# Explanation wanted for a demonstration in MIT phyiscs lec #6

1. Jul 12, 2009

### cfung

The demonstration: A mass of 2kg is hanged to a string that is hanged vertically from a fixture. Another string is attached to the bottom of the mass and then this bottom string is given a downward pull. Note that the strings used are assumed identical.

This demonstration can be found here (starting at time 43:00, or the fourty-third mins):

The initial question was, would the top string or the bottom string break first?
The answer, according to the demonstration, is apparently that either string could break depending on the way you pull down on the bottom string.

The explanation for the top string breaking first isn't too difficult; it is simply due to the fact that no matter how hard you pull down on the bottom string (thus giving it tension), the tension of the top string is always greater than the bottom string by 20N (contributed by the 2kg hanging mass).

As for the breaking of the bottom string. I've read through the video comments but none seems to have a correct answer rigorously examined. I've spent a week thinking about the demonstration thoroughly yet I still don't have a confident answer myself.

I wish to understand the truth of the matter - Of what cause does the bottom string breaks? - and I need someone to enlighten me here.

2. Jul 12, 2009

### cfung

Just thinking about the problem empirically, it seems impossible for the bottom string to break (Note it can be assumed that both the strings break at an identical tension value). This is because no matter how much force you exert on the bottom string, the top string will always experience 20N more tension. For example, if we knew that the string will break at 50N, it would only takes me to pull down on the string with a force of 30N. Consequently, there is NO WAY the bottom string can break before the top string does (keeping in mind the key idea that no matter what you do to the bottom string, the top always feels 20N more tension).

If it were possible for the bottom to break before the top, then it must be the case that there's a way of pulling the string such that the bottom string experience a tension of 50N+ while the top string feels a tension of less than 50N. This would violates my common sense as stated above.

However, the fact is so, the demonstration could make that to happen, apparently by giving a significant acceleration in the act of pulling the string. This hints for a potential explanation in the concept impulse or acceleration.

Last edited: Jul 12, 2009
3. Jul 12, 2009

### TVP45

So, did you actually try this? It's a simple experiment.

4. Jul 12, 2009

### negitron

It's just inertia.

5. Jul 12, 2009

### Fenn

If the strings break at a particular tension, then a gradually increasing pulling force on the bottom string will introduce a nice graduale increase in both strings' tensions. But what if you increase the force very rapidly?

Imagine that the strings are not ideal, in that they tend to stretch a bit when they are under tension. In my mind, the rapidly increasing force will introduce a tension directly to the lower string. The lower string will pull on the weight, which will act as a kind of buffer for the upper string. The weight is a massive object, initially at rest. It must be accelerated and ultimately displaced in order to introduce more tension on the upper string.

This transferring process can conceivably take some amount of time. If the rapidly increasing pulling force is sufficiently fast, there will be a surge of tension on the lower string, enough to break the string, before the massive weight has had enough of a chance to move and exert the additional tension on the upper string.

It might be easier to visualize this if you replace the strings with identical springs. Imagine a slow, gradual pull, where both springs stretch equally. Now, reset the system, and yank down on the lower string rapidly. The lower string will stretch considerably before the mass has a chance to move, and thus (eventually) stretch the upper spring.

6. Jul 12, 2009

### negitron

Yes, we call that "inertia."

7. Jul 12, 2009

### Redbelly98

Staff Emeritus
That's only true when the mass's acceleration is zero, and approximately true only when the acceleration is negligible compared to g.

8. Jul 13, 2009

### cfung

Now, you stated "If the rapidly increasing pulling force is sufficiently fast, there will be a surge of tension on the lower string, enough to break the string," which is the farthest I could arrive and it's exactly the part where I can't be confident.

Because, that cause which is "enough to break the string" is vague and not obvious and it can refer to the following possibilities:

1.) The act of "rapidly increasing pulling force"?
2.) The Inertia of the 2kg mass body?
or,
3.) The 50N tension that the string feels (assuming 50N is the value of breaking)?

