Why Does Tension Act in Two Directions at a Point in a String?

[Chestermiller]

I see, the 2T tension in the upper string would also reverse its direction in that case. Either way, either the two tensions acting on the pulley are down or u but not both. Am I right? And the 2T tension above adjusts its direction accordingly. Right?

You are aware that when you have an action-reaction pair at the contact point between two bodies A and B, one member of the pair is the force exerted by body B on body A, and the other member of the pair is the force exerted by body A on body B, correct? When you are doing a force balance on body A, the two don't cancel. The force balance on body A includes only forces exerted by other bodies on body A, and not forces exerted by body A on other bodies. If both members of an action-reaction pair always had to be included, then no body could ever exert a force on any other body.

Can you see how this all relates to your rope and pulley?

Chet

 

[andyrk]

You are aware that when you have an action-reaction pair at the contact point between two bodies A and B, one member of the pair is the force exerted by body B on body A, and the other member of the pair is the force exerted by body A on body B, correct? When you are doing a force balance on body A, the two don't cancel. The force balance on body A includes only forces exerted by other bodies on body A, and not forces exerted by body A on other bodies. If both members of an action-reaction pair always had to be included, then no body could ever exert a force on any other body.

Can you see how this all relates to your rope and pulley?

Chet

So is the action reaction logic the reason why a point in a rope has tensions in two opposite directions all throughout the rope? And if that is the case then why does the end of the string still have tension T while the block is pulling the string down with force m₁g on one end and m₂g on the other end?

 

[Chestermiller]

So is the action reaction logic the reason why a point in a rope has tensions in two opposite directions all throughout the rope?

Suppose you have a rope under tension T (with both of its ends fixed), and you decide to cut the rope at a certain location. Say the rope is horizontal, and, after you cut it, you have a left section and a right section. You decide what you are going to do is, by hand, replace the force that the right section was exerting on the left section, so that the tension within the left section is the same as it was before. What force would you have to apply by hand to accomplish this? What direction would this force be pointing?

Now you decide to do the same thing with the right section, rather than the left section. What force would you have to apply by hand to the right section so that it is under the same tension as before? What direction would this force be pointing?

Chet

 

[andyrk]

Suppose you have a rope under tension T (with both of its ends fixed), and you decide to cut the rope at a certain location. Say the rope is horizontal, and, after you cut it, you have a left section and a right section. You decide what you are going to do is, by hand, replace the force that the right section was exerting on the left section, so that the tension within the left section is the same as it was before. What force would you have to apply by hand to accomplish this? What direction would this force be pointing?

Now you decide to do the same thing with the right section, rather than the left section. What force would you have to apply by hand to the right section so that it is under the same tension as before? What direction would this force be pointing?

Chet

For the left section, I would have to apply the same force that the right section was pulling it with when it was not cut. And for the right section, I would have to apply the same force that the left section was pulling it with when it was not cut. But whether the two forces are equal or not is still unclear.

 

[Chestermiller]

For the left section, I would have to apply the same force that the right section was pulling it with when it was not cut. And for the right section, I would have to apply the same force that the left section was pulling it with when it was not cut. But whether the two forces are equal or not is still unclear.

If you go around to the other side of the string, the left side becomes the right side, and right side becomes the left side. So the forces have to be the same, since the place where you personally are located can't affect the string.

Chet

 

[andyrk]

If you go around to the other side of the string, the left side becomes the right side, and right side becomes the left side. So the forces have to be the same, since the place where you personally are located can't affect the string.

Chet

What? What do you mean by left side becomes right side and vice-versa? I still don't get why the forces would have to be equal? Oh. Okay. I get it. But what about the ends of the string when it is connected to blocks at the two ends? If the block pulls the string down by a force equal to its weight then shouldn't the string pull it back with the same force leading to a tension at the ends which is equal to the weight of the block attached to the end of the string? But that is not the case as we know

 

[Chestermiller]

What? What do you mean by left side becomes right side and vice-versa? I still don't get why the forces would have to be equal? Oh. Okay. I get it. But what about the ends of the string when it is connected to blocks at the two ends? If the block pulls the string down by a force equal to its weight then shouldn't the string pull it back with the same force leading to a tension at the ends which is equal to the weight of the block attached to the end of the string? But that is not the case as we know

If the block is accelerating, then it does not pull the string down by a force equal to its weight. It pulls the string down by a force less than its weight:

Newton's 2nd law:

T - mg = -ma

If the block pulled the string down with a force equal to its weight, then it would be in equilibrium, and couldn't be accelerating.

Chet

 

[andyrk]

Right. How could I forget that.

 

[Chestermiller]

Right. How could I forget that.

If it were me, I would blame it on a "Senior Moment." But you're much younger than I am.

Chet

 

[andyrk]

If it were me, I would blame it on a "Senior Moment." But you're much younger than I am.

Chet

What's a senior moment? And what about the other end? Why does the tension at the other end also equal T? ( The block is accelerating upwards here). Why can't it be something other than T which would impart the same acceleration to the lighter block as the heavier block on the other end?

 

[Chestermiller]

What's a senior moment?

A senior moment refers to what happens to your memory when you become a senior citizen.

And what about the other end? Why does the tension at the other end also equal T? ( The block is accelerating upwards here). Why can't it be something other than T which would impart the same acceleration to the lighter block as the heavier block on the other end?

This follows from a moment balance on the pulley. Unlike the blocks, the pulley is assumed to be massless, so that its moment of inertial is also zero. So, $T_1R-T_2R=I\alpha = 0$, and thus T₂ = T₁.

