A body is in equilibrium under concurrent forces when the vector sum of all forces acting on it is zero — it has zero acceleration. This sounds simple, but NEET exploits two specific traps in this topic that cost marks year after year.
Trap 1: The Atwood pulley — system shortcut loses the tension. When two masses hang over a frictionless pulley, many aspirants write a single F = ma for the whole system. That gives acceleration correctly (a = (m₁ − m₂)g/(m₁ + m₂)), but the tension T disappears from the equation. When the question asks for T, they're stuck. The reliable method: draw a free-body diagram for each mass separately and write Newton's second law for each. Two equations, two unknowns (a and T). Solve simultaneously. T = 2m₁m₂g/(m₁ + m₂) — always less than the weight of the heavier mass and more than the weight of the lighter one when the system accelerates. NCERT does not print this formula; it builds it the same way, one second-law equation per body, in its block-and-trolley pulley example (NCERT Class 11 Physics Chapter 4, page 62) and sets this exact two-mass pulley as Exercise 4.16 (page 70).
Trap 2: The ladder against a smooth wall — pivot choice. A uniform ladder leaning against a smooth wall with a rough floor involves four forces: weight at the centre, normal and friction at the floor, and normal reaction at the wall. Taking torques about the centre of mass introduces all four forces with non-zero moment arms. Instead, take torques about the floor contact point — friction and normal at the floor contribute zero torque, leaving only weight and wall reaction. The limiting condition simplifies to μ_min = 1/(2 tan θ) (NCERT Class 11 Physics Chapter 4, page 64).
The equilibrium conditions for concurrent forces are: ΣF_x = 0, ΣF_y = 0, and for extended bodies, Στ = 0 about any chosen point. The pivot choice is free — pick the one that eliminates the most unknowns.
Watch-out: Newton's third law pairs act on different bodies. The book's weight and the table's normal force on the book are NOT an action-reaction pair — both act on the book. That's equilibrium, not Newton's third law.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Two masses m₁ = 4 kg and m₂ = 2 kg are connected by a light inextensible string over a frictionless pulley. What is the tension in the string? (Take g = 10 m/s²)
Answer: A. Write F = ma for each mass. Heavier (4 kg, moving down): 40 − T = 4a. Lighter (2 kg, moving up): T − 20 = 2a. Adding the two removes T: 20 = 6a, so a = 10/3 m/s². Then T = 20 + 2 × 10/3 = 80/3 ≈ 26.7 N, which is what T = 2m₁m₂g/(m₁ + m₂) = 160/6 gives. (Second law: NCERT Class 11 Physics Chapter 4, page 54; this two-mass pulley is Exercise 4.16, page 70.)
Why B is wrong: B. 20 N assumes T equals the weight of the lighter mass (m₂g), ignoring that T must be greater than m₂g to accelerate m₂ upward. (trap: confusing tension with single weight)
Why C is wrong: C. 40 N assumes T equals the weight of the heavier mass (m₁g), which is true only when acceleration is zero. (trap: treating pulley as static equilibrium)
Why D is wrong: D. 30 N comes from averaging the two weights (m₁g + m₂g)/2 = 30 N, which has no physical basis for an Atwood machine. (trap: arithmetic averaging instead of harmonic-mean-like formula)
A uniform ladder of length L and mass M leans against a smooth vertical wall making angle θ with the floor. The floor is rough with coefficient of static friction μ_s. The minimum value of μ_s for the ladder to remain in equilibrium is:
Answer: D. Taking torques about the base eliminates floor friction and normal force. The wall reaction N_w × L sin θ = Mg × (L/2) cos θ, giving N_w = Mg cos θ/(2 sin θ). Horizontal equilibrium: f = N_w. Vertical: N_floor = Mg. So μ_min = f/N_floor = cos θ/(2 sin θ) = 1/(2 tan θ) (NCERT Class 11 Physics Chapter 4, page 64).
