Weight on an Elastic thread question.

[Firefox123]

From http://star.tau.ac.il/QUIZ/ website:

A weight is hanging on an elastic thread. An additional stretching force F is applied and is gradually (slowly) increased. When the force reaches value Fo the thread breaks. What should be the minimal size of a force that breaks the thread, if such a force is applied instantaneously and remains unchanged.

(1/06) This problem has been solved correctly (13/9/05) by Qiu Shi Wang and Ying Cun Luo, freshmen at Peking University, China (e-mail inklings@163.com), and (13/9/05) by Chetan Mandayam Nayakar, a student at India Institute of Technology, Madras, India (e-mail mn_chetan@yahoo.com). There are many, essentially equivalent ways to solve the problem. We will present what appears to be the simplest solution.


The answer: The thread will break if F=Fo/2.

The solution:
Before the force is applied the weight of the object hanging on the thread is balanced by the tension force of the thread. Once the additional force F is applied downwards the TOTAL force becomes F, and the weight starts executing harmonic oscillation under the influence of the forces. It starts the oscillation at the top point of the period. After a quarter of the period it reaches the midpoint of the oscillation at which the total force vanishes. After half of the period it reaches the bottom point of the oscillation, at which, by symmetry, the total force is F UPWARDS. This total force is result of the applied external force F pointing downwards, and the increase in the thread tension, which must be 2F and point upwards. Thus, the maximal thread tension is TWICE larger than the applied force. Consequently, F=Fo/2 suffices to break the thread.

This problem can also be found on page 32 http://kvant.mirror1.mccme.ru/1990/05/resheniya_zadach_f1206_-_f1212.htm , 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?

 

[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.

 

[Firefox123]

K^2...

Thanks for the reply...

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

Okay...

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.

Makes sense...

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.

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

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.

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

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.

That seems reasonable...

 

[K^2]

Start with equation of motion.
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.

 

[Firefox123]

Start with equation of motion.
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.

Got it...makes perfect sense.

Thanks.

 

[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 vacuum (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 ?

 

[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.

 

[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 x_max results in the following in other texts I've looked at...

x_max = \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.

 

[Firefox123]

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 x_max results in the following in other texts I've looked at...

x_max = \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.

Can anyone confirm or deny what femr2 is saying here?