This MCQ module is based on: Third Law Equilibrium
TOPIC 16 OF 33
Third Law Equilibrium
🎓 Class 11
Physics
CBSE
Theory
Ch 4 – Laws of Motion
⏱ ~14 min
🌐 Language: [gtranslate]
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Third Law Equilibrium
4.7 Newton's Third Law of Motion
The first two laws of motion tell us how an applied force changes the motion and provide us with a method of determining the force. Newton's third law of motion gives yet another important property of force.
To every action, there is always an equal and opposite reaction. In simple terms, the law states that forces in nature always occur between pairs of objects. The two forces are equal in magnitude and opposite in direction.
Newton's Third Law: If body A exerts a force \(\vec{F}_{AB}\) on body B, then body B simultaneously exerts a force \(\vec{F}_{BA}\) on body A, such that:
\[\vec{F}_{AB} = -\vec{F}_{BA}\]
Crucial points about action-reaction pairs:
- Always act on DIFFERENT bodies — if both forces acted on the same body, they would cancel and nothing would ever accelerate!
- Equal in magnitude — \(|F_{AB}| = |F_{BA}|\) always.
- Opposite in direction — along the line joining the two bodies (for contact forces).
- Simultaneous — appear and disappear together. There is no "first" force.
- Same type of force — if action is gravitational, reaction is also gravitational.
⚠️ Common Misconception: The pair "weight of book (W) and normal force (N) on book" is NOT a Newton's Third Law pair! Both act on the SAME body (book). They balance each other (Newton's First Law). The TRUE third-law pair for W is the gravitational pull of the book on the Earth.
Application: Conservation of Momentum (Two-Body)
Consider two bodies A and B colliding. The third law says \(\vec{F}_{AB} = -\vec{F}_{BA}\). During the collision (time Δt):
\[\vec{F}_{AB} \cdot \Delta t = -\vec{F}_{BA} \cdot \Delta t\]
\[\Delta\vec{p}_B = -\Delta\vec{p}_A\]
\[\Delta\vec{p}_A + \Delta\vec{p}_B = 0\]
The total momentum of the (A+B) system is conserved. This is the foundation of the law of conservation of momentum.
4.8 Equilibrium of a Particle
Equilibrium of a particle in mechanics refers to the situation when the net external force on the particle is zero. According to the first law, this means that, the particle is either at rest or in uniform motion.
If two forces \(\vec{F}_1\) and \(\vec{F}_2\) act on a particle, equilibrium requires:
\[\vec{F}_1 = -\vec{F}_2\]
i.e., the two forces on the particle must be equal and opposite. Equilibrium under three forces requires:
\[\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0\]
That is, the three forces must form a closed triangle when added head-to-tail. In component form, for equilibrium:
\[\sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0\]
Worked Examples
Worked Example 1: Identifying Action-Reaction Pairs
A horse pulls a cart and the cart moves. By Newton's third law, the cart pulls the horse back with equal force. Then how does the cart move at all?
This is a famous puzzle that often confuses students. The resolution lies in identifying which forces act on which body.
Forces on the cart: (i) horse pulls cart forward (F), (ii) friction with ground (f_cart, backward).
Net force on cart = F − f_cart. If F > f_cart, cart accelerates forward.
Forces on the horse: (i) cart pulls horse backward (F'), (ii) friction with ground PUSHES HORSE FORWARD (because the horse pushes the ground backward through its hooves; reaction = forward).
Net force on horse = friction − F'. If friction > F', horse accelerates forward.
Key insight: The action-reaction pair (F, F') doesn't cancel because they act on DIFFERENT bodies. The horse-cart system as a whole moves forward because the EXTERNAL force (ground friction on horse) is forward.
Forces on the cart: (i) horse pulls cart forward (F), (ii) friction with ground (f_cart, backward).
Net force on cart = F − f_cart. If F > f_cart, cart accelerates forward.
Forces on the horse: (i) cart pulls horse backward (F'), (ii) friction with ground PUSHES HORSE FORWARD (because the horse pushes the ground backward through its hooves; reaction = forward).
Net force on horse = friction − F'. If friction > F', horse accelerates forward.
