Slide or topple

A tall filing cabinet, pushed at the top, goes over. A brick on the same floor, pushed with the same hand, skids. Both are held down by their own weight and held back by the same coefficient of friction, and the difference between them is not that one is heavier or that one was pushed harder. It is a ratio of two lengths.

Two conditions, and what they have in common #

Set a rectangular block of width $b$ and height $h$ on a level floor, and push it horizontally with a force $P$ applied at a height $y$. The weight $W$ acts down through the centre of gravity. Three things then have to balance: the vertical forces, the horizontal forces, and the moments.

Vertically, the floor supplies $N = W$ and there is nothing else to say. Horizontally, friction supplies $F = P$, and it can do that only up to its ceiling: static friction is an inequality rather than a value, supplying exactly what equilibrium demands until it is asked for more than $\mu_s N$. So the sliding condition is

$$P \ge \mu_s W.$$

The moments are the less familiar half. Take moments about the leading bottom edge — the corner the block would rotate about if it went over. The push has a moment $Py$ trying to turn the block over that edge; the weight has a moment $W(b/2)$ holding it down, because the centre of gravity sits half a width back from that corner. So the tipping condition is

$$P y \ge W\frac{b}{2}, \qquad\text{that is}\qquad P \ge W\frac{b}{2y}.$$

Now put the two conditions side by side. Both have $W$ on the right, so divide by it:

$$\text{slides at } \frac{P}{W} = \mu_s, \qquad\qquad \text{tips at } \frac{P}{W} = \frac{b}{2y}.$$

The weight has gone. Whichever of those two numbers is smaller is reached first as the push is increased, so the block tips rather than slides exactly when

$$\frac{y}{b} > \frac{1}{2\mu_s},$$

and there is no force anywhere in that inequality. A cabinet 300 mm deep pushed at 900 mm has $y/b = 3$; on a floor with $\mu_s = 0.42$ the threshold is $1/0.84 = 1.19$; three is larger, and it goes over. Its mass is not relevant, and neither is how hard it is shoved.

Where the floor is actually pushing back #

The moment condition above was written about a corner, which is a way of getting the answer without asking where the reaction acts. Asking is more instructive.

The floor does not push at a point. It pushes over the whole contact patch, with some distribution of pressure that depends on how stiff both bodies are and that nobody knows exactly. What is known is the resultant of that distribution — the single force it could be replaced by, and where it acts. Taking moments about the centre of the base, with friction acting at floor level and therefore contributing nothing,

$$N e = P y \quad\Longrightarrow\quad e = \frac{P y}{W},$$

with $e$ the distance from the centre of the base to the line of the resultant. At zero push it is in the middle. As the push rises it walks toward the leading edge, in exact proportion.

That is what tipping is. Not a threshold written into the problem, but the reaction running out of base. It is the same statement as the corner argument — $e = b/2$ gives $Py = Wb/2$ immediately — and it is the same reasoning that puts the resultant of a dam’s water pressure a third of the way up rather than halfway. It says something the corner argument does not: that the object is already leaning on its front edge, in a well-defined sense, long before it goes anywhere.

Change where the push is applied and the whole picture changes with it, because $e$ is proportional to $y$.

Setting $e = b/2$ and $P = \mu_s W$ at the same instant gives the height at which the two failures arrive together:

$$y^\ast = \frac{b}{2\mu_s}.$$

Push below $y^\ast$ and it slides; push above and it goes over. For the cabinet on that floor, $y^\ast$ is 357 mm — about a foot off the ground.

The one figure with no force in it #

The condition $y/b > 1/2\mu_s$ has two quantities in it, and neither is a force, a mass or a material property except through $\mu$. That makes it drawable as a single boundary in a plane, with every object in the world somewhere on it.

That is worth stating on its own, because it inverts the usual intuition about friction. Roughening the floor raises the sliding threshold and leaves the tipping threshold exactly where it was. So the effect of a better grip is not to make an object safer but to change how it fails — and for anything tall, the failure it changes to is the worse one.

The same inequality, at speed #

The comparison has one application large enough that a regulator computes it for every model of car sold, and it is the same two lengths against the same coefficient.

Take a vehicle cornering. The lateral force it needs is supplied by the tyres, and it is limited by friction at $\mu$ times the weight, exactly as the sliding condition above. The overturning moment is that lateral force acting at the height of the centre of gravity; the restoring moment is the weight acting at half the track width. So the vehicle rolls rather than slides when

$$\frac{t}{2h} < \mu,$$

with $t$ the distance between the wheels and $h$ the height of the centre of mass — which is the cabinet’s inequality with the base replaced by the track.

