Friction

Chapter Snapshot - Friction

Friction is among the most practically tested chapters in NEET mechanics. Static friction (self-adjusting, 0 to μsN), kinetic friction (constant at μkN), angle of friction, and angle of repose anchor every MCQ in this chapter. The key insight is that static friction is not μsN — it equals the applied force and only reaches μsN at the point of impending motion. Two-block problems (block on block, Atwood with friction) and inclined-plane dynamics are the dominant NEET question formats. Students who master tan(angle of repose) = μs = tan(angle of friction) and the incline acceleration formula a = g(sinθ − μcosθ) will answer 70% of this chapter's questions almost instantly.

✓ Use This To Plan Your First 2–3 Hours

Expected Questions (Typical)

2-3

Friction is tested every year — typically 2 questions and sometimes 3. The three recurring formats are: (1) angle of repose / angle of friction numericals, (2) a = g(sinθ − μcosθ) or two-block friction acceleration problems, and (3) minimum force / stopping-distance conceptual MCQs.

Time Required (Practical)

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6-8 hrs

Theory and types of friction 1.5 hrs; angle of friction/repose and incline formulas 2 hrs; two-body friction problem practice 2 hrs; stopping distance, cart/rotor applications, and MCQ bank 1.5 hrs.

Difficulty Level

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Moderate

Formulas are compact, but NEET exploits conceptual gaps: whether friction equals μN or equals applied force, whether to use static or kinetic coefficient, and how to identify friction direction when the applied force is oblique. The mathematics is straightforward once the free-body diagram is set up correctly.

Most Asked Style: Numerical MCQ: find acceleration of body sliding down rough incline; find minimum mass to move block connected by string over pulley; angle of repose given μ; stopping distance of moving block; minimum force direction for moving body on horizontal surface.Biggest Trap: Saying friction on a stationary body equals μsN. Static friction is self-adjusting and equals only the applied external force until impending motion — it reaches μsN only at the limiting (maximum) condition. NEET structures 1 MCQ per paper around this exact misconception.Fast Win: Memorise: angle of repose = angle of friction = tan⁻¹(μs). From this single equality, tan(angle) = μ follows immediately and 30% of angle-related problems become one-liners. Also memorise: a = g(sinθ − μcosθ) for sliding down and stopping distance S = u²/(2μg) on horizontal surface.Revision-Friendly: Yes. All key results fit on two flashcards: Card 1 — friction types, μs > μk, angle of friction = angle of repose; Card 2 — incline acceleration, two-block conditions, stopping formulas. A 40-minute pre-exam review covers over 80% of testable content.

Subtopics - Friction (NEET)

Four major blocks: classification and laws of friction (static, limiting, kinetic, rolling; graph of friction vs applied force); angle of friction and angle of repose (their equality and consequences); dynamics on inclined planes and force calculations (accelerations, work done, minimum force problems); and multi-body friction systems (block-on-block, Atwood with friction, stopping problems, cart and rotor applications).

Revision tip: Before any friction numerical: (1) draw the free-body diagram and mark the direction of impending motion, (2) identify which friction acts — static (adjustable) or kinetic (μkN), (3) write normal reaction from perpendicular equilibrium, then substitute into friction force. This ritual eliminates 90% of free-body errors. For two-block problems always check whether F < F_limiting first — if yes, both bodies move together.

NCERT Lines MCQs Quick Test

1) Types of Friction, Laws, and the F-vs-Applied Force Graph

Defines and distinguishes all three types of friction (static, kinetic/dynamic, rolling) and their coefficients. States the two empirical laws: friction is proportional to normal reaction and independent of apparent area of contact. Derives the inequality μk < μs and explains the graph of friction force vs applied force showing the OA (static increasing), A (limiting), BC (kinetic, constant) regions. Rolling friction is the smallest of the three types.

Static: self-adjustingKinetic: μkN constantRolling < Kinetic < StaticLaws: F ∝ R, area-independent

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Static Friction and Limiting FrictionStatic friction: the opposing force when a body tends to move but has not yet started. It is self-adjusting — it equals the applied force exactly as long as the body remains stationary and has not yet reached its maximum value. Maximum static (limiting) friction: F_l = μsR where μs is the coefficient of static friction (dimensionless, μs = F_l / R) and R is the normal reaction. Static friction direction: always opposite to the direction of impending motion (not necessarily opposite to applied force). F_l ∝ R always; F_l is independent of apparent area of contact and relative velocity (since body is stationary). Coefficient μs depends only on the nature and material of the surfaces in contact.

