Trouble with the concept of tension....

[Achintya]

Homework Statement

Two persons A & B are pulling the two ends of a rope (As people do in tug war).Person A pulls rope with a force=10N and the person B is pulling another end of the rope with a force =20N. According to Newton's 3rd, rope will also pull person A with force=10N and B with a force=20N. It means tension at one end of the rope is 10N and at the other end tension is 20N(Because of the definition of tension).Is it correct? If yes ,then pulling force on A by the rope will only depend on force exerted by A on rope and will not depend upon the force applied by B on the rope. In that case , if B is pulling the rope with a force=1000N and A is pulling the other end of rope with a force=10N, then the tension at that end(where A is applying force) will be only 10N.It means the force by which rope is pulling A will be only 10N. Is it possible?

Relevant Equations

Newton's laws and fundamental of mechanics

...

 

[PeroK]

This is very confused. There can only be one tension along the rope between the two people. If there is an unbalanced force on the rope, then the rope will accelerate (in which case the tension will vary slightly along the length of the rope depending on its mass). If we assume the mass of the rope is small compared to the forces involved, then the tension is still approximately constant.

In a tug of war, the two people are also pushing on the ground.

 

[Achintya]

This is very confused. There can only be one tension along the rope between the two people. If there is an unbalanced force on the rope, then the rope will accelerate (in which case the tension will vary slightly along the length of the rope depending on its mass). If we assume the mass of the rope is small compared to the forces involved, then the tension is still approximately constant.

In a tug of war, the two people are also pushing on the ground.

even if the rope is light(but not completely massless), then what forces will A experience...will it be just the Newton's 3rd law reaction pair force of his own force or the force exerted by B from the other side...pleasez it will be really helpful if you could explain with help of an example and on the basis of Newtonic mechanics...
i would really appreciate the help!

 

[archaic]

 

[Adesh]

A will experience the reaction force + the force applied by B (in the direction opposite to which A pulls)

 

[PeroK]

even if the rope is light(but not completely massless), then what forces will A experience...will it be just the Newton's 3rd law reaction pair force of his own force or the force exerted by B from the other side...pleasez it will be really helpful if you could explain with help of an example and on the basis of Newtonic mechanics...
i would really appreciate the help!

The simplest way to think about it is that there is a tension $T$ in the rope. And both people are pulling and being pulled with the same force $T$. For an idealised, massless rope there can be no unbalanced force. The forces at either end must be the same.

The difference in a tug-of-war comes from the forces exerted on the ground.

Alternatively, analyse a tug-of-war in space.

 

[Achintya]

A will experience the reaction force + the force applied by B (in the direction opposite to which A pulls)

then what will be the tension in the rope? Because we know that for a massive rope the tension won't be constant throughout the string rather it would be a function of the distance x from the end of the rope...by Newton's 3rd law if A pulls on the rope with 30 N , the rope will also pull A with 30 N ...so the tension of the rope segment connected to A will be 30N , then what about the force that B is exerting on the rope ...will it be of no significance to A?

 

[hutchphd]

If the rope has no mass then then tension in the rope will be the same everywhere (regardless of its state of acceleration). The force exerted by the rope at each end will be T in the obvious directions. If the rope has mass then the tension will depend upon the acceleration of the system of rope + people and will vary according to mass distribution in the rope (linear in a uniform rope) .
PLEASE ask one particular question at a time...not an incoherent jumble.

 

[PeroK]

then what will be the tension in the rope? Because we know that for a massive rope the tension won't be constant throughout the string rather it would be a function of the distance x from the end of the rope...by Newton's 3rd law if A pulls on the rope with 30 N , the rope will also pull A with 30 N ...so the tension of the rope segment connected to A will be 30N , then what about the force that B is exerting on the rope ...will it be of no significance to A?

The two ends of the rope cannot be pulled with different forces without the rope itself accelerating. And, if the rope is light, the difference in forces (tension) at each end will be small.

In an ideal analysis, you have $T$ constant throughout the rope.

