Friction: dry friction, wedges and belt friction

Why it matters

Friction decides whether a vehicle can brake, climb a gradient or transmit engine torque through a clutch or a belt. It is also what keeps a wedge, a cotter or a screw jack from slipping back under load. Most friction problems in exams are equilibrium problems with one extra condition, F ≤ μN, so a clean free-body diagram and a clear idea of which way motion is impending are the whole game.

Key ideas

Dry (Coulomb) friction. When two dry solid surfaces are pressed together, the contact force has a normal part N and a tangential part F.

• Before sliding, F is whatever is needed for equilibrium, from zero up to a limit. Static friction is self-adjusting.
• At impending motion, F reaches its limiting value F_max = μ_s·N.
• Once sliding starts, F = μ_k·N, roughly constant and usually a little smaller than μ_s·N.
• F always opposes the relative motion (or the impending relative motion) of the surfaces.

Coulomb's empirical laws. Limiting friction is proportional to the normal force, independent of the apparent area of contact, and (for kinetic friction) roughly independent of sliding speed at moderate speeds. These are approximations valid for dry, clean surfaces; lubricated contacts follow fluid-film behaviour instead. Typical μ values depend on the materials and surface condition, so take them from a data book or the problem statement.

Angle of friction and angle of repose. The resultant R of N and F_max makes an angle φ with the normal, where tan φ = μ. A block on an incline starts to slide under its own weight when the incline angle reaches the angle of repose, which equals φ. Thinking of R as a single force at angle φ to the normal makes wedge and screw problems much shorter.

Deciding the case. For any "will it slip?" problem: assume equilibrium, find the friction force F needed, and compare with μ_s·N. If F ≤ μ_s·N the body stays put and the friction force is F (not μN). If not, it slides and F = μ_k·N. When a body can either slide or tip, check both and take the one requiring the smaller force.

Wedges. A wedge converts a small push into a large force at right angles. Friction acts on every sliding face and always opposes the wedge's motion relative to that face. Two consequences:

• Raising a load needs more force than the frictionless value, because friction adds to the resistance.
• A wedge is self-locking (stays in place when the push is removed) when the wedge angle is small compared with the friction angles; for a wedge with friction angles φ₁ and φ₂ on its two sliding faces, self-locking needs α ≤ φ₁ + φ₂.

Belt (rope) friction. Consider a flat belt in contact with a pulley over an angle θ (the angle of lap). Equilibrium of a small element, with limiting friction everywhere along the contact, gives the capstan relation T₁/T₂ = e^(μθ), where T₁ is the tight-side tension and T₂ the slack side. The tension ratio depends only on μ and θ, not on the pulley radius. Larger wrap angle and larger μ give exponentially more grip. This is why a rope wrapped a few turns round a bollard holds a ship, and why idlers are used to increase the lap on small pulleys.

• V-belts: wedging in the groove raises the effective coefficient to μ/sin(β/2), where β is the groove angle.
• At high speed the belt's own mass creates a centrifugal tension T_c = m·v² that adds to both sides without contributing to grip. Maximum power occurs when T_c = T_max/3.

Formulas

F_max = μ_s·N, F_k = μ_k·N

• F: friction force (N); N: normal reaction (N); μ_s, μ_k: static and kinetic coefficients (dimensionless). F_max is a limit, not the actual force, unless motion is impending.

tan φ = μ

• φ: angle of friction (rad or degrees). The angle of repose of a block on an incline equals φ.

P_up = W·(sin α + μ·cos α), P_hold = W·(sin α − μ·cos α)

• P: force parallel to an incline of angle α needed to start the block moving up, or to just prevent it sliding down; W: weight (N). P_hold applies only when tan α > μ; otherwise no force is needed.

P = W·[tan(α + φ₁) + tan φ₂]

• Horizontal force to drive a wedge of angle α under a load W that is guided to move vertically by smooth guides; φ₁ on the wedge–load face, φ₂ on the wedge–floor face.

T₁ / T₂ = e^(μ·θ)

• T₁, T₂: tight- and slack-side tensions (N); θ: angle of lap in radians. Flat belts or ropes on the point of slipping.

