Concept Card: Newton’s Third Law of Motion & Action-Reaction Pairs
1. Formal Statement of Newton’s Third Law:
To every action, there is always an equal and opposite reaction; or the mutual interactions of two bodies upon each other are always equal in magnitude and directed along opposite directions.
Vector representation:
$$\vec{F}_{AB} = -\vec{F}_{BA}$$
where $\vec{F}_{AB}$ represents the force exerted on body $A$ by body $B$, and $\vec{F}_{BA}$ represents the force exerted on body $B$ by body $A$.
2. Core Axioms & Essential Characteristics of Action-Reaction Pairs:
- Simultaneity: Action and reaction arise and cease at the exact same instant. There is strictly no time delay or cause-and-effect relationship; either force can be considered the action and the other the reaction.
- Act on Different Bodies: $\vec{F}_{AB}$ acts exclusively on body $A$, while $\vec{F}_{BA}$ acts exclusively on body $B$. Because they act on two completely distinct physical entities, action and reaction forces NEVER cancel each other out!
- Identical Fundamental Origin: Both forces in a pair always belong to the same fundamental interaction:
- Gravitational pull of Earth on an apple $\iff$ Gravitational pull of the apple on Earth.
- Electromagnetic normal push of a table on a book $\iff$ Electromagnetic normal press of the book on the table.
- Electromagnetic friction of a road on a shoe $\iff$ Electromagnetic friction of the shoe on the road.
- Collinearity: In classical central-force mechanics, $\vec{F}_{AB}$ and $\vec{F}_{BA}$ act along the straight line joining the centers of the interacting particles.
3. Critical JEE Conceptual Pitfall: Normal Reaction ($N$) vs. Weight ($mg$):
For a block of mass $m$ resting on a horizontal surface in static equilibrium ($N = mg$):
$\vec{N}$ and $m\vec{g}$ are NOT an action-reaction pair!
- Both forces act on the same body (the block).
- They have completely different physical natures ($m\vec{g}$ is gravitational, whereas $\vec{N}$ is electromagnetic contact repulsion).
- If the supporting surface is removed, $\vec{N}$ vanishes immediately while $m\vec{g}$ remains unchanged.
True Action-Reaction Pairs:
– Pair 1 (Gravitational): Downward pull of Earth on the block ($m\vec{g}$) $\iff$ Upward gravitational pull of the block on the Earth ($m\vec{g}$ at Earth’s center).
– Pair 2 (Contact): Upward normal support of the table on the block ($\vec{N}$) $\iff$ Downward normal push of the block on the table ($\vec{N}’$).
4. Internal vs. External Forces & Conservation of Momentum:
When multiple interacting bodies are treated as a combined system:
$$\sum \vec{F}_{\text{internal}} = \sum_{i \neq j} \vec{F}_{ij} = \vec{0}$$
Because all internal interactions occur as equal and opposite action-reaction pairs between constituents within the system boundary, their vector sum is identically zero.
Consequently, only net external forces accelerate the center of mass:
$$\vec{F}_{\text{ext, net}} = \frac{d\vec{P}_{\text{total}}}{dt}$$
If $\vec{F}_{\text{ext, net}} = \vec{0}$, then $\frac{d\vec{P}_{\text{total}}}{dt} = \vec{0} \implies \vec{P}_{\text{total}} = \text{constant}$.
Newton’s Third Law is the foundational guarantor of the Law of Conservation of Linear Momentum.
Solved Examples
Example 1 (Force Pair Identification in a Multi-Body Stack):
A textbook of mass $m_1 = 2.0\text{ kg}$ rests on a wooden block of mass $m_2 = 5.0\text{ kg}$, which sits on a horizontal lab bench. Taking $g = 9.8\text{ m/s}^2$:
(a) Identify all vertical forces acting on the book, the block, the bench, and the Earth.
(b) Explicitly pair each force with its Newton’s third law reaction force.
(c) Calculate the magnitude of all contact forces in static equilibrium.
Solution:
(a) & (b) Force inventory and pairs:
– Pair 1: Earth’s downward gravitational pull on book $\vec{W}_1$ ($19.6\text{ N}$ down on book) $\iff$ Book’s upward gravitational pull on Earth $\vec{W}_1’$ ($19.6\text{ N}$ up on Earth’s center).
– Pair 2: Earth’s downward gravitational pull on block $\vec{W}_2$ ($49.0\text{ N}$ down on block) $\iff$ Block’s upward gravitational pull on Earth $\vec{W}_2’$ ($49.0\text{ N}$ up on Earth’s center).