If you add the mass of the rope, then you have $T + am$ at one end and $T$ at the other. The difference in tension along the rope is only $am$, where $a$ is the acceleration of the rope/system and $m$ is the mass of the rope. In the idealised scenario, you neglect the $am$ term, as it is small.

Your scenario of an approximately static, light rope pulled by $30N$ at one end and $20N$ at the other is physically impossible. That isn't a valid free-body diagram for a stationary rope.

 

[Adesh]

then what will be the tension in the rope? Because we know that for a massive rope the tension won't be constant throughout the string rather it would be a function of the distance x from the end of the rope...by Newton's 3rd law if A pulls on the rope with 30 N , the rope will also pull A with 30 N ...so the tension of the rope segment connected to A will be 30N , then what about the force that B is exerting on the rope ...will it be of no significance to A?

If the forces on the rope don’t cancel each other then the rope will accelerate. Tension exists in the static situation .

 

[hutchphd]

If the forces on the rope don’t cancel each other then the rope will accelerate. Tension exists in the static situation .

Let us be precise. Tension will exist for any (non-slack) rope. It will be uniform for a massive rope only if not accelerated.

 

[Achintya]

The two ends of the rope cannot be pulled with different forces without the rope itself accelerating. And, if the rope is light, the difference in forces (tension) at each end will be small.

In an ideal analysis, you have $T$ constant throughout the rope.

If you add the mass of the rope, then you have $T + am$ at one end and $T$ at the other. The difference in tension along the rope is only $am$, where $a$ is the acceleration of the rope/system and $m$ is the mass of the rope. In the idealised scenario, you neglect the $am$ term, as it is small.

Your scenario of an approximately static, light rope pulled by $30N$ at one end and $20N$ at the other is physically impossible. That isn't a valid free-body diagram for a stationary rope.

But i never said that the rope is stationary...it is obvious that if there are unbalanced forces on a system then the system would accelerate according to Newton's 2nd law

 

[Achintya]

If the forces on the rope don’t cancel each other then the rope will accelerate. Tension exists in the static situation .

so you mean once the system accelerates there will be no tension in the rope?

 

[Adesh]

so you mean once the system accelerates there will be no tension in the rope?

Take a rope which is not attached to any thing, if you pull it then it will certainly going to move and the stretch in the rope will be due to molecular attraction/repulsion.

 

[PeroK]

But i never said that the rope is stationary...it is obvious that if there are unbalanced forces on a system then the system would accelerate according to Newton's 2nd law

Okay, but what about the case where two people are anchored to the ground and one pulls the rope with $30N$ and the other pulls the rope with $20N$?

That's impossible.

It's also invalid for a massless rope, since any unbalanced force would lead to an infinite acceleration of the rope.

It's also invalid for a light rope ($1kg$, say), unless the rope and the whole system is accelerating at $10m/s^2$.

Fundamentally, you cannot have different forces pulling two ends of a rope without the rope accelerating under those forces. If there are other objects in the system stopping it from accelerating, then those cannot be the forces on the rope.

 

[Achintya]

Let us be precise. Tension will exist for any (non-slack) rope. It will be uniform for a massive rope only if not accelerated.

yes i have understood the fact that if system is in static equilibrium(0 acceleration), then tension would be constant throughout the string whether it be massless or massive...but what happens when the system accelerates like in my original question(20N and 30N forces are acting at each end of the string, making the system accelerate)...i hope i am clear.

 

[PeroK]

yes i have understood the fact that if system is in static equilibrium(0 acceleration), then tension would be constant throughout the string whether it be massless or massive...but what happens when the system accelerates like in my original question(20N and 30N forces are acting at each end of the string, making the system accelerate)...i hope i am clear.

That's physically impossible.

 

[Achintya]

Take a rope which is not attached to any thing, if you pull it then it will certainly going to move and the stretch in the rope will be due to molecular attraction/repulsion.

yes i could feel this

 

[Achintya]

That's physically impossible.

it would be really helpful if you can explain me why is it physically impossible.

 

[PeroK]

it would be really helpful if you can explain me why is it physically impossible.

Newton's second law: $F = ma$.

