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The most basic form of my question is: "Can the normal force from the ground be enough to let a chain's upper end fall at constant velocity?" I suspect the answer is no, but to elaborate, here is the actual question:

The upper end of a uniform chain of linear mass density $\mu$ is lowered with a speed $v$. The reaction force offered by the ground after the chain "after the chain falls through a distance $y$" is?

(In the diagram, a part of the chain is shown to already rest on the ground. Unfortunately, the quantity $y$ has not been marked in it.)

Since the original length is not given, i suspect the paper setter meant "when a length $y$ of the chain is yet to fall."

With this assumption, let us consider a force $F$ (variable if needed), that is applied on the upper end. Also, i assume that the speed of all the parts of chain still in motion is $v$, and thus (is this a correct conclusion?) the tension throughout the chain in motion is zero. Now considering the part of the chain in motion, for its equilibrium, the normal force $\mathcal N$ and the force $F$ together must balance the weight of the chain in motion, thus giving: $$F + \mathcal N = \mu gy $$ Also, considering a small element $\mathrm d y$ coming to rest in time $\mathrm d t$ is only due to the normal force $\mathcal N$, thus giving: $$\mu v \, \mathrm dy = \mathcal N \mathrm dt$$ $$\mathcal N = \mu v^2$$ But they give the answer as $\mathcal N = \mu (v^2 + gy)$. The only way i could get it by assuming that the normal force by the ground also balanced the weight of the remaining chain. Is this true? If not, is my previous reasoning (and my answer) correct? If not, where am i going wrong?

Edit: After reading George Smyridis's answer, i still have a doubt. How can the normal force alone be responsible for the chain falling with constant speed? If it were so, what makes it different from an ordinary chain falling on the ground? Shouldn't there be an extra force $\vec F$ acting on it (as shown in the following diagram), that acts on the chain, so that it falls with constant velocity?

chain image

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    $\begingroup$ A chain cannot transfer any compressive forces, so $\mathtt{N}$ is felt by one link only. Is this correct? But it is more like a rope than a chain in this situation, so where is $\mathtt{N}$ applied on? $\endgroup$ – ja72 Mar 8 '16 at 19:55
  • $\begingroup$ Well, the question asks for the reaction force by the ground, which is meant to be the $\mathcal N$. So it must be applied on the bottom end of the chain only, i suppose? And if this is true, then it must be the same in real life too, right? About transfer of compressive forces, i am of the opinion that the chain wouldn't transfer any compressive force, and should be treated as a rope. $\endgroup$ – FreezingFire Mar 9 '16 at 3:43
  • $\begingroup$ I had referred the previous comment to @ja72. $\endgroup$ – FreezingFire Mar 9 '16 at 4:16
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The chain can be considered a system with variable mass, and the mass of the parts that touch the ground is the mass that it loses. The general equation of variable mass motion is written as:

$$ \vec F_{external} + \vec v_{rel} \frac{dm}{dt}=m \frac{d \vec v}{dt}$$

Since the velocity of the chain is constant:

$$\frac{d \vec v}{dt}=0$$

Therefore,

$$\vec F_{external}=- \vec v_{rel} \frac{dm}{dt}$$

The chain is falling down and the part of the chain that falls on the ground stays still, therefore:

$$\vec v_{rel}=-\vec v$$

where $\vec v $ is the velocity of the chain.

Now since the chain is uniform, its mass can be calculated by

$$m=μy$$

The rate at which the chain loses its mass is:

$$\frac{dm}{dt}=μ \frac{dy}{dt}=μv$$

The force applied on the chain by the ground is not the only external force acting on it, gravitational force also acts on it. Therefore:

$$\vec F_{external}=\vec N + \vec W$$

As a result: $$N= μv^2+W \Rightarrow N=μ(v^2+gy)$$

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  • $\begingroup$ But one fundamental doubt remains unclear, how can the normal force by the ground be sufficient for the chain to fall with constant speed? Shouldn't there be another external force $\vec F$, that should be applied on the upper end of the chain (so that $\vec{F}_{external} = \vec N + \vec W + \vec F$) ? Otherwise what difference is there between a chain dropped simply, and one falling with a constant speed (in terms of acting on it)? $\endgroup$ – FreezingFire Mar 8 '16 at 4:41
  • $\begingroup$ * I meant "in terms of force acting on it". $\endgroup$ – FreezingFire Mar 9 '16 at 17:15
  • $\begingroup$ There is a quantity vdm/dt in the equation that has dimensions of force and it occurs from the change of mass but it is not external force. I think this is the force that you are looking for. $\endgroup$ – George Smyridis Mar 10 '16 at 0:00
  • $\begingroup$ That term equals to $\mu v^2$. When i solved a similar problem (I.E. Irodov problem 1.157) involving a chain dropped freely with its lower end just touching the ground, the normal force exerted by the falling part of the chain turned out to be $\mu v^2$ (there $v$ was variable and increased due to gravity). Now i feel that the extra force $\vec F$ would be needed only to balance the weight of the chain, and the normal force by the ground $\vec N$ would balance the $\mu v^2$ part! $\endgroup$ – FreezingFire Mar 10 '16 at 3:49

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