M040 · Weight plus arriving momentum

Falling Chain on a Scale

Why can the pan force exceed the weight already on it? The pan must also stop incoming downward momentum. Multiply the mass arriving each second by its incoming speed, then add the deposited weight. For an ideal continuous chain released from rest with its lower end touching the pan, the arrival force is three deposited weights; after the last link, only the whole-chain weight remains.

Drop a continuous chain onto a fixed pan. Compare deposited weight, arriving mass flow, stopping momentum flux and the force drop after the last link.

Deposited weight $5.19\,\mathrm{N}$

Arriving mass per second $2.5\,\mathrm{kg\,s^{-1}}$

Stopping momentum flux $10.4\,\mathrm{N}$

Total pan force $15.6\,\mathrm{N}$

Model assumptions

A uniform continuous flexible chain falls vertically under uniform gravity. Its lower end starts at the chosen gap above a fixed receiving plane, with an optional downward initial speed. Incoming material stops in a perfectly inelastic capture; the collected part rests at zero height. No upstream bending tension, pile height, rebound, finite-link collision or detector response is modeled. The threefold force relation requires release from rest with the lower end initially touching the pan. A gap or initial speed adds arriving momentum and changes that ratio. Real chains can develop a bottom tension and fall faster than free fall, as Hamm and Géminard observed; that effect is outside this ideal model. Link sizes, support hardware and stored-link positions are schematic; the stored layout is not a calculated coiling trajectory. The support arrow is translated beside the pan for visibility and does not specify a torque. Ideal force jumps are retained at the start and end of arrival. The complete trial repeats in eight seconds; the clock and curves use physical SI time.

Start with the essentials

Focus question: Why can the reading exceed the weight already on the pan?
One-sentence intuition: Multiply the mass arriving each second by the speed it loses at capture, then add the deposited weight.

Core mathematical model

Follow the arriving material

m_d=\lambda d,\qquad W_d=\lambda gd,\qquad q_{\rm in}=\lambda v

The chain is uniform. Here speed is positive downward, while signed velocities in the exported momentum balance are positive upward.

Weight plus stopping force

N=W_d+q_{\rm in}v=W_d+\lambda v^2

This applies while material reaches the fixed pan and immediately comes to rest, with no upstream bending tension. It is the pan force, not a second weight.

The special threefold case

H=v_0=0:\quad v^2=2gd,\quad N=3W_d;\qquad t\ge t_{\rm end}:\quad N=\lambda Lg

The ideal force drops after the final material arrives. A nonzero gap or initial speed adds incoming kinetic energy, so the threefold relation no longer applies.

Check the whole chain as well

N-\lambda Lg=\frac{d}{dt}\left[-\lambda(L-d)v\right]

Only the uncollected part moves. This closed-system check agrees with the deposited-part control volume, which has a mass inflow and zero stored momentum.

Common difficulties

An apparent weight is a support force

Typical misconceptionThe chain suddenly has three times as much gravitational mass.

Better mental modelMass is unchanged. The additional support force stops downward momentum; it vanishes when nothing more arrives.

A real chain can pull its upper part down

Typical misconceptionEvery physical chain must follow the threefold curve exactly.

Better mental modelA finite minimum bending radius can create bottom tension and a faster-than-free-fall motion. This Lab isolates the freely flexible, zero-tension ideal limit.

Run the experiment

01
Watch the final link arrive
Start with Three deposited weights, then replay through the final arrival.
What to observe: The arrival term ends abruptly and the readout drops to the whole-chain static weight.
02
Add a gap below the chain
Drag the release clamp upward, then replay the new trial.
What to observe: Material reaches the pan with speed gained before any chain has been deposited. The peak exceeds the zero-gap limit.
03
Follow the energy loss
Open Where the mechanical energy goes in the data row.
What to observe: Remaining kinetic and gravitational energy decrease as completely inelastic capture removes mechanical energy. Their sum with the capture loss stays fixed.