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Kinetic Friction, Rolling Friction, and the Friction–Force GraphKinetic (dynamic) friction: once the body moves, friction = F_k = μkR where μk < μs always. Kinetic friction is constant regardless of applied force magnitude or velocity of the body — this is why the F_k region (BC) on the graph is horizontal. Rolling friction: F_rolling = μr × R/r where r is the radius of the rolling body; μr has dimensions of length (metres). Rolling friction is much smaller than sliding friction — hence wheels are used for heavy loads. Graph of friction vs applied force: segment OA is linear (static, self-adjusting); point A is limiting friction (peak); beyond A, friction drops to the lower constant kinetic value — this slight drop from F_l to F_k is because once motion starts, fewer surface irregularities interlock. Hierarchy: rolling friction < kinetic friction < limiting friction.

2) Angle of Friction, Angle of Repose, and Resultant Surface Force

Defines angle of friction (λ) as the angle between the resultant surface force S and the normal reaction R at limiting condition: tan λ = F_l/R = μs. Defines angle of repose (α) as the inclination at which a body just begins to slide: tan α = μs. Proves their equality α = λ — one of the most tested equalities in NEET friction. Derives the magnitude of the resultant force S = mg√(μ² + 1) and its range mg ≤ S ≤ mg√(μ² + 1). Derives minimum-force formulas for pulling, pushing, and motion on inclined planes.

tan λ = μsAngle of repose = angle of frictionS = mg√(μ²+1)Min-force: P = W sinθ / cos(α−θ)

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Angle of Friction and Angle of Repose — Derivation and EqualityAngle of friction λ: defined as the angle which the resultant of limiting friction (F_l) and normal reaction (R) makes with R. At limiting condition: tan λ = F_l / R = μs; therefore λ = tan⁻¹(μs). Block stays static if the applied force angle from normal is at most λ; slides if angle exceeds λ. Angle of repose α: the inclination angle of a plane at which a body just begins to slide (impending motion condition). At that angle: F_l = mg sin α and R = mg cos α giving tan α = F_l / R = μs. Since both equal tan⁻¹(μs), angle of repose = angle of friction always. This is a direct NEET equality: from μ alone, determine both angles simultaneously. The resultant surface force S = √(F_l² + R²) = mg√(μ² + 1); without friction (μ=0) S = mg (minimum); with friction S can be up to mg√(μ² + 1).

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Minimum Force Calculations — Horizontal Surface and Inclined PlaneFor body on horizontal surface, force P applied at angle α from horizontal: equilibrium gives F = P cos α and R = W − P sin α. Substituting F = μR gives P = W sin θ / cos(α − θ) for pulling (where θ = angle of friction). For pushing at angle α below horizontal: R = W + P sin α; P = W sin θ / cos(α + θ) — greater force needed than pulling because normal reaction (and hence friction) increases. Minimum pulling force on horizontal surface: minimum P occurs when α = θ (angle of pull equals angle of friction); P_min = μmg / √(1 + μ²). For moving body up a rough incline: P = W sin(θ + λ) / cos(α − θ). For preventing sliding down incline: P = W [sin(λ − θ) / cos(θ + α)]. The common thread: the angle of friction θ = tan⁻¹(μ) appears in every formula.

3) Dynamics on Inclined Plane and Work Done Against Friction

Derives the net acceleration for a block sliding down (a = g(sin θ − μcos θ)) and the retardation for a block moving up (a = g(sin θ + μcos θ)) a rough inclined plane. Calculates stopping distance on horizontal (S = u²/2μg) and on incline (S = u²/2g(sin θ + μcos θ)) from energy or kinematics. Derives work done against friction on incline: W = mgs(sin θ + μcos θ). Covers coefficient-of-friction measurement from time-ratio on smooth vs rough wedge: μ = tan θ (1 − 1/n²).

a = g(sinθ − μcosθ)S = u²/(2μg)W = mgs(sinθ + μcosθ)Wedge n-ratio: μ = tanθ(1−1/n²)

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Acceleration on Rough Incline — Down and Up MotionBlock sliding DOWN rough incline (θ > angle of repose): along incline: mg sin θ − μmg cos θ = ma; so a = g(sin θ − μcos θ). If θ < angle of repose, block stays stationary — static friction adjusts to balance gravity component. Block moving UP rough incline (retardation): both gravity component and kinetic friction act down the incline; ma = mg sin θ + μmg cos θ; retardation a = g(sin θ + μcos θ). For frictionless incline (μ = 0): a = g sin θ in both cases. On horizontal surface (θ = 0): a = μg (retardation). The incline acceleration formula is the most numerically tested result in this chapter — every NEET paper has at least one calculation using it.