The net force on the rope (tension at one end minus tension at the other) must equal the mass of the rope times its acceleration. In this case:

$F = T_2 - T_1 = ma \approx 0$

 

[Adesh]

yes i could feel this

Consider this scenario:

We got a sponge, I push it from east and you push it from west with same force. Would the sponge going to be squeezed?

 

[hutchphd]

Are the people in your original question rigidly attached to the earth? Are they holding the ends of a fixed length string? So they must also be accelerating. Please succinctly rephrase your question because it is inconsistent as stated.

 

[PeroK]

Consider this scenario:

We got a sponge, I push it from east and you push it from west with same force. Would the sponge going to be squeezed?

I fail to see the relevance of this.

 

[Adesh]

I fail to see the relevance of this.

I’m trying to explain to him with that sponge example because deformation in sponge is very intuitive. If we have unequal forces on sponge it will move with some net deformation in it.

 

[hutchphd]

unless your rope stretches I don't see the relevance. Stick to original

 

[Adesh]

unless your rope stretches I don't see the relevance. Stick to original

Tension will be caused only when there is a stretch, in my opinion.

 

[hutchphd]

Your opinion is incorrect. Tension implies nothing about stretch inherently. An "ideal" rope is absolutely inexstensible but still has tension. Not a contradiction. In practice all things stretch a little (although Carbon Fiber is remarkably resistant) but the approximation is fine. Rigid bodies aren't absolutely rigid either!
So restate your original question in a concise and precise way please, and it will be addressed..

 

[Adesh]

Your opinion is incorrect. Tension implies nothing about stretch inherently.

It seems that you’re using this definition of tension :
The tension force is the force that is transmitted through a cable, rope, wire or string when it is pulled tight by forces acting from opposite ends. It is directed along the length of the cable and pulls equally on the objects on the opposite ends of the wire..

What I meant by the tension was the restoring force that the rubber band exerts when we pull (keeping one it’s ends fixed).

 

[hutchphd]

The two are not mutually exclusive. You may use what you like but the rest of the world uses the more generic one you quote. If you wish to describe an elastic rope then you need to indicate that and characterize it as to details. Otherwise everyone will be confused as you have witnessed. Hopefully lesson learned!

 

[Adesh]

The two are not mutually exclusive. You may use what you like but the rest of the world uses the more generic one you quote. If you wish to describe an elastic rope then you need to indicate that and characterize it as to details. Otherwise everyone will be confused as you have witnessed. Hopefully lesson learned!

Yes! I should use those things the way everyone uses it (if I’m posting it in public).

 

[Achintya]

Newton's second law: $F = ma$.

The net force on the rope (tension at one end minus tension at the other) must equal the mass of the rope times its acceleration. In this case:

$F = T_2 - T_1 = ma \approx 0$

here you are considering massless string but i am asking you about massive string(m not equals 0)

 

[Achintya]

Are the people in your original question rigidly attached to the earth? Are they holding the ends of a fixed length string? So they must also be accelerating. Please succinctly rephrase your question because it is inconsistent as stated.

no they are not rigidly attached to the earth...they are simply standing on the Earth's surface and are free to accelerate.. yes they are holding the ends of the fixed length massive string and are applying forces of different magnitudes(20N and 30N in my case)...which would make the system accelerate with a net force of 10N...so i want to know the tension produced in the string in this case...it would be better if you could explain me what is exactly happening to the string at the molecular level with the help of Newton's laws

 

[PeroK]

here you are considering massless string but i am asking you about massive string(m not equals 0)

Okay. Let's assume the rope has a mass of $1kg$ and is accelerating at $1m/s$. Then:

$T_2 - T_1 = 1N \ne 10N$

You cannot have $T_2 = 30N$ and $T_1 = 20N$ unless the rope is accelerating at $10m/s^2$. Assuming the mass of th e rope is $1kg$.

If you focus on the rope, the forces on the rope (and only the forces on the rope) determine its acceleration. If you have a large unbalanced force on the rope it must be accelerating rapidly.