Chet

 

[andyrk]

A senior moment refers to what happens to your memory when you become a senior citizen.This follows from a moment balance on the pulley. Unlike the blocks, the pulley is assumed to be massless, so that its moment of inertial is also zero. So, $T_1R-T_2R=I\alpha = 0$, and thus T₂ = T₁.

Chet

What? Wait, the tensions at the pulley can be equal but why do they also need to be equal at the ends connected to the 2 blocks?

 

[Chestermiller]

What? Wait, the tensions at the pulley can be equal but why do they also need to be equal at the ends connected to the 2 blocks?

If you do a force balance on the section of a massless string between the pulley and a block, you get

$$T_{by \ pulley}-T_{by \ block}=(string \ section \ mass)\times (string \ section \ acceleration)=0$$

Chet

 

[andyrk]

If you do a force balance on the section of a massless string between the pulley and a block, you get

$T_{by \ pulley}-T_{by \ block}=(string \ section \ mass)\times(string \ section \ acceleration)=0$

Chet

So do you mean to say that the block is pulling the string at the end with a tension T and not its weight?

 

[Chestermiller]

So do you mean to say that the block is pulling the string at the end with a tension T and not its weight?

Yes. I though we already settled that.

Chet

 

[andyrk]

Yes. I though we already settled that.

Chet

Yup but I was thinking that what stops the string from having tension equal to the weight? Is Newton's Second Law a sufficient enough reason? That is, the block accelerates so their has to be a net force. Is that a good enough reason why T would not and should not equal the weight of the block?

If you do a force balance on the section of a massless string between the pulley and a block, you get

$T_{by \ pulley}-T_{by \ block}=(string \ section \ mass)\times(string \ section \ acceleration)=0$

Chet

And we can calculate this unknown T by solving the equations we make with the help of Newton's Second Law, right?

 

[Chestermiller]

Yup but I was thinking that what stops the string from having tension equal to the weight? Is Newton's Second Law a sufficient enough reason? That is, the block accelerates so their has to be a net force. Is that a good enough reason why T would not and should not equal the weight of the block?

Sure. We either accept Newton's 2nd law, or we don't. Personally, I accept it.

And we can calculate this unknown T by solving the equations we make with the help of Newton's Second Law, right?

Sure. That's what this is all about.

Chet

 

[andyrk]

You are aware that when you have an action-reaction pair at the contact point between two bodies A and B, one member of the pair is the force exerted by body B on body A, and the other member of the pair is the force exerted by body A on body B, correct? When you are doing a force balance on body A, the two don't cancel. The force balance on body A includes only forces exerted by other bodies on body A, and not forces exerted by body A on other bodies. If both members of an action-reaction pair always had to be included, then no body could ever exert a force on any other body.

Can you see how this all relates to your rope and pulley?

Chet

One final question, why does the rope have to pull the pulley down at the point it touches the pulley and not up? Since tension can be in either direction, so why does it have to be down only?

 

[Chestermiller]

One final question, why does the rope have to pull the pulley down at the point it touches the pulley and not up? Since tension can be in either direction, so why does it have to be down only?

If the string were exerting an upward force on the pulley, the string would have to be under compression. Have you ever seen what happens to a string when you try to put it under compression? But that still doesn't prevent you from drawing an upward arrow for the force of the string acting on the pulley at the contact point, and it doesn't prevent you from drawing a downward arrow for the force of the pulley on the string. However, when you solve the problem, the value of this force would come out negative, indicating that the arrows should be pointing in the opposite direction.

Chet

 

[andyrk]

However, when you solve the problem, the value of this force would come out negative, indicating that the arrows should be pointing in the opposite direction.

You mean when I apply Newton's law on the blocks and strings and get the tension, I would get a specific direction for tension? But as I understood it, tension is always acting in 2 directions. So how can I get a single direction? Lastly, why doesn't the string have tension pointing in 2 directions at the point it is pulling the pulley down? (Even though it is unclear as to why it pulls it down..that is another question altogether at the moment).

 

[Chestermiller]

You mean when I apply Newton's law on the blocks and strings and get the tension, I would get a specific direction for tension? But as I understood it, tension is always acting in 2 directions. So how can I get a single direction? Lastly, why doesn't the string have tension pointing in 2 directions at the point it is pulling the pulley down? (Even though it is unclear as to why it pulls it down..that is another question altogether at the moment).

See posts 31-37.

Chet

 

[andyrk]

See posts 31-37.

Chet

Yes, but I still don't understand as to why does the string pulls the pulley down and not up? I don't think compression as an explanation which I need to go through because it is way beyond what my curriculum demands of me. Could you explain it more high school-ish terms which are easy to comprehend and fall within the realm of logical capacity for high school students?

And the reason you provided for why a point is pulley equally in 2 directions is not rigorous enough. Its too qualitative. Could you provide a more quantitative explanation? Also, why does a massless string have constant tension at all points? Can you explain this quantitatively rather than qualitatively?

 

[Chestermiller]

Yes, but I still don't understand as to why does the string pulls the pulley down and not up? I don't think compression as an explanation which I need to go through because it is way beyond what my curriculum demands of me. Could you explain it more high school-ish terms which are easy to comprehend and fall within the realm of logical capacity for high school students?

It doesn't matter whether you assume that the string is pulling down on the pulley or pushing up. When you set up Newton's 2nd law equations and solve the problem, the sign of the tension will take care of itself.

And the reason you provided for why a point is pulley equally in 2 directions is not rigorous enough. Its too qualitative. Could you provide a more quantitative explanation?

If body A exerts a contact force F on body B in the positive x direction, what force do you think that body B is exerting on body A (at their contact point)?

Also, why does a massless string have constant tension at all points? Can you explain this quantitatively rather than qualitatively?

See post #43. Just apply the same approach over a shorter segment of string.