The quantity $t/2h$ has a name in the trade: the static stability factor, and it is published for every model. An ordinary saloon comes out around 1.35 — a wide track and a low centre of mass. A tall vehicle with a high floor comes out nearer 1.05. Dry tyres on dry asphalt give $\mu$ around 0.85.

So on a flat road both slide before either rolls, and the saloon has far more margin. What changes the answer is anything that raises the effective $\mu$ at the moment of the manoeuvre — and the thing that does that most decisively is not a road surface but an obstruction. A wheel striking a kerb sideways provides a lateral force at ground level that is not limited by friction at all, and against that the sliding escape route is closed and only the geometric threshold remains. Most rollovers happen that way, which is the practical content of the observation that high friction removes the alternative.

The other thing that changes the answer is loading, and it acts on the other side of the inequality. A roof load raises $h$; nothing on a car lowers it. The effect is not small, because $h$ is in the denominator of a ratio that was only a third above the threshold to begin with.

The squat case, and the quantity that is not in it #

A brick laid flat is the same problem with the ratio the other way round. Its base is 215 mm and its height 65 mm, so pushed at the top $y/b = 0.30$, and the tipping threshold is $b/2y = 1.65$ — a push of nearly twice the brick’s weight. Friction gives way at 0.65 long before that, so it slides, and it would slide on any floor whose coefficient is under 1.65, which is every floor.

One quantity is conspicuously absent from all of this, and its absence is not an oversight. The area of the base does not appear. A wide flat brick and a narrow tall one of the same mass press with the same total force, and friction’s ceiling depends on that total and not on how it is spread — which is the same independence of area that makes μ a single number in the first place, and which holds for the same reason: the real contact area is set by the load and the hardness of the asperities, not by the outline of the block.

So the two failures depend on disjoint properties. Sliding is decided by the surfaces and is blind to the shape; tipping is decided by the shape and is blind to the surfaces. Their comparison is the only place either quantity meets the other.

Tilting the floor instead of pushing the object #

The same comparison appears again if the object is left alone and the floor is tilted. On a slope of angle $\theta$ the weight has a component $W\sin\theta$ along the surface and presses with $W\cos\theta$, so sliding begins when $\tan\theta = \mu_s$ — the angle of repose, and again independent of the weight. Meanwhile the line of the weight, dropped from the centre of gravity, moves toward the downhill edge of the base and leaves it when $\tan\theta = b/2h$, with $h$ now the height of the centre of gravity.

$$\text{slides at } \tan\theta = \mu_s, \qquad\qquad \text{tips at } \tan\theta = \frac{b}{2h}.$$

The comparison is the same one, with $y$ replaced by $h$ — which is what a push applied at the centre of gravity would give. A tilt is a push at the centre of gravity, because that is where gravity is applied.

The two thresholds are read off the same free-body diagram and differ only in which balance is written. For a cargo container, $b/2h$ is around 0.35 and $\mu_s$ on a steel deck is around 0.35 too, which is why lashing is calculated rather than guessed. For a wine glass the two differ by a factor of ten and there is no calculation to do.

A body with no edge to reach #

The reaction walking toward the edge of the base is not confined to solids on floors. Float the object instead, and the upward force is buoyancy, whose resultant acts through the centre of the displaced volume. Heel the hull and the displaced volume changes shape, so that point moves sideways — exactly as the reaction under the cabinet does.

The difference is instructive. Under a rigid block the reaction can move only as far as the base extends, and the failure is the arrival at that limit. Under a hull the support point moves smoothly and can go on moving for a long way, so the righting moment rises, peaks and falls, and the failure is the peak being passed rather than a limit being reached. Both are the same question — how far can the upward force move before it stops being able to hold the body up — asked of a support that has an edge and of one that does not.

Pushing at an angle #

The horizontal push is a convenient case and an unusual one. A hand pushing a piece of furniture is almost never horizontal, and the difference is worth working out, because the two thresholds respond to it in opposite directions.

Push with a force $P$ at an angle $\phi$ below the horizontal. The horizontal component is $P\cos\phi$ and the vertical component presses down, so the normal force becomes $W + P\sin\phi$ and the sliding condition is

$$P\cos\phi \ge \mu_s\,(W + P\sin\phi).$$

Pressing down helps friction as much as it loads it, so the threshold rises — and if $\tan\phi$ exceeds $1/\mu_s$ it becomes impossible to slide the object by pushing at that angle at all, however hard. There is a cone of directions within which no amount of force makes anything move, and the object is wedged by its own friction.

The tipping condition moves the other way. The downward component’s line of action passes in front of the centre of gravity if the push is applied at the front face, so it adds a small restoring moment; but the horizontal component’s moment arm is unchanged. Net, the tipping threshold rises much less than the sliding one does.