If the rope is attached to things that stop it accelerating, then that influences the tension. The tension at either end must balance if the rope is not accelerating.

 

[Achintya]

Okay. Let's assume the rope has a mass of $1kg$ and is accelerating at $1m/s$. Then:

$T_2 - T_1 = 1N \ne 10N$

You cannot have $T_2 = 30N$ and $T_1 = 20N$ unless the rope is accelerating at $10m/s^2$. Assuming the mass of th e rope is $1kg$.

If you focus on the rope, the forces on the rope (and only the forces on the rope) determine its acceleration. If you have a large unbalanced force on the rope it must be accelerating rapidly.

If the rope is attached to things that stop it accelerating, then that influences the tension. The tension at either end must balance if the rope is not accelerating.

but the net force on the rope is 10N..so why did you take it 1N.

 

[PeroK]

but the net force on the rope is 10N..so why did you take it 1N.

It can't be. It's physically impossible, Which part of $F = ma$ don't you understand? The tension in the rope and its acceleration are governed by Newton's laws. You can have a difference of $10N$ but if you do a $1kg$ rope must accelerate at $10m/s$. You can't have arbitrary tension at either end. Newton's laws forbid it.

If you tie a rope to a wall and pull, then whatever force you apply is equalled by the wall. You can't have an unbalanced force in that case, If you ask: how come the wall pulls back with the same force?, then you have Isaac Newton to thank for the answer: $F = ma$. There is no acceleration, hence the force (and tension) at the wall equals the force you apply at the other end.

If you ask how does nature achieve this? Then, the answer is because she's clever!

 

[hutchphd]

If I may reiterate the answer to the original question:
The answer is yes. For a massive rope all of those possibilities exist; the bigger the net unbalanced force from A and B the bigger the acceleration. The tension in the uniform rope will smoothly match the external forces at each end and taper in magnitude linearly down the (accelerating) rope between.

 

[Achintya]

It can't be. It's physically impossible, Which part of $F = ma$ don't you understand? The tension in the rope and its acceleration are governed by Newton's laws. You can have a difference of $10N$ but if you do a $1kg$ rope must accelerate at $10m/s$. You can't have arbitrary tension at either end. Newton's laws forbid it.

If you tie a rope to a wall and pull, then whatever force you apply is equalled by the wall. You can't have an unbalanced force in that case, If you ask: how come the wall pulls back with the same force?, then you have Isaac Newton to thank for the answer: $F = ma$. There is no acceleration, hence the force (and tension) at the wall equals the force you apply at the other end.

If you ask how does nature achieve this? Then, the answer is because she's clever!

i totally agree with you with that wall example...but in that case basically we have equal forces at the two ends of the rope because of Newton's third law and thus the tension equaled the magnitude of the two equal and opposite action-reaction pairs...but where my confusion lies is that what would happen if the system somehow accelerates(which is only possible when the forces at the two ends of the ropes are unequal)...so for that case what would be the tension in the string...see according to me at the end of A(exerting 20N force), the tension of the rope segment joining the man A should be 20N (Newton's 3rd law)...correct me if i am wrong here...then same is for other end B(where the force exerted is 30N)...then the tension of the rope segment lying between these two segments will range from 20 to 30N ...also correct me if i am wrong here

 

[Achintya]

If I may reiterate the answer to the original question:
The answer is yes. For a massive rope all of those possibilities exist; the bigger the net unbalanced force from A and B the bigger the acceleration. The tension in the uniform rope will smoothly match the external forces at each end and taper in magnitude linearly down the (accelerating) rope between.

so yes that's what my basic question was all about ...will the tension experienced by A is not effected by the force exerted by B...[[PLEASE REFER TO THE MSG I SENT ABOVE]]

 

[PeroK]

i totally agree with you with that wall example...but in that case basically we have equal forces at the two ends of the rope because of Newton's third law and thus the tension equaled the magnitude of the two equal and opposite action-reaction pairs...but where my confusion lies is that what would happen if the system somehow accelerates(which is only possible when the forces at the two ends of the ropes are unequal)...so for that case what would be the tension in the string...see according to me at the end of A(exerting 20N force), the tension of the rope segment joining the man A should be 20N (Newton's 3rd law)...correct me if i am wrong here...then same is for other end B(where the force exerted is 30N)...then the tension of the rope segment lying between these two segments will range from 20 to 30N ...also correct me if i am wrong here

Now, replace the wall with another person holding the rope. Unless they move, they are no more able than the wall to pull with a lesser force. And, if they pull with a greater force, then you have no choice but to move or pull with the same force. Or, let go of the rope.

