The force that is whatever it needs to be

A crate stands on a concrete floor with somebody leaning on it, and nothing moves. Ask what sideways force the floor is applying and the only honest answer is: exactly as much as is being pushed, and not a newton more. Push harder and the floor pushes back harder; stop and the floor stops. No property of the concrete, the crate or the contact fixes the number — it is set by whatever else is happening.

That is a strange kind of force in a subject whose whole method is to write equations and solve them for the unknowns. Everywhere else in statics, a reaction is what the equations say it is: a roller under a beam carries whatever $\Sigma M = 0$ hands it, and the arithmetic produces one number. Friction produces no number at all. It produces a bound.

The force with an inequality where the others have an equation #

The whole of what follows sits in one line:

$$|F| \le \mu N$$

Read it as a constraint rather than as a formula and the consequences arrive immediately. A constraint does not determine anything; it rules things out. The friction force at a contact is decided by the rest of the structure, and the coefficient is consulted only afterwards, to ask whether the value equilibrium demanded is one the contact can supply.

So a friction problem holds two questions. The first is what force is being asked for, which is ordinary statics; the second is can the contact supply it, which is an inspection. And where a structure has more than one friction contact, the first question has no single answer either.

This is friction as a constraint on equilibrium, which is a different subject from friction as physics. Why $\mu$ has the value it has — asperities, real contact area growing with pressure, adhesion at the junctions — belongs with the contact and is argued elsewhere. What is argued here is what an inequality does to a count of equations, and the answer turns out to be more disruptive than the physics.

A cone of directions, with no weight anywhere in it #

The geometry that makes the inequality legible comes from adding the two forces the contact supplies. A normal force $N$ perpendicular to the surface and a friction force $F$ along it sum to one reaction, leaning away from the normal by $\arctan(F/N)$, and the inequality caps that lean at

$$\phi = \arctan \mu$$

so the admissible reactions fill a cone of half-angle $\phi$ about the normal, and equilibrium is possible exactly when the reaction the structure demands points inside it. In the figure above the demand leans 15.0° against a cone of 19.3°, and the block stands.

Nothing in that comparison is a force. Both quantities are angles: one belongs to the geometry and the loading, the other to the surfaces. The weight cancels before the comparison is made, because it appears in $F$ and in $N$ alike and the test is on their ratio. That is why the angle of repose of a heap of dry sand is $\arctan \mu$ and nothing else — 19.3° for the coefficient drawn here — and why a bigger heap has exactly the same slope as a small one.

That is the loss stated in the language of the drawing that finds the reactions. Concurrency is a genuine theorem and friction does not touch it. What friction removes is the construction’s second ingredient — a known direction — and with a range of directions in, a range of answers comes out.

What friction does to the count #

The counting rule asks whether statics can answer at all: unknowns against equations, with the verdict read off the difference.

Counting unknowns against equations is the first thing done to any structure, and its verdict on a redundant frame is that the missing information is elastic: bring in stiffness, solve compatibility, and the indeterminacy resolves. That route is what one support too many is about, and it always works, at the price of needing to know a modulus.

Friction is short in a direction where that route is closed. There is no constitutive law relating a friction force to anything — no $F = k\delta$, no strain, nothing to be compatible with. The Coulomb model supplies a bound and declines to supply a value, so a friction indeterminacy is not resolved by a stiffness calculation; it is not resolved at all. What can be answered is whether any admissible state exists, and that is a different question from what the forces are.

The ladder does not have an answer #

The standard problem in every statics course is the ladder against the wall, and it is the cleanest demonstration there is, because it is set up as a determinate problem and is not one. Four unknowns: a normal and a friction force at the floor, and the same pair at the wall. Three equations, because a plane body has three. The system is short by one, and the missing information is the kind friction does not carry. There is therefore no ladder problem with an answer. There is a one-parameter family of equilibrium states, and the useful question is whether any member of it satisfies both cones.