Key insight: The action-reaction pair (F, F') doesn't cancel because they act on DIFFERENT bodies. The horse-cart system as a whole moves forward because the EXTERNAL force (ground friction on horse) is forward.
Worked Example 2 (NCERT Example 4.6): Equilibrium of Three Forces
A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the midpoint P of the rope. What is the angle the rope makes with the vertical in equilibrium? (Take g = 10 m/s². Neglect mass of the rope.)
Setup: The mass hangs from the lower half of rope; tension T_1 in upper half, T_2 in lower half (= mg = 60 N).
At point P, three forces:
Dividing: \[\tan\theta = \frac{50}{60} = \frac{5}{6}\] \[\theta = \tan^{-1}(0.833) \approx \boxed{40.6°}\] The rope makes ~40.6° with the vertical.
At point P, three forces:
- T_1 along upper rope (up & away from vertical, at angle θ from vertical)
- T_2 = 60 N down (lower rope tension = weight of mass)
- F = 50 N horizontal
Dividing: \[\tan\theta = \frac{50}{60} = \frac{5}{6}\] \[\theta = \tan^{-1}(0.833) \approx \boxed{40.6°}\] The rope makes ~40.6° with the vertical.
Worked Example 3: Rocket Thrust
A rocket of mass 200 kg expels gas backward at a rate of 5 kg/s with a relative velocity of 800 m/s. Find the thrust on the rocket.
By Newton's Third Law, the force the rocket exerts on the gas (action) equals the force the gas exerts on the rocket (thrust).
Thrust = (rate of mass ejection) × (exhaust velocity) \[T = \frac{dm}{dt} \times v_{rel} = 5 \times 800 = \boxed{4000 \text{ N}}\] This thrust is what propels the rocket forward.
Thrust = (rate of mass ejection) × (exhaust velocity) \[T = \frac{dm}{dt} \times v_{rel} = 5 \times 800 = \boxed{4000 \text{ N}}\] This thrust is what propels the rocket forward.
🎯 Interactive: Three-Force Equilibrium
A weight hangs from a string. A horizontal force pulls it sideways. Vary the force and see the equilibrium angle.
📐 Activity 4.3 — Tug of War & Newton's Third Law
Setup: Two students of similar mass on a smooth (low-friction) floor each hold one end of a rope. They pull as hard as they can.
Variation 1: Both students wear shoes with good grip.
Variation 2: Student A is on smooth ice (no friction); Student B has good grip.
Predict: In each case, who wins, and why?
Variation 1: The rope tension is THE SAME on both (Newton's Third Law). Whoever pushes the ground harder via friction (better grip + more leg strength) "wins" — actually moves the other one.
Variation 2: Student A (on ice) cannot generate friction on the ground. So when B pulls, A simply slides toward B. The rope tension is still equal on both sides, but there is no external force on A to oppose the motion. B easily wins.
Insight: Tug of war is NOT about who pulls the rope harder. It's about who can push the GROUND harder (and the ground pushes back via friction).
🎯 Competency-Based Questions
Q1. A football player kicks a ball with force 200 N. The reaction force on the player's foot is: L2 Understand
Answer: (c) 200 N. By Newton's Third Law, action and reaction are equal in magnitude. The ball pushes back on the foot with the same 200 N (in opposite direction).
Q2. Why can swimmers push themselves forward in water? L3 Apply
Answer: The swimmer pushes water BACKWARD with their hands and feet. By Newton's Third Law, the water pushes the swimmer FORWARD with equal force. This forward push from the water (the reaction force) propels the swimmer.
Q3. A 3-kg mass and 5-kg mass collide elastically. The 3-kg mass exerts 30 N on the 5-kg mass during contact. What force does the 5-kg mass exert on the 3-kg mass? Find the accelerations of each. L4 Analyse
Answer: Force on 3-kg from 5-kg = 30 N (Newton's Third Law).
Acceleration of 3-kg: a = 30/3 = 10 m/s² (in opposite direction to its velocity).
Acceleration of 5-kg: a = 30/5 = 6 m/s² (in direction of force from 3-kg).
