Pulling up a uniform chain with constant velocity

Question

A uniform chain lies in a heap on a table. If one end is raised vertically with uniform velocity $v$, show that the upward force exerted on the end of the chain is equal to the weight of a length $z + v^2/g$ of the chain, where $z$ is the length that has been uncoiled at any given instant.

Let $\lambda$ be the linear density of the chain. The standard solution applies Newton's Second Law ($F = \mathrm{d}p/\mathrm{d}t$) to the uncoiled part to obtain: $$F - \lambda z g = \frac{\mathrm{d}p}{\mathrm{d}t} = v \frac{\mathrm{d}m}{\mathrm{d}t} = v \cdot \lambda v = \lambda v^2$$ hence $$F = \lambda g \left( z + \frac{v^2}{g} \right).$$

However, this approach seemed less intuitive to me, so I attempted to use the work-energy theorem. Specifically, I reasoned that if the end of the chain rises by $\mathrm{d}z$, the work done by the force $F$ should be $F\mathrm{d}z$, which should equal the change in mechanical energy of the entire chain. The mechanical energy (gravitational potential energy and kinetic energy) of the chain is given by: $$E = \lambda z g \frac{z}{2} + \frac{1}{2} \lambda z v^2.$$ Taking the differential of $E$ yields: $$F\mathrm{d}z = \lambda g z \mathrm{d}z + \frac{1}{2} \lambda v^2 \mathrm{d}z.$$ From this, I arrive at: $$F = \lambda g \left( z + \frac{v^2}{2g} \right).$$

Could someone please point out where my reasoning went astray? Any insights would be greatly appreciated.

Answer 1

I think the lesson here is that conservation of momentum is less likely to lead you astray than conservation of energy.

Since the work done by the force $F$ is less than the gain in energy of the chain, some energy must be "lost" to the environment. Basically, lifting the chain is more like an inelastic collision than an elastic collision.

Answer 2

Chains on tables are notoriously tricky because as a link rises of the table the force on the link is the upward pull from the chain above it plus the force exerted by the table on the part of the link that still touches it as it rotates upwards. Consider the example of the chain fountain for example. Because of this I'd trust your work energy calculation more than the momentum calculation. The force from the table complicates the momentum calculation, but, as it does no work, it does not complicate the work-energy theorem.

There are similar problems with chain held at both ends, and then one is realeased. The traditional momentum-based textbook solution is not consistent with the experimental results which show that energy is conserved. For a paywall-free discussion of this latter problem (with references) see my online notes

Answer 3

I think I've found the answer to your conundrum in this question. Please read it, but the upshot is that for a variable mass system, the usual KE expression $\frac{1}{2}mv^2$ is apparently incorrect. I'll try to summarize the issue in a way more relevant to your problem. We know that the change in KE is the work done by a force: $$ \Delta\text{KE} = \int \frac{\mathrm{d}p}{\mathrm{d}t} \mathrm{d}z $$ In our problem however $\mathrm{d}p/\mathrm{d}t = v \mathrm{d}m/\mathrm{d}t $ because speed is constant. This gives: \begin{align*} \Delta\text{KE} &= \int \frac{\mathrm{d}p}{\mathrm{d}t} \mathrm{d}z \\&= v\int\frac{\mathrm{d}m}{\mathrm{d}t}dz \\&= v\int\frac{\mathrm{d}m}{\mathrm{d}t}\frac{\mathrm{d}z}{\mathrm{d}t}\mathrm{d}t\\&=v^2\int \mathrm{d}m \end{align*} Which for the case of your problem, we can evaluate from $m=0$ to $m=\lambda z$ to give: $$ \Delta\text{KE} = \lambda z v^2 $$

It seems like it accounts for the missing factor of two in your second method. Of course, I hope I'm not missing something important here, please tell me if I am :)

Edit: some supplementary observations

I was interested to find out whether any published papers solved this problem by starting from energetic considerations, and also whether any experimental tests were done to compare the results with the standard solution.

Sadly, I didn't find a lot of information. However, I was sidetracked for a while by a very similar problem I came across, which seems to be a kind of "forefather" to the one in this question. It is known as "Buquoy's Problem", named after the mathematician Jiří Buquoy.

The difference in Buquoy's version is that the condition of uniformity is on the upward force applied to the chain, and not on the resulting upward speed:

Consider an ideally flexible fibre lying reeled on a horizontal plane. Determine its motion when a constant vertical force (directed upward) is exerted on the end of the fibre.

This makes the problem more involved, because it implies we will have a non-uniform motion. We will have $\ddot{z}=\dot{v}$ non vanishing, which means both terms from $\dot{p}$ don't vanish. For one thing, in that version of the problem the chain doesn't even necessarily move up. Even if it does, it will in general attain some maximal height and then fall back down once it reaches some weight that equals the magnitude of the constant force.

For what it's worth, there is one small observation I gleaned from looking at analyses of Buquoy's problem, such as this one (Šíma, 2005): they all seem to start from application of the modern form of Newton's 2nd law, $F=dp/dt$, rather than from energy considerations!

To summarize, my little research to find alternative approaches to this problem didn't amount to much, but I still hope that this little addendum will be of interest.

Answer 4

The chain as a whole is submitted to three external forces, the only forces to be taken in account in Newton's Second Law :

1. The force F by which it is lifted
2. Its weight $\lambda Lg$
3. The vertical reaction $R$ of the table which support it

The part of the chain lying on the table has no movement and is submitted to $R$ and its own weight, that is $\lambda(L-z)g$. Those two forces compensate one another so that Newton's Law leads indeed to your first equation $$\lambda v^2 = F - \lambda g z.$$ If you use the energy method, you have to take in the internal forces which provide an internal non conservative work so that $$\frac{\mathrm{d}E}{\mathrm{d}t}=Fv+P_{\mathrm{int}}$$ as to obtain $P_{\mathrm{int}}=-\frac12 \lambda v^3$.