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Stopping Distance, Work Done, and Rough Wedge Time RatioStopping distance on HORIZONTAL surface: using v² = u² − 2aS with v = 0 and a = μg gives S = u² / (2μg). Alternatively, S = P² / (2μm²g) where P = mu is momentum. Time to stop: from v = u − μgt = 0 gives t = u/(μg). Stopping distance on INCLINE: S = u² / [2g(sin θ + μcos θ)] since retardation = g(sin θ + μcos θ). Work done against friction on horizontal: W = F × S = μmgs. On rough incline: W = mgs(sin θ + μcos θ) — includes both the gravitational component and friction. Rough wedge time experiment: if smooth wedge takes time t for descent and rough wedge takes nt (n > 1) for same path: both have same distance S; equating half × g sin θ × t² = half × g(sin θ − μcos θ) × (nt)² gives μ = tan θ × (1 − 1/n²). Used to measure μ experimentally.

4) Two-Body Friction Problems and Special Applications

Analyses the block-on-block system (force applied to upper or lower block) under four sub-cases: no friction, friction between blocks only (F < F_l, move together; F > F_l, different accelerations), friction also between lower block and floor. Derives Atwood-type conditions: mass on horizontal table vs hanging mass (m2 = μm1 at limiting equilibrium); mass on rough incline vs hanging mass. Covers maximum hanging chain length (μ = hanging length / table length), sticking of block to accelerating cart (a_min = g/μ), and rotor wall sticking (ω_min = √(g/μr)).

Two-block: check F vs μ_s mgAtwood: m2 = μm1 horizontalCart sticking: a_min = g/μRotor: ω_min = √(g/μr)

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Block-on-Block and Atwood-Type Friction SystemsBlock A (mass m) on block B (mass M), force F applied to UPPER block A: if F < F_l = μs mg (limiting friction between A and B), both move together with common acceleration a = F/(M+m). If F > F_l, blocks separate: aA = (F − μk mg)/m; aB = μk mg/M; relative acceleration = aA − aB = [MF − μk mg(M+m)]/(mM); time for A to fall off length L of B: t = √(2mML / [MF − μk mg(M+m)]). Force applied to LOWER block B: move together if F < μs(M+m)g; if F > μs mg (limiting), B moves faster; aA = μk g; aB = (F − μk mg)/M. Atwood on horizontal: block m1 on table, hanging mass m2 — at limiting equilibrium T = m2g and T = F_l = μs m1g; so m2_min = μs m1. On rough incline: m2_min = m1(sin θ + μcos θ).

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Hanging Chain, Accelerating Cart, and Rotor ProblemsMaximum hanging chain: for uniform chain of length l with length l' hanging over edge, at limiting condition μ = (hanging mass)/(table mass) = l'/(l − l'); so maximum fraction that can hang without sliding: l'_max = μl/(1 + μ). Accelerating cart and block: vertical block pressed against front face of horizontally accelerating cart of mass M; pseudo-force on block = ma (horizontal, into wall); normal reaction R = ma; friction force = μma must support weight mg; condition: μma ≥ mg → a_min = g/μ; minimum cart force = (M+m) × g/μ. Rotor wall sticking: person stands inside rotating cylindrical drum; centrifugal force outward = mω²r = normal reaction R; upward static friction = μR = μmω²r must balance mg; condition μω²r ≥ g → ω_min = √(g/μr). These are favourite NEET applied-friction problems because they require physical reasoning, not just formula substitution.

Friction Download Notes & Weightage Plan

For each topic in the Friction chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.

2 Downloads

Types of Friction, Laws, and the F-vs-Applied Force Graph

The definitional and conceptual foundation: three friction types, their coefficients, the proportionality laws, and the graph that encodes almost every conceptual MCQ on friction.

1-2 Q/yearConceptual + definitionalGraph interpretationμs > μk always

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

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Topic Notes (Condensed)Static friction: self-adjusting = applied force, max value = μs R. Limiting friction: maximum static friction = μs R. Kinetic friction: F_k = μk R, constant, independent of velocity, μk < μs. Rolling friction: F = μr R/r, smallest of all three. Laws: F_l ∝ R (normal force); independent of area. Friction graph: OA linear (static), A = F_l peak, BC horizontal (kinetic, constant, lower than A). Friction as cause of motion: walking, vehicle motion, object in accelerating vehicle — all caused by friction.

