Coefficient of Friction Between Body and Wedge – Complete Notes, Revision, Important Questions & Downloads
Coefficient of Friction Between Body and Wedge covers the TOC subtopic Wedge Friction Formula. A body slides on a smooth wedge of angle θ in time t. On the rough version of the same wedge, it takes nt (n > 1 times longer). From these two conditions, the coefficient of friction is μ = tanθ[1 − 1/n²]. NEET tests this as a direct substitution problem: given θ, t, nt, find μ. The key steps are setting equal distances from both smooth and rough cases and solving for μ.
NEET Weightage — Coefficient of Friction Between Body and Wedge
Friction (Chapter 5)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 0 | 0 | |
| 2023 | 1 | 4 | |
| 2022 | 0 | 0 | |
| 2021 | 1 | 4 | |
| 2020 | 0 | 0 | |
| 2019 | 0 | 0 | |
| 6-Year Total (2019–2024) | 0–2 | 0–8 |
DERIVATION: Equate S: ½ gsinθ · t² = ½ g(sinθ − μcosθ) · n²t² → gsinθ = n²g(sinθ − μcosθ) → sinθ = n²sinθ − n²μcosθ → n²μcosθ = n²sinθ − sinθ = sinθ(n²−1) → μ = sinθ(n²−1)/(n²cosθ) = tanθ(1 − 1/n²).
RESULT: μ = tanθ[1 − 1/n²]. Special cases: n=1 (same time = smooth wedge): μ = 0 ✓. n→∞ (body barely moves on rough wedge): μ → tanθ (coefficient equals tangent of angle — body on verge of not sliding). For n=2 and θ=30°: μ = tan30°(1−1/4) = (1/√3)(3/4) = 3/(4√3) = √3/4 ≈ 0.433.
How to Solve the Wedge Friction Coefficient Problem
Step 1 — Write accelerations for both cases Smooth: a₁ = gsinθ. Rough: a₂ = g(sinθ − μcosθ). Both motions start from rest along the same incline length S.
Step 2 — Equate distances S = ½a₁t² and S = ½a₂(nt)². Therefore a₁t² = a₂n²t² → a₁ = n²a₂ → ratio a₁/a₂ = n².
Step 3 — Solve for μ gsinθ = n²g(sinθ − μcosθ) → sinθ = n²sinθ − n²μcosθ → n²μcosθ = sinθ(n²−1) → μ = tanθ(n²−1)/n² = tanθ[1−1/n²].
Study Materials — Coefficient of Friction Between Body and Wedge
PDF · Cheat Sheet · MCQ Set · PYQSubtopics — Coefficient of Friction Between Body and Wedge
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Rapid Revision — Coefficient of Friction Between Body and Wedge
Concept → Trap → Example1) Wedge Friction Formula — Deriving μ = tanθ(1−1/n²)
Wedge Friction FormulaPROBLEM: A body slides from rest on a frictionless wedge (angle θ) in time t, reaching the bottom. On the same wedge with friction (coefficient μ), starting from rest, the body takes n times longer (time = nt). Derive μ in terms of θ and n. SMOOTH CASE: Acceleration a₁ = gsinθ. Distance S = ½a₁t². ROUGH CASE: Forces along incline: gravity component down = mgsinθ; friction (resisting motion, up the incline) = μN = μmgcosθ. Net force = mg(sinθ − μcosθ). Acceleration a₂ = g(sinθ − μcosθ). Same distance S = ½a₂(nt)² (starts from rest, takes time nt). EQUATING S: ½a₁t² = ½a₂(nt)² → a₁ = n²a₂ → gsinθ = n²g(sinθ − μcosθ) → sinθ = n²sinθ − n²μcosθ → n²μcosθ = n²sinθ − sinθ = sinθ(n²−1) → μ = [sinθ(n²−1)]/(n²cosθ) = tanθ × (n²−1)/n² = tanθ[1 − 1/n²]. RESULT: μ = tanθ(1 − 1/n²). Alternatively: μ = tanθ(n²−1)/n².
- PHYSICAL INTERPRETATION: For n=1 (rough wedge takes same time as smooth): μ = tanθ(1−1) = 0 — consistent, no friction. For n→∞ (body barely slides on rough wedge): μ → tanθ — the coefficient approaches tanθ, which is exactly the angle-of-repose condition (body on the verge of not sliding). So the formula captures both extremes correctly.
- INVERSE PROBLEM: Given μ and θ, find n. From μ = tanθ(1−1/n²): μ/tanθ = 1−1/n² → 1/n² = 1 − μ/tanθ → n² = 1/(1 − μ/tanθ) = tanθ/(tanθ − μ) → n = √[tanθ/(tanθ − μ)]. This is the time ratio. Condition for body to slide: μ < tanθ (otherwise body doesn't slide at all on rough wedge, n→∞).
- NEET TRAP: Students mix up which wedge is smooth and which is rough, or forget the n² factor from squaring the time ratio. The key algebraic step is: same distance S means a₁t² = a₂(nt)² → a₁ = n²a₂. Write this ratio immediately after stating both accelerations. The n² comes from squaring the time ratio (not from linear ratio).
US Curriculum Gaps — Coefficient of Friction Between Body and Wedge
Topics in this section are in NEET but may be framed differently in US physics courses.Kinematic Time Comparison on Inclines in AP Physics 1
AP Physics 1 covers friction on inclined planes but does not test the specific comparison between smooth and rough descent times as a method to compute friction coefficient. The formula μ = tanθ(1−1/n²) does not appear in AP Physics curriculum. NRI students with strong kinematics can derive it but need to recognise the 'equate distances' approach.
- AP Physics 1: friction on inclines covered; time-ratio method to find μ not tested
- NEET: μ = tanθ(1−1/n²) is a direct-formula question — memorise and apply in under 30 seconds
- Key step: a₁t² = a₂(nt)² → a₁/a₂ = n² — the n² ratio from squaring time is the critical algebraic manipulation
Angle of Repose Connection in University Physics
Halliday & Resnick introduces angle of repose (tanθ = μ) as the incline angle at which an object just begins to slide. The wedge formula μ = tanθ(1−1/n²) shows that the limiting case (n→∞) recovers the angle-of-repose condition. This connection between dynamic and static friction on inclines is typically not made explicit in US university courses at the level of NEET problems.
- US university: angle of repose defined but smooth/rough time comparison not a standard problem type
- NEET: the formula μ=tanθ(1−1/n²) directly connects to angle-of-repose when n→∞
- Memorisation: write μ = tanθ(1−1/n²); verify units — μ is dimensionless, tanθ is dimensionless ✓
NEET-Style Practice — Coefficient of Friction Between Body and Wedge
4 QuestionsPractice Problems — Coefficient of Friction Between Body and Wedge
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Physics — Friction Revision Checklist
Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ — Coefficient of Friction Between Body and Wedge
Notes · Downloads · Revision · Important QuestionsWhat is the formula for coefficient of friction on a wedge when time ratio is n?
Why does the formula use n² and not n?
What does the formula predict for n = 1?
What does n → ∞ mean physically?
What is the condition for the body to slide on the rough wedge at all?
Can I apply this formula if the body is being pushed up the rough wedge instead of sliding down?
If the body descends in 2s (smooth) and 4s (rough), does the rough wedge have exactly 4× greater resistance?
What if I know μ and want to find the time ratio n?
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