The pile that lands heavier than it weighs

Hold a chain vertically with its lowest link just touching a kitchen scale, and let go. The scale’s reading during the fall is not the weight of the heap that has formed. It is three times that weight, at every instant, and the factor is exact.

The reason the factor is a whole number is worth doing slowly, because it is the same argument that runs through every one of the situations below.

Where the three comes from #

Let the chain have mass $\lambda$ per unit length and let $x$ be the length that has landed. Two things press on the pan.

The pile itself weighs $\lambda x g$. That much is obvious.

The second term is the force required to stop the links that are arriving. In a time $dt$ a length $v\,dt$ of chain joins the pile, bringing momentum $\lambda v^2\,dt$ downward and leaving with none, so the pan must supply $\lambda v^2$ of force to destroy it. That is a momentum flux, and it is a force in exactly the sense the second law means: a rate of change of momentum.

Now the arithmetic. The chain is in free fall until it lands, so $v^2 = 2gx$, and the flux is $2\lambda g x$ — precisely twice the weight of the pile, at every stage of the fall, because both terms are proportional to $x$ and the constant of proportionality differs by two. The reading is $3\lambda g x$, it reaches three times the chain’s whole weight at the last link, and it drops to one times it immediately afterwards, because the flux stops all at once.

Nothing about the chain’s material enters, and nothing about the height it was dropped from — only that it started from rest with its end touching. Drop it from higher and the arriving links come in faster, but they also start arriving later, and the factor changes.

The law that does not apply #

The thing that trips this problem up is not the arithmetic. It is that the moving part of the chain is not a legal subject for Newton’s second law.

$F = ma$ is a statement about a definite set of matter. The falling portion of the chain is not a definite set of matter; it is a region, and matter is leaving it. Writing $F = d(mv)/dt$ and letting $m$ change is not a generalisation of the second law but a different equation, and it gives the wrong answer here for a reason that becomes obvious once stated: it silently assumes the departing matter leaves with the velocity the remaining matter has, which is exactly what does not happen — the links that leave the moving part do so by being brought to rest.

The correct procedure is the one used for a rocket, and it is the same procedure in reverse: draw a boundary, be explicit about the momentum crossing it, and apply the second law to a fixed set of matter for an instant.

A rocket is the same ledger read the other way. There, matter leaves the moving part carrying momentum away with it, and the thrust is the rate at which that momentum is exported; here, matter joins the resting part and has its momentum destroyed against the floor. Both are problems in which the moving mass changes, and in neither is $F = ma$ a statement about the changing part. What is conserved is the momentum of the whole set of matter, boundary and all, and the force on the floor is the rate at which the falling branch’s momentum is being removed from it.

The bookkeeping, done three ways #

A claim as clean as “exactly three” invites a check, and there are two independent ones.

The impulse the floor delivers over the whole fall must equal the momentum the arriving links brought. Integrating the momentum gives $\tfrac{2}{3}\lambda L\sqrt{2gL}$; integrating the force over the time gives the same number; evaluating the integral in closed form gives the same number a third time. They agree to five decimal places, which is the sampling error and not a physical difference.

The energy ledger is the striking one. The chain’s centre of mass falls from $L/2$ to zero, releasing $\tfrac12\lambda g L^2$. The energy destroyed in the impacts — every link arriving at $\sqrt{2gx}$ and leaving at rest — integrates to $\tfrac12 \lambda g L^2$ as well.

So all of it is lost. Not most of it: all of it. The chain finishes at rest in a heap, with nothing stored anywhere, and every joule the fall released was destroyed link by link at the moment of arrival. That is what makes the situation different in kind from a collision between two bodies, where the loss depends on how elastic the bodies are and can be anything from nothing to everything. Here it is fixed at everything by the geometry, whatever the chain is made of.

A chain lowered gently releases exactly the same potential energy and loses none of it — all of it comes out as work on whatever is doing the lowering. The two processes have identical endpoints and completely different ledgers, which is a good short demonstration that energy released is not the same quantity as energy usefully obtained.

Picking it back up #

Run the process backwards and the mirror image is just as sharp, and rather more useful.

Lift the free end of a heaped chain at a constant speed $v$. The force is $\lambda g y + \lambda v^2$: the weight of what is hanging, plus the momentum being handed to links that were lying still. The second term does not depend on how much has been lifted, so the curves are the weight line shifted upward by a constant.

The energy ledger of that is the part to carry away. The power delivered is $\lambda g y v + \lambda v^3$. The power that ends up as height and speed is $\lambda g y v + \tfrac12\lambda v^3$. The difference is $\tfrac12 \lambda v^3$, which is exactly half of the kinetic energy being handed to the chain — at every speed, at every height, whatever the chain is made of.

