Static Friction

The Coulomb Friction Model

Dry (Coulomb) friction resists the tendency of surfaces to slide. While a body is stationary, the static friction force is only as large as needed to prevent motion, up to a ceiling:

$f_s \le \mu_s N$

where μs is the coefficient of static friction and N is the normal force. The friction force is the equilibrium value below the limit; only at the verge of motion (impending slip) does it equal fs,max = μsN. Once the body slides, kinetic friction takes over:

$f_k = \mu_k N, \qquad \mu_k < \mu_s$

Because kinetic is smaller, a block is harder to start than to keep moving (the familiar "jerk then glide").

Trap: Friction force is not automatically μsN. If the applied load is below the slipping threshold, friction equals only the applied tangential force. Use fs = μsN only when the problem states motion is impending or slipping.

Inclined Planes and the Friction Angle

For a block of weight W resting on an incline at angle θ, resolve along and perpendicular to the surface:

• Perpendicular: N = W cos θ
• Along the surface (driving the block downhill): W sin θ

The block is on the verge of sliding down when the driving component equals the maximum friction, W sin θ = μs(W cos θ), which simplifies to the clean result:

$\tan\theta = \mu_s \quad\Longrightarrow\quad \theta = \phi = \tan^{-1}(\mu_s)$

φ is the angle of friction (repose). If the incline angle θ < φ the block holds; if θ > φ it slides; at θ = φ motion impends. This is why a coefficient of 0.577 corresponds to a 30° threshold — tan 30° = 0.577.

Worked Example — Force to Hold on an Incline

A 500 N block sits on a 25° incline with μs = 0.30. Is it about to slide, and if held, what friction acts?

Driving force = W sin θ = 500 sin 25° = 211 N. Max friction = μs W cos θ = 0.30 × 500 × cos 25° = 0.30 × 453 = 136 N.

Since 211 N > 136 N available friction, the block cannot stay — it slides, and an external up-slope force of at least 211 − 136 = 75 N is required to hold it. (Equivalently, θ = 25° exceeds φ = tan⁻¹0.30 = 16.7°.)

Wedges

Wedges are short inclined planes used to lift heavy loads with small forces. Solve by drawing a separate FBD for the wedge and for the lifted body, drawing every friction force opposite the impending motion of that surface, and applying equilibrium to each body. All normal forces must come out positive (surfaces push, never pull).

Belt (Capstan) Friction

When a flat belt or rope wraps around a fixed drum, the tensions on the two sides differ because friction along the contact arc carries part of the load:

$\frac{T_2}{T_1} = e^{\mu\beta}$

where T₂ is the tight (larger) side, T₁ is the slack side, μ is the belt-drum friction coefficient, and β is the wrap angle in RADIANS. The exponential growth means even modest wrap angles produce large tension ratios — the principle behind capstans and bollards.

Worked Example — Belt Tension

A belt wraps 180° around a drum with μ = 0.30; the slack side holds T₁ = 100 N. Find the maximum tight-side tension before slipping.

Convert the wrap: 180° = π rad = 3.1416 rad.

$T_2 = T_1\,e^{\mu\beta} = 100\,e^{(0.30)(3.1416)} = 100\,e^{0.9425} = 100(2.566) = 256.6\text{ N}$

If instead the wrap were 270° = 4.712 rad, T₂ = 100 e^(0.30·4.712) = 100 e^1.414 = 411 N — the extra wrap sharply raises capacity.

Top trap: plugging the wrap angle in degrees instead of radians. Always convert (multiply degrees by π/180) before exponentiating. A second check: confirm you placed the larger tension as T₂ on top — the ratio is always ≥ 1.

Tipping vs. Sliding

Many FE problems ask whether a block slides or tips first. Sliding impends at P = μsN; tipping impends when the resultant normal force reaches the block's edge (ΣM about that edge = 0). Compute both thresholds; the smaller force wins and determines the actual failure mode.

Worked Example — Slide or Tip?

A uniform crate 0.6 m wide and 1.2 m tall weighing 400 N rests on a floor with μs = 0.35. A horizontal force P is applied at the top. Sliding impends at P = μsW = 0.35(400) = 140 N. Tipping impends when ΣM about the leading bottom edge = 0: P(1.2) = W(0.3), so P = 400(0.3)/1.2 = 100 N. Because 100 N < 140 N, the crate tips before it slides — the answer is a tipping failure at 100 N. A tall, narrow object tips; a short, wide one slides.

Quick-Reference Friction Formulas

Self-check before answering any friction question: (1) Is motion actually impending, or is friction merely resisting a sub-threshold force? (2) Are all angles for belt problems in radians? (3) Is the normal force correct — it changes when a force has a vertical component pressing the block down or lifting it? Getting N right is half of every friction problem, because friction scales directly with it.