A chain with tiny links falls onto a scale.Angels55 - 6 Answers
Ok, this another attempt at the answer. The solution is similar to how the force of a jet hitting a plate at right angles is worked out.
First define some variables:
M is the mass of the chain
L is the length of the chain
l is the distance of a point on the chain from the scale on which
it falls. This ranges from 0 to L
Ftotal is the total force on the scale
v is the velocity of the falling chain at a point which is at the distance l from the scale
Mr is the mass flow rate of the chain
W is the weight of the chain
F is the force due to the change in momentum of the chain as it falls
onto the scale
The total force is equal to the static weight of the chain W plus the force due to the change in momentum of the chain
Ftotal = W + F
W = Mg
The force due to the change in momentum of the chain is
F = Mrv
From the equations of motion, the velocity v of any point on the falling chain at a distance l from the scale, at the moment of impact after it has fallen by a distance l = square root (2gl)
Mass flow rate Mr = M/L x velocity of chain = M/L(square root(2gl))
So F = Mrv = M/L (square root (2gl)) x square root(2gl) = 2Mgl/L
This is a maximum as the last piece of chain hits the scale and l = L
So F = 2MgL/L = 2Mg
So Ftotal = W + F = Mg + 2Mg = 3Mg or 3W
Q: X days = a few days ? ?...solve or X
(hint X appears to be greater than 33!)
I only ask because I own 2 antique wooden scales that use pure gold chains to use as a weighing mechanism. Enclosed in beautifully preserved wooden/glass (and the other in a silver metal/glass case) these elaborate scales weigh out to 7 and 9 decimal places! (wooden =9) and I have confirmed the accuracy before carefully packing them up very well, by measuring the difference between my fingerprint and an ex-girlfriend?s where using standard imprint techniques, and only the grease from out fingers, they consistently showed a noticeable difference with considerable precision! The scales easily predate my father!
Either way....I would love to ponder the question for days, but I don't have the time at the moment, but the answer would be very interesting to learn.
I could easily outsmart the question and say it is a trick where a vital component is lacking from the original assumptions made by everyone in the first place. the term Mass is used rather than weight. unless you have a magic scale that can account for gravity automatically, the scale would measure the standard mass differently on the moon, or under water, or even in more nitrogen rich air than normal. The mass would be the same for chain regardless of location or gravity, but the scale would be operating under some "unknown" set of operating conditions compered to what it was made to operate under. (using weight doesn't help either, actually makes matters worse!)
I would imagine that the 3x reading would indicate that this is not the solution. So I will nullify that concept. When the chain was released and the scale "felt " the full weight of force of gravity pushing against it (since all four of the fundamental forces push, not pull) The weight would be effectively zero when the scale is pressing up with equal force as the chain is being pushed down by gravity. Then the effect of inertia would start to become counteracted by the opposing forces of the 1 unit of weight chain and the virtually "limitless" upward force of the scale's weighing mechanism whatever it might be. (it's considered limitless just for ease, if you chose a scale that couldn?t weight the chain properly, that would be an evil thing to do! =)
So at the zero equalization point, the inertia of the chain would depress the scale the amount equal to it's own mass (as the inertia is a direct result of the mass of an object an it speed,which we are negating here)
This inertia and scale mechanism would finally completely zero-point when this point is reached which is equal to the actual weight of the chain, the additional weight of the chain depressing the scale innards using gravity, and then the inertia of the chain falling pressing against he scale even further.
At this point there is no more inertia present in the chain's downward vector component and like a ball thrown in the air, the end-point or peak is reached for only a brief moment, before being overcome by the opposing forces (the scales internal mechanism - spring, or whatever) when an final equilibrium is reached the scale is calibrated to show the mass of the item on it, which by definition in the given question is one unit.
So the final point is M and the maximum depression is 3M.
This cannot be true, because if the mass was small, say M=1 ug or something trivial like that. then the length was 100km the time it takes for the chain to fall would create a serious problems with the effect of the inertia. (although it is being experienced one unit of length at a time, and the overall summation of depression would equal out to he same as the maximum depression described before, it would be spread out over time, creating a difficult state(s) to explain in such simplistic terms.
(that was my "not really thinking it through until my fingers were actually typing" attempt, and I wasn?t even going to try....remember, so does it hold any merit as a on the spot thought experiment?)
Since the chain is uniform and of length L, the lower part of the chain has no energy since it has zero distance from the scales. The upper point of the chain is a distance of L from the scale. Assuming SI units, the potential energy of the chain averages out at 1/2 MgL joules (if the chain was concentrated at a distance L from the scales, the energy would be MgL).
The way i would tackle this would be to equate the potential energy of the chain to the energy stored in the spring. I don't know whether that is correct, but anyway here goes!
If the chain falls and all the energy is stored in the spring then
Energy in spring = 1/2kx^2 where x is the displacement of the spring and k is the stiffness.
Equating potential energy to spring energy gives
1/2MgL = 1/2kx^2
x = square root( MgL/k)
When everything is in equilibrium the weight of the chain creates a displacement d so
Mg = kd or d = Mg/k
So the ratio of the initial reading to the equilibrium reading would be x/d which works out at square root(Lk/Mg)
The above answer is obviously incomplete although on the right track. The correct answer is that as the last link of the chain falls on the scale (which might be a spring scale or some other kind, it is assumed to be fairly sensitive) the reading will be three times the weight of the chain, 3Mg. This result is independent of the scale. When the scale becomes stable, the reading will be Mg as Mebournian has stated.
I will give you a few more days to come up with an answer describing what is happening, then will give the answer myself.
Ok let's get down to this:
Initially the scale will read zero, assuming that the just touching is negligible.
As the chain falls onto the scale, the reading will increase, and will reach maximum right as the last piece of chain hits it. At this point, it will show the mass M + a value determined by the spring constant of the scale's spring. When stable, the scale will show M.
That is OK for a simple weight falling on a spring scale, which is not actually the question. Try again. the answer I gave is correct, and there is no L or k in it.
This is not a trivial problem to solve, although easy to state. It took me a long time to figure it out when I first came across it some 40 years ago in a physics textbook on classical mechanics (when I was eleven years old:)).