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In my text book the following system was given

A massless inextensible string wearing a bead of mass $m$ is suspended from point A and B. At the given instant they said that the tension on both sides of mass $m$ will be equal but I don't understand how we can prove that?

EDIT: The system is not in equilibrium. It is like when a bead is dropped and the string suddenly becomes taut giving an impulse on the bead.

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  • $\begingroup$ It's too easy to understand when you think what's happening in microscopic level. $\endgroup$
    – Debanjan Biswas
    Commented Jul 25 at 16:29
  • $\begingroup$ First we need to figure it out why tension in a string appears. It's due to the fact when one part of the string is pulled, the distance between the molecules of the string increases and due to its perfectly elastic nature, the intermolecular attraction rises between the molecules which is called tension. $\endgroup$
    – Debanjan Biswas
    Commented Jul 25 at 16:36
  • $\begingroup$ Now two atoms next to each other in any string pulled, will exert same and opposite force or tension on each other. In a string the force is distributed symmetrically throughout its length so that it satisfies the relation $T=\frac{F(l-x)}{l}$, where $F$ is the pulling force, $l$ is the length of string from the part which is pulled to the end and $x$ being the distance of the point at which tension is determined from the pulled part. $\endgroup$
    – Debanjan Biswas
    Commented Jul 25 at 16:47
  • $\begingroup$ The above holds for the strings with mass. For a massless string, every point attracts every point in it with same force because none of the points have mass and no force exerted on them is wasted to pull them. Quite simple! $\endgroup$
    – Debanjan Biswas
    Commented Jul 25 at 16:51

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$\dots$ the tension on both sides of mass $m$ will be equal $\dots$ if there is no friction between the bead and the string. If there are no frictional forces then the fact that the string is massless means that there can be no net force on any segment of the string.

Think of an extreme case where the string is glued to the bead and then the tensions are not necessarily equal.

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    $\begingroup$ If the system is not at equilibrium, the key assumption is that the contact between bead and string is frictionless. This is the only acurate answer, IMHO. $\endgroup$
    – vals
    Commented Nov 15, 2023 at 15:12
  • $\begingroup$ But there can be net force on the bead which is not massless so how can we comment that tension in string on both sides of bead will also be same? $\endgroup$
    – Garv Chaudha
    Commented Nov 15, 2023 at 16:17
  • $\begingroup$ If the tensions are at different angles to the horizontal as per your diagram then the bead is not in equilibrium. $\endgroup$
    – Farcher
    Commented Nov 15, 2023 at 17:13
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"A massless inextensible string wearing a bead of mass m"

From this line I assume there is only one string used throughout, and is passed through the hollow centre of the bead.

As long as only one string is used of massless nature, the tension throughout the string must be constant when it is in static equilibrium.

Let's assume the tension was not uniform in this string. If the tension was not uniform throughout, and we take a segment of this string (let's say dl). This segment would have different magnitudes of tension on either side, and therefore this would be indicative of a net tension in a particular direction, causing the string as a whole to elongate in that direction till it attains static equilibrium.

Hence, there's no need to prove that tension is equal on both sides. It's the inherent property of any ideal massless tensed body to have uniform tension in equilibrium.

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  • $\begingroup$ I have edited my question to clarify the system which is not in equilibrium. $\endgroup$
    – Garv Chaudha
    Commented Nov 15, 2023 at 7:22
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The phrase "inextensible string" and the fact it's tied down at a point, is what determines that the tension at every point along the string is the same:

"Inextensible" doesn't just meant that the whole length of the string doesn't change - it means it that no part of it can stretch at all. It's not a rubber-band that can stretch out more at one part and less on other parts. No part of the string can stretch at all. If you assume this, and that the string is tied down at one point (e.g., A) we can prove that the tension must be the same throughout the string:

If the tension was not the same throughout the string, imagine a short piece of the string - we can assume it's straight if it's short enough - where the tension on one end is different from the tension on the other part. In this case, this short straight piece of string is experiencing net force, and therefore accelerating in the direction of the string. Since we know one end of the string (e.g., A) is tied in place and not accelerating, it would mean that the string is stretching, which we assumed is not possible. So we can conclude that this acceleration is zero and therefore that the tension must be the same throughout the string.