1.) One thing that is worth pointing out first is that the "rapidly increasing pulling force" is not the essential element in breaking the bottom string. Simply, it should be anything in line with a huge force. Imagine if you can magically replace your pulling hand, in an instant, with a second mass (its mass comparable to a car) that hangs to the end of the bottom string, then the bottom string will break no problem, even without an external force acting on this second mass to create the rapidly increasing pulling force.

2.) Now is it the inertia, or the hanging mass's resistance to motion? In saying this, one risks implying that the top string doesn't feel any increasing tension (identical to the bottom string) when the bottom string is being rapidly pulled to its breaking point. In this scenario, one is really saying that since the bottom string is being pulled too fast, it doesn't have enough time to fully transfer its effects to overcome the hanging mass' resistance to motion, or, its inertia, and therefore the effect doesn't reach the top string at all.

I disagree. Because an object possessing inertia, even if it is enormous relative to an external force, is still susceptible to motion. And this is merely a restatement of the Newton's Second Law, F = ma. In this case, the object will experience a very small acceleration since F is small and m is huge.

Likewise, the idea of the above, F = ma, is being violated if we explain the demonstration by saying that the inertia of the hanging mass stops the tension (created by rapidly pulling the bottom string to its breaking point) from reaching the top string.

3.) This is likely the cause of breaking the bottom string. Since we knew that the string breaks when it experience a tension of 50N, it is logically safe to conclude that, in the demonstration, the string breaks because it has indeed experienced a tension of 50N.

In short, it seems as if 3. should be the only cause of breaking the bottom string. But the bottom string can not feel any tension when we pull down on it rapidly unless the hanging mass pulls back with a force. This is in accordance with the Newton's 3rd Law.

So there is a moment when the 2 kg mass exerts a force to the top string equal to the tension of the bottom, plus its own g force.

And this goes back to the original problem, the top string experience more tension than the bottom, but the bottom break first. I feel that the physics has ultimately got to do with inertia but it just doesn't make complete sense to me, perhaps my above analysis is flawed.

Last edited: Jul 13, 2009
9. Jul 13, 2009

### cfung

True and worthwhile to point out. In general, the top string feels 20N more tension is always true when the mass is in any uniformly accelerated frame of reference at 10m/s^2, as this demonstration has assumed.

Last edited: Jul 13, 2009
10. Jul 13, 2009

### Fenn

There is no violation of a law here. You are correct that, no matter how enormous an object is, it will still accelerate under the influence of a net force. But with a sufficiently large mass, the acceleration will be sufficiently small such that the rate of change of motion is slow compared to the force exerted on the bottom string.

Imagine unbreakable, but stretchable strings. When a new force instantly pulls down on the mass, this tension exists only in the second string. If you draw a free body diagram for the mass, there will be a net force acting downward, and thus an acceleration. As time goes on, the mass is accelerates and causes the upper string to stretch, which increases the tension in the upper string. Eventually, the tension in the upper string will increase to the point where it will stop the motion. With dampening effects, the system will eventually settle down, where both string have an increased tension due to the applied force.

Now, back off the requirement that the strings are unbreakable. If the applied force is sufficiently large and the mass is sufficiently heavy, the bottom string will break. Like I said earlier, it could help to imagine replacing strings with springs, and picture how each will stretch when you rapidly apply a force. The amount the spring stretches is proportional to the amount of tension on that spring.

11. Jul 14, 2009

### cfung

If it were possible that only some percentage of the force applied becomes effective to the system, then this really explains. And it's somewhat convincing because it's very ordinary to lose energy in a system. But it's difficult to imagine a nearly massless string would lose significant amount of energy to heat or sound.

Then that rapid force applied would travel through the spring as longitudinal waves (streches, compresses and so) and these stretches and compressions are strong enough to cause the material to break midway.

That sounds imaginable and practical, though it only raises another question. Do all materials in the world act like a spring in the sense that any applied force would travel through the material as longitudinal waves?

Suppose we replace the top string and the bottom string with strong alloy rods, and have someone pulling the lower rod with sufficiently large force so that it breaks, just as in the original demonstration. Would an explanation to this demonstration no more different than the original one?

