# Newtonian mechanics question

I was given this problem to solve (first year maths undergraduate) in an introductory mechanics course, but I got the wrong answer:

A flexible homogeneous rope, whose length is $L$ and total mass is $m$, lays on the ground. At some moment you start to vertically lift the rope by one of its edges at a constant speed $v$. Calculate the force you need to apply and the power you need to invest as a function of the height of the rope’s edge from the ground.

My initial guess was to consider $F= m\times g$ (as acceleration would be $0$), and then just compute the power as an integral of $F\times v$ with respect to time, but this is wrong. The solution proceeds by choosing a very specific set of particles into a system (the length of the vertical rope + a length $=$ $v\times \mathrm{d}t$ on the floor), calculates the momentum at $t$ and $t+\mathrm{d}t$ and then take the derivative.

I reckon that my problem has something to do with not being rigorous about what ''system'' of points I choose, but i don t really get it. Any hints:D? Thank you

• I'm a first year student too. If I have time later I'll try this problem out because it seems interesting. My hunch is to begin by inspecting an infinitesimal portion of rope much like is done here: math.ubc.ca/~feldman/m256/wave.pdf. From an analysis of this sort you can figure out the net downwards force on the rope at any moment of time assuming constant force is applied. – theideasmith Mar 15 '17 at 1:00
• The force you have to apply is not constant, since there is a portion of the rope in the air, but also a portion on the floor. The latter does not does not contribute to the force now, but it will contribute later when parts of it are lifted off the floor. – NickD Mar 15 '17 at 1:16

You might find it easier to think about a slightly different problem first. Replace the continuous heavy rope with a set of beads of mass $m$ attached to a massless rope, at distance $d$ apart.

When each bead starts to lift off the ground, its velocity suddenly changes from $0$ to $v$. So its momentum suddenly changes, and the change in momentum requires an impulsive force (transmitted through the rope) as given by Newton's second law.

Going back to the continuous rope, you have an "infinite number of infinitesimal impulses", which is equivalent to a steady force at the bottom of the vertical section of the rope.

Your hunch is correct : you have not identified what is "the system". The mass $m$ is not the whole mass of the rope; it is only the mass which is hanging in the air.

Even so, the force applied at any instant is not simply the weight of the rope hanging in the air. The momentum of the rope is constantly increasing as element after element is accelerated from rest to speed $v$ as it lifts off the ground. This rate of change of momentum requires force additional to the weight of hanging rope. The same consideration is required if you are given the reverse problem : to calculate the reading of a weight scale as sand is dropped onto it at a constant rate of mass.

Some assumptions are necessary :

1. The rope uncoils vertically. Then the rate at which mass leaves the ground is constant and the change in momentum is vertical, like the weight. If the rope had been laid out in a horizontal line then one end raised vertically, the rate at which mass leaves the ground and the direction in which it is moving would be far more difficult to find.

2. The rope has no stiffness at all, so that it offers no support to the rope which has been lifted off the ground. A better model than the rope is a chain with very many small links. Chains have no stiffness.