Centroids, Area Moments and Friction

Centroids, second moments of area, parallel-axis theorem, friction, wedges and belts.

Centroids and centers of gravity

x̄ = Σ(Ai xi) ÷ ΣAi    ȳ = Σ(Ai yi) ÷ ΣAi

Second moment of area

I = ∫ y² dA    rectangle: I = b h³ ÷ 12    circle: I = π d⁴ ÷ 64    parallel-axis: I = Ic + A d²

Worked example: a T-section

Worked example: centroid and I of a T-section

A flange 100 mm wide by 20 mm thick sits on top of a web 20 mm wide by 80 mm deep (total height 100 mm). Measure heights from the bottom.

  • Areas: flange 2 000 mm² at y = 90 mm; web 1 600 mm² at y = 40 mm; total 3 600 mm².
  • Centroid: ȳ = (2 000 × 90 + 1 600 × 40) ÷ 3 600 = 67.78 mm above the bottom.
  • Flange: I = 100 × 20³ ÷ 12 = 66 667 mm⁴, plus A d² = 2 000 × 22.22² = 987 654. Web: I = 20 × 80³ ÷ 12 = 853 333, plus A d² = 1 600 × -27.78² = 1 234 568.
  • Total I = 3.142 × 10⁶ mm⁴. Section moduli: Z (top) = I ÷ 32.22 = 97.5 × 10³ mm³ and Z (bottom) = I ÷ 67.78 = 46.4 × 10³ mm³ (the flange side is stressed less in bending because it is nearer the neutral axis).
  • For comparison, a solid 20 × 100 mm bar has I = 1.667 × 10⁶ mm⁴ with 2 000 mm² of area: the T-section uses 1.8 times the material for 1.89 times the stiffness.

Dry friction

F ≤ μs N (static)    F = μk N (sliding)    angle of friction φ = arctan μ

Worked example: a block on a slope

Worked example: a 50 kg block, μ = 0.4

  • The block slides when the slope angle reaches arctan 0.4 = 21.8°, independent of its mass.
  • On a 30° slope the weight component along the slope is 245 N but friction can resist only 170 N, so it slides.
  • Force to push it up the slope (parallel to the slope): F = m g (sin θ + μ cos θ) = 415 N. Force needed just to hold it from sliding down: m g (sin θ − μ cos θ) = 75 N.

Wedges, ladders and belts

  • Ladder against a smooth wall with floor friction μ: it stays put if tan θ ≥ 1 ÷ (2 μ) for a load at mid-length. With μ = 0.4 that is θ ≥ 51.3° from the horizontal; with a person near the top the angle must be steeper still.
  • Belt or rope friction (capstan): T1 ÷ T2 = eμ β. With μ = 0.3 and half a turn (β = π) the ratio is 2.57; three half turns (1.5 turns) give 17, so a 1 000 N pull is held with only 59 N. That is why a few turns of rope round a bollard hold a ship, and why belt drives slip first on the small pulley.
  • Wedges and screws are inclined planes: a wedge or screw is self-locking when the friction angle exceeds the wedge or lead angle (see Power Screws and Lead Screws).

Related

FAQ

Why is I measured about the centroid?

The neutral axis of a beam in bending passes through the centroid, so bending stress and curvature use I about that axis.

Does friction depend on contact area?

To a good approximation not; it depends on the normal force and the surface pair. Soft rubber and very clean metals are exceptions.

Is static friction always μ N?

No. It is whatever is needed to hold the body, up to the limit μ N. Use the equality only when the body is about to slip.

References

  • Hibbeler, R. C., Engineering Mechanics: Statics, Pearson.
  • Gere, J. M., Goodno, B. J., Mechanics of Materials, Cengage (tables of section properties).
  • Meriam, J. L., Kraige, L. G., Engineering Mechanics: Statics, Wiley.

Last reviewed: October 2026. Friction coefficients are typical; measure the pair for any design that depends on them.