# Weight on an Elastic thread question.

1. Jul 26, 2010

### Firefox123

From http://star.tau.ac.il/QUIZ/ [Broken] website:

This problem can also be found on page 32 http://kvant.mirror1.mccme.ru/1990/05/resheniya_zadach_f1206_-_f1212.htm [Broken], but it's in Russian so might be hard to read...

Can someone explain this solution? Is it always true that F=Fo/2? Don't we need to know some details about the thread?

I would think this is just a nonhomogeneous differential equation with a force "F".

What do you guys think about the solution? Does it always hold true in general?

Last edited by a moderator: May 4, 2017
2. Jul 26, 2010

### K^2

The wording is a little bit weird, but yes, it's exactly F0/2.

Simplest way to see this. If you gradually apply a force and increase it to F, you displace the mass by F/k. This new point is the new equilibrium point.

If you instantly add point F, the equilibrium is still shifted by F/k, but mass has not moved yet. It will accelerate towards equilibrium point, overshoot it, and pass it by another F/k. So now the mass is 2F/k from point where it was before you applied force, and that means total force on the thread is 2F.

So by applying a force instantly, you cause maximum force to be twice the applied force, and that means you can break the thread by applying only 1/2 of the force you need to break it.

Edit: It's not clear from wording of the problem, but the Quant's solution starts out by stating that they treat the thread as obeying Hook's Law, which is fair for a tight thread with low elasticity.

3. Jul 26, 2010

### Firefox123

K^2....

Okay....

Makes sense...

How do we know it will overshoot by exactly F/k?

That makes sense assuming the overshoot is exactly F/k.....

That seems reasonable....

4. Jul 26, 2010

### K^2

$$m\ddot{x} + kx - F - mg = 0$$

Suppose the following is a solution.
$$x = A cos(\omega t) + D$$

Substituting...
$$- \omega^2 m A cos(\omega t) + k A cos(\omega t) + kD - F - mg = 0$$

Since it must hold for all t, there are actually two equations there.
$$k = \omega^2 m$$
and
$$kD = F + mg$$

Amplitude A is determined by initial condition. The oscillations have frequency ω=sqrt(k/m), no surprise. And the oscillations take the mass to ±A from point x=D=(mg+F)/k, which is the equilibrium point. Since the object starts at old equilibrium point (mg/k) and new equilibrium point is (mg+F)/k, amplitude A is trivially given by A=(mg+F)/k-mg/k = F/k.

So the maximum is achieved at x=D+A=mg/k+2F/k, the total force there is mg+2F. If mg+2F = mg+F0, the thread breaks. So F=F0/2.

5. Jul 27, 2010

### Firefox123

Got it....makes perfect sense.

Thanks.

6. Jul 27, 2010

### femr2

Hi :)

I have no issue with the solution Fo/2 as the theoretical limiting case.

However, the example is simplified and omits many real-world factors such as damping ratio of the thread, energy dissipation of any kind (such as generation of heat), external resistance such as conducting the experiment in treacle rather than a vaccuum (drag coefficient), etc etc...

Are all here agreed that the solution Fo/2 is the theoretical limiting case, and that in any real-world scenario, solutions can tend to Fo/2, but cannot reach it exactly ?

7. Jul 27, 2010

### K^2

For a tight thread, all of these coefficients are not going to play a big role in a single oscillation. I suppose, you can "break" the solution by introducing a material that has a lot of viscous give before it snaps.

So yeah, it's an approximation, but it's an extremely good one.

8. Jul 27, 2010

### femr2

Thanks.

An example I've used in other venues is using natural rubber as the material for the elastic thread.

Natural rubber has a damping ratio range of 0.01 to 0.08.

Incorporating damping ratio for xmax results in the following in other texts I've looked at...

xmax = $$\stackrel{F}{k}$$(1+e-$$\zeta$$$$\pi$$)

Not used to the formatting here I'm afraid - xmax=F/k*(1+e^-zeta*pi)

Where $$\zeta$$ (zeta) is the damping ratio.

Applying 0.08 to (1+e-$$\zeta$$$$\pi$$) = 1.78

Using damping ratio of zero results in 2 as expected.

I agree that Fo/2 is a reasonable approximation for general use, but do think that 1.78 is a quite significant difference to 2.

Any issues there ?

ETA: It would also be useful to have opinion on Fo/2 being an unattainable limiting case in any real-world scenario. I cannot think of any real-world situation which would allow for exactly Fo/2 to be reached. Splitting hairs of course, but it would be useful to have additional confirmation. Would imply a material with a damping ratio of exactly zero.

Last edited: Jul 27, 2010
9. Jul 27, 2010

### Firefox123

Can anyone confirm or deny what femr2 is saying here?