andyrk
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If we take one end of a mass-less string and pull it with force F, would the string have any tension in it? Would it have any tension when we pull it with force F from both the ends?
Massless string pulled by a force
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Yes.
andyrk
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Why? And what would it be? Can we derive it?
AdityaDev
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If I take a small element dx of the string, and if T1 acts on one side and T2 acts on the other, net force is ##T_1-T_2=ma=0##. So T1=T2.
This means that net force on every element of string is zero right?
This means that net force on every element of string is zero right?
andyrk
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Yes, but what I am asking is that why would Tension arise in the first place?AdityaDev said:
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Why would it not? What is your understanding of tension?
AdityaDev
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From the equation of in post 4, the force on every segment of the thread is zero. Then why does it need tension?Orodruin: 5053035 said:
andyrk
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That tension arises only when something is pulling at some particle. So the point which is being pulled with F has some pulling force on it. But what is pulling on the particles behind that first point at which force is being applied?Orodruin said:
The fact that you are using T1 and T2 on dx element itself means that tension is present.AdityaDev said:
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AdityaDev said:
The net force is zero, yes. This is not the definition of tension. The definition of tension is the force which one part of the string exerts on the other part of the string. What you have shown in #4 is that the tension is constant.
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andyrk said:
This is wrong. Tension (at point x) is the force that acts from one part of the string on the other part (for a given cut at the point x of the string).
andyrk
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So how would you explain that different parts of the string impart forces on other parts so that tension exists throughout the string?Orodruin said:
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It has to, otherwise the string would be accelerating as per Newton's 2nd law as shown in post #4 by Aditya. If you let one of the ends of the element you consider be an actual end of the string (where you are pulling with force F), the force on the other side of the element must also be F (but in the opposite direction). Thus, the tension in the string must be F.
AdityaDev
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@Orodruin , so tension is not the net force on each part of the string but the force that is exerted by a nearby segment of the same string on it.
also, when the string has some mass and when equal forces are applied on both ends, since acceleration of body is zero, again the tension in each segment of the string has to be F right?
also, when the string has some mass and when equal forces are applied on both ends, since acceleration of body is zero, again the tension in each segment of the string has to be F right?
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AdityaDev said:
Assuming that acceleration is zero, yes.
andyrk
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So how would you explain that this same tension exists at all points of the string besides the first? We know that F exists on the very first element. But what about the rest?Orodruin said:
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andyrk said:
One part of the string pulls on the other part of string and vice versa. If it did not, the different parts of the string would accelerate, as post #4 clearly shows. This is the very definition of what tension is.
andyrk
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How do you know that one part pulls the other part with the same force as the first part was pulled by? Why can't they be different?Orodruin said:
AdityaDev
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Third law.andyrk said:
when one part pulls the nearby part, then that nearby part will pull the first part with the same force.
andyrk
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There are 2 parts. One is right at the end of the string, which is being pulled by force F. The second part which is adjacent to this part but not in direct contact with the point at which the force is being applied. So how does this second part get pulley by force F?AdityaDev said:
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By the internal forces of the string, how else?andyrk said:
andyrk
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Because otherwise the string would accelerate (in fact, if the string is massless, it would have infinite acceleration).
andyrk
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And why should the string not accelerate? The string is being pulled from only one direction. So it should accelerate.
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andyrk said:
You are referring to the case when you only pull one end? You simply cannot do that without giving the string infinite acceleration (if it is massless). I hope you realize that what breaks down in this case is your assumption that the string is massless. If the string has even the tiniest mass, the acceleration will not be infinite and the tension will change along the string depending on the string's density.
andyrk
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Right, but what I was asking was would tension exist in such a case or not? (String is massless and being pulled from only one direction)Orodruin said:
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andyrk said:
I just told you this situation cannot exist. If you are pulling one end of the string only and there is no other masses involved, you cannot approximate the string as massless. The tension in the (non-massless, i.e., massive) string will depend on how its mass is distributed.
andyrk
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Oh..you mean if there exists an unbalanced force...then their has to be some mass...mass-less case can't exist side by side with unbalanced force?
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Not unless you like infinite accelerations. Generally, it just means you neglected a mass that you probably should not have neglected and the acceleration is large, but not infinite.
andyrk
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And if I say that I am fine with infinite acceleration, then what would the tension be?Orodruin said:
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You should not be fine with infinite acceleration apart from for an infinitely short moment as an approximation, which does not allow you to apply force considerations. It is unphysical, you cannot make a massless string.