Friction & Its Applications

Friction problems appear in GATE ME both in Engineering Mechanics (~1–2 marks) and inside machine elements (brakes, clutches, belts, screw jacks). The governing laws are simple; the marks come from applying them to the right contact.

Laws & key relations

Coulomb friction: limiting (maximum static) friction F = μₛN, where μₛ = coefficient of static friction, N = normal reaction. Once sliding, kinetic friction F = μₖN with μₖ < μₛ. Friction is independent of contact area and acts opposite to (impending) motion.

Angle of friction φ: tan φ = μ. Angle of repose = the incline angle at which a block just slides = φ (for a block on an incline, impending slip when tan θ = μ).

Applications:

• Screw/screw jack: effort torque relates to tan(α + φ) for raising, tan(φ − α) for lowering; self-locking if φ ≥ α (lead angle).
• Belt friction: T₁/T₂ = e^(μθ), where θ = wrap angle in radians, T₁ = tight side, T₂ = slack side.
• Wedge, ladder, block-on-incline: draw FBD with friction along each contact.

Exam Tricks & Tips

• 🎯 Angle of repose = angle of friction: if a block is on the verge of sliding on an incline, tan θ = μ directly — a one-line answer.
• 🎯 Belt ratio T₁/T₂ = e^(μθ) — remember θ is in radians and larger wrap (or lapping) raises grip exponentially.
• 🎯 Self-locking screw when friction angle φ ≥ helix/lead angle α — a standard conceptual mark in power screws.
• 🎯 Friction magnitude ≤ μN, not always = μN: below impending motion, friction only equals the applied tangential force (static, self-adjusting).
• 🎯 Independent of area and speed (idealised): flags in an MCQ that "larger area → more friction" as false.
• ❌ Common mistake: using F = μN when the body is NOT on the verge of sliding — until motion impends, friction is whatever is needed for equilibrium, up to the μN limit.

Expected exam pattern

A 1-mark block-on-incline or angle-of-repose calculation, and a 2-mark belt-friction (T₁/T₂ = e^(μθ)) or screw-jack effort/self-locking problem. Conceptual MCQs test the "independent of area, F ≤ μN" subtleties.

Quick recap

Limiting friction F = μN, tan φ = μ, angle of repose = φ. Static friction self-adjusts up to μₛN; kinetic = μₖN (< static). Belt: T₁/T₂ = e^(μθ) (θ in radians). Power screw self-locks when φ ≥ α. Friction is independent of area.

Cover the answer, recall, then check. 11 cards on friction for GATE ME.

Q1. State the law of limiting (maximum static) friction.
A1. F = μₛN, proportional to the normal reaction N, independent of contact area, and directed opposite to impending motion.

Q2. How do static and kinetic friction coefficients compare?
A2. μₖ < μₛ. Once sliding starts, friction drops from μₛN to μₖN.

Q3. Relate the angle of friction φ to the coefficient μ.
A3. tan φ = μ. φ is the angle the total contact reaction makes with the normal at impending slip.

Q4. What is the angle of repose?
A4. The incline angle at which a resting block just begins to slide; it equals the angle of friction φ (tan θ = μ).

Q5. State the belt-friction equation.
A5. T₁/T₂ = e^(μθ), with T₁ tight side, T₂ slack side, θ = angle of wrap in radians.

Q6. When is a power screw self-locking?
A6. When the friction angle φ ≥ the helix (lead) angle α — the load cannot drive the screw back on its own.

Q7. Is friction always equal to μN?
A7. No. Static friction self-adjusts to balance the applied tangential force, up to a maximum of μₛN. It equals μN only at impending motion.

Q8. Does friction depend on the apparent area of contact?
A8. No (Coulomb idealisation) — it depends only on the normal force and the surface pair, not on contact area or sliding speed.

Q9. Torque to raise a load on a square-thread screw?
A9. Proportional to tan(α + φ) (mean-radius torque). Lowering uses tan(φ − α).

Q10. A 5 kg block rests on a 30° incline, μ = 0.7. Does it slide?
A10. tan 30° = 0.577 < 0.7 = μ, so required friction < μN → it stays at rest (no sliding).

Q11. How does increasing the wrap angle affect belt grip?
A11. Grip (T₁/T₂) rises exponentially with θ, since the ratio is e^(μθ) — more wrap dramatically increases holding capacity.

