The ladder that lets go of the wall

Lean a ladder against a wall, let its foot slip on a polished floor, and the ladder slides down with its top scraping the wall and its foot skidding out — for a while. Partway down, the top comes away from the wall, and the ladder finishes its fall clear of it, the foot still sliding, the top dropping in an arc through empty air. Anyone who has watched it happen has seen the gap open, and probably assumed it was the top bouncing off a bump in the wall.

It is not a bump. A ladder with a frictionless foot sliding against a frictionless wall must leave the wall, and it does so at a definite point: when its top has fallen to two-thirds of the height it started at. Two-thirds whatever the ladder’s length, whatever its mass, whatever angle it was leaning at when it started. The result is one of the cleanest uses of energy in mechanics, because it needs no force to be known in advance — only the energy and the one geometrical fact about where the centre of a ladder between a wall and a floor must be.

The circle the centre has to stay on #

For the mechanics, the ladder against the wall is a uniform rod: its steps and side rails matter only through how its mass is spread, and a uniform ladder spreads it as a rod does. While both ends of the rod touch, the foot is on the floor and the top on the wall. Draw the rod at any moment, mark its centre, and drop perpendiculars to the floor and the wall: the centre is at half the foot’s distance from the wall and half the top’s height above the floor. A little geometry shows that its distance from the corner where wall meets floor is always exactly half the rod’s length. So while the rod slides, its centre moves on a circle of radius $L/2$ centred on the corner.

That is a fact about geometry, not about forces, and it does all the work. The rod’s state at any moment is fixed by one number — its angle $\theta$ from the vertical — and its energy can be written in terms of that angle and how fast it is changing. The centre moves round its circle at speed $(L/2)\dot\theta$, and the rod also turns about its centre at rate $\dot\theta$, so its kinetic energy is

$$\tfrac12 m\left(\tfrac L2\right)^2\dot\theta^2 + \tfrac12\cdot\tfrac{mL^2}{12}\,\dot\theta^2 = \frac{mL^2}{6}\,\dot\theta^2,$$

the first term for the centre’s motion and the second for the turning about it, which the mass, and where it sits found always has to be counted for anything that turns as it moves. The potential energy is $mg(L/2)\cos\theta$. Neither the floor nor the wall does any work — each pushes at right angles to the surface the rod’s end slides along, and with no friction there is no force along it — so, as the floor that does no work insisted, the energy is conserved:

$$\dot\theta^2 = \frac{3g}{L}\left(\cos\theta_0 - \cos\theta\right).$$

Everything about the motion follows from that one line.

A wall that can only push #

The wall’s job is to keep the top against it, which means keeping the centre on its circle. Moving in a circle needs a force towards the circle’s centre, and the circle’s centre is the corner. Early in the fall, the rod is nearly upright and moving slowly, the centre is high on its circle and moving down it, and holding it on the circle needs a push away from the wall — the centre is being steered outwards as it goes down. The wall supplies that push.

The push is the rod’s mass times its centre’s horizontal acceleration, and the energy equation gives that acceleration at every angle. The result is

$$N_w = \tfrac34\,mg\sin\theta\,\left(3\cos\theta - 2\cos\theta_0\right).$$

It starts at zero, grows, and then falls back. It reaches zero again when $\cos\theta = \tfrac23\cos\theta_0$ — when the top’s height, $L\cos\theta$, has fallen to two-thirds of its starting value $L\cos\theta_0$. Below that the formula goes negative. To keep the top against it, the wall would have to pull the rod towards itself.

A wall cannot pull. At two-thirds of the starting height the push reaches zero, and from then on the top moves away from the wall. Every one of the curves reaches zero at the same place, whatever the starting angle — the starting angle sets how large the push grows, not where it vanishes. The length of the rod appears nowhere, and neither does its mass, which cancels from both sides. Less obviously, neither does the way the mass is spread along the rod. A rod with heavy ends turns more sluggishly than a uniform one, and one with all its mass at its middle turns not at all about its centre, and both fall at different rates from the uniform rod — but the extra turning energy is a fixed fraction of the centre’s at every angle, and redoing the calculation with any moment of inertia gives the same zero at the same place. The two-thirds belongs to the circle the centre is held on, and to nothing else.

Why does the centre need to be pulled inwards later in the fall? Because it is moving faster. Moving round a circle at speed $v$ needs an inward force of $mv^2$ divided by the radius, and as the rod falls and speeds up that need grows. Early on the outward steering dominates; later the inward need for the circular motion does, and at two-thirds height the two cancel. Past that point, a circle would need an inward pull from the corner, and the corner can push only outwards. The rod’s centre leaves the circle by going outside it.