Why A is wrong: A. cos θ / 2 omits the division by sin θ from the wall reaction's moment arm. (trap: incomplete torque calculation)
Why B is wrong: B. tan θ / 2 inverts the trigonometric ratio. This would mean steeper ladders need MORE friction, which contradicts physics — a nearly vertical ladder needs very little friction. (trap: swapping sin and cos in the torque equation)
Why C is wrong: C. 1/(2 sin θ) drops the cos θ from the weight's moment arm, replacing it incorrectly with 1/sin θ. (trap: wrong component resolution of the weight's moment arm)
A body is in static equilibrium under three concurrent forces. If two of the forces are 5 N and 12 N acting perpendicular to each other, the magnitude of the third force is:
Answer: A. For equilibrium, the third force must be equal and opposite to the resultant of the first two. Resultant = √(5² + 12²) = √(25 + 144) = √169 = 13 N. The third force is 13 N directed opposite to this resultant (NCERT Class 11 Physics Chapter 4, page 52).
Why B is wrong: B. 17 N adds the two forces arithmetically (5 + 12 = 17), which applies only when forces are parallel and in the same direction. (trap: scalar addition instead of vector addition)
Why C is wrong: C. 7 N subtracts the two forces (12 − 5 = 7), which applies only when forces are antiparallel, not perpendicular. (trap: wrong vector addition rule)
Why D is wrong: D. 8.5 N is the average of 5 and 12, which has no physical basis in force equilibrium. (trap: averaging instead of vector sum)
A book rests on a table. A student says: "The weight of the book and the normal reaction from the table form a Newton's third law action-reaction pair." This statement is:
Answer: C. Both weight and normal reaction act on the same body (the book). Newton's third law pairs always act on two different bodies. The reaction to the book's weight is the gravitational pull the book exerts on the Earth; the reaction to the table's normal force on the book is the force the book pushes down on the table (NCERT Class 11 Physics Chapter 4, page 58).
Why A is wrong: A confuses equilibrium (equal-opposite forces on the SAME body) with Newton's third law (equal-opposite forces on DIFFERENT bodies). Equal magnitude and opposite direction is necessary but not sufficient for a third-law pair. (trap: action-reaction same-body confusion)
Why B is wrong: B makes the same error as A — forces cancelling on one body is equilibrium, not the third law. Newton's third law pairs never cancel because they act on different objects. (trap: conflating equilibrium with Newton's third law)
Why D is wrong: D is factually wrong — the normal force equals the weight for a body at rest on a horizontal surface. This option introduces a false premise. (trap: incorrect physics claim)
In an Atwood machine with masses m₁ = 5 kg and m₂ = 3 kg over a frictionless pulley, the acceleration of the system is: (Take g = 10 m/s²)
Answer: B. a = (m₁ − m₂)g/(m₁ + m₂) = (5 − 3) × 10/(5 + 3) = 20/8 = 2.5 m/s². This follows from writing separate F = ma equations for each mass and solving simultaneously, the method NCERT uses for its pulley example (NCERT Class 11 Physics Chapter 4, page 62).
Why A is wrong: A. 10 m/s² uses g directly, as if one mass were in free fall. This ignores the constraint that the string connects both masses through the pulley. (trap: treating the system as a single falling body)
Why C is wrong: C. 6.25 m/s² comes from m₁g/(m₁ + m₂) = 50/8, using only the heavier mass's weight as the driving force instead of the difference of the two weights. (trap: using one weight instead of the net force)
Why D is wrong: D. 3.75 m/s² comes from m₂g/(m₁ + m₂) = 30/8, using only the lighter mass's weight as the driving force. (trap: using one weight instead of the net force)
The actual value of static friction on a block resting on a rough horizontal surface when a horizontal force of 3 N is applied to it is: (Given: mass = 5 kg, μ_s = 0.4, g = 10 m/s²)
Answer: B. Maximum static friction = μ_s × mg = 0.4 × 50 = 20 N. Since the applied force (3 N) is well below this maximum, static friction self-adjusts to exactly match the applied force: f_s = 3 N. The block does not move (NCERT Class 11 Physics Chapter 4, page 58).
Why A is wrong: A. 20 N uses f_s = μ_s × N regardless of the applied force. This is the MAXIMUM possible static friction, not the actual value when the applied force is small. (trap: always using μ_s × N for static friction)
Why C is wrong: C. 10 N has no physical basis — it is neither the applied force nor μ_s × N nor mg. (trap: arbitrary intermediate value)
Why D is wrong: D. 0 N would mean no friction acts, but that would leave a net force of 3 N and the block would accelerate, contradicting the observation that it stays still. (trap: ignoring friction entirely)
Which of the following is the correct Newton's third law reaction pair to the gravitational force the Earth exerts on a book lying on a table?