Download Notes Printable PDF

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NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.

NCERT Lines Flashcards

Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."

MCQ Set PYQs

How to revise: Draw the OA→A→BC graph from memory and label each region with the type of friction, its formula, and whether it depends on velocity. Then write a 3-row table: Type | Formula | Velocity dependent? | Relative magnitude. This visual review captures everything NEET tests on this topic in under 20 minutes.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One conceptual MCQ per paper — usually on whether friction equals applied force or μN, or comparing rolling vs sliding friction, or reading the graph to identify the kinetic friction value.

Time Required1.5 hrs45 min theory for three types, laws, and graph; 45 min practice MCQs on self-adjusting friction, graph reading, and rolling friction facts.

DifficultyEasy-ModerateDefinitions are straightforward; the trap is conceptual — static friction is NOT always μsN. Once this one insight is secured, almost every MCQ in this topic becomes easy.

Scoring Focus: Key testable facts: (1) static friction is self-adjusting and ≤ μsN — it equals applied force below the limit. (2) kinetic friction < limiting friction, both proportional to R but NOT area. (3) Rolling friction < kinetic friction — hence wheels.
High-risk Area: Asserting that static friction always equals μsN. This is wrong — friction equals applied force up to a maximum of μsN. Every NEET paper contains a trap option exploiting this error.
Best Practice Style: Make a flashcard: one face shows the graph (OA, A, BC); the other face lists which friction equation applies in each region. Test yourself by hiding one face. This 5-minute drill before exam day is high-return.

Priority rule: Medium. 1 reliable mark. Master the self-adjusting nature of static friction first. Allocate the first 20% of chapter time here — it is the conceptual prerequisite for every subsequent topic.

Angle of Friction, Angle of Repose, and Resultant Surface Force

The angular language of friction: the two key angles derived from μs, their equality, and the minimum-force derivations for horizontal surface and inclined plane problems.

1-2 Q/yeartan λ = μsAngle equalities criticalMin-force formulas

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

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Topic Notes (Condensed)Angle of friction λ = tan⁻¹(μs). Angle of repose α = tan⁻¹(μs). Therefore α = λ always. Resultant surface force S = mg√(μ² + 1); range: mg ≤ S ≤ mg√(μ² + 1). Minimum pulling force at angle α: P = W sin θ / cos(α−θ). Minimum force is least when α = θ (pull at angle of friction): P_min = μmg/√(1+μ²). Pushing always requires more force than pulling because normal reaction increases with push. On incline: min pull UP = W sin(θ+λ)/cos(α−θ).

Download Notes Printable PDF

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NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.

NCERT Lines Flashcards

Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."

MCQ Set PYQs

How to revise: Write the proof that angle of repose = angle of friction in 4 steps: (1) at limiting condition F_l = mg sin α, R = mg cos α; (2) tan α = F_l/R = μs; (3) angle of friction λ defined by tan λ = μs; (4) therefore α = λ. Memorise this chain — if you forget the result, you can re-derive it in 60 seconds under exam conditions.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2One MCQ on finding the angle of repose given μ (or vice versa); occasionally a second on minimum force direction for pulling a block on a horizontal surface.

Time Required2 hrs45 min proving α = λ and range of S; 45 min minimum-force derivations for horizontal and inclined cases; 30 min numericals on angle and minimum force.

DifficultyModerateThe equality α = λ is straightforward once derived; the minimum-force formulas require careful trigonometric manipulation. NEET rarely asks for derivation — it tests numerical substitution and the direction of minimum force.

Scoring Focus: α = λ = tan⁻¹(μs) is the single most tested equality here. Minimum force direction on horizontal surface = angle of friction from horizontal (α = θ). These two results provide instant answers with no computation.
High-risk Area: Confusing angle of friction with angle of repose conceptually — they are defined on different setups (one is the angle of the resultant with normal; other is the incline angle) but they are numerically equal. NEET sometimes exploits hesitation about this.
Best Practice Style: Solve 3 problems each: given μ find angle of repose; given angle of repose find μ; find minimum force to pull block. Cross-check with tan of the angle — should always equal μ. 30-minute targeted drill gives 100% confidence.

