Friction: dry friction, wedges and belt friction

Why it matters

Friction decides whether a ladder slips, a screw jack holds its load, a wedge stays in place, a brake stops a drum and a belt drive transmits power. Designers sometimes want more of it (brakes, belts, bolted joints) and sometimes less (bearings, guides), so they must be able to calculate it reliably.

Key ideas

Dry (Coulomb) friction acts between unlubricated solid surfaces, tangent to the contact and opposing the relative motion or the tendency of motion.

• While the body is at rest, the friction force F is whatever equilibrium requires, from zero up to a limit: F ≤ μs·N.
• At impending motion (limiting friction), F = μs·N.
• Once sliding, F = μk·N, with μk usually a little less than μs.
• Coulomb's laws (approximately true for dry surfaces): limiting friction is proportional to the normal reaction, independent of the apparent contact area, and kinetic friction is nearly independent of sliding speed at moderate speeds.

Angle of friction φ. At limiting friction the resultant of N and F makes angle φ with the normal, where tan φ = μs. A block on an incline is on the point of sliding under its own weight when the incline angle equals φ; this is the angle of repose for a block.

Solving friction problems. Draw the FBD with friction opposing the likely motion. Then either (a) assume equilibrium, solve for F and check F ≤ μs·N; or (b) assume impending motion, put F = μs·N and solve. When there are several contacts, check which one slips first, and whether the body slides or tips.

Wedges. A wedge converts a small push into a large normal force on the surfaces it bears against. Friction acts on every sliding surface. Treat each part (wedge, load block) as a separate FBD; at each sliding face replace N and F by a single reaction inclined at φ to the normal, turned so its friction component opposes the relative sliding. A wedge is self-locking (stays in place when the push is removed) if the wedge angle is small enough compared with the friction angles; for a wedge with friction only on its inclined face, the condition is α ≤ φ.

Belt and rope friction. A flat belt or rope wrapped round a drum through an angle θ can resist a much larger tension on one side than the other. The tension grows exponentially with the wrap angle, so a few turns of rope round a bollard hold a ship. In a belt drive the tight side T1 and slack side T2 differ by the effective pull that transmits torque. For a V-belt with groove angle 2β, the wedging action raises the effective coefficient to μ/sin β.

Limits. These relations assume dry contact, rigid bodies and (for belts) a flexible, inextensible belt with no centrifugal effect. At high belt speed, centrifugal tension mv² must be subtracted (see machine design). Lubricated contacts follow fluid-film laws, not Coulomb's.

Formulas

F_max = μs · N; F_k = μk · N

• F = friction force (N), N = normal reaction (N), μ = coefficient of friction (dimensionless, from tests or a data book).

tan φ = μs

• φ = angle of friction (rad or degrees).

P = W (sin α + μ cos α) (push up an incline, P parallel to the incline) P = W (sin α − μ cos α) (least force parallel to the incline to hold a block that would otherwise slide down, valid when tan α > μ)

• W = weight (N), α = incline angle.

P = W tan(α + φ) (wedge of angle α with friction only on its inclined face, load guided vertically without friction, wedge on a frictionless floor)

T1 / T2 = e^(μθ) (flat belt or rope, on the point of slipping)

• T1 = tight-side tension (N), T2 = slack-side tension (N), θ = angle of wrap in radians.

T1 / T2 = e^(μθ / sin β) (V-belt, 2β = groove angle)

Power = (T1 − T2) · v

• v = belt speed (m/s), power in W.

Worked examples

Example 1 (standard). A block of weight 200 N rests on a plane inclined at 30°. μs = 0.25. Find (a) the force parallel to the plane needed to start it moving up, and (b) the least force parallel to the plane that prevents it sliding down.

1. Will it slide on its own? tan 30° = 0.577 > μs = 0.25, so yes; a holding force is needed.
2. Normal reaction: N = W cos α = 200 × 0.866 = 173.2 N; limiting friction μs·N = 43.3 N.
3. (a) Impending motion up, friction acts down the slope: P = W sin α + μs N = 100 + 43.3 = 143.3 N.
4. (b) Impending motion down, friction acts up the slope: P = W sin α − μs N = 100 − 43.3 = 56.7 N.

Answer: (a) 143.3 N, (b) 56.7 N. Any force between these keeps the block at rest.

Example 2 (GATE level, belt). A flat belt runs over a pulley with an angle of contact of 165°. μ = 0.3, the maximum allowable belt tension is 1500 N and the belt speed is 20 m/s. Neglecting centrifugal tension, find the slack-side tension and the maximum power transmitted.

1. Angle of wrap in radians: θ = 165 × π/180 = 2.880 rad.
2. Tension ratio: T1/T2 = e^(0.3 × 2.880) = e^0.864 = 2.372.
3. T2 = 1500 / 2.372 = 632.2 N.
4. Power = (T1 − T2) v = (1500 − 632.2) × 20 = 17 356 W.

Answer: T2 ≈ 632 N, power ≈ 17.4 kW.

Example 3 (wedge). A load of 1000 N rests on a 10° wedge and is guided so that it can move only vertically, without friction at the guides. The floor under the wedge is frictionless; μ = 0.25 at the inclined face between wedge and load. Find the horizontal push needed to raise the load, and state whether the wedge is self-locking.

1. Friction angle: φ = tan⁻¹ 0.25 = 14.04°.
2. Load block: the reaction from the wedge is inclined at α + φ = 24.04° to the vertical. Vertical equilibrium: R cos 24.04° = 1000, so R = 1094.9 N.
3. Wedge: horizontal equilibrium P = R sin 24.04° = W tan(α + φ) = 1000 × 0.4460 = 446.0 N.
4. Self-locking check: α = 10° < φ = 14.04°, so the wedge stays in place when P is removed.

Answer: P ≈ 446 N; the wedge is self-locking.

Common mistakes

• Writing F = μN for a body that is not on the point of slipping. Below the limit, F comes from equilibrium.
• Taking N = W on an incline or when another force has a vertical component.
• Using degrees instead of radians in e^(μθ).
• Getting the friction direction wrong on a wedge; draw the relative sliding of each face first.
• Assuming the body slides when it may tip first (check the line of action of the normal force).
• Saying friction depends on contact area. For dry friction it does not, to a first approximation.

For GATE PI

Typical questions: limiting force on an incline, ladder and block problems with two rough contacts, wedge forces and self-locking, band brakes and belt tension ratios with power, and V-belt effective friction. Practise converting wrap angles to radians quickly and setting up separate FBDs for wedge and load.

Quick check

1. If μs = 0.5, what is the angle of friction?
2. A 100 N block on a horizontal floor (μs = 0.4) is pushed horizontally with 20 N. What is the friction force?
3. For μ = 0.25 and a wrap of 180°, what is T1/T2?
4. When is a wedge with friction only on its inclined face self-locking?

Answers: 1. 26.6°. 2. 20 N (below the 40 N limit, so friction equals the push). 3. e^(0.25π) ≈ 2.19. 4. When the wedge angle is no more than the friction angle (α ≤ φ).