Half is the same fraction that a perfectly inelastic collision between equal masses loses, and that is not a coincidence: each link joining the moving part is an inelastic collision between a piece of moving chain and a piece at rest, and the mass ratio at the boundary is what fixes the fraction. Anybody hauling anything flexible at speed is paying that toll continuously.

The chain that falls faster than gravity #

Fold a chain in half, hold both ends together, and release one. The falling branch shortens as it goes, and what happens next has been argued about for a hundred and thirty years.

There are two models and they disagree about something visible.

If the fold behaves like the pile — every link joining the stationary branch has its momentum destroyed — then the falling branch is in free fall and the free end accelerates at exactly $g$ the whole way.

If instead the fold dissipates nothing, energy conservation gives $\dot{x}^2 = gx(2L-x)/(L-x)$, and differentiating that gives an acceleration that starts at $g$, passes $2g$ at 0.423 of the way down, and diverges at the end. The mechanism is not mysterious: the moving branch is getting shorter, so the same released energy is carried by less and less matter.

Measurements settle it. The free end of a real folded chain arrives early and is observed accelerating past $g$, which puts the energy-conserving fold in the lead. It is one of the few dissipation questions in mechanics that a stopwatch decides.

The divergence at the end is not to be believed, and the reason is worth stating: the model assumes the moving branch is straight and moves as one piece, which stops being true when only a few links are left. What the calculation establishes is the sign of the effect and its size in the middle of the fall, and it hands the last five per cent back to the experiment.

What decides which model #

The two accounts differ in whether the fold is a place where energy goes. That is not a question about mechanics; it is a question about the chain.

A chain of rigid links joined by loose pins is full of small impacts, and each one is inelastic. A smooth flexible cord has no such joints. The dissipative model is the honest description of the first and the energy-conserving one of the second, and a real chain sits somewhere between — which is why the experimental answer is “more than $g$” rather than “the divergence”.

The two models differ in exactly the way two equal masses colliding differ from two equal masses meeting on a spring. A perfectly inelastic collision between equal masses loses half the kinetic energy — a fraction fixed by the mass ratio and by nothing else, not by how hard the impact was or how long it took — and the fold in a link chain is that collision repeated once per link. An energy-conserving fold is the same encounter with the spring, in which the same momentum is transferred and nothing is lost. The chain is not choosing between two mechanics; it is reporting which of the two encounters its links are having.

The general rule that comes out of the pair is this: wherever matter joins or leaves a moving part, ask what velocity it has on each side of the boundary, and the energy ledger follows. The chain landing goes from moving to still, and everything is lost. The chain being lifted goes from still to moving, and half is lost. The rocket’s exhaust goes from moving with the rocket to moving backwards, and the loss depends on the exhaust speed. Same equation, three answers.

Where the same arithmetic turns up #

A hose has to be held. A jet of water leaving at speed $v$ carries momentum away at $\rho A v^2$, and that is the force on whoever is holding the nozzle. It is the same $\lambda v^2$ as the chain’s, with $\lambda = \rho A$, and it is why a fire hose is a two-person job.

A conveyor belt loading sand costs more power than the sand’s weight suggests. Material dropped onto a moving belt has to be accelerated to the belt’s speed, and half the energy handed to it is lost in the slipping — the lifting result, with the direction rotated.

Rain increases the weight of a moving vehicle by more than the weight of the water. Drops arriving with a horizontal velocity relative to the vehicle deliver a flux as well as a mass, and the effect is measurable on a train.

A pile of granular material poured onto a floor loads it the same way, and the excess disappears the moment the pouring stops — which is a small part of why what a silo’s floor feels has almost nothing to do with what the silo holds.

And a falling chimney breaks in mid-air. The rigid analogue of the folded chain: a rod pivoted at its base has a tangential acceleration at the tip exceeding $g$ once it has fallen far enough, so the top of a toppling chimney is falling faster than the free fall it would follow if broken, and the compression that requires eventually snaps it.

What the whole family shares is a division of energy made at a boundary rather than within a body. A ball rolling down a slope divides its energy between going and spinning, and the split is fixed by the shape rather than by the slope: a hoop arrives slowest because the largest share of what it was given had to go into rotation. The chain’s problem is the same accounting with the division made at the fold — what matters is where the energy has to go, and how much of it has nowhere to go but heat.

The measurement nobody quite believes #

The factor of three is easy to state and surprisingly hard to see, and the reason is instructive.

A kitchen scale measures force by letting something deflect, and the deflection takes time. Dropping a metre of chain takes about 0.45 seconds, during which the reading is supposed to sweep from zero to three times the chain’s weight and then fall to one; a scale whose own response time is a tenth of a second turns that sweep into a smeared hump with a peak well below three. What is being measured is the convolution of the physics with the instrument, and the instrument wins.

