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).
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.
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.
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).
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)).
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.
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) 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.
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.
- 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.
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) 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.
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.
- 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.
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) 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.
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.
- 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.
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) 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.
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.
- 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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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).
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.