By the way, if the string were not tied down at one end, it is possible for the tension to be different throughout the string! This can happen in a moving pulley (see for example https://xaktly.com/AtwoodsMachine.html), and also in a accelerating train being pulled at one end by a locomotive - the tension is not the same between each car, even though the cars are all moving in unison.

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  • $\begingroup$ I'm not quite convinced by this argument, presented as-is. Let's consider a massive inextensible string which is tied down at both ends, which is not the scenario in the original question, but your argument should apply to this scenario equally well. Intuitively, it seems easy to imagine a situation where the tension is not uniform along the string (perhaps the string was laid out in a serpentine shape, and just a moment ago, I suddenly moved one of the anchors away from the other, and the resulting wave has not reached the other anchor yet). (Cont'd...) $\endgroup$
    – Sophie Swett
    Commented Nov 15, 2023 at 22:03
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    $\begingroup$ Then we do indeed have short pieces of string where the tension at one end is different from the tension at the other end, and those pieces of the string do indeed experience a net force and accelerate in a direction parallel to their own length. Furthermore, both ends of the string are tied in place and not accelerating. However, this information isn't enough to conclude that the string is stretching (as shown by the fact that in this scenario, all the premises are true yet the conclusion is false—the string is not stretching). $\endgroup$
    – Sophie Swett
    Commented Nov 15, 2023 at 22:06
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The string is massless and inextensible; so if there is a net force on any part of the string it will cause its acceleration. Since m approaches zero, acceleration for even a small unbalanced force will approach infinity. So we will just have to assume tension to be the same on both sides.

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Let me put it one more way:

If the tensions at two points of the string were different, the string piece between these two points would experience a net force from the tension difference.

There are no other forces acting upon this piece of string:

  • As the string is massless, there is zero force from gravity.
  • Nothing else is mentioned that can cause forces on that string piece (friction, magnetism, ...).

Any net force (from tension difference in our case) will cause an acceleration, and with a zero-mass piece of string, any non-zero net force will cause an acceleration of its center of mass with an infinite magnitude, which never happens in the real world.

Nitpicking: But the example contains a few assumptions that contradict the real world: the string is massless, and it is inextensible. And this creates one situation with infinite acceleration (or deceleration, to be more specific). That is, when the string becomes taut, it instantaneously decelerates from some falling speed to zero. The force needed to infinitely accelerate a zero-mass object is undefined (zero times infinity), so in this very instant, there can be a net force on the string, meaning a tension difference.

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    $\begingroup$ Just note that it's not accurate to say there are no forces at all acting on the string - after all, the string will be moving because of the bead. As an example, consider a pendulum string (string tied to the ceiling in one end, with a massive ball on the other end) - it definitely feels the force (exerted by the massive ball, itself feeling gravity). It's just that the force isn't along the direction of the string (otherwise the string would stretch) - it's perpendicular to the direction of the string. $\endgroup$
    – Nadav Har'El
    Commented Nov 15, 2023 at 10:16
  • $\begingroup$ @NadavHar'El As I said, if any non-zero net force were acting on the string (in whatever direction), it would react with an infinite-magnitude acceleration (as it has a mass of zero). With zero mass, it also feels zero gravity force itself. $\endgroup$
    – Ralf Kleberhoff
    Commented Nov 15, 2023 at 14:21
  • $\begingroup$ This is true, but in practice, you can have a massive ball swinging on a string of arbitrarily small mass, not zero, but maybe a picogram. How does this work? Well, because the string is inextensible, the moving ball will pull it with the same acceleration as the ball's - which because of the tiny mass of the string will translate to a very tiny force. Nothing works with zero mass, but it can be as small as you'd like and everything works out. $\endgroup$
    – Nadav Har'El
    Commented Nov 15, 2023 at 17:11
  • $\begingroup$ I can replace "zero" with "tiny" and "infinite" with "huge". Then the net force acting on the tiny-mass string will be tiny, compared to the force on the ball, meaning that the tension difference will be tiny. And a non-tiny net force will result in a huge acceleration. $\endgroup$
    – Ralf Kleberhoff
    Commented Nov 16, 2023 at 9:30

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