Trouble with the concept of tension....

[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.

 

[haruspex]

if the system somehow accelerates(which is only possible when the forces at the two ends of the ropes are unequal)

If the forces at the ends are F1 and F2 then there is a net force F1-F2 on the rope. If the rope has mass m then the acceleration of the rope is (F1-F2)/m. The tension will vary along the rope from F1 at one end to F2 at the other.
No rope is truly massless, but if the mass is very small then either the force difference is very small or the acceleration is very large.

 

[PeroK]

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

I draw your attention to Newton's third law.

If you pull me with a force of 100N, then there is an equal an opposite reaction and I miust pull you with a force of 100N. We both accelerate (depending on our masses) in response to a 100N force. If we both have masses of 100kg, then we both accelerate towards each other at $1m/s^2$.

Your analysis, that neither of us would accelerate "because forces balance" is wrong. You are confusing Newton's second and third laws.

If you pull me one way with $100N$ and someone else pulls me in the opposite direction with $80N$, then the net force on me is $20N$.

In order for this to happen, I must be able to sustain these forces and the resulting internal strain.

 

[Achintya]

The same as the force at each end.

you mean the tension at each end is equal to the force applied at each end (by Newton's third law?)

 

[Achintya]

I draw your attention to Newton's third law.

If you pull me with a force of 100N, then there is an equal an opposite reaction and I miust pull you with a force of 100N. We both accelerate (depending on our masses) in response to a 100N force. If we both have masses of 100kg, then we both accelerate towards each other at $1m/s^2$.

Your analysis, that neither of us would accelerate "because forces balance" is wrong. You are confusing Newton's second and third laws.

If you pull me one way with $100N$ and someone else pulls me in the opposite direction with $80N$, then the net force on me is $20N$.

In order for this to happen, I must be able to sustain these forces and the resulting internal strain.

yes you are right...

 

[Achintya]

If the forces at the ends are F1 and F2 then there is a net force F1-F2 on the rope. If the rope has mass m then the acceleration of the rope is (F1-F2)/m. The tension will vary along the rope from F1 at one end to F2 at the other.
No rope is truly massless, but if the mass is very small then either the force difference is very small or the acceleration is very large.

That's what i want to know that in this case, if A is exerting force of F1 and B is exerting force of F2,,,then the tension at the end of A and B will be F1 and F2 respectively..(acc. to Newton's 3rd law?)...if YES then my last question is will the tension at the end of A remain unaffected by the force exerted by B?

 

[haruspex]

if A is exerting force of F1 and B is exerting force of F2,,,then the tension at the end of A and B will be F1 and F2 respectively.

Yes, but remember that:
- if we are to take the rope as massless then F1and F2 must be the same.
- if we take the mass of the rope as very small yet F1 and F2 differ significantly then the rope will have a great acceleration, so the bodies at the ends of the rope will also have a great acceleration.
Your basic misunderstanding is that you seem to think that, say, two people can pull on opposite ends of a rope each with whatever force they care to apply, independently of the other. This is simply not true. Take an extreme case: there is no-one on the other end. You can try to pull hard on your end but you will not succeed. You will fly backwards and create very little tension in the rope.

will the tension at the end of A remain unaffected by the force exerted by B?

The tension at end A will equal the force F1. The ability of the body at end A to exert a force on the rope may be limited by the body's own inertia - as in my one person tug of war example above. If the force F2 is increased it may allow F1 to increase as well, so whether F2 affects the tension at end A depends on what you mean by that.

 

[Achintya]

Your basic misunderstanding is that you seem to think that, say, two people can pull on opposite ends of a rope each with whatever force they care to apply, independently of the other. This is simply not true. Take an extreme case: there is no-one on the other end. You can try to pull hard on your end but you will not succeed. You will fly backwards and create very little tension in the rope.

The tension at end A will equal the force F1. The ability of the body at end A to exert a force on the rope may be limited by the body's own inertia - as in my one person tug of war example above. If the force F2 is increased it may allow F1 to increase as well, so whether F2 affects the tension at end A depends on what you mean by that.

basically what i meant is that if A exerts a force F1 and B is not able to exert the same force F1...so naturally B will accelerate towards A...so how is Newton's third law satisfied here...as B is not able to exert the same force(F1)...what my basic confusion is that whether the Newton's 3rd law reaction of B (in response to the force F1 by A ) is same as the force exerted by B(F2)... and if B is experiencing the tension equal to its own applied force(F2)by the adjacent rope segment then why would B accelerate towards A...i might sound confusing but it will be great if you could help me come out of this misconception..thanks in advance...

 

[PeroK]

basically what i meant is that if A exerts a force F1 and B is not able to exert the same force F1...so naturally B will accelerate towards A...

That's the same mistake again. By Newton's laws that's not what will happen at all. If you look at a free-body diagram for that:

$A$ is subject to a reaction force of $F_1$ from the rope and will accelerate towards $B$ at an acceleration of $F_1/m_A$.

$B$ will accelerate towards $A$ at $F_2/m_B$.

The rope will accelerate in the direction of $B$ to $A$ at an acceleratation of $(F_1 - F_2)/m$, where $m$ is the mass of the rope.

If $A$ and $B$ are anchored to the ground, then only the rope will accelerate. In this case, the rope must be moving through their hands.

 

[Achintya]

$A$ is subject to a reaction force of $F_1$ from the rope and will accelerate towards $B$ at an acceleration of $F_1/m_A$.

$B$ will accelerate towards $A$ at $F_2/m_B$.

okay, so basically this reaction force appears as the tension respectively at the two ends?

 

[PeroK]

okay, so basically this reaction force appears as the tension respectively at the two ends?

Yes. There's a third law pair of forces at either end of the rope.