Priority rule: High. 1-2 marks per paper. Angle of repose calculation is almost guaranteed every year. Allocate 25% of chapter time here.

Dynamics on Inclined Plane and Work Done Against Friction

Quantitative mechanics: acceleration formulae for sliding down and retardation for moving up a rough inclined plane, stopping distances using kinematics, and work-energy considerations including the rough wedge time-ratio experiment.

1-2 Q/yeara = g(sinθ−μcosθ)S = u²/2μg horizontalWedge: μ = tanθ(1−1/n²)

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

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Topic Notes (Condensed)Sliding down: a = g(sin θ − μ cos θ); condition for motion: θ > angle of repose. Sliding up (retardation): a = g(sin θ + μ cos θ). Horizontal (θ=0): a = μg. Stopping distance horizontal: S = u²/(2μg). Time to stop: t = u/(μg). Stopping on incline: S = u²/[2g(sin θ + μcos θ)]. Work against friction on incline: mgs(sin θ + μcos θ). Wedge time-ratio: μ = tan θ × (1 − 1/n²).

Download Notes Printable PDF

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NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.

NCERT Lines Flashcards

Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."

MCQ Set PYQs

How to revise: Draw the incline free-body diagram twice: (1) body sliding down — mark mg sin θ down the slope, friction F_k up the slope, N perpendicular. (2) body moving up — mark both mg sin θ AND F_k down the slope. Write Newton's second law for each and derive a. This physical derivation from free body is faster than memorising two separate formulas.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2At least one numerical per paper on incline acceleration — either find a given θ and μ, or find stopping distance, or find the angle at which body just begins to slide (angle of repose). The wedge n-ratio occasionally appears as a medium-difficulty MCQ.

Time Required2 hrs45 min deriving and practising incline acceleration formulas; 45 min stopping distance and work-done problems; 30 min wedge time-ratio and roughness measurement problems.

DifficultyModerateIncline acceleration formula is the most formula-intensive part of this chapter; errors arise when students flip the direction of friction (should be opposite to motion, not just opposite to applied force). Once the free-body diagram is drawn correctly, calculation is straightforward.

Scoring Focus: a = g(sin θ − μcos θ) for sliding down is the most numerically tested formula in this chapter. Stopping distance S = u²/(2μg) provides quick answers to horizontal motion questions. Memorise both as fundamental results.
High-risk Area: Using kinetic friction formula when the body is stationary on incline (below angle of repose). If θ < angle of repose, body stays put — static friction adjusts; kinetic formula a = g(sin θ − μcos θ) does not apply. NEET gives answer choices where this wrong calculation produces a plausible-looking value.
Best Practice Style: Practise 5 incline problems in this order: (1) find a for sliding down, (2) find range of μ for body to remain stationary, (3) find retardation going up, (4) find stopping distance from v=u, (5) find velocity at bottom from energy. This sequence enforces all formula variants.

Priority rule: Highest priority in this chapter. 1-2 marks every year from incline dynamics. Allocate 30% of chapter time here. Master before attempting two-body problems.

Two-Body Friction Problems and Special Applications

Composite systems: block-on-block dynamics (two key modes — together vs separate), Atwood-type problems (horizontal table + incline variants), maximum hanging chain, and applied-friction problems (accelerating cart block sticking and rotor wall sticking).

1 Q/yearTwo-block: check F vs μsmgCart: a_min = g/μChain: μ = l'/（l−l'）

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓

Topic Notes (Condensed)Two-block (F on upper): if F < μs mg → common acceleration F/(M+m). If F > μs mg → aA = (F − μk mg)/m, aB = μk mg/M. Atwood on table: m2_min = μs m1. Chain: μ = l'/(l − l'). Wedge: μ = tan θ (1 − 1/n²). Cart sticking: a_min = g/μ. Rotor sticking: ω_min = √(g/μr). Stopping of two compressed blocks: S ∝ 1/m² (same P and μ); S1/S2 = (m2/m1)².

Download Notes Printable PDF

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NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.

NCERT Lines Flashcards

Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."

MCQ Set PYQs

How to revise: Two-block problems: always check the limiting condition first (F vs μs mg). Draw the FBD for both blocks stating which friction acts (kinetic between them, static to floor). For the cart problem, identify the normal force as mA_cart (horizontal pseudo-force provides the support), then apply friction condition. The rotor problem is the same structure with centripetal acceleration. Practice recognising the 'same structure' across different physical setups.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1Approximately 1 question per paper — usually on the two-block problem critical force condition or the cart-sticking minimum acceleration. The Atwood-table configuration is also asked frequently.