– Pair 3: Normal upward contact force exerted by block on book $\vec{N}_{12}$ ($19.6\text{ N}$ up on book) $\iff$ Normal downward press exerted by book on block $\vec{N}_{21}$ ($19.6\text{ N}$ down on block).
– Pair 4: Normal upward contact force exerted by bench on block $\vec{N}_{23}$ ($68.6\text{ N}$ up on block) $\iff$ Normal downward press exerted by block on bench $\vec{N}_{32}$ ($68.6\text{ N}$ down on bench).
(c) Quantitative equilibrium calculations:
For book ($m_1$): $N_{12} = m_1 g = (2.0)(9.8) = 19.6\text{ N}$.
For block ($m_2$): $N_{23} = N_{21} + m_2 g = 19.6 + (5.0)(9.8) = 19.6 + 49.0 = 68.6\text{ N}$.
Example 2 (Resolving the Classic Horse-Cart Paradox):
A horse of mass $m_H = 400\text{ kg}$ is harnessed to a cart of mass $m_C = 600\text{ kg}$. The horse pushes backward on the ground with horizontal static friction force $F_{\text{ground}} = 3000\text{ N}$. The rolling resistance opposing the cart’s wheels is $f_r = 600\text{ N}$.
(a) Explain why the equal and opposite forces between the horse and cart do not prevent acceleration.
(b) Determine the common forward acceleration $a$ of the horse and cart.
(c) Find the tension $T$ in the harness connecting the horse to the cart.
Solution:
(a) The pull of the horse on the cart ($\vec{T}$) acts on the cart. The pull of the cart on the horse ($-\vec{T}$) acts on the horse. Because they act on two different bodies, they cannot cancel each other. Acceleration is determined by the net external force on each individual body or the combined system.
(b) For the combined (Horse + Cart) system:
Net external horizontal force: $F_{\text{net, ext}} = F_{\text{ground}} – f_r = 3000 – 600 = 2400\text{ N}$.
Total system mass: $M = m_H + m_C = 400 + 600 = 1000\text{ kg}$.
Acceleration: $a = \frac{F_{\text{net, ext}}}{M} = \frac{2400\text{ N}}{1000\text{ kg}} = 2.4\text{ m/s}^2$.
(c) Harness tension from cart’s equation of motion:
$T – f_r = m_C a \implies T = f_r + m_C a = 600 + (600)(2.4) = 600 + 1440 = 2040\text{ N}$.
Verification on horse: $F_{\text{ground}} – T = 3000 – 2040 = 960\text{ N} = (400)(2.4)$ (Consistent!).
Example 3 (Weighing Scale Action-Reaction in an Accelerated Lift):
A student of mass $M = 60\text{ kg}$ stands on a spring weighing machine fixed to the floor of an elevator cabin of mass $m_{\text{lift}} = 400\text{ kg}$. The elevator is accelerated upward by a hoisting cable with tension $T_0 = 5520\text{ N}$. Taking $g = 9.8\text{ m/s}^2$:
(a) Calculate the upward acceleration $a$ of the elevator.
(b) Determine the reading of the weighing machine in Newtons and in kilograms-force ($\text{kgf}$), and identify the corresponding action-reaction pair.
(c) What does the scale read if the hoisting cable suddenly snaps?
Solution:
(a) For the entire system ($M_{\text{tot}} = 400 + 60 = 460\text{ kg}$):
$T_0 – M_{\text{tot}} g = M_{\text{tot}} a \implies 5520 – (460)(9.8) = 460 a \implies 5520 – 4508 = 460 a \implies 1012 = 460 a \implies a = \frac{1012}{460} = 2.2\text{ m/s}^2$.
(b) FBD of the student: $N – M g = M a \implies N = M(g + a) = 60(9.8 + 2.2) = 60(12.0) = 720\text{ N}$.
Scale reading in $\text{kgf}$: $R = \frac{N}{g} = \frac{720}{9.8} \approx 73.47\text{ kgf}$.
Action-Reaction Pair: Upward normal push of the scale on the student’s feet ($720\text{ N}$ up on student) $\iff$ Downward normal press of the student’s feet on the scale ($720\text{ N}$ down on scale). The scale measures the magnitude of this contact interaction.
(c) In free fall ($a = g$ downwards): $N = M(g – g) = 0\text{ N} \implies \text{Scale reads zero (weightlessness)}$.