Answer: C. Newton's third law: if Earth pulls the book downward with force mg, the book pulls the Earth upward with force mg. The two forces act on different bodies (Earth and book), are equal in magnitude, opposite in direction, and of the same type (gravitational). This is the defining requirement for a third-law pair (NCERT Class 11 Physics Chapter 4, page 58).
Why A is wrong: A is the most common wrong answer. The normal force and the weight both act on the book — they form an equilibrium pair, not a third-law pair. Third-law pairs must act on different bodies. (trap: action-reaction same-body confusion)
Why B is wrong: B. The weight of the table is a force on the table from the Earth, unrelated to the force on the book. Different object, different interaction. (trap: mixing unrelated forces)
Why D is wrong: D. Friction between the book and the table is a contact force unrelated to the gravitational interaction between Earth and the book. (trap: confusing force types)
A uniform rod of mass 10 kg and length 4 m leans against a smooth wall at 60° to the floor. If the floor is rough, the friction force at the base is: (Take g = 10 m/s²)
Answer: A. A is correct. The wall is smooth, so it pushes horizontally only. Taking torques about the floor contact removes both floor forces from the equation: N_wall × L sin 60° = Mg × (L/2) cos 60°, since the weight acts at the mid-point of a uniform rod. The length cancels: N_wall = Mg cos 60°/(2 sin 60°) = Mg/(2 tan 60°) = 100/(2√3) = 50/√3 N ≈ 28.9 N. Horizontal equilibrium then gives friction f = N_wall = 50/√3 N (NCERT Class 11 Physics Chapter 4, page 64).
Why B is wrong: B. 50√3 N ≈ 86.6 N inverts the ratio, putting tan 60° = √3 in the numerator instead of the denominator. Notice the shape of the result: a steeper rod should press LESS hard on the wall, so the answer must fall as the angle grows, and only 50/√3 does. (trap: inverting the trigonometric ratio)
Why C is wrong: C. 100/√3 N omits the factor of 2, by putting the weight's moment arm at L cos 60° instead of (L/2) cos 60°. The rod is uniform, so its weight acts at the mid-point. (trap: forgetting the centre-of-mass is at half-length)
Why D is wrong: D. 25/√3 N carries an extra factor of 2 in the denominator, as though the L/2 applied to both sides of the torque equation rather than to the weight alone. (trap: double-counting the factor of 2)
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Pattern: Ladder leaning against a smooth wall (NEET pattern: ladder leaning wall, anchored to PYQ 2025 47 Q42)
Given
A uniform ladder of mass M = 20 kg and length L = 5 m leans against a smooth vertical wall at angle θ = 45° to the horizontal floor. The floor is rough. Take g = 10 m/s².
Required
Find the minimum coefficient of static friction μ_s at the floor for the ladder to remain in equilibrium.
Concept
For a rigid body in static equilibrium: ΣF_x = 0, ΣF_y = 0, and Στ = 0 about any point. The key insight is choosing the pivot point wisely to eliminate unknowns.
Formula
Torque equation about the base contact point:
N_wall × L sin θ = Mg × (L/2) cos θ
Horizontal equilibrium: f = N_wall
Vertical equilibrium: N_floor = Mg
μ_min = f / N_floor
Substitution
N_wall × L sin 45° = Mg × (L/2) cos 45°
The length cancels from both sides, leaving N_wall sin 45° = Mg cos 45° / 2, so
N_wall = Mg / (2 tan 45°) = 200 / (2 × 1) = 100 N
f = N_wall = 100 N
N_floor = Mg = 200 N
μ_min = f / N_floor = 100/200 = 0.5
Calculation
μ_min = 1/(2 tan 45°) = 1/(2 × 1) = 0.50
Exact constants note: g = 10 m/s² is a problem-defined exact value. The integers M = 20, L = 5 are exact counting/given values. These do not limit significant figures. tan 45° = 1 (exact mathematical constant).
Final answer
μ_min = 0.50
The minimum coefficient of static friction is 0.50. Below this, the ladder slides.
