Friction: dry friction, wedges and belt friction

Why it matters

Friction decides whether a part on an inclined conveyor stays put, whether a lead screw holds a vertical axis when the motor is switched off, and how much torque a flat or V-belt can carry before it slips. Mechatronic designers both fight friction (it costs motor torque and causes stick-slip in positioning) and rely on it (self-locking screws, brakes, belt drives). The Coulomb model below is simple, but it predicts all of these well enough for design.

Key ideas

Coulomb (dry) friction. When two dry solid surfaces are pressed together by a normal force N, the tangential friction force F opposes the relative motion, or the impending motion if the bodies are still at rest.

• While at rest, F is whatever equilibrium requires, up to a limit: F ≤ μs·N. It is not automatically μs·N.
• At impending slip, F = μs·N (limiting friction).
• Once sliding, F = μk·N, with μk usually a little smaller than μs. The drop from μs to μk is the root of stick-slip in slow servo axes.
• To first order, F is independent of the apparent contact area and of sliding speed.

Angle of friction. At limiting friction the resultant of N and F makes the angle φ with the normal, where tan φ = μs. A block on an incline starts to slide on its own when the incline angle reaches φ (the angle of repose). The total reaction can lie anywhere inside a cone of half-angle φ; this "friction cone" view makes wedge and screw problems quick.

How to solve a friction problem. Draw the FBD with F opposing the likely motion. Either assume equilibrium, solve for F and check F ≤ μs·N, or, if the question asks for the force to start motion, put F = μs·N directly. When several surfaces could slip (for example stacked blocks or a wedge), test each possible slip mode; the one needing the smaller force happens first.

Wedges. A wedge turns a small push into a large lifting or clamping force. Each sliding face carries a normal force and a friction force, so friction adds to the push needed to drive the wedge in. A wedge with friction angle φ on both sliding faces stays in place when the push is removed (self-locking) if its angle α ≤ 2φ. Small wedge angles give large force multiplication and self-locking; large angles are easy to withdraw.

Screws. A square-thread power screw is an inclined plane wrapped round a cylinder, with lead angle λ, where tan λ = L/(π·d_m). Raising a load is like pushing a block up an incline of angle λ. The screw is self-locking (will not back-drive under load) when φ ≥ λ, and a self-locking square-thread screw has efficiency below 50 %. This is why ball screws, which have very low friction, need a brake on vertical axes.

Belt friction (capstan relation). For a flat belt or rope wrapped through angle θ on a drum, friction lets the tension grow from T₂ (slack side) to T₁ (tight side). At impending slip T₁/T₂ = e^(μθ). The ratio grows exponentially with lap angle, not with contact area, which is why a few turns of rope on a bollard can hold a ship. A V-belt wedges into its groove; the normal force rises by 1/sin β, so the effective coefficient becomes μ/sin β, where β is half the groove angle. The torque transmitted is (T₁ − T₂)·r and the power is (T₁ − T₂)·v. At high belt speed the centrifugal tension m·v² (m = mass per metre) must be subtracted from both tensions before applying the ratio.

Formulas

F ≤ μs·N, F = μk·N (sliding)

• F: friction force (N); N: normal force (N); μs, μk: static and kinetic coefficients (dimensionless). Take μ from a data book or test; it depends on materials, finish and lubrication.

tan φ = μs

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

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

• Force parallel to an incline of angle θ to push a block of weight W (N) up it, or the minimum force to stop it sliding down (only needed when tan θ > μ).

tan λ = L / (π·d_m), T_raise = W·(d_m/2)·tan(λ + φ), T_lower = W·(d_m/2)·tan(φ − λ)

• L: lead (m); d_m: mean thread diameter (m); W: axial load (N); T: torque (N·m). Square threads, collar friction neglected. T_lower > 0 means self-locking.

η = tan λ / tan(λ + φ)

• Screw efficiency, square thread, no collar friction.

T₁ / T₂ = e^(μθ) (flat belt), T₁ / T₂ = e^(μθ / sin β) (V-belt)

• T₁, T₂: tight- and slack-side tensions (N); θ: angle of lap (rad, never degrees); β: half the groove angle.