Example 4 (Spring Balances in Series & Connected Strings):
(a) Two identical light spring balances $A$ and $B$ are connected in series. The top hook of balance $A$ is fixed to a ceiling, while a mass of $m = 10\text{ kg}$ hangs from balance $B$. Determine the readings of balance $A$ and balance $B$ ($g = 10\text{ m/s}^2$).
(b) A light string passes over a frictionless pulley supported by a spring balance. Two masses $m_1 = 3\text{ kg}$ and $m_2 = 5\text{ kg}$ hang vertically from the ends of the string. Find the reading of the spring balance while the masses accelerate under gravity.
Solution:
(a) Balance $B$ supports $mg = 100\text{ N}$ at its lower hook. For balance $B$ to be in equilibrium, balance $A$ pulls upward on it with $100\text{ N}$. By Newton’s Third Law, balance $B$ pulls downward on balance $A$ with $100\text{ N}$. Thus, both balance $A$ and balance $B$ experience a restoring force of $100\text{ N}$ and read $10\text{ kg}$ (or $100\text{ N}$).
(b) Acceleration of Atwood’s machine: $a = \frac{(5 – 3)(10)}{3 + 5} = \frac{20}{8} = 2.5\text{ m/s}^2$.
Tension in hanging string: $T = \frac{2 m_1 m_2 g}{m_1 + m_2} = \frac{2(3)(5)(10)}{8} = \frac{300}{8} = 37.5\text{ N}$.
Downward force on pulley: $2T = 2(37.5) = 75\text{ N}$.
By Newton’s Third Law, the pulley pulls downward on the spring balance with $75\text{ N}$.
Hence, the spring balance reads $F = 75\text{ N}$ (or mass equivalent $7.5\text{ kg}$, which is strictly less than $m_1 + m_2 = 8\text{ kg}$).
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A person walks forward on a horizontal road. The force of friction exerted by the road on the person’s shoe acts:
(A) In the forward direction
(B) In the backward direction
(C) Perpendicular to the road surface
(D) Is zero since walking requires frictionless contact
Problem 2 (JEE Main – Single Correct):
A block of mass $m$ rests on a horizontal table. The action-reaction pair to the weight $mg$ of the block is:
(A) The upward normal reaction force exerted by the table on the block
(B) The downward normal force exerted by the block on the table
(C) The upward gravitational force exerted by the block on the Earth
(D) The frictional force between the block and the table
Problem 3 (JEE Main – Single Correct):
Two blocks of masses $m_1 = 4.0\text{ kg}$ and $m_2 = 6.0\text{ kg}$ are kept in contact on a smooth horizontal surface. A horizontal force $F = 50\text{ N}$ is applied on $m_1$ pushing it towards $m_2$. The contact force between the two blocks is:
(A) $50\text{ N}$
(B) $30\text{ N}$
(C) $20\text{ N}$
(D) Zero
Problem 4 (JEE Main – Single Correct):
In a tug-of-war match between Team $A$ and Team $B$, Team $A$ wins by pulling Team $B$ across the centerline. Team $A$ won because:
(A) Team $A$ exerted a greater tension force on the rope than Team $B$ did on the rope
(B) Team $A$ exerted a greater backward horizontal force on the ground than Team $B$ did on the ground
(C) The rope exerted a greater force on Team $B$ than on Team $A$
(D) The mass of Team $A$ was greater than Team $B$
Problem 5 (JEE Main – Single Correct):
A swimmer pushes the water backwards with hands and feet to move forward in water. This is an immediate illustration of:
(A) Newton’s First Law of Motion
(B) Newton’s Second Law of Motion
(C) Newton’s Third Law of Motion
(D) The Principle of Conservation of Energy
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are TRUE regarding Newton’s Third Law of Motion?
(A) Action and reaction forces act simultaneously on two different bodies.
(B) Action and reaction forces always belong to the same fundamental interaction.
(C) Because action and reaction are equal and opposite, their net vector sum is zero, meaning internal forces cannot change the total linear momentum of an isolated system.
(D) In non-inertial frames, pseudo-forces obey Newton’s Third Law and have corresponding reaction forces.
Problem 7 (JEE Advanced – One or More Correct):
A block of mass $m$ is at rest on a rough inclined plane of angle $\theta$ with the horizontal. Which of the following statements is/are correct?
(A) The contact force exerted by the incline on the block has magnitude $mg$ directed vertically upward.
(B) The normal force $N = mg\cos\theta$ and the component $mg\cos\theta$ of gravity form a Newton’s third law pair.
(C) The static friction force $f_s = mg\sin\theta$ and the component $mg\sin\theta$ of gravity form a Newton’s third law pair.
(D) The force exerted by the block on the inclined plane is $mg$ directed vertically downward.
