A simple, perhaps unsolvable, question: lifting a chain?

Question

This was asked on here and it has me stumped:

Imagine a long chain, linear mass density ρ, that you pick up with constant force F.

How many meters of chain are off the ground in 5 seconds?

I started with:

F - gρL = ρL(dv/dt) where L is the length of suspended chain.

Multiplied by 1/v = dt/dL to get

(F/L - ρg)dL = ρvdv and integrated to

Fln(L/L₀) - ρg(L - L₀) = ρ/2(v^2 - v₀^2)

Problem is, this doesn't integrate easily to find L in terms of t.  Moreover, it has some nasty behavior that I hoped integrating would smooth out.  For instance, it blows up initially, if there is no initial chain length.  This is due to an infinite acceleration at the beginning, since the suspended mass is 0.

Help?

oldprof · Answer

We have S = 1/2 aT^2; where a = 2F/(dm/dT * T) = 2F/(rho dL/dT * T) is the average acceleration over T.  dL/dT = v = aT is the average speed of hoist.  So a = 2F/(rho aT^2) and a^2 = 2F/(rho T^2) so that a = sqrt(2F/rho)/T

And now we have S = 1/2 aT^2 = 1/2 sqrt(2F/rho)T.

We check the units.  F ~ kg.m/s^2 and rho = kg/m so sqrt ~ sqrt((m/s)^2) ~ m/s a speed metric.  And then S ~ m/s * s ~ m.  Ta da.  The units are consistent with distance, the height pulled up in T = 5 seconds.

We check the physics.  Clearly we get more height over T when we apply more F.  The derivation shows that.  And further we get less height for a fixed F when the density of the chain increases.  Again, this is shown.

So for the made assumptions, S = 1/2 sqrt(2F/rho)T is probably correct.

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A simple, perhaps unsolvable, question: lifting a chain?

1 Answer