That single frame carries the argument. Equilibrium is a line, not a point, because three equations in four unknowns describe a line. Admissibility is a region, because each inequality cuts a pair of half-planes and the intersection of the two pairs is a triangle. What is both admissible and in equilibrium is the chord the line cuts across the triangle — a segment 46.2 N long, measured along the floor’s friction force.

The textbook answer is the point where that line meets the axis, obtained by declaring the wall frictionless so the count comes right. It satisfies all three equations exactly, and it is 4.1 N outside the triangle: the floor would have to supply 404.1 N against a capacity of $0.4 \times 1000 = 400$ N, a ratio of 1.010. The determinate answer to this problem does not exist, and the ladder stands.

Which free body produced the number #

Every number above comes from one free body and three equations, and the family falls out of them in two lines.

Cut the whole ladder free — floor contact at A, wall contact at B, self-weight $W = 200$ N at mid-length, a climber of 800 N at 0.75 of the length, leaning at $\theta = 60°$. Write the load moment about the base as $Q = W/2 + W_p \cdot a = 100 + 600 = 700$ N in units of the ladder’s horizontal projection. Then:

$$\Sigma F_x: \; F_{floor} = N_{wall} \qquad \Sigma F_y: \; N_{floor} = W_{total} - F_{wall} \qquad \Sigma M_A: \; N_{wall}\tan\theta + F_{wall} = Q$$

Choose $F_{wall}$ and everything else follows: $N_{wall} = (Q - F_{wall})/\tan\theta$, then $F_{floor}$ from the first equation and $N_{floor}$ from the second. Nothing has been assumed and nothing has been solved — the choice was free.

Check the drawn state by hand. $F_{wall} = 63.4$ gives $N_{floor} = 1000 - 63.4 = 936.6$, the 937 in the figure. $N_{wall} = (700 - 63.4)/\tan 60° = 367.6$, and $\Sigma F_x$ makes the floor’s friction the same 367.6, whose ratio to capacity is $367.6 / (0.4 \times 936.6) = 0.981$ — comfortably inside. The residuals the generator reports are $6 \times 10^{-14}$, arithmetic noise rather than approximation.

Two readings come out of that sweep that a single answer cannot give. The floor’s demand falls monotonically as the wall takes more friction, so the frictionless-wall assumption is the worst case for the base — conservative wherever it is admissible at all, which is why it survives as a teaching device. And the band is bounded at one end by the floor and at the other by the wall, so which contact binds changes with the angle, the coefficients and where the climber stands.

The only determinate ladder is the one about to slide #

Imposing $F = \mu N$ at both contacts adds two equations to three, and five equations in four unknowns are inconsistent in general — consistent here at exactly one angle.

At 58.93° the family collapses to a point, the indeterminacy vanishes and the forces become determinate. Below it no equilibrium state exists and the ladder slides. So the one configuration in which the classical friction calculation returns a genuine unique answer is the one in which the structure is on the point of failing. Setting every friction force to its limit is not a modelling convenience; it is a statement that everything is simultaneously about to slip, and a count that returns a number for every input will return one for the cases it was never valid on.

Two machines that are made out of the inequality #

Friction is not only a nuisance to be checked. Two devices exist because the bound can be arranged to be un-exceedable, and in both the condition contains nothing but angles.

A wedge stays where it is driven while $\alpha \le 2\phi$, equivalently while $\mu \ge \tan(\alpha/2)$ — a condition on the coefficient alone, with no weight, no size and no material strength in it. A screw jack is the same inclined plane wrapped round a cylinder, and it locks while its helix angle is below the friction angle. The efficiency of a self-locking screw is capped at $(1 - \mu^2)/2$ at the boundary $\alpha = \phi$ and so never reaches one half: half the work is the price of staying put with the spanner removed.

The capstan is the one place the bound turns into an exponential, and its free body is one element of rope wide. Over an angle $d\beta$ the tension changes by $dT$, the bollard presses with $dN = T\,d\beta$, and the friction available is $\mu\,dN$, so $dT/d\beta = \mu T$ and

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

The radius is not in it. A rope round a thin pin holds exactly as well as the same rope round a fat one, because the smaller radius squeezes harder over a shorter length of contact and the two cancel exactly. The radius does not cancel out of the pressure, which is why a rope burns on a thin pin and not on a bollard.

