# Homework: Rope falling on scale [closed]

I think I have a solution for the following question, but it seems suspiciously simple to me: Suppose we have fixed a rope of Length $L$ and Mass $M$ at the ceiling such that the tip of the rope just touches a scale positioned vertically underneath the rope's suspension point. Let's suppose we now cut the rope. The question is what the scale will show at the exact moment when a length of $x \leq L$ is resting on it.

My thoughts were that the gravitational force will push the portion of the rope that is resting on the scale downwards, which is what the scale will measure, while the gravitational force acting on the portion of the rope still falling down will only accelerate its fall. So the solution would be ($w$ stands for the weight the scale measures)

$$w = x \frac{M}{L}g$$

Am I missing something here?

EDIT: Apparently, I forgot to take the impact of the falling rope into account. In a time interval $dt$, a mass $dm$ will fall on the scale with momentum $dp = dm \cdot v$. If we have a length of $x$ resting on the scale, the segment of the rope impacting the scale at this instant will have been hung up at distance $L-x$ from the ceiling and will have moved to a distance $L$ at the time of impact, so conservation of energy gives $$\frac{dm v^2}{2} = dmgx \Rightarrow v = \sqrt{2gx}$$

Also, we have $dm = dx \cdot \frac{M}{L}$ which yields

$$F_{impact} = \frac{dm}{dt}\cdot v = \frac{dx}{dt} \cdot \frac{M}{L} \cdot v = \frac{M v^2}{L}$$

Altogether, we get

$$w = x \frac{M}{L}g + M\frac{v^2}{L} = 3\frac{Mg}{L}x$$

## closed as off-topic by ACuriousMind♦, John Duffield, Kyle Kanos, John Rennie, user36790 Oct 23 '15 at 13:10

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• This derivation doesn't take into account settling after the rope has finished falling. Yes it should be $Mg$ after the rope has come to rest, but at the last instant when the last piece of the rope hits, it is still impacting the scale, and thus has an impact beyond $Mg$ – tmwilson26 Oct 22 '15 at 19:24
• There is also a problem in this derivation. Your conservation of energy equation assumes that the rope has fallen from a height $L$ at all times, which is not the case. Parts of the rope that impact earlier fall from a lower height than $L$ – tmwilson26 Oct 22 '15 at 19:29
• The conservation of energy term is still a bit off. Think about this: If there is no rope on the scale, $x=0$, yet your velocity of the part of the rope $dm$ is high. Also, when all of the rope is finally on the scale, you'd expect the last bit of the rope to have a high velocity, yet your equation gives a velocity of $v=0$ – tmwilson26 Oct 22 '15 at 19:49

What you have there is the weight from just the rope that is laying on the scale at that instant. You also need to consider the force of the falling rope impacting the scale, which will have force $F = \frac{dp}{dt}$ associated with it.
In a time $dt$ a certain amount of mass $dM$ will fall on the scale at velocity $v$, where $v$ can be computed based on how far that particular piece of rope has fallen by that time.