P = (T₁ − T₂)·v, T_c = m·v²

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

Worked examples

Example 1 (standard): part on an inclined conveyor. Given: a 50 kg tote on a stationary conveyor bed inclined at 20°, μs = 0.35, g = 9.81 m/s². Will it slide down on its own? What force parallel to the bed is needed to push it up, and what is the smallest force that will hold it?

1. Weight: W = 50 × 9.81 = 490.5 N.
2. Slip test: tan 20° = 0.364 > μs = 0.35, so the tote slides down unaided.
3. Push up (friction acts down the slope): P_up = W·(sin θ + μ·cos θ) = 490.5 × (0.342 + 0.35 × 0.940) = 329 N.
4. Hold (friction acts up the slope): P_hold = W·(sin θ − μ·cos θ) = 490.5 × (0.342 − 0.329) = 6.4 N. Any force between 6.4 N and 329 N keeps the tote at rest; friction adjusts itself within ±μs·N.

Example 2 (GATE level): flat-belt drive power. Given: a flat belt on a motor pulley with lap angle 165°, μ = 0.3, maximum allowable tension 1.2 kN, belt speed 15 m/s; neglect centrifugal tension. Find the maximum power.

1. Lap angle in radians: θ = 165 × π/180 = 2.880 rad.
2. Tension ratio: T₁/T₂ = e^(μθ) = e^(0.3 × 2.880) = e^0.864 = 2.372.
3. T₁ = 1200 N, so T₂ = 1200/2.372 = 505.8 N.
4. Power: P = (T₁ − T₂)·v = (1200 − 505.8) × 15 = 10 413 W. Answer: P ≈ 10.4 kW. With a 40° V-groove (β = 20°) the ratio for the same μ and θ would be e^(0.864/0.342) ≈ 12.5, which is why V-belts carry far more power for the same tension.

Example 3 (GATE level): will a lead screw hold a vertical axis? Given: square-thread lead screw, mean diameter 20 mm, lead 4 mm, μ = 0.15, axial load 2 kN; collar friction neglected.

1. Lead angle: tan λ = 4/(π × 20) = 0.0637, so λ = 3.64°.
2. Friction angle: φ = tan⁻¹ 0.15 = 8.53°.
3. Raising torque: T_raise = W·(d_m/2)·tan(λ + φ) = 2000 × 0.010 × tan 12.17° = 4.31 N·m.
4. Lowering torque: T_lower = W·(d_m/2)·tan(φ − λ) = 2000 × 0.010 × tan 4.89° = 1.71 N·m. It is positive, so the screw is self-locking (φ > λ): the load will not drive the screw down when the motor is unpowered.
5. Efficiency: η = tan 3.64° / tan 12.17° = 0.295 (29.5 %).

Common mistakes

• Writing F = μN for a body that is not about to slip. Below the limit, find F from equilibrium.
• Drawing friction in the wrong direction. It opposes the relative motion (or tendency) of that surface, which may differ for each body in a stack.
• Using the lap angle in degrees in e^(μθ).
• Labelling the tensions the wrong way round: T₁, the larger, is the span being pulled onto the driving pulley.
• Forgetting the normal force on an incline is W·cos θ, not W, or forgetting that a pull at an angle changes N.
• Treating a wedge as frictionless on one face when the problem says all faces are rough.

For GATE ME

Expect blocks on inclines and stacked blocks (which surface slips first), ladders with friction, wedge and screw self-locking conditions, screw torque and efficiency, and flat- or V-belt tension ratio and power, sometimes with centrifugal tension. Practise drawing friction forces with the right sense, converting lap angles to radians, and checking whether equilibrium is actually possible before assuming limiting friction.

Quick check

1. A block rests on a horizontal floor with μs = 0.4 and weight 200 N, and a horizontal pull of 50 N acts on it. What is the friction force?
2. What is the angle of repose for μs = 0.4?
3. For a flat belt with μ = 0.25 and θ = π rad, what is T₁/T₂?
4. When is a square-thread screw self-locking?
5. Does doubling the contact area of a dry block double the limiting friction?

Answers: 1. 50 N (below the limit of 80 N, so no slip). 2. tan⁻¹ 0.4 = 21.8°. 3. e^(0.785) = 2.19. 4. When the friction angle is at least the lead angle (φ ≥ λ). 5. No, limiting friction depends on N, not on apparent area.