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 either mass when the system accelerates (NCERT Class 11 Physics Chapter 4, page 10).
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 16).
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)
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:
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:
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:
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)
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)
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?
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)
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 × 5 × sin 45° = 20 × 10 × (5/2) × cos 45° Since sin 45° = cos 45°, they cancel: N_wall × 5 = 200 × 2.5 N_wall = 500/5 = 100 N Wait — let me redo carefully: N_wall × L sin 45° = Mg × (L/2) × cos 45° N_wall × sin 45° = Mg × cos 45° / 2 N_wall = Mg / (2 × tan 45°) = Mg / (2 × 1) = 200/2 = 100 N f = N_wall = 100 N N_floor = Mg = 200 N μ_min = 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. 10Approach 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. 16Figure 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. 4An object moving in a circle of radius r at constant speed v has acceleration of magnitude v^2/r (or equivalently omega^2 * r) directed toward the centre. This is centripetal (radially inward), not tangential.
| Symbol | Quantity | SI Unit |
|---|---|---|
| a_c | Centripetal acceleration | m/s^2 |
| v | Tangential speed | m/s |
| r | Radius of circle | m |
| omega | Angular speed | rad/s |
On a road banked at angle theta from horizontal with tyre-road friction coefficient mu_s, this is the maximum speed for safe negotiation of a curve of radius r.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v_max | Maximum safe speed | m/s |
| g | Gravitational acceleration | m/s^2 |
| r | Radius of the curve | m |
| mu_s | Coefficient of static friction | (dimensionless) |
| theta | Banking angle | rad/deg |
The net inward force required to keep a body of mass m moving in a circle of radius r at speed v is m*v^2/r. This 'centripetal' force is NOT a new fundamental force — it is whichever real force (tension, friction, gravity, etc.) provides the inward acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F_c | Centripetal force | N |
| m | Mass of the body | kg |
| v | Tangential speed | m/s |
| r | Radius of the circle | m |
| omega | Angular speed | rad/s |
If the net external force on a system of particles is zero, the total linear momentum of the system is conserved (vector equality of total p before and after any internal interaction). Basis of all collision and recoil analysis.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F_ext | Net external force on system | N |
| p_total | Sum of m_i*v_i over all particles | kg*m/s |
Impulse equals the product of force and the time interval over which it acts. By Newton's Second Law, impulse equals the change in linear momentum during that interval.
| Symbol | Quantity | SI Unit |
|---|---|---|
| J | Impulse (vector) | N*s = kg*m/s |
| F | Force (treated as average) | N |
| Delta_t | Time interval | s |
| Delta_p | Change in momentum | kg*m/s |
When two surfaces slide relative to each other, kinetic friction opposes the motion with magnitude proportional to the normal force.
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_k | Kinetic friction force | N |
| mu_k | Coefficient of kinetic friction | (dimensionless) |
| N | Normal force | N |
Static friction is the only force available to provide centripetal acceleration on a level road. Setting mu_s*N = m*v^2/r and N = m*g gives this maximum-safe speed bound.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v_max | Maximum safe speed (no skid) | m/s |
| mu_s | Coefficient of static friction (tyre vs road) | (dimensionless) |
| g | Gravitational acceleration | m/s^2 |
| r | Radius of the circular path | m |
The 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 |
The maximum value of static friction between two surfaces in contact equals the coefficient of static friction times the normal force. Below f_s_max, static friction self-adjusts to whatever value is needed to prevent relative motion.
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_s_max | Maximum static friction | N |
| mu_s | Coefficient of static friction | (dimensionless) |
| N | Normal force | N |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student claims velocity is constant in uniform circular motion (it's not — direction changes).
Question asks 'in uniform circular motion at constant speed, which is also constant?'
In UCM: SPEED constant; KE constant. VELOCITY (vector) NOT constant. ACCELERATION (centripetal, magnitude v²/r) constant in MAGNITUDE but NOT in direction.
Category: Sign Convention
Student picks the direction of MOTION as the direction of the net force, instead of the direction of the change in momentum (Δp). When velocity changes direction at constant speed, the force is perpendicular to BOTH the initial and final velocity vectors (in the limit) — it's the direction of Δv.
Question describes a body changing direction (e.g. turning) and asks for the force direction or the direction of Δp.
F ∝ Δp = m Δv = m (v_f − v_i). Always draw v_i and v_f as vectors and subtract; the result (v_f − v_i) is the direction of net force.