The same factor on a hemisphere #

The factor $3\cos\theta - 2\cos\theta_0$ is familiar from a quite different problem. A small block released at rest on the top of a smooth hemisphere, a little off the summit at angle $\theta_0$, slides down its curved surface and leaves it at a definite point. The surface pushes outwards on the block with whatever force keeps it on the curve; the block’s speed, from energy, is set by how far it has dropped; and moving on a circle needs an inward force of mass times speed squared over radius, which gravity’s inward component must supply with the surface’s push subtracted. The push comes out as $mg(3\cos\theta - 2\cos\theta_0)$, and the block leaves when $\cos\theta = \tfrac23\cos\theta_0$ — at two-thirds of its starting height above the hemisphere’s centre.

The rod and the block share the factor because they share the structure. In both, a point is kept on a circle by a constraint that can push and cannot pull; in both, energy alone fixes how fast it moves at each angle; and in both, gravity’s pull along the circle grows as the point descends while the force needed to keep it on the circle grows faster, until the constraint has nothing left to give. The rod adds its turning — its centre does not carry all its kinetic energy — but the rotation’s share is a fixed fraction of the centre’s at every angle, so it rescales the speeds without moving the point where the push vanishes. The hill that gives it back drew energy as a landscape a body rolls on; on both the hemisphere and the circle of the rod’s centre, the landscape stops describing the motion at two-thirds of the way down, because past that point the body is no longer on it.

Turning is an acceleration, and constant speed does not help made the general point that moving on a curve takes a force towards the curve’s centre whether or not the speed changes. Here the speed does change, and the force needed grows with its square. What the two problems add to that general point is the moment the force runs out: the constraint can supply the inward force only by pushing less, and when it has stopped pushing altogether, the body leaves the curve along a path of its own.

Free of the wall, the sliding is frozen #

Once the top has left the wall, the only horizontal force on the rod was the wall’s push, and it is gone. The floor pushes up; gravity pulls down. Nothing acts sideways. So the rod’s centre keeps the horizontal speed it had at the moment of separation, exactly as the point that keeps moving as if nothing had happened found for any system with no external force along a direction — its centre of mass moves uniformly along it, whatever its parts do.

The energy figure shows the freeze. While the top is on the wall, the wall’s push is feeding the sideways motion, and the share of energy in sliding climbs to about a tenth of the total. At the moment of separation it stops dead, and from then until the rod hits the floor the sliding share is constant. Everything the rod still has to lose goes into falling and turning. By the time it lies flat, two-thirds of its energy is in its centre’s downward motion and a fifth in its spin; only the tenth it collected while the wall was pushing it is in sliding.

That freeze is also why the separation is a real event and not an approximation. A wall that could pull would hold the rod on the circle and feed more and more energy into sliding; the floor would see the foot shoot out faster and faster. The real ladder’s foot slides at a speed that stops growing at separation. A high-speed film of a rod released on a smooth floor shows it: the foot accelerates, then cruises.

The top that lands harder than a stone #

A natural guess is that no part of a falling ladder can strike the floor faster than a stone dropped from the same height, because the rod’s energy is shared among all its parts, while the stone keeps all of its own. It is wrong for the top.

At first the top lags: while it slides down the wall, the rod’s whole mass is only beginning to turn, and the top moves slowly. Once the rod is well over, the top is the far end of a rod turning about a point near its foot, and its speed is the rod’s turning rate times nearly its whole length — while the energy that drives that turning comes from the fall of the centre, which is only halfway up. The top is being flung. It overtakes the stone at about 2.3 metres and lands at 10.3 metres a second against the stone’s 8.7.

The same arithmetic explains why a falling chimney breaks in the air and why a tree being felled snaps its top when it hits. A rigid rod pivoted at its base and falling from upright has a top that accelerates downwards, near the end of the fall, at up to one and a half times $g$: the inner part of the rod is falling more slowly than it would freely and holds the top back; the top is falling faster than it would freely and is pulled along by the rest. A chimney is not rigid enough to transmit those forces and breaks where the bending is greatest, partway up. The pile that lands heavier than it weighs found a falling chain hitting a table harder than its weight for a related reason: the parts of a connected body are not each falling freely, and the connection can speed some parts beyond free fall.

