# Uniform chain rising with constant velocity

1. Nov 14, 2016

### Dustgil

1. The problem statement, all variables and given/known data
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 that mus be 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 instant.

2. Relevant equations
Newton's second comes in handy here.

3. The attempt at a solution

$$\frac {dp} {dt} = T - Mg$$

dp = p(t+dt) - p(t)
= (M+dM)v + dm(v-u) - MV ,where v - u gives the absolute speed of the grounded part of the chain (= 0)
= dMV

so

$$\frac {dM} {dt} v = T - Mg$$

$$\frac {dM} {dz} \frac {dz} {dt}v = T - Mg$$
$$\frac {dM} {dz} v^{2}=T-Mg$$

separating variables and integrates gives me

$$-\frac {1} {g} ln(1 - \frac {Mg} {T}) = \frac {z} {v^2}$$

and here's where i get stuck. I tried doing a log expansion, but that didn't pull out the correct expression. Any help?

2. Nov 14, 2016

### ehild

It is correct so far.
Note that T is not constant, so you integral is wrong. You have to determine T, and you can get it from the equation in red. If the linear density of the uniform chain is λ, M=λz. What is T in terms of z?
Finding the equivalent length which has the same weight that T, you have to divide T with gλ.

3. Nov 15, 2016

### Dustgil

okay, thanks i've got it now.

$$T = Mg + \frac {dM} {dz} v^{2}$$

Since

$$M = \lambda z$$
$$\frac {dM} {dz} = \lambda$$

So

$$T = \lambda z g + \lambda v^{2}$$
$$\frac {T} {\lambda g} = z + \frac {v^{2}} {g}$$

4. Nov 15, 2016

Good work!