Category: Unit Conversion
Student computes μmg (friction FORCE in Newtons) when asked for the friction-limited ACCELERATION. The two differ by a factor of m: a = μg, F = μmg.
Question gives μ, g, and a body's mass and asks for the maximum acceleration of the supporting surface OR the friction force on the body.
Read carefully: 'maximum acceleration of the vehicle so the body stays still' = μg (no mass). 'Friction force on the body' = μmg (mass present). Units expose the error: N for force, m/s² for acceleration.
Category: Sign Convention
Student forgets that velocity DIRECTION reverses on bounce; computes |v₁ - v₂| instead of |v₁ + v₂|.
Question describes ball dropped from height h₁, rebounding to height h₂; asks for impulse on ball.
Impulse J = Δp = m(v_after - v_before). Take down as positive: v_before = +v₁, v_after = -v₂. So J = m(-v₂ - v₁) = -m(v₁ + v₂); magnitude = m(v₁ + v₂).
Category: Sign Convention
Student writes a = g sin θ for a rough incline (which is the smooth-incline answer); forgets to subtract μ g cos θ.
Question contrasts rough vs smooth inclines, or asks for acceleration on a rough incline.
On a rough incline (block sliding down): a = g(sin θ - μ_k cos θ). On a rough incline (block sliding up): a = -g(sin θ + μ_k cos θ). Smooth case (μ = 0): just g sin θ.
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: Sign Convention
Student adds magnitudes of momenta of fragments instead of vectors; ignores cancellation when fragments fly perpendicular or in opposite directions.
Question describes a body at rest exploding into multiple fragments with given mass ratios and partial velocity info.
Total momentum is a VECTOR. Initial p = 0; therefore Σ m_i v_i = 0 as a vector equation. Decompose along chosen axes (often natural symmetry axes); sum = 0 in each.
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.
Category: Overthinking
Student applies the full external force F to a single block instead of recognising the system needs to be analysed for the contact (internal) force.
Question gives horizontal force F on block A which pushes block B; asks for contact force between A and B or acceleration.
System acceleration: a = F / (m_A + m_B). Contact force on B from A = m_B × a. The full F acts on the system, not on each block independently.
Category: Overthinking
Student computes P = Mgv (just lifting against gravity) and ignores the friction-opposing-motion term.
Question describes a lift moving at constant speed with explicit friction force on cable or guides.
At constant speed, net force = 0, so cable tension T = Mg + f_friction. Power = T × v = (Mg + f) × v. Always add friction when stated.
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.
Root cause: concept gap
Centripetal force is NOT a new fundamental force. It is the NET inward radial component of the real forces (tension, friction, normal, gravity, etc.). On a free-body diagram, draw only the real forces; their net inward component must equal m*v^2/r.
Distractor sums tension + 'centripetal force' as separate inward forces.
Root cause: unit error
Impulse J has dimensions of momentum (kg*m/s) and equals F*Delta_t. Force has dimensions kg*m/s^2. Confusing them inflates or deflates an answer by a factor of seconds. Always check units before declaring an answer.
Distractor reports an answer in newtons where the correct answer is in N*s (or vice versa).
Root cause: concept gap
Linear momentum is conserved only when the net EXTERNAL force is zero. Gravity over a finite time changes momentum. For collision problems we use conservation because the collision happens over a brief time interval where external impulses are negligible compared to internal collision forces.
Distractor sets initial momentum = final momentum for a free-fall problem where gravity has acted for several seconds.
Root cause: formula misuse
Static friction is SELF-ADJUSTING: f_s exactly cancels the applied tangential force up to a ceiling f_s_max = mu_s * N. Below the ceiling, f_s = applied force. At the ceiling, motion is impending. Substituting mu_s * N too early over-estimates the friction.
Distractor uses mu_s * N for the static friction force in a no-slip scenario where the applied force is well below threshold.
treats velocity as scalar
Conflates speed (scalar) with velocity (vector)
scalar sum of momenta
Treating momentum as scalar; ignores vector cancellation
force along final velocity
Default to thinking force points in direction of motion
force along initial velocity
Newton-1 misread: object 'wants' to keep moving in original direction
subtracts v2 from v1 instead of adding
Forgetting velocity DIRECTION reversal at bounce
ignores mu cos theta term
Forgetting the friction-along-incline component
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)
uses mu times g times mass
Confusing force (μ·N = μmg) with acceleration
treats each block with full F
Forgetting Newton's 3rd law internal force decomposition
forgets friction term
Ignoring opposing force
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