# Heavy chain moving upward at constant velocity

1. Nov 6, 2015

### AHashemi

1. The problem statement, all variables and given/known data
We have a heavy chain with length of L and weight of m placed on a table. we take an end of it and move it with constant velocity of v upwards. so each moment it gets heavier. If the force needed for this uniform speed is called F, and length of chain which is higher than the table is called y, write F as a function of y.

2. Relevant equations
F=ma
y=vt​

3. The attempt at a solution
equation of chain's weight at each moment is
$$M=ym/L=vtm/L$$
and delta M is equal to
$$\delta M=vm/L$$
But I can't calculate the force needed to make this motion with uniform speed.

2. Nov 6, 2015

### stockzahn

In the first instant, when all the chain is lying on the table, what is the necessary force to accelerate to a certain velocity (assuming your arm has no mass)?
After then what is the necessary condition to keep the velocity constant?

3. Nov 6, 2015

### AHashemi

When it's all on the table so it has no weight to resist upward movement. chain gets heavier each moment so we need more force (F) to deal with the added weight (Mg). so we have an acceleration here (because F=Ma) but I can't figure out how it increases.
$$F-Mg=Ma$$

4. Nov 6, 2015

### stockzahn

In your first post you've already shown the connectedness of the weight of the chain as function of the time, hence the necessary force to overcome the gravitation.

If the velocity is constant also the mass of the chain which is lifted of the table (vs. time) is constant → constant necessary force depending on the velocity.

5. Nov 6, 2015

### AHashemi

But I need to remove the time parameter here and describe F as a function of y (the length of chain not lying on the table)
I've got a bit confused. need more explanation.

6. Nov 6, 2015

### stockzahn

You need the velocity, which is constant and therefore no function of the time (v ≠ f(t)). If the v is constant, then you can substitute: t = y / v. So in the end you would have a function F(y) = Fa + Fg(y) instead of F(t) = Fa + Fg(t), but you've derived the equation for the time-dependent part of the force, so you can substitute the time already. What's missing is the constant (and therefore time-independent) part of the acceleration of the chain leaving the table, so no "t" in your formula. However, I hope I don't misread the statement.

7. Nov 6, 2015

### J Hann

Newton's second law says F = dP/dt
In this case F = m dv/dt + v dm /dt
For this problem dv / dt = 0
So you have F = v dm / dt
Also dm is just p * dL where p is the density of the chain
The rest of the solution should follow.

8. Nov 6, 2015

### haruspex

That works, but I would add a word of caution.
The formulation F = m dv/dt + v dm /dt can be misleading. Strictly speaking, it implies mass being created out of nothing. In reality, the mass must be entering the system (the system referred to by the P of dP/dt) from somewhere else, and it does so with some initial velocity. Thus, this formulation only works if the added mass is known to have zero initial velocity in the reference frame. In this case, it is.