You must match their force, move or let go.

In other word, the people have the freedom to pull with different forces, but if they do, then motion must result. There cannot be what you demand: people pulling with different forces yet no motion. That's physically impossible.

 

[Achintya]

Now, replace the wall with another person holding the rope. Unless they move, they are no more able than the wall to pull with a lesser force. And, if they pull with a greater force, then you have no choice but to move or pull with the same force. Or, let go of the rope.

You must match their force, move or let go.

In other word, the people have the freedom to pull with different forces, but if they do, then motion must result. There cannot be what you demand: people pulling with different forces yet no motion. That's physically impossible.

but i never that motion is not occurring ...please don't misinterpret my question
forget about the rope ..just imagine me and you ...if i pull you with 100N force,but your maximum capacity is 80N...so a net 20N force will act on you...m i correct here?

 

[haruspex]

where my confusion lies is that what would happen if the system somehow accelerates(which is only possible when the forces at the two ends of the ropes are unequal)...so for that case what would be the tension in the string...see according to me at the end of A(exerting 20N force), the tension of the rope segment joining the man A should be 20N (Newton's 3rd law)...correct me if i am wrong here...then same is for other end B(where the force exerted is 30N)...then the tension of the rope segment lying between these two segments will range from 20 to 30N ...also correct me if i am wrong here

If we ignore the mass of the rope, you are simply posing an impossible scenario. The two will always apply the same force to the rope.
Take a more obvious situation. You hold the upper end of a rope and on the other hangs a mass m. The mass, via gravity, exerts a force mg and so do you.
Now you pull with a force 2mg. The mass accelerates upwards at rate g and now exerts a force 2mg on the rope to match yours.

 

[Achintya]

If we ignore the mass of the rope, you are simply posing an impossible scenario. The two will always apply the same the same force to the rope.
Take a more obvious situation. You hold the upper end of a rope and on the other hangs a mass m. The mass, via gravity, exerts a force mg and so do you.
Now you pull with a force 2mg. The mass accelerates upwards at rate g and now exerts a force 2mg on the rope to match yours.

so what's the tension in the rope here?

 

[PeroK]

okay forget about the rope ..just imagine me and you ...if i pull you with 100N force,but your maximum capacity is 80N...so a net 20N force will act on you...m i correct here?

No. My mass is, give or take, $100kg$. If you pull me with $100N$, then I must be capable of a force of $100N$. I then accelerate at $1m/s$ under a net force of $100N$.

If I am not capable of generating a force of $100N$, then you cannot pull me with $100N$. I either lose my grip on the rope or my arms get pulled off or whatever.

 

[Achintya]

No. My mass is, give or take, $100kg$. If you pull me with $100N$, then I must be capable of a force of $100N$. I then accelerate at $1m/s$ under a net force of $100N$.

If I am not capable of generating a force of $100N$, then you cannot pull me with $100N$. I either lose my grip on the rope or my arms get pulled off or whatever.

so you mean if you are not able to produce 100 N then Newton's third law will not be valid?(i mean in that case i won't feel any reactive force of my applied force?)

 

[etotheipi]

The simplest way to think about it is that there is a tension $T$ in the rope. And both people are pulling and being pulled with the same force $T$. For an idealised, massless rope there can be no unbalanced force. The forces at either end must be the same.

The difference in a tug-of-war comes from the forces exerted on the ground.

Alternatively, analyse a tug-of-war in space.

I might be a little late to the party, but I thought I'd add that the above is undoubtedly the best way to think of it.