T₁ / T₂ = e^(μ·θ / sin(β/2))

• β: V-groove angle. V-belts and ropes in grooved pulleys.

T_c = m·v², P = (T₁ − T₂)·v, with T₁ = T_max − T_c

• m: belt mass per unit length (kg/m); v: belt speed (m/s); P: power transmitted (W).

Worked examples

Example 1 (standard): block on an incline. Given: block weight W = 500 N on a 25° incline, μ_s = 0.3. Find the force parallel to the incline to (a) start it moving up, (b) just prevent it sliding down.

1. N = W·cos α = 500 × 0.9063 = 453.15 N.
2. Component down the slope: W·sin α = 500 × 0.4226 = 211.31 N.
3. Limiting friction: μ·N = 0.3 × 453.15 = 135.95 N.
4. (a) Motion impending upward, so friction acts down the slope: P = 211.31 + 135.95 = 347.3 N.
5. (b) tan 25° = 0.466 > 0.3, so the block would slide. Friction now acts up the slope: P = 211.31 − 135.95 = 75.4 N. Answer: (a) 347.3 N, (b) 75.4 N. Any force between these keeps the block at rest.

Example 2 (GATE level): power through a flat belt with centrifugal tension. Given: angle of lap θ = 160°, μ = 0.25, maximum allowable belt tension T_max = 2000 N, belt mass m = 0.5 kg/m, belt speed v = 20 m/s. Find the maximum power transmitted.

1. Centrifugal tension: T_c = m·v² = 0.5 × 20² = 200 N.
2. Tight-side tension available for driving: T₁ = T_max − T_c = 1800 N.
3. Lap angle in radians: θ = 160 × π/180 = 2.7925 rad, so μθ = 0.6981.
4. T₂ = T₁ / e^(μθ) = 1800 / 2.0100 = 895.5 N.
5. P = (T₁ − T₂)·v = (1800 − 895.5) × 20 = 18 090 W. Answer: P ≈ 18.1 kW.

Example 3 (wedge). Given: a load W = 5 kN, guided by smooth vertical guides, is to be raised by driving a 15° wedge under it; μ = 0.2 on both wedge faces. Find the horizontal push P and check self-locking.

1. φ = tan⁻¹(0.2) = 11.31°.
2. P = W·[tan(α + φ) + tan φ] = 5 × [tan 26.31° + 0.2] = 5 × (0.4944 + 0.2) = 3.47 kN.
3. Self-locking check: φ₁ + φ₂ = 22.6° > 15°, so the wedge stays in place when P is removed. Answer: P ≈ 3.47 kN; the wedge is self-locking.

Common mistakes

• Writing F = μN when the body is not on the point of slipping. Below the limit, friction comes from equilibrium.
• Drawing friction in the wrong direction. It opposes the relative motion of the surfaces, which for a wedge is different on each face.
• Using degrees in e^(μθ). θ must be in radians.
• Taking the normal force as W on an incline, or ignoring the normal-force change when the applied force has a component perpendicular to the surface.
• Swapping T₁ and T₂ (the ratio must be greater than 1) or forgetting centrifugal tension at high belt speed.
• Assuming a block slides when it might tip first; check both when the block is tall.

For GATE ME

Typical questions: blocks on inclines and connected blocks, ladders with friction at one or both ends, minimum force and its best angle (pull at angle φ above the incline), wedge forces and self-locking, band brakes and belt tension ratios, and maximum power with centrifugal tension. Practise deciding the direction of impending motion first, then solving with R at angle φ to the normal where possible.

Quick check

1. A block rests on a 20° incline with μ_s = 0.5. Does it slide under its own weight?
2. A 200 N block on a level floor (μ_s = 0.4) is pushed horizontally with 50 N. What is the friction force?
3. What happens to T₁/T₂ if the angle of lap doubles?
4. State the condition for maximum belt power when centrifugal tension is included.

Answers: 1. No, tan 20° = 0.364 < 0.5. 2. 50 N (limit is 80 N, so friction just balances the push). 3. The ratio is squared. 4. T_c = T_max/3.