Friction is where "ideal" mechanics meets reality — it holds ladders up, drives belts and brakes, and decides whether a block slides or tips. GATE loves it because a small conceptual slip (impending motion vs. moving, slide vs. tip) changes the whole answer.

Core concept: dry (Coulomb) friction opposes relative sliding with a force limited by F ≤ μN; at the verge of motion it equals μₛN and points opposite to impending slip.

Deep explanation

Beginner — laws of dry friction

• Friction force F = μN only at impending or actual sliding; below that, F takes whatever value equilibrium needs (static friction is self-adjusting up to the limit μₛN).
• Kinetic friction μ_k N acts once sliding, with μ_k < μₛ.
• The angle of friction φ satisfies tan φ = μ; the resultant of N and F leans at φ from the normal.
• Angle of repose = the incline angle at which a block just slides = φ (tan θ = μ).

Intermediate — inclined plane and self-locking

On an incline of angle θ, a block is on the verge of sliding down when tan θ = μ. To just move a body up a plane needs force accounting for both the weight component and μN. A screw/wedge is self-locking if its lead angle < friction angle (it will not back-drive) — the basis of jacks and power screws.

Advanced — belt friction and slide-vs-tip

• Belt/rope friction (capstan): T₁/T₂ = e^(μβ), where β is the wrap angle in radians and T₁ > T₂ (tight side). This governs belt drives, band brakes and rope on a bollard.
• Slide vs. tip: compare the pushing force needed to slide (μN) with the force that makes the normal reaction reach the tipping edge. Whichever happens at the lower force governs — always check both.

Worked example

A belt wraps 180° (β = π rad) around a pulley with μ = 0.25. If the slack-side tension is 400 N, find the maximum tight-side tension before slipping.
T₁/T₂ = e^(μβ) = e^(0.25·π) = e^(0.7854) = 2.193.
T₁ = 2.193 × 400 = 877 N (approx). So the belt can transmit up to (T₁ − T₂) = 477 N of effective pull before it slips.

GATE relevance

Friction appears in Engineering Mechanics (blocks, wedges, ladders) and directly powers Design/TOM topics: belt drives (e^(μβ)), band and block brakes, clutches, and power screws (self-locking). The e^(μβ) relation is a recurring high-value formula.

Exam tricks & shortcuts

• Use F = μN only at impending/actual sliding; otherwise solve friction as an unknown from equilibrium.
• For "just slides" incline problems, jump to tan θ = μ.
• Mnemonic: "tan of the friction angle IS mu (tan φ = μ)."

Key formulas

• Static/kinetic friction: f_s ≤ μ_sN; f_k = μ_kN; angle of friction tanφ = μ.
• Angle of repose = angle of friction: tanθ = μ_s.
• Ladder, wedge, screw problems use ΣF = ΣM = 0 with friction.
• Screw jack (square thread): effort torque T = W r tan(α + φ); efficiency η = tanα/tan(α + φ).
• Belt friction: T₁/T₂ = e^(μθ).

Friction opposes relative motion; belt drives and screw threads follow exponential and lead-angle relations.

Worked Example

Problem: A block of weight W = 100 N rests on an inclined plane; the coefficient of static friction is µ_s = 0.3. (a) Find the maximum incline angle before the block starts to slide. (b) At an incline of 20°, find the friction force needed and determine whether the block slides.

Solution:
(a) On the verge of sliding, the driving component W·sinθ equals the maximum friction µ_s·N = µ_s·W·cosθ:
W·sinθ = µ_s·W·cosθ ⇒ tanθ = µ_s.
θ_max = arctan(µ_s) = arctan(0.3) ≈ 16.7° (this is the angle of friction).

(b) At θ = 20°:
Friction force required to hold the block = W·sinθ = 100·sin20° = 100 × 0.342 = 34.2 N.
Maximum friction available = µ_s·W·cosθ = 0.3 × 100 × cos20° = 0.3 × 93.97 ≈ 28.2 N.

Since the required 34.2 N exceeds the available 28.2 N, static friction cannot hold the block — it slides. (Also 20° > θ_max = 16.7°, confirming motion.)

Answer: (a) θ_max ≈ 16.7°; (b) needs 34.2 N but only 28.2 N is available, so the block slides.