The problem isn't helped by the fact that students are introduced to Hooke's law by considering only an external force acting on a spring (i.e. due to a hanging mass). You might then ask what the force exerted on the spring by the hook (no pun intended) on which it is hung is? Then the obvious question is what does the $F$ in $F=ke$ refer to? Only one of these forces? Both? This can cause problems.

You must think of tension as a force exerted by the string/spring etc. on anything attached to/in contact with it. A taut horizontal string will have tension (which, if the string is massive and accelerating will be a function of position along its length, or if it is massless will be constant along its length). The force exerted on any object tied to the ends of the spring will be $T$. Indeed, even if the string is bent around a pulley, we can take two little tension forces in the directions of the string.

So returning to the example of the spring, the resolution becomes simple (I'll assume the spring is massless, just to make things easier!). The spring exerts a force of $T$ on both the hanging mass and the hook, and by $\text{N III}$ these will both exert forces $T$ on the spring.

 

[PeroK]

so you mean if you are not able to produce 100 N then Newton's third law will not be valid?

No. It means you can't pull me with a force of $100N$. Force is not a fundamental factor in these cases. It's really power.

Let's say you throw a cricket ball as hard as you can. Let's say it weights $160g$. You might be able to accelerate it at $20m/s^2$. I'm guessing. So, a force of $3.2N$.

If we take a table tennis ball. You throw it as fast as you can. Let's say it weights $3g$. You might be able to accelerate it at about the same rate. You cannot generate a force of $3.2N$ in this case.

Similarly, if you have a solid cable attached to a car. You may be able to pull with a force of $500N$.

But, if you pull a toy car, with a mass of only $1kg$, you couldn't possible generate a force of $500N$.

Your ability to generate a force on an object depends on the object; not just on you.

 

[PeroK]

PS another example is pulling a person against pulling a papier mache dummy. If you try to pull the dummy with too much force you pull its arms off. You may momentarily generate a greater force than it can sustain, but only for the short time that it takes to pull the object apart.

 

[Achintya]

No. My mass is, give or take, $100kg$. If you pull me with $100N$, then I must be capable of a force of $100N$. I then accelerate at $1m/s$ under a net force of $100N$.

If I am not capable of generating a force of $100N$, then you cannot pull me with $100N$. I either lose my grip on the rope or my arms get pulled off or whatever.

NO , if you are not able to generate 100 N force..then your whole body will accelerate with the net force(100- the max. force that you are capable of generating), in no case your arm gets pulled of

 

[Achintya]

I might be a little late to the party, but I thought I'd add that the above is undoubtedly the best way to think of it.

The problem isn't helped by the fact that students are introduced to Hooke's law by considering only an external force acting on a spring (i.e. due to a hanging mass). You might then ask what the force exerted on the spring by the hook (no pun intended) on which it is hung is? Then the obvious question is what does the $F$ in $F=ke$ refer to? Only one of these forces? Both? This can cause problems.

You must think of tension as a force exerted by the string/spring etc. on anything attached to/in contact with it. A taut horizontal string will have tension (which, if the string is massive and accelerating will be a function of position along its length, or if it is massless will be constant along its length). The force exerted on any object tied to the ends of the spring will be $T$. Indeed, even if the string is bent around a pulley, we can take two little tension forces in the directions of the string.

So returning to the example of the spring, the resolution becomes simple (I'll assume the spring is massless, just to make things easier!). The spring exerts a force of $T$ on both the hanging mass and the hook, and by $\text{N III}$ these will both exert forces $T$ on the spring.

well i agree with you but where my confusion lies is that what would happen if the system somehow accelerates(which is only possible when the forces at the two ends of the ropes are unequal)...so for that case what would be the tension in the string...see according to me at the end of A(exerting 20N force), the tension of the rope segment joining the man A should be 20N (Newton's 3rd law)...correct me if i am wrong here...then same is for other end B(where the force exerted is 30N)...then the tension of the rope segment lying between these two segments will range from 20 to 30N ...also correct me if i am wrong here

 

[haruspex]

so what's the tension in the rope here?

The same as the force at each end.