## Weighty Problem

Posted by Chris on July 2, 2010 – 5:21 pm

Pinched from the old ToM site.

A rope of length L metres and mass M kg is suspended vertically by one end with the other end just touching a weighing scale (calibrated in kg). The rope is released. At the moment just before the entire rope has fallen onto the scale what does the scale read?

Assume the rope is mathematically perfect, the weighing scale has 0 response time, etc.

July 3rd, 2010 at 1:51 am

To easy Kilograms and Grams! no weight is given for the rope and you already said it is calibrated to 0 kilogrammes so it reads in kilogrammes and gramms

July 3rd, 2010 at 3:28 am

Well, I’m going to go with a variety of answers….

At the moment just prior to release, I will assume the answer will be 0.

At the moment of release, before gravity starts to work, the answer will still be 0.

The moment gravity drags the rope down, whilst it is still straight, before the rope accelerates towards the scales, I think the scales will accurately show the full weight of the rope.

As the rope accelerates towards the scale, assuming all the rope lands on the scale the weight shown on the scales will increase until half the rope is on the scales and then start to decrease, until all the rope is resting on the scales when the true weight will again be shown.

Now to prove the theory… I need rope and scales….

Back soon.

July 3rd, 2010 at 4:22 am

Most weighing machines are calibrated in kg (or stones and pounds). If I stood on one, it’d read around 75 kg. I haven’t seen one that would say 735 newtons. The confusion arises because (almost) everyone talks about weight (a force) in units of mass.

I used rope because that (or a chain) is typically used in this and related problems. It might be better if you think of a cylinder of sand (or water) being used. In this case the sand is somehow (magically) initially in a state of suspended animation, and then suddenly it is free to fall, with no constraints other than the usual laws of physics. The salient feature is that the object falling isn’t rigid, and is unable to transmit forces within itself.

Don’t worry about the height of the pile of rope (or sand or water) building up on the pan. The rope (or sand or water) doesn’t fall off the pan.

To solve this problem you will need (simple) calculus, or equivalent.

July 4th, 2010 at 12:37 am

Hi Rewman. The weight of the rope is Mg, where g is the acceleration due to gravity. However, most people would say that it had a weight M (kg) because that’s what a bathroom scale would say. The fact is that people say “my weight is …” rather that “my mass is …”. It causes difficulties for some people when they are learning physics.

The scale will display grammes as thousandths of a kg.

July 4th, 2010 at 12:42 am

I have a mass of 78 kg (12 stone 4 lbs). So my weight is 765 newtons on the Earth and it’ll be roughly 130 newtons on the Moon, but my mass will still be 78 kg on the Moon.

The apple that Newton allegedly pondered on would have had a weight of about 1 newton.

July 6th, 2010 at 3:56 am

It looks like no-one is up for this. So…

Newton’s second law is F = dp/dt where p = mv is momentum and F is force.

So calculus suggests that F = m dv/dt + v dm/dt. This derivative needs to be interpreted with care. NB I’m assuming that the speeds involved are sufficiently small that air resistance (drag) effects are negligible.

Now apply this to the rope (or sand or water) at the scale pan. Consider that over an infinitesimally short time, dt, that an infinitesimally small mass, dm, arrives at the pan and comes to rest. dm dv/dt is infinitesimally small (dv/dt is finite as infinitely large accelerations are not possible). But v dm/dt is finite, so we may neglect the dm dv/dt term.

Now consider an infinitesimal portion of the rope that originally was at a height h above the pan. When it gets to the pan it will have a speed v = sqrt(2gh), where g is the acceleration due to gravity. Over a time dt a length v dt of the rope will fall onto the pan and it will have a mass, dm = (M/L)v dt. So v dm/dt = 2gh(M/L). NB (M/L) is the mass per unit length of the rope.

The length of rope that has come to rest on the pan is h (obviously) and it must have a weight gh(M/L). So the total force on the pan is 3gh(M/L). For the very last bit of the rope, h = L, so the total force is 3Mg. But the scales are calibrated in kg, so the reading is 3M. Moments later the reading will drop to M.