Common trap
Taking torques about the centre of mass introduces all four forces (floor normal, floor friction, weight, wall reaction) with non-zero moment arms — four unknowns in one equation. Taking torques about the base eliminates two forces instantly (floor friction and floor normal have zero moment arm about the base contact point). Always choose the pivot that eliminates the most unknowns.
Similar NEET-style question
A uniform rod of mass 15 kg and length 6 m rests against a smooth wall at 30° to the floor. Find μ_min at the floor.
Answer: μ_min = 1/(2 tan 30°) = 1/(2 × 1/√3) = √3/2 ≈ 0.87.
---
A particle is in equilibrium when the net (vector) external force on it is zero. Equivalently, the sum of forces along every coordinate axis is zero (ΣFx = 0, ΣFy = 0, ΣFz = 0). Equilibrium of concurrent forces means three or more forces all passing through a single point sum to zero.
-- NCERT Class 11 Physics, Ch. 4, p. 58Approach to mechanics problems: isolate each body, draw all external forces on it as vectors (free-body diagram), apply Newton's Second Law along chosen axes. The choice of axes is free; aligning one axis with the direction of acceleration usually simplifies the algebra.
-- NCERT Class 11 Physics, Ch. 4, p. 65Figure 4.2 illustrates a book at rest on a horizontal table: weight W = mg acts downward; the table exerts an upward normal reaction R. With the book at rest (in equilibrium), R = W. This is NOT Newton's third law — both forces act on the SAME body (the book). Action-reaction pairs always act on DIFFERENT bodies.
-- NCERT Class 11 Physics, Ch. 4, p. 52The net external force on a body equals the rate of change of its linear momentum. For a body of constant mass, this reduces to F = m*a — net force equals mass times acceleration. Both F and a are vectors; the acceleration is in the direction of the net force.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | Net external (vector) force | N |
| m | Mass of the body | kg |
| a | Acceleration (vector) | m/s^2 |
| p | Linear momentum (= m*v) | kg*m/s |
| t | Time | s |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student takes torque about the centre of mass (introducing all 4 forces with non-zero moment arms) instead of about the floor contact (where 2 forces have zero moment arm and the equation simplifies).
Question gives a uniform rod or ladder leaning against a wall; asks for friction coefficient or limiting condition.
Pick the pivot to ELIMINATE unknown forces from the torque equation. Floor-contact pivot: normal force and friction at floor contribute zero torque; only weight (mid-length) and wall normal (top) appear. Result: μ_min = 1 / (2 tan θ).
Category: Overthinking
Student tries to apply F=ma to the system as a whole (using net force = (m1-m2)g and total mass m1+m2) but loses track of the tension. The correct approach writes Newton's Second Law SEPARATELY for each mass and treats T as an unknown in two simultaneous equations.
Question contains a frictionless pulley with two unequal masses tied to a string. Asks for tension T or acceleration a (or both).
Draw a free-body diagram for EACH mass. Write F=ma per body, treating T as the same magnitude on both sides of the string. Solve simultaneously: a = (m1 - m2) g / (m1 + m2); T = 2 m1 m2 g / (m1 + m2).
Category: Negative Marking
Multi-mass pulley problem requires computing acceleration first, then tension. T = 2 m1 m2 g/(m1+m2). Sign errors in m1−m2 propagate.
Atwood machine or pulley system with multiple masses.
Compute a = (m1-m2)g/(m1+m2) first with m1 the heavier mass and downward as positive. Then T from F=ma on either mass. Always re-check by plugging back into both Newton's 2nd Law equations.
Root cause: concept gap
Action-reaction pairs ALWAYS act on DIFFERENT bodies. The pair to the book's weight is the gravitational pull the book exerts on the Earth. The pair to the normal force from table on book is the force the book exerts on the table. Equal-and-opposite forces on the SAME body are an equilibrium statement, not third-law statement.
Distractor labels two forces on the same body as a Newton's-third-law pair.
More in Laws of Motion: 12 exam traps and mistakes · 8 formulas · 8 question patterns from its other lessons.
torque about wrong pivot
Default pivot at center of mass adds complexity
uses g where a belongs
Forgetting that the system accelerates, so weight is balanced by net force minus T
confuses tension with weight
Treating T = m·g for one of the masses (which is true only when a=0)
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →8 real NEET Laws of Motion questions (2022–2026) with mechanics you can watch move (22 min)
Watch on YouTube