Time Required2 hrs45 min two-block problem analysis with all four sub-cases; 45 min Atwood-type and hanging-chain problems; 30 min cart and rotor applications.

DifficultyModerate-HardThe two-block cases require careful identification of which friction is static and which is kinetic, and for which body each acts. Rotor and cart problems demand physical reasoning about which surface provides the normal force — this is non-obvious and a common source of error.

Scoring Focus: For Atwood on table: remember m2_min = μm1 directly from limiting equilibrium. For cart: R = mA_cart horizontally — static friction upward = μmA_cart ≥ mg gives minimum A. For rotor: R = mω²r — same structure gives ω_min = √(g/μr). Recognising the structural similarity makes these three problems a single pattern.
High-risk Area: In two-block problems, students often apply the same acceleration to both blocks even when F exceeds the limiting friction between them. Always check the condition first: if F > μs mg (force on upper block), they move separately and must have different accelerations with free body diagrams for each.
Best Practice Style: For every two-block problem: (1) compute F_limiting = μs mg. (2) Compare applied force F with F_limiting. (3) If F < F_l: write one equation for combined system. If F > F_l: write two separate Newton's law equations. This checklist catches every case without memorising separate results.

Priority rule: Medium. 1 reliable extra mark per year. Cover after mastering incline dynamics. Allocate 25% of chapter time — disproportionate return from Atwood-table and cart problems if understood structurally.

Friction Chapter NEET Traps & Common Mistakes (Topic-Wise)

Each subtopic below is of the Friction chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.

! Avoid Easy Negatives

Static Friction Is Self-Adjusting — It Does NOT Always Equal μsN

NEET 2019NEET 2022Static frictionSelf-adjustingMost common misconception

Mistake Snapshot (What Students Do Wrong)

Saying friction on a stationary body equals μsN when the applied force is less than μsN:: Static friction = applied force as long as the body has not reached limiting condition. Friction = μsN is only the MAXIMUM value of static friction at the point of impending motion. If a 100 N box has μs = 0.5 and is pushed with 30 N, friction = 30 N — not 0.5 × (100 × 10) = 490 N. NEET frequently presents options where the wrong digit comes from using μN on a stationary body before the limit is reached.
Claiming friction direction is always opposite to applied force:: Friction opposes relative motion or tendency of relative motion between surfaces — it is not necessarily opposite to the applied force. On a block on an incline where gravity component acts down the slope, static friction acts up the slope. If an external push is applied along the incline, friction direction depends on net tendency — it could still be different from the direction of the applied push.

2–3 Line Example (Typical Error)

Block of mass 5 kg on rough horizontal floor (μs = 0.4, g = 10 m/s²). Applied force = 10 N. Limiting friction = μs × mg = 0.4 × 50 = 20 N. Since 10 N < 20 N, body stays at rest; friction = 10 N (NOT 20 N). NEET option trap: 20 N (wrong — limiting value, not actual friction); 10 N (correct). If force is increased to 25 N, friction = 20 N (limiting) and body just starts to slide.

How NEET Frames The Trap

NEET presents a stationary block with an applied force less than μmg and asks 'what is the force of friction?' Distractor option is always the limiting value μmg. Correct answer is the applied force itself.

NEET-Style Trap Question Format

Q. A block of mass 4 kg rests on a rough horizontal surface (μs = 0.5, μk = 0.4, g = 10 m/s²). A horizontal force of 12 N is applied. What is the force of friction acting on the block?
A. 12 N B. 16 N C. 20 N D. 8 N
Trick: Limiting friction = μs × N = 0.5 × 40 = 20 N. Applied force = 12 N < 20 N, so body is stationary. Static friction = applied force = 12 N. Option A is correct. Option C (20 N) is the limiting friction — body has NOT reached impending motion so friction adjusts to 12 N, not the maximum.

Quick rule: Static friction = applied force as long as body is at rest. It equals μsN ONLY at the limiting (impending motion) condition. Always compare applied force with μsN first; if applied < μsN, friction = applied force, not μsN.