Problem 8 (JEE Advanced – One or More Correct):
Two interacting particles $1$ and $2$ with masses $m_1$ and $m_2$ are isolated from external forces. Let $\vec{F}_{12}$ be the force exerted on $1$ by $2$, and $\vec{F}_{21}$ be the force exerted on $2$ by $1$. Which of the following must be true at all times?
(A) $\vec{F}_{12} + \vec{F}_{21} = \vec{0}$
(B) $m_1 \vec{a}_1 + m_2 \vec{a}_2 = \vec{0}$
(C) $\vec{a}_1 = -\vec{a}_2$
(D) The velocity of the center of mass of the two particles is strictly constant.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
Two blocks of masses $m_1 = 3.0\text{ kg}$ and $m_2 = 2.0\text{ kg}$ are in contact on a smooth horizontal floor. A horizontal force $F = 20\text{ N}$ is pushed from the left against $m_1$. The magnitude of the normal contact force between $m_1$ and $m_2$ in Newtons is $N_c$. Find $N_c$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A person of mass $m = 70\text{ kg}$ stands in an elevator cabin accelerating downward with uniform acceleration $a = 2.8\text{ m/s}^2$. Taking $g = 9.8\text{ m/s}^2$, the normal reaction force exerted by the elevator floor on the person in Newtons is $N$. Find the integer value of $N$.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (C) | 3. (B) | 4. (B) | 5. (C) | 6. (A, B, C) | 7. (A, D) | 8. (A, B, D) | 9. 8 | 10. 490
Solution 1:
When walking, the shoe pushes backward on the road. By Newton’s Third Law, the road pushes forward on the shoe. This forward friction force accelerates the walker forward.
Correct Answer: (A)
Solution 2:
The weight $mg$ is the Earth’s gravitational pull on the block. The reaction force must act on the Earth and be gravitational—namely, the gravitational pull exerted by the block on the Earth.
Correct Answer: (C)
Solution 3:
Common acceleration: $a = \frac{F}{m_1 + m_2} = \frac{50}{4.0 + 6.0} = 5.0\text{ m/s}^2$.
Contact force accelerating $m_2$: $N_c = m_2 a = 6.0 \times 5.0 = 30\text{ N}$.
Correct Answer: (B)
Solution 4:
Tension is uniform across a light rope. The winning team exerts a greater horizontal force against the ground, which by Newton’s Third Law produces a greater forward ground reaction friction force.
Correct Answer: (B)
Solution 5:
Pushing water backward results in the water pushing the swimmer forward with equal and opposite force, directly demonstrating Newton’s Third Law.
Correct Answer: (C)
Solution 6:
– (A) Action and reaction act simultaneously on different bodies (True).
– (B) Both forces share the same physical nature (True).
– (C) Internal pairs sum to zero and cannot change total momentum (True).
– (D) Pseudo-forces are non-inertial frame corrections without physical reaction counterparts (False).
Correct Answer: (A, B, C)
Solution 7:
– (A) Contact force is the vector sum $\vec{N} + \vec{f}_s$, which balances gravity $m\vec{g}$ vertically (True).
– (B) & (C) Normal force and friction are contact forces, whereas gravity is gravitational; they act on the same body and are not action-reaction pairs (False).
– (D) By Newton’s Third Law, the force on the incline is equal and opposite to the total contact force on the block, having magnitude $mg$ directed downward (True).
Correct Answer: (A, D)
Solution).
Correct Answer: (A, D)
Solution 8:
– (A) $\vec{F}_{12} + \vec{F}_{21} = \vec{0}$ by Newton’s Third Law (True).
– (B) $m_1\vec{a}_1 + m_2\vec{a}_2 = \vec{F}_{12} + \vec{F}_{21} = \vec{0}$ (True).
– (C) $\vec{a}_1 = -\frac{m_2}{m_1}\vec{a}_2 \neq -\vec{a}_2$ unless $m_1 = m_2$ (False).
– (D) Zero net external force implies $\vec{v}_{\text{cm}} = \text{constant}$ (True).
Correct Answer: (A, B, D)
Solution 9:
$a = \frac{20}{3.0 + 2.0} = 4.0\text{ m/s}^2$.
$N_c = m_2 a = 2.0 \times 4.0 = 8.0\text{ N}$.
Correct Answer: 8
Solution 10:
In downward accelerating elevator: $mg – N = ma \implies N = m(g – a) = 70(9.8 – 2.8) = 70 \times 7.0 = 490\text{ N}$.
Correct Answer: 490