Note: forces equal, but accelerations differ because masses differ.
Acceleration of 3-kg: a = 30/3 = 10 m/s² (in opposite direction to its velocity).
Acceleration of 5-kg: a = 30/5 = 6 m/s² (in direction of force from 3-kg).
Note: forces equal, but accelerations differ because masses differ.
Q4. A bird sits on a tree branch. List all action-reaction pairs involving the bird. L4 Analyse
Answer: The bird is in equilibrium with two forces: gravity (Earth pulling bird down) and normal (branch pushing bird up).
Action-Reaction Pairs:
(i) Earth pulls bird DOWN ↔ bird pulls Earth UP (gravitational, equal magnitude).
(ii) Branch pushes bird UP (normal) ↔ bird pushes branch DOWN (normal, equal magnitude).
Note: The "balance" of weight and normal acting on the BIRD is NOT a third-law pair — it's the first-law equilibrium condition.
Action-Reaction Pairs:
(i) Earth pulls bird DOWN ↔ bird pulls Earth UP (gravitational, equal magnitude).
(ii) Branch pushes bird UP (normal) ↔ bird pushes branch DOWN (normal, equal magnitude).
Note: The "balance" of weight and normal acting on the BIRD is NOT a third-law pair — it's the first-law equilibrium condition.
Q5. HOT: A box of mass 5 kg sits on the floor of a lift. The lift accelerates upward at 2 m/s². Find: (a) the normal force on the box, (b) what the box exerts on the floor, (c) any third-law pair acting on the floor. L6 Create
(a) N − mg = ma ⇒ N = m(g + a) = 5 × 11.8 = 59 N upward.
(b) By Newton's Third Law, the box exerts 59 N DOWNWARD on the floor (reaction).
(c) Third-law pair on the floor: the box-on-floor force (59 N down). The floor's reaction (floor-on-box, 59 N up) is the third-law partner.
Insight: The lift's upward acceleration increases the apparent weight by ma = 5×2 = 10 N. The box "feels heavier" (10 N more than 49 N at rest).
(b) By Newton's Third Law, the box exerts 59 N DOWNWARD on the floor (reaction).
(c) Third-law pair on the floor: the box-on-floor force (59 N down). The floor's reaction (floor-on-box, 59 N up) is the third-law partner.
Insight: The lift's upward acceleration increases the apparent weight by ma = 5×2 = 10 N. The box "feels heavier" (10 N more than 49 N at rest).
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: Action and reaction forces always cancel each other.
R: They are equal in magnitude and opposite in direction.
Answer: (D). A is FALSE — they don't cancel because they act on DIFFERENT bodies. R is TRUE. If they cancelled, no motion would ever be possible. The "non-cancelling on the same body" is what enables motion.
A: A rocket can be launched in space (vacuum) where there is no air to push against.
R: A rocket works on the principle of Newton's Third Law — exhaust gases pushed back, rocket pushed forward.
Answer: (A). Both true; R correctly explains A. The rocket needs no air — it carries its own propellant. The third-law reaction pushes the rocket forward.
A: When two bodies are in equilibrium, all forces between them must form a closed polygon.
R: For equilibrium of a body, the vector sum of all forces on that body must be zero.
Answer: (D). A is FALSE/incomplete — the closed-polygon rule applies to forces on a SINGLE body, not "between two bodies". R is TRUE — Σ F = 0 for equilibrium.
Frequently Asked Questions - Third Law Equilibrium
What is the main concept covered in Third Law Equilibrium?
In NCERT Class 11 Physics Chapter 4 (Laws of Motion), "Third Law Equilibrium" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Third Law Equilibrium useful in real-life applications?
Real-life applications of Third Law Equilibrium from NCERT Class 11 Physics Chapter 4 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Third Law Equilibrium?
Key formulas in Third Law Equilibrium (NCERT Class 11 Physics Chapter 4 Laws of Motion) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 4?
NCERT Class 11 Physics Chapter 4 (Laws of Motion) is structured so each part builds on the previous one. Third Law Equilibrium connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Third Law Equilibrium?
CBSE board questions from Third Law Equilibrium typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Third Law Equilibrium lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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