Which reverses the practical advice. Pushing downward at a shallow angle, which feels like the safe thing to do with something heavy, makes sliding harder and tipping comparatively easier — so a downward-angled shove on a tall object is more likely to put it over than a level one. Pulling upward does the opposite: it unloads the friction, and the object skids sooner. Removal firms tilt and drag rather than push for exactly that reason, and the tilt is doing two things at once — it lowers the effective push height and it takes weight off the friction.

One threshold is a cliff and the other is not #

The two failures also differ in what happens the instant after they are reached, and the difference is not symmetric.

Sliding is a cliff. The coefficient that has to be overcome to start the motion is larger than the one that resists it afterwards — typically by a fifth to a third for dry solids — so the moment the object breaks away, the force required to keep it going drops. A push that was just enough to start it is now more than enough to accelerate it, and the object leaves suddenly. That is why furniture skids rather than creeps, and why a push applied slowly enough to be controlled is exactly the push that produces the least controlled result.

Tipping has no such discontinuity. Past the threshold the restoring moment falls smoothly to zero as the object rotates onto its edge and then reverses; there is no second coefficient waiting to change value, and nothing about the object’s resistance changes as it starts to move. The failure is progressive, and it is progressive in a way that gives a person about a quarter of a second to react.

Which is the opposite of the intuition most people have, and it matters for the practical question this essay is really about. A heavy object that slides is more likely to injure somebody than one that begins to tip, because the sliding is the one that happens without warning — and the standard advice to fasten tall furniture to a wall is aimed at the failure that announces itself, while the one that does not is left to the flooring.

How little it takes to move the answer #

The comparison is a ratio of two lengths against a coefficient, and both sides are known to very different accuracy. That is worth saying plainly, because it decides what is worth worrying about.

The aspect ratio is a length over a length and can be measured to a per cent. The coefficient is uncertain by tens of per cent, varies with the state of both surfaces, and can change by a factor of two when a floor is polished or a foot is worn. So the boundary in the regime figure is a sharp curve with every object’s horizontal position badly known, and the useful question is never “which side of the line is this object on” but “how far from it”.

The sensitivity to the centre of gravity is the part that catches people out. The threshold is $b/2h$, so it depends on the reciprocal of the height — and a centre of gravity is not where most people put it. Loading the top shelf of a bookcase and leaving the bottom empty moves it upward by a large fraction of the height, and halving $b/2h$ takes an object that would have slid and makes it fall. The standard instruction to load heavy things low is not about the total weight, which cancels out of everything here; it is about one length in a ratio.

What the pictures cannot show #

Everything above treats the contact as a line and the object as rigid. Neither is true, and each failure matters somewhere.

The pressure under a real base is not uniform and is not even a smooth function: it concentrates at the edges, and for a stiff object on a stiff floor most of the load is carried within a few millimetres of the perimeter. That does not move the resultant, which is fixed by moment balance whatever the distribution, so every threshold above survives. It does mean that the material at the leading edge is loaded far beyond the average, which is why a heavy cabinet pushed on a soft floor leaves a mark at the front foot and not under the middle.

Nothing here shows the transient. The conditions are static: they say when equilibrium becomes impossible, not what happens next. A block just past the tipping threshold begins to rotate slowly, and whether it falls over or drops back depends on whether the push continues and on how much energy it has — a question about a hill and a well rather than about balance, and one whose answer looks like the separatrix a spinning body sits on.

And the coefficient itself is the weakest number in the whole account. $\mu_s$ is not a material constant: it varies with cleanliness, with how long the surfaces have rested together, with humidity, and — contrary to the usual statement of the law — slightly with the load. None of that is a defect in the argument above, which needs only that a coefficient exists; it is a statement about how precisely any particular object can be placed on the boundary.

Where the ladder goes next #

The step taken here is small and general: a support force is a distribution, its resultant is located by moment balance, and a body fails when that resultant reaches the boundary of the region it is allowed to occupy. That reappears whenever something rests on something else. It decides which way a ladder slips, whether a stack of blocks can overhang, and — with the region a whole cross-section instead of a base — how far a compressive load can be moved off-centre before a joint opens. It is also the reason a torque is a force multiplied by a distance and not by a force: the distance is the only thing that distinguishes the two conditions here.

Two rungs follow directly. One is dynamic: the same block on a floor that accelerates, where the effective gravity tilts and the whole comparison happens with a modified vertical — which is the argument of a rotating frame applied to a static problem. The other is the stack: several blocks, each resting on the last, where the resultant under each has to lie inside the one below, and the answer to how far a pile can lean turns out to be a divergent series.

The habit worth taking away is the first move. Two failure conditions were written down, both proportional to the weight, and the weight was divided out — leaving a comparison between a property of the surfaces and a property of the shape. A great deal of mechanics is that move: find the quantity that appears on both sides of a comparison, and remove it, and see what the question was actually about.