12. Jul 14, 2009

### Fenn

Well, consider a solid as being made up of some arrangement of atoms or molecules. The atoms and molecules are held in place by interatomic and intermolecular forces. In a sense, you can picture this structure as being a bunch of atoms held together by a lattice of springs. The more rigid the material, the more rigid these springs holding them together.

If you apply an external force to a solid, this force must be applied to the atoms within that material. Imagine pushing on one side of a lattice of metal atoms. The atoms which you press on are compressed into neighbouring atoms, which causes subsequent compression and pushing of atoms further down. A chain reaction of this action eventually leads to the material being moved.

Of course, when considering a rigid material, such as a rod, you can expect such an action to take place rather fast, and on length scales that you would never be able to witness. But what if you had a big chunk of flimsy rubber instead of metal?

As another example, consider a metal tuning fork. When you strike the fork, it vibrates and hums. Those vibrations are just waves of displacement travelling throughout the rigid solid.

13. Jul 14, 2009

### Civilized

If he goes from 0 to beyond the maximum tension in a shorter time than it takes for the signal to propagate up to the top string, then the bottom string breaks. Otherwise, the top string breaks. I found this professor irritating!

14. Jul 15, 2009

### cfung

Cool, the conclusion is to attribute the break of the string to the overwhelming stretch in the string's molecular structure, albeit rapid. It's a sensible explanation I should accept. Thanks.

15. Jul 15, 2009

### sganesh88

No energy loss is assumed here. Just because of the relatively high mass, the acceleration is low; not of the magnitude you'd expect from a lighter one. That doesn't mean there are losses. Distribute 100 chocolates among 10 children. Each gets 10. Among 100? Just one each. No losses here either.

16. Jul 15, 2009

### rcgldr

I think this is covered now, but note that every real world object has a stress strain (deformation) relationship. In order to increase the overall tension in a string, that string must be stretched. If the bottom string is pulled quickly enough, the momentum of the relatively large mass between the strings reduces the rate of stretch of the upper string, so the bottom string breaks first if it's jerked hard enough. If the bottom string is jerked hard enough to break quickly, the large mass senses a large but very momentary force, so the work done is very small, and the large mass is only moved downwards a small amount, increasing the tension in the upper string by only a small amount.

This would probably be easier to visualize if the upper string were replaced by a spring that didn't break, and then noting the amplitude of spring and mass oscillation versus the jerk and force applied to the lower string.

17. Jul 15, 2009

### cfung

According to (Force)*(Time) = (Impulse), if 50 Newton is exerted on the 2kg mass through a string in 0.5 seconds, then the hanging mass's change of momentum would be 25kgm/s.

Likewise, if in an experiment one pulls on the string with 50N tension and the string takes 0.5 seconds to completely transfer that force to the 2kg hanging mass, then one would expect the 2kg mass to have an impulse of 25kgm/s.

What if in this experiment the string broke and the measured impulse from the 2kg mass turns out to be something less, say 10kgm/s? We know that the whole process took 0.5 seconds, so the total force that the 2kg mass absorbed is actually (impulse)/(Time) = (10kgm/s)/(0.5s) = 20N.

Clearly, in the event that the string broke, not all of its tension would get transferred to the attached mass. In the above example, a difference of 30N vanished and it has to had gone somewhere.

18. Jul 15, 2009

### sganesh88

We can't say that. You're missing the other forces in the picture. If a body is subjected to a constant net force of F newton for t seconds, its change in momentum will be, magnitude of F multiplied by t, yes. Here the 50 N isn't the net force.

Force isn't a substance to be transferred. It is applied.
And regarding the situation you mentioned where the string breaks, if the string had indeed maintained a constant force of 50 N for the time 0.5 s on the body under the influence of no other forces before breaking, then yes. the body should have a change in momentum of 25 Kgm/s.

19. Jul 15, 2009

### cfung

I may have used the wrong diction, pardon me for that, but in that context I meant force as being "propagated".

Though thank you for your clarification on the subject of impulse.