Weight on an Elastic thread question.

In summary: Perfect summary. In summary, the solution to the given problem is that the thread will break if F=Fo/2. This is because when an additional force F is applied, the total force on the thread becomes 2F, which is double the applied force. This can be proven by considering the equation of motion and using the fact that the thread follows Hook's Law. This solution was correctly solved by Qiu Shi Wang and Ying Cun Luo, freshmen at Peking University, China, and by Chetan Mandayam Nayakar, a student at India Institute of Technology, Madras, India. The solution can also be found on a Russian website, but it may be difficult to understand.
  • #1
Firefox123
183
1
From http://star.tau.ac.il/QUIZ/ [Broken] 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 [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?
 
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  • #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
K^2...

Thanks for the reply...

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

Okay...

K^2 said:
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...

K^2 said:
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?

K^2 said:
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...

K^2 said:
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...
 
  • #4
Start with equation of motion.
[tex]m\ddot{x} + kx - F - mg = 0[/tex]

Suppose the following is a solution.
[tex]x = A cos(\omega t) + D[/tex]

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

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

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
K^2 said:
Start with equation of motion.
[tex]m\ddot{x} + kx - F - mg = 0[/tex]

Suppose the following is a solution.
[tex]x = A cos(\omega t) + D[/tex]

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

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

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.
 
  • #6
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 ?
 
  • #7
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
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 = [tex]\stackrel{F}{k}[/tex](1+e-[tex]\zeta[/tex][tex]\pi[/tex])

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

Where [tex]\zeta[/tex] (zeta) is the damping ratio.

Applying 0.08 to (1+e-[tex]\zeta[/tex][tex]\pi[/tex]) = 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.
 
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  • #9
femr2 said:
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 = [tex]\stackrel{F}{k}[/tex](1+e-[tex]\zeta[/tex][tex]\pi[/tex])

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

Where [tex]\zeta[/tex] (zeta) is the damping ratio.

Applying 0.08 to (1+e-[tex]\zeta[/tex][tex]\pi[/tex]) = 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?
 

1. What is the purpose of the weight on an elastic thread experiment?

The purpose of this experiment is to demonstrate the relationship between the weight of an object and the elongation of an elastic thread. It also helps to understand the concept of Hooke's Law, which states that the force applied to an elastic material is directly proportional to the amount of stretch or compression of the material.

2. How is the weight on an elastic thread experiment set up?

The experiment involves attaching a weight to an elastic thread and measuring the amount of elongation of the thread. The weight should be heavy enough to cause a noticeable elongation, but not heavy enough to break the thread. The thread should be securely attached at both ends, and a ruler or measuring tape can be used to measure the elongation.

3. What factors can affect the results of the weight on an elastic thread experiment?

The results of this experiment can be affected by several factors, such as the type and quality of the elastic thread used, the weight of the object, and the accuracy of the measurements. Other factors, such as temperature and humidity, can also have an impact on the elasticity of the thread and should be kept constant for accurate results.

4. How does Hooke's Law apply to the weight on an elastic thread experiment?

Hooke's Law states that the force applied to an elastic material is directly proportional to the amount of stretch or compression of the material. In the weight on an elastic thread experiment, this means that the weight applied to the thread will cause a proportional amount of elongation in the thread. This relationship can be observed and measured in the experiment.

5. What are the practical applications of the weight on an elastic thread experiment?

The weight on an elastic thread experiment has practical applications in various fields, such as engineering, materials science, and physics. It helps to understand the behavior of elastic materials and their ability to withstand forces. This knowledge is essential in designing structures and materials that can withstand various loads and stresses.

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