isolated system, watch the internal pairs cancel, and total momentum stands still.
7. Frequently Asked Questions
State Newton's third law of motion.
If body A exerts a force on body B, then body B simultaneously exerts an equal and opposite force on body A along the same straight line: F_AB = -F_BA. The two forces are called an action-reaction pair. Action and reaction are equal in magnitude, opposite in direction, collinear, simultaneous, act on two different bodies, and are always the same kind of force (both contact, both gravitational, both magnetic, and so on).
If action and reaction are always equal and opposite, why don't they cancel out?
Because the two forces of the pair act on different bodies. Forces cancel only when they act on the same body. The action accelerates the second body while the reaction accelerates the first — each body feels exactly one force of the pair. For example, when you push a wall, your push acts on the wall and the wall's reaction acts on you; the wall's motion is decided by forces on the wall alone, yours by forces on you alone. Adding an action to its own reaction is adding forces on different bodies — physically meaningless.
Are the normal force on a book and its weight an action–reaction pair?
No. Both act on the same body (the book), so they cannot be a third-law pair — they merely balance in equilibrium. The true pairs are: Earth pulls the book down (weight) and the book pulls Earth up with an equal gravitational force; and the table pushes the book up (normal N) while the book presses the table down with an equal contact force. Recognising this distinction is the most frequently tested third-law skill in NEET and JEE Main.
How does walking work according to the third law?
While walking you push the ground backward and slightly downward with your foot (action); the ground, by the third law, pushes you forward and slightly upward (reaction). It is this frictional reaction from the ground that accelerates you forward — without friction (on polished ice) you cannot push effectively and cannot walk. The forward force on you is exerted by the ground, not by your own muscles directly; your muscles only supply the push on the ground.
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