Coulomb friction

Push a heavy crate across a floor and, up to a point, nothing happens — the floor itself supplies exactly the resisting force needed to keep you at a standstill. Push harder and eventually the crate breaks free and slides. That everyday experience is governed by one of the oldest and most useful rules in engineering mechanics: Coulomb's law of dry friction. It is the friction model behind almost every statics problem you will meet in this chapter — blocks on inclines, ladders against walls, wedges driven into gaps, clamps holding a workpiece — and it builds directly on the free-body-diagram and equilibrium skills from trusses and frames. This article compresses the static/kinetic friction laws, the angle of friction, and the impending-motion procedure into one exam-ready reference, with two worked problems and a quick-recall card at the end.

The core insight

Friction at a dry contact can rise to a maximum of $\mu_s N$ before slip begins, and once slip occurs it holds steady at $\mu_k N$ ; the borderline case $F=\mu_s N$ (impending motion) is what most equilibrium problems actually solve for.

The core idea

Where this sits. Coulomb friction is the third major topic of the "Structures and Friction" chapter in ME10001 Mechanics, following plane trusses and frames/machines. You are expected to know the static and kinetic friction laws, the concept of impending motion, the angle of friction, and how to fold friction into a full 2-D equilibrium solution.

The two laws. At any point of contact between two dry surfaces, the total reaction splits into a normal component $N$ (perpendicular to the surface) and a friction component $F$ (tangent to the surface, opposing relative sliding or the tendency to slide):

F \le \mu_s N \quad \text{(static: no sliding, or motion impending)}

F = \mu_k N \quad \text{(kinetic: surfaces actually sliding)}

where: $\mu_s$ is the coefficient of static friction, $\mu_k$ the coefficient of kinetic friction, and $\mu_k$ is usually slightly smaller than $\mu_s$ (it is generally easier to keep a body sliding than to start it moving). The maximum available static friction is $\mu_s N$ ; as long as the friction required by equilibrium stays below this ceiling, the body sits still. Motion is impending exactly when the equality $F = \mu_s N$ is reached — the borderline condition that most limiting-case problems analyze.

The angle of friction. At impending slip, the total reaction (the resultant of $N$ and $F$ ) makes an angle $\phi$ with the normal, called the angle of friction:

\tan\phi = \frac{F}{N} = \mu_s \quad \Rightarrow \quad \phi = \tan^{-1}\mu_s

where: $\phi$ depends only on the surface pair, not on the weight or size of the body. This gives a fast geometric shortcut: a block resting under gravity alone on an incline of angle $\theta$ stays put as long as $\theta \le \phi$ ; once $\theta$ exceeds $\phi$ , no normal force can supply enough friction to prevent sliding. This special case ( $\theta = \phi$ ) is sometimes called the angle of repose.

Figure

Free body diagram of a block on a horizontal surface pulled by a force P at an angle above the horizontal showing weight W downward normal force N upward friction force F opposing sliding and the applied force P resolved into horizontal and vertical components

Procedure. Draw the FBD with $N$ and $F$ as unknowns at every frictional contact. Then either (a) set $F = \mu_s N$ if you are asked for a limiting condition (maximum load, minimum angle, and so on), or (b) solve for the required $F$ purely from equilibrium and only afterward check it against the ceiling $\mu_s N$ to decide whether the assumed equilibrium is actually possible.

Derivation AThe friction-angle test for a block resting on an incline

The goal is to show, starting only from the two equilibrium equations for a block on a rough incline and Coulomb's static law, that the block stays at rest under gravity alone precisely when the incline angle does not exceed the angle of friction — turning a two-equation equilibrium problem into a single instant angle comparison you can apply without re-deriving it each time.

Step 1. Set up axes along and perpendicular to the incline for a block of weight $W$ on a surface inclined at $\theta$ , with no applied force besides gravity. Resolving weight into these axes gives a down-slope component $W\sin\theta$ and a perpendicular component $W\cos\theta$ .

Step 2. Apply equilibrium perpendicular to the incline: $\sum F_{\perp}=0$ gives $N = W\cos\theta$ . Apply equilibrium along the incline: $\sum F_{\parallel}=0$ gives the friction force actually required to hold the block, $F_{req} = W\sin\theta$ (up-slope).

Step 3. Compare the required friction to the maximum available, $F_{max}=\mu_s N = \mu_s W\cos\theta$ . Equilibrium is possible only if $F_{req}\le F_{max}$ , that is $W\sin\theta \le \mu_s W\cos\theta$ . The weight $W$ cancels from both sides — the result is independent of how heavy the block is.

Step 4. Divide both sides by $\cos\theta$ : $\tan\theta \le \mu_s$ . Since $\tan\phi=\mu_s$ by definition of the angle of friction, this is exactly $\theta \le \phi$ . The block remains in equilibrium as long as the incline angle does not exceed the angle of friction; beyond that angle no value of $N$ can generate enough friction to balance the down-slope pull of gravity, and the block slides regardless of its mass.

