What is the friction force?

How it works

• Friction force: f = μ × N, with μ the coefficient for the two surfaces and N the normal force. For kinetic friction (sliding) this is the friction force; for static friction it is the largest force before the object slips.
• Coefficient: μ = f ÷ N from a measured friction force.

The normal force is typed, or worked out from a mass m and gravity g:

• Level ground: N = m × g.
• Slope at θ above level: N = m × g × cos θ.

On a slope the page also shows the weight along the slope, m × g × sin θ, and:

• with static friction, whether it slides: it holds while sin θ ≤ μ × cos θ (μ_s ≥ tan θ). The answer reads No: static friction can hold it or Yes: it slides.
• with kinetic friction, the acceleration of an object sliding down: a = g × (sin θ − μ × cos θ). A negative value means a sliding object slows down.

The page also shows arctan μ in degrees: with static friction the steepest slope that holds; with kinetic friction the slope on which an object slides at a steady speed (μ_k = tan θ).

Surfaces (OpenStax Table 6.1, static μ_s and kinetic μ_k): rubber on dry concrete 1.0 and 0.7; wood on wood 0.5 and 0.3; waxed wood on wet snow 0.14 and 0.1; metal on wood 0.5 and 0.3; steel on steel, dry 0.6 and 0.3; steel on steel, oiled 0.05 and 0.03; Teflon on steel 0.04 and 0.04; bone lubricated by synovial fluid 0.016 and 0.015; shoes on wood 0.9 and 0.7; shoes on ice 0.1 and 0.05; ice on ice 0.1 and 0.03; steel on ice 0.04 and 0.02. The table gives a range for rubber on wet concrete (0.5 to 0.7 and 0.3 to 0.5): type your own μ for it.

Units (exact): 1 kN = 1,000 N; 1 lbf = 4.4482216152605 N; 1 kgf = 9.80665 N; 1 g = 0.001 kg; 1 t = 1,000 kg; 1 oz = 0.028349523125 kg; 1 lb = 0.45359237 kg; 1 short ton = 907.18474 kg. The angle is in degrees and g in m/s² (9.80665 standard).

Rules

• μ from 0 to 10; normal force from 10⁻¹² N to 10¹⁵ N; measured friction force from 0 to 10¹⁵ N; mass from 10⁻⁶ kg to 10¹² kg; slope from 0° to 90°; g from 0.001 to 1,000 m/s². Limits are in SI units and include their ends; a value typed in another unit is converted first.
• On a 90° slope (a wall) the normal force is 0, so the friction force is 0, and the coefficient has no answer.
• A result that is exactly 0 shows as 0. Any other result too large or too small to show as a number gives no answer.

Output format

The answer line reads f = a N (b lbf) for a friction force or μ = a for a coefficient. It and every number show to 6 significant figures, rounded half up, with trailing zeros dropped and thousands commas, and below 10⁻⁶ or from 10¹⁵ up as a × 10ⁿ. Typed values are read back as the exact decimals you typed, and every unit size is exact. sin θ and cos θ are exact where they are rational (0, ½ or 1, at 0°, 30°, 60° and 90°), so those results are exact fractions rounded once and the slide test compares exact values; other angles use 64-bit floats.

Worked examples by hand

Static friction on a crate. OpenStax Example 6.10: 20.0 kg on a level floor with g = 9.80 m/s², so N = 20 × 9.8 = 196 N. With μ_s = 0.700, static friction holds up to 0.7 × 196 = 137.2 N (30.8438 lbf).

Kinetic friction. The same crate sliding, μ_k = 0.600: f = 0.6 × 196 = 117.6 N.

A coefficient from a slide. OpenStax Example 6.11: 45.0 N of friction on a 62 kg skier sliding down a 25° slope (g = 9.80 m/s²). N = 62 × 9.8 × cos 25° = 550.673 N, so μ_k = 45 ÷ 550.673 = 0.0817182.

Sliding down a slope. OpenStax Example 6.14: a snowboarder on a 13° slope with μ_k = 0.20 and g = 9.80 m/s²: a = 9.8 × (sin 13° − 0.2 × cos 13°) = 9.8 × (0.224951 − 0.194874) = 0.294755 m/s².

Does it hold? A 10 kg wooden block on a 30° wooden ramp (μ_s = 0.5): tan 30° = 0.57735 is more than 0.5, so it slides. The weight along the slope is 10 × 9.80665 × 0.5 = 49.0333 N, and the steepest slope that holds is arctan 0.5 = 26.5651°.