The way round it is to measure the impulse rather than the force. The integral of the reading over the whole fall does not care about the instrument’s response time, provided the instrument is linear and settles eventually, because a convolution preserves an integral. That is a general and rather useful trick: when the shape of a fast signal is beyond an instrument, its area often is not. It is the same reason a slow detector can still count the photons in a pulse it cannot resolve — the total is recoverable when the profile is not.

The same chain, over an edge #

There is a variant of this problem that looks like the same apparatus and is not, and the pair of them together is the sharpest statement of what the momentum flux is doing.

Arrangement one: a chain lies heaped at the edge of a table with one end hanging over. The hanging part pulls, links are drawn out of the heap, and each is set moving from rest. Writing the momentum balance for the moving part — rate of change of $\lambda x \dot x$ equal to the weight $\lambda g x$ — and looking for a constant acceleration gives

$$\ddot x = \frac{g}{3}.$$

A third of gravity, at every stage, whatever the chain and whatever the length already hanging.

Arrangement two: the same chain laid out straight on a frictionless table, running over a smooth peg at the edge. Now nothing is picked up from rest — every link is already moving, and the whole chain is one body of fixed mass. The equation is $\ddot x = gx/L$, and the motion grows exponentially rather than uniformly.

Two chains, two tables, one difference: whether the moving part is gaining members. The first arrives at the floor measurably slower than the second, and the shortfall is energy destroyed at the heap as each link is jerked into motion. It is the lifting result of the previous section, running downhill instead of up.

The distinction is easy to lose in practice, which is why the problem has a long history of published wrong answers. The tell is always the same question: is the matter that joins the moving part already moving? If it is, this is ordinary mechanics; if it is not, there is a flux term and there is a loss.

Weighing a stream by what it destroys #

The scale under the falling chain is doing something an instrument can be built out of, and it is.

Pour a steady stream of sand onto a plate at rate $\dot m$, having fallen a height $h$. The grains arrive at $\sqrt{2gh}$ and leave at rest, so the plate carries the weight of whatever is resting on it plus $\dot m\sqrt{2gh}$ — a force proportional to the mass flow rate, with a coefficient the drop height fixes. Mount the plate on a load cell, subtract the standing weight, and the reading is a flow meter with no moving parts, no pipe to obstruct and nothing to wear out. Impact flow meters of that kind are standard on grain, cement and mineral conveyors.

A falling stream of water behaves differently in an instructive way. Sand grains keep their spacing, so the mass arriving per second is the mass leaving the nozzle per second. A liquid stream accelerates as it falls and therefore thins, by continuity, so the same mass per second arrives through a narrower cross-section at a higher speed. The flux is $\dot m v$ either way — the momentum arriving per second — and the thinning is a statement about the stream’s shape rather than about the force.

Which is the general lesson of the whole page, stated once more in the way that transfers. What a surface feels from something arriving is the momentum it destroys per second, and getting it right is a matter of knowing the speed on arrival and the mass rate, and nothing else about the material at all.

What the picture cannot show #

The discontinuity is not real. A real chain has links of finite size, so the flux stops over the time one link takes to land rather than instantaneously, and a real scale has a response time of its own. What is measured is a peak of about three times the weight followed by a settling transient, and the vertical drop in the figure is the limit of an idealisation rather than an observation.

The links are treated as arriving perfectly inelastically. A link that bounces off the pile takes some momentum back upward, which reduces the flux, and a chain dropped onto a hard surface does bounce a little. The factor of three is an upper bound for a real chain and a good one for a chain landing on a heap of itself.

The chain is treated as perfectly flexible and inextensible. Real chains stretch, and the tension wave that runs up a chain when its end is caught travels at a finite speed; nothing in the model above knows about that wave. For a light chain landing gently it does not matter, and for a heavy cable arrested suddenly it is the whole problem.

And the folded-chain result is model-dependent in exactly the way the figure says. Two defensible accounts of one fold give accelerations differing by a factor that grows without limit near the end. The experiment picks one of them for real chains; it does not follow that the same choice is right for a rope, a cable or a strip of tape, and the honest position is that the fold has to be characterised rather than assumed.

The ladder from here #

Later rungs on this anchor: the general variable-mass equation and the cases where the departing matter’s velocity is the whole answer; the falling chimney worked out properly, with the bending moment along its length and the height at which it breaks; the water-rocket and the hose reaction as one problem; the chain fountain, where a chain lifted out of a beaker rises above the rim and the explanation needs the beaker to push back; and the tension wave along a struck cable, which is where the inextensible assumption is finally paid for.

The neighbouring ladders are collisions, which is the fixed-mass version of every argument here, and the rocket, which is the same boundary bookkeeping with the matter going the other way. The centre of mass is what the whole of it is really about: it never stops falling at $g$, whatever the individual links are doing.