Worked example

A 50 kg block rests on a horizontal surface with $\mu_s = 0.30$ and $\mu_k = 0.25$ . (a) A purely horizontal force $P$ is applied — what is the maximum $P$ for which the block remains at rest? (b) The same force is instead applied at $20°$ above the horizontal — find the new maximum $P$ for equilibrium. (c) If in case (b) a slightly larger force of $P = 145$ N is actually applied, so the block starts sliding, find its acceleration.

Step 1 — Part (a), horizontal pull. Weight $W = mg = 50(9.8) = 490$ N. Vertical equilibrium gives $N = W = 490$ N. At impending motion, $F = \mu_s N = 0.30(490) = 147$ N, and horizontal equilibrium gives $P_{max} = F = 147$ N.

Step 2 — Part (b), pull at $20°$ . Now $P$ has components $P\cos20°$ horizontal and $P\sin20°$ vertical (upward), which reduces the normal force: $\sum F_y=0 \Rightarrow N = 490 - P\sin20°$ . At impending motion, $F=\mu_s N$ , so horizontal equilibrium gives $P\cos20° - \mu_s(490-P\sin20°)=0$ , i.e. $P(\cos20°+\mu_s\sin20°) = \mu_s(490)$ . Substituting numbers, $P(0.9397 + 0.3\times0.3420) = 147 \Rightarrow P(1.0423)=147 \Rightarrow P_{max}=141.0$ N. Tilting the pull upward slightly lowers the force needed to move the block because it reduces $N$ (and hence the friction ceiling) faster than it reduces the useful horizontal pull — up to the optimal angle $\phi=\tan^{-1}(0.3)=16.7°$ , close to the $20°$ used here.

Step 3 — Part (c), kinetic phase. Since $P=145$ N exceeds the $141.0$ N threshold found in Step 2, the block is now sliding, so friction switches from the static law to the kinetic law $F=\mu_k N$ . The normal force still depends on $P$ 's vertical component: $N = 490 - 145\sin20° = 490 - 49.6 = 440.4$ N, so $F = \mu_k N = 0.25(440.4) = 110.1$ N. Newton's second law along the horizontal then gives the net force and acceleration: $P\cos20° - F = ma \Rightarrow 145(0.9397) - 110.1 = 50\,a \Rightarrow 136.3-110.1=50\,a \Rightarrow a = 0.52\ \text{m/s}^2$ .

The block slides once $P$ exceeds 141.0 N, and at $P=145$ N it accelerates at $a \approx 0.52\ \text{m/s}^2$ .

Common traps

Assuming $F = \mu_s N$ always. This equality holds only at impending motion. In ordinary equilibrium, friction takes on whatever value up to $\mu_s N$ is needed — solve for the required $F$ first, then compare it against the ceiling.
Confusing $\mu_s$ and $\mu_k$ . Use $\mu_s$ to decide "will it start moving"; switch to $\mu_k$ (typically smaller) only once the body is confirmed sliding.
Getting the friction direction wrong. Friction opposes the relative or impending sliding motion of the contact — determine the direction from the physical tendency to move, never by a fixed default.
Misusing the $\theta$ -versus- $\phi$ shortcut. The rule $\theta\le\phi$ applies only to a block resting under gravity alone on an incline with no other force. Add an applied force, a cable, or a second contact, and you must return to the full equilibrium equations.

Ask the tutor: If you're unsure whether a problem needs the impending-motion equality $F=\mu_sN$ or requires checking a computed $F$ against $\mu_sN$ after solving equilibrium, ask the AI tutor to walk through the distinction with a fresh example.

Why is the kinetic coefficient usually smaller than the static one?

Once two surfaces are sliding, their microscopic high points have less time to interlock and cold-weld together than when the surfaces are sitting still relative to each other, so on average the resisting force is a little lower. That is why $\mu_k$ is typically slightly less than $\mu_s$ for the same pair of surfaces, and why a body often "jerks" forward the instant it breaks free — the resisting force drops the moment sliding starts.

If required friction is less than the maximum available, is the actual friction force still μₛN?

No. Coulomb's law only sets an upper bound, $F\le\mu_s N$ , on how much static friction a contact can supply — it does not say friction is always at that maximum. In ordinary equilibrium (not on the verge of slipping) the friction force is whatever value equilibrium demands, found from $\sum F=0$ , and that value can be anywhere from zero up to $\mu_s N$ . You only set $F=\mu_s N$ when the problem explicitly asks about the limiting, impending-motion condition.

Does the angle of friction depend on the size or weight of the block?

No, and that is exactly why it is such a useful shortcut. The derivation shows the weight $W$ cancels out of the inequality $\tan\theta\le\mu_s$ , leaving a condition, $\theta\le\phi=\tan^{-1}\mu_s$ , that depends only on the coefficient of static friction between the two surfaces in contact. A heavier block needs more friction force to hold it, but it also presses down harder, generating proportionally more available friction — the two effects cancel.

Coulomb friction

The core idea

Worked example

Common traps

Before this

Where this leads