The same shortage of equations, one field over #

The inequality is not confined to bodies resting on things. It sits inside structural elements, deciding which of two mechanisms carries the load.

A slip-critical joint carries nothing until it slips is the same argument in a connection: below the bound the plates are held by friction and the shear in the bolt shanks is genuinely zero; above it the joint has moved into a different load path where the hole goes oval. Nothing about the applied load says which regime applies. The inequality does.

The generalisation worth carrying reaches into masonry. A friction problem asks whether some admissible state exists rather than what the state is, and that is exactly the form of the safe theorem behind the line that must stay inside: an arch is indeterminate, its thrust line is not unique, and the proof of safety is the exhibition of one line that fits within the masonry rather than the discovery of the real one. Plastic collapse analysis makes the same move with its lower-bound theorem. All three abandoned the search for the actual internal forces and replaced it with an existence question, for the same reason — the equations ran out and no constitutive law was going to supply the missing ones. That the ladder in the first chapter of every statics book belongs to that family is not usually mentioned.

The other escape from the same shortage is to fix the force by a displacement instead of bounding it, which is what retained soil does: the pressure a wall receives depends on how far the wall has moved.

Where the model stops #

Coulomb friction is a model with one constant, and a rough one. A single $\mu$, independent of area, pressure and speed, is a fit rather than a law, and dirt, moisture and finish move it by more than most people would accept in a material property. Everything here is a statement about the structure of the problem, which survives a different $\mu$ — the numbers do not.

The cone assumes the contact is a point. A real ladder foot has area, over which pressure and friction are distributed, and a resultant leaning inside the cone is consistent with parts of that area having already slipped — which is what a rubber foot rolling at its edge is doing.

The bodies are rigid, and admissible does not mean occupied. A real ladder bends, and bending brings in a stiffness — so a real ladder does have a determinate answer, decided by elastic compatibility at the contacts and by the order in which the load arrived. Whether the climber walked up or was there when the ladder was placed changes which member of the family is occupied, and no equilibrium equation contains that. Elasticity does not remove the indeterminacy; it replaces it with a dependence on things nobody measures.

The inequality is checked at one instant. Vibration walks a contact through its cone repeatedly, and a mean force well inside the bound can still creep — which is why machinery bases loosen, and why a load that will not hold still is a different subject.

The pictures on this page have a limitation worth naming. Every one of them draws a set — a cone, a triangle, a chord, a band — and a set is exactly what a picture of a structure cannot show. The ladder figure has to draw one state, and it draws a member of the family chosen for legibility rather than by any physical argument: any single elevation implies a determinacy that is not there. That is why the argument’s centrepiece is a plot of the two friction forces against each other. The elevation is what a reader recognises; the plane of the two forces is where the answer lives.

The ladder from here #

Later rungs on this anchor: the cone derived from Coulomb’s experiments, and what its constant really summarises. Friction in three dimensions, where the admissible set stops being a plane region. Impending motion, and static against kinetic coefficients as a stability question. The order of loading, and structures whose friction forces depend on how they were assembled. Wedges and screw jacks in full, including the mechanical advantage a self-locking device gives up. Belt and band brakes, where the capstan exponential acquires a direction of rotation. Friction dampers, a bound placed in a structure deliberately to cap the force it can attract. Rocking against sliding, where the geometry decides whether a body tips before it slides. And the limit theorems for frictional systems, which are the general statement of what the ladder shows in miniature.

Coulomb reached the memoir that carries his name, in 1785, through a prize question on the rigging of ships — where the capstan, the wedge and the rope over a bollard were the practical problems of the day. The inequality has not changed since. What has changed is what is asked of it: Coulomb wanted to know whether ships’ tackle would hold, and the same three lines are now asked to say what the forces are, which they have never claimed to do.