Angle of Friction vs Angle of Repose — Different Definitions, Same Value

NEET 2018NEET 2023Angle of frictionAngle of reposeConceptual equality trap

Mistake Snapshot (What Students Do Wrong)

Treating angle of friction and angle of repose as different values:: Students see two different definitions — angle of friction is defined via the resultant force on horizontal surface; angle of repose is defined via the inclined plane at which body just slides — and assume they must be different. Both equal tan⁻¹(μs). They are numerically identical and NEET tests whether students know this equality.
Confusing angle of repose with the angle at which a body slides OFF a frictionless incline:: On a frictionless incline any angle > 0 allows sliding. The angle of repose only exists for a rough surface (μ > 0) and is the exact angle at which static friction reaches its limit. Below angle of repose the body stays stationary; above it the body slides. This boundary condition must be correctly applied when checking if a body moves.

2–3 Line Example (Typical Error)

μs = 0.577 (= tan 30°). Angle of friction = tan⁻¹(0.577) = 30°. Angle of repose = tan⁻¹(0.577) = 30°. They are the same angle. A block on a 25° incline with μs = 0.577 stays at rest (25° < 30°). On a 35° incline it slides (35° > 30°). NEET question: 'Find angle of repose if angle of friction is 30°' — answer is 30°, not a different value.

How NEET Frames The Trap

NEET states angle of friction λ and asks for angle of repose, or vice versa. Distractor options include λ/2, 2λ, tan⁻¹(2μ), or μ itself — all plausible if the equality is not memorised.

NEET-Style Trap Question Format

Q. The coefficient of static friction between a block and an inclined surface is 1/√3. What is the angle of repose and what is the angle of friction?
A. 30° and 60° B. 30° and 30° C. 45° and 30° D. 60° and 30°
Trick: Angle of repose α = tan⁻¹(μs) = tan⁻¹(1/√3) = 30°. Angle of friction λ = tan⁻¹(μs) = 30°. Both are 30°. Option B is correct. Option A (30° and 60°) incorrectly treats angle of friction as tan⁻¹(√3) = 60° — a common error from using the wrong μ formula.

Quick rule: Angle of friction = angle of repose = tan⁻¹(μs). They are two names for the same numerical angle derived from the same coefficient. If the problem gives one, the other is automatically the same value — no computation needed.

Using Kinetic Friction When Body Is Stationary on Incline

NEET 2020NEET 2021Inclined planeStatic vs kineticWrong friction type applied

Mistake Snapshot (What Students Do Wrong)

Applying a = g(sinθ − μk cosθ) for a body at rest on a rough incline:: The formula a = g(sin θ − μcos θ) applies ONLY when the body is actually sliding (kinetic friction active). If θ < angle of repose (θ < tan⁻¹(μs)), the body is stationary and static friction adjusts to balance mg sin θ. Applying the kinetic formula gives a non-zero acceleration, which is physically impossible — the body is not moving.
Assuming friction always acts down the slope on an incline:: Friction direction on incline depends on the tendency of motion. For a body tending to slide DOWN, friction acts UP the slope. For a body being pushed UP the slope, friction acts DOWN the slope. The default 'incline friction acts up' is wrong for bodies being pushed upward.

2–3 Line Example (Typical Error)

Block on incline with θ = 20°, μs = 0.5 (angle of repose = tan⁻¹(0.5) ≈ 26.6°). Body is stationary since 20° < 26.6°. Wrong approach: a = g(sin 20° − 0.5 × cos 20°) = 10(0.342 − 0.47) = −1.28 m/s² — negative sign, yet students report this magnitude as acceleration. Correct answer: a = 0 (stationary). Friction = mg sin 20° = mg × 0.342 (self-adjusting, not 0.5 × mg cos 20°).

How NEET Frames The Trap

NEET provides μ, θ and asks for friction force or acceleration on a stationary block. The computation g(sin θ − μcos θ) gives a number; NEET includes that number as a distractor. The correct answer of 0 for acceleration (or friction = mg sin θ) is often overlooked.

NEET-Style Trap Question Format

Q. A block rests on an inclined plane of angle 30°. The coefficient of static friction is μs = 0.6 (angle of repose ≈ 31°). What is the acceleration of the block?
A. 1.25 m/s² B. 2.5 m/s² C. 0 m/s² D. g sin 30°
Trick: Angle of repose = tan⁻¹(0.6) ≈ 31°. Since θ = 30° < 31°, the body is ON OR BELOW the angle of repose — it does NOT slide. Acceleration = 0. Option C is correct. Option A uses a = g(sin 30° − 0.6 cos 30°) = 10(0.5 − 0.52) = −0.2 (negative, physically impossible for sliding motion) — this trap catches students who mechanically apply the kinetic formula.