The same fall at every size #

The length of the rod cancels from where it leaves the wall, and it nearly cancels from everything else. The only length in the problem is $L$ and the only acceleration is $g$, so every time in the motion is a multiple of $\sqrt{L/g}$ and every speed a multiple of $\sqrt{gL}$, with the multiples fixed by the starting angle alone. A 4-metre ladder released at 15° hits the floor 1.34 seconds later; a 1-metre broom handle released at the same angle against a polished wall falls in exactly half that time, 0.67 seconds, and leaves the wall at the same fraction of its starting height. A tall ladder looks as if it falls in slow motion because it does: its fall takes longer as the square root of its length, the same scaling that makes a tall tree topple majestically and a pencil snap over. It is the kind of argument the jump that does not get higher with size used for animals: when the only quantities in a problem are a length and $g$, size changes the clock and leaves the shape of the motion alone.

That is also why the broom handle is the way to see the effect. Stand one against a smooth wall on a smooth floor, let it go, and watch its top: it scrapes down the wall, then lifts clear, and the handle falls the last stretch in the air. The gap between its top and the wall at the moment it lands is about a twentieth of its length, set by the sliding speed it had collected at separation and kept thereafter.

Standing before it slides #

None of this happens if the ladder against the wall does not start sliding, and a real one has friction where it meets the floor. Whether it stands is a statics question, and slide or topple already found the form of the answer: the friction needed is set by moments, and the friction available by the floor.

With the wall smooth, the wall’s push is the only horizontal force at the top, and it must be balanced by friction at the foot. Taking moments about the foot, the wall’s push is set by how far the ladder’s weight acts from the foot, and the ladder stands if

$$\tan\alpha \ge \frac{1}{2\mu},$$

where $\alpha$ is the angle from the floor and $\mu$ the friction coefficient at the foot. A person standing on the ladder changes the balance according to how high they stand.

A climber halfway up changes nothing, because their weight acts where the ladder’s own does. A climber near the top changes a great deal, because their weight acts with a longer lever about the foot, pressing the wall harder and so the foot outwards harder. The working rule — a foot out from the wall for every four feet up, 76° — leaves a margin for that: with a climber near the top it is safe on any foot with a friction coefficient above 0.21, which rubber on concrete easily exceeds and a metal foot on wet tiles does not. The accidents that kill people on ladders are mostly the foot sliding out, when the climber reaches the top of a ladder set too shallow on too slippery a floor.

A ladder that has started to slide is in the dynamic problem. With friction at the foot, energy is no longer conserved — friction works against the foot’s sliding — and the place where the top leaves the wall moves; with enough friction the ladder stops before it leaves. The frictionless case is the clean limit, and it is the limit approached by a ladder on ice or a broom handle on a polished floor.

What the figures leave out #

The figures treat the ladder as a uniform rigid rod, the wall and floor as perfectly smooth and rigid, and the motion as confined to a vertical plane. A ladder on the floor whose centre of mass is not at its midpoint — one heavier at its foot, say — keeps its centre on an ellipse rather than a circle and leaves the wall at a different fraction of its height; the two-thirds belongs to a ladder balanced at its middle. A real ladder flexes, its feet bounce, and its top may rattle against the wall before leaving it. And the separation is drawn as a clean event: in reality the push approaches zero smoothly, and the slightest roughness of the wall can hold the top on or knock it off a little early. The domain is a slender uniform ladder on surfaces smooth enough for friction to be negligible, released from rest.

Still open: what a rough wall and floor do #

With friction at both ends the problem has no tidy answer. Whether the top slides down the wall or stops, where it leaves the wall if it does, and whether the foot stops sliding before the top does, depend on both friction coefficients and on the starting angle, and the motion can switch between sliding and sticking at either end during the fall. Treatments that follow all the cases find regions in which the equations of rigid-body motion with dry friction have no consistent solution, or more than one — the same Painlevé paradoxes that arise when a rigid rod is dragged across a rough surface — and resolving them requires giving up rigidity at the contacts. What a real ladder does in those regions is decided by the flexibility of its feet and the floor, and it is a question for experiments rather than for the rigid equations.

The frictionless result is exact and short. A ladder sliding between a smooth wall and a smooth floor keeps its centre on a circle of radius $L/2$ about the corner, so energy alone gives $\dot\theta^2 = (3g/L)(\cos\theta_0 - \cos\theta)$; the wall’s push, $\tfrac34 mg\sin\theta(3\cos\theta - 2\cos\theta_0)$, reaches zero when the top has fallen to two-thirds of its starting height — 2.58 m of 3.86 m for a 4 m ladder at 15° — and since the wall cannot pull, the ladder lets go, keeps its sliding speed, and lands its top faster than a stone. The wall decides where the ladder leaves by being able to do only half of what a constraint usually does.