Quick rule: Before using a = g(sinθ − μcosθ), check: is θ > angle of repose? If θ ≤ tan⁻¹(μs), body is stationary and a = 0. Static friction adjusts to mg sin θ — it does NOT equal μs N in this case.

Two-Block Condition: Checking F vs Limiting Friction Before Applying Acceleration Formulas

NEET 2017NEET 2024Two-body problemBlock on blockCondition check before formula

Mistake Snapshot (What Students Do Wrong)

Applying individual accelerations aA and aB without first checking whether F exceeds limiting friction between A and B:: The two-block separation formulas (aA = (F − μk mg)/m, aB = μk mg/M) apply ONLY when F > μs mg (limiting friction between A and B). If F < μs mg, both blocks move together with a = F/(M+m). Using the separation formulas when F < F_limiting gives wrong accelerations — aA would come out less than aB, violating the no-separation condition.
Forgetting that kinetic friction (μk mg) acting on B from A is the CAUSE of B's motion (not a retarding force on B):: In two-block (force on upper block A), kinetic friction from A acts FORWARD on B — it is the only horizontal force on B, causing B to accelerate. Students often draw this friction as opposing B's motion (backwards on B) which reverses the sign of aB. Friction on B is in the direction of A's motion (forward) because A tends to slide forward relative to B.

2–3 Line Example (Typical Error)

Block A (2 kg) on block B (8 kg) on frictionless floor. μs = 0.4 (between A and B). F = 6 N applied to A. Limiting friction = μs × mA × g = 0.4 × 2 × 10 = 8 N. Since F = 6 N < 8 N, blocks move together: a = 6/(2+8) = 0.6 m/s². If F = 12 N > 8 N (limiting), then aA = (12 − μk × 2 × 10)/2 and aB = μk × 2 × 10/8 separately.

How NEET Frames The Trap

NEET provides the two masses, μ, and an applied force, asking for the acceleration of the lower block B. Students who skip the limiting-friction check calculate separate accelerations even when F < F_limiting and report aB = μk mg/M instead of F/(M+m).

NEET-Style Trap Question Format

Q. Block A (3 kg) sits on block B (7 kg) on a smooth floor. μs = 0.3, μk = 0.2 between A and B. A horizontal force F = 6 N is applied to A. What is the acceleration of block B? (g = 10 m/s²)
A. 0.6 m/s² B. 0.2 m/s² C. 0.8 m/s² D. 0 m/s²
Trick: Limiting friction between A and B = μs × mA × g = 0.3 × 3 × 10 = 9 N. Since F = 6 N < 9 N, both blocks move TOGETHER. Common acceleration = 6/(3+7) = 0.6 m/s². Block B also accelerates at 0.6 m/s². Option A is correct. Option B (0.2 m/s²) incorrectly uses aB = μk × mA × g / mB = 0.2 × 3 × 10 / 7 ≈ 0.86 m/s² — the formula for when blocks slide separately, which does not apply here.

Quick rule: For two-block problems: FIRST compute F_limiting = μs × m_upper × g. THEN compare F with F_limiting. If F < F_limiting → a = F/(M+m) for both. If F > F_limiting → use separate FBD equations. Never skip the comparison step.

Chapter Snapshot - Friction

Subtopics - Friction (NEET)

1) Types of Friction, Laws, and the F-vs-Applied Force Graph

2) Angle of Friction, Angle of Repose, and Resultant Surface Force

3) Dynamics on Inclined Plane and Work Done Against Friction

4) Two-Body Friction Problems and Special Applications

Friction Download Notes & Weightage Plan

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

Friction Chapter NEET Traps & Common Mistakes (Topic-Wise)

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

NEET > Physics > Laws of Motion Chapters

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Friction

Overview content

Chapter Snapshot - Friction

Subtopics - Friction (NEET)

1) Types of Friction, Laws, and the F-vs-Applied Force Graph

2) Angle of Friction, Angle of Repose, and Resultant Surface Force

3) Dynamics on Inclined Plane and Work Done Against Friction

4) Two-Body Friction Problems and Special Applications

Friction Download Notes & Weightage Plan

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

1) Download Packs For This Topic (And How To Use Them)

2) Importance, Weightage & Time Allocation (Practical)

Friction Chapter NEET Traps & Common Mistakes (Topic-Wise)

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

Mistake Snapshot (What Students Do Wrong)

How NEET Frames The Trap

NEET > Physics > Laws of Motion Chapters

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