Trouble with the concept of tension....

[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.

 

[Achintya]

OKAY THANKS A LOT EVERYONE FOR BEING WITH ME AND HELPING ME OUT...i REALLY APPRECIATE YOU GUYS FOR YOUR LOVE FOR PHYSICS AND THE WAY YOU GUYS ANSWERED MY NEVER ENDING QUESTIONS WITH SUCH PATIENCE...THANKS FOR YOUR TIME AND NOW I CAN FINALLY SAY THAT U HAVE REDUCED MY TENSION REGARDING TENSION.! Hope to see you guys again with some other box of doubts.
REGARDS,
Achintya(From INDIA)

 

[PeroK]

OKAY THANKS A LOT EVERYONE FOR BEING WITH ME AND HELPING ME OUT...i REALLY APPRECIATE YOU GUYS FOR YOUR LOVE FOR PHYSICS AND THE WAY YOU GUYS ANSWERED MY NEVER ENDING QUESTIONS WITH SUCH PATIENCE...THANKS FOR YOUR TIME AND NOW I CAN FINALLY SAY THAT U HAVE REDUCED MY TENSION REGARDING TENSION.! Hope to see you guys again with some other box of doubts.
REGARDS,
Achintya(From INDIA)

Finally, to go back to a question I asked many post ago about a tug-of-war in space. In that scenario, each astronaut accelerates according to how much he/she pulls the rope and independent of how much the other pulls the rope.

 

[Achintya]

Finally, to go back to a question I asked many post ago about a tug-of-war in space. In that scenario, each astronaut accelerates according to how much he/she pulls the rope and independent of how much the other pulls the rope.

YES! got that.. but on Earth surface we don't accelerate with our own force because we get a equal and opposite reaction from the ground...right?

 

[jbriggs444]

YES! got that.. but on Earth surface we don't accelerate with our own force because we get a equal and opposite reaction from the ground...right?

We do not accelerate due to the force of the rope on our hands because there is an equal and opposite force from the ground, yes.

But that force from the ground is not a "reaction" force.

Conventionally we reserve the term "reaction force" to refer to the third law partner of a given force. The force of the ground on your feet is not the third law partner to the force of the rope on your hands. The third law partner of the force of the rope on your hands is the force of your hands on the rope.

The third law assures us that momentum is conserved for the system as a whole. Any momentum lost by A due to the force on it from B is equal and opposite to the momentum gained by B due to the force on it from A.

 

[sophiecentaur]

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.

No. It will stretch because it has mass if it is accelerating. (N2)

 

[kuruman]

Now that the answer has been sorted out, I should point out that having 20 N at one end of a string and 30 N at the other is not as unphysical as it may look. Imagine an Atwood machine with 20 N and 30 N at each end. The 20 N weight is tied to the floor with a second string that is under tension. Clearly the tension in the Atwood string is 30 N and the tension in the floor string is 10 N; the system is in equilibrium and the acceleration of each mass is $a=0.$ At $t=0$ the floor string is burnt with a blowtorch. What is the acceleration at $t=0?$ If we agree that some finite time is required for the acceleration to jump from zero to the steady-state common value, $a=\dfrac{m_2-m_1}{m_2+m_1}g$, then it is reasonable to conclude that the tension is 20 N at one end and 30 N at the other at $t=0.$

 

[Stephen Tashi]

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

Imagine the rope modeled as a series of small masses connected by hook's law springs. A change in the force exerted at one end of the rope is not instantly transmitted to small mass at the other end. If unequal forces act on opposite ends of the rope at time t = 0, the motion of the masses can be quite complicated. the tensions in the springs can be different and change as a function of time.

So if you want to talk about "the" tension in such a model, you have to imagine the system getting into a steady state where all the tensions in each spring are constant. However, if we are using idealized springs that don't dissipate energy, this will never happen. The system can oscillate forever as it (as a whole) accelerates.

 

[kkps]

restating the question. A is pulling a massless rope with 10N, B is pulling in opposite direction of A with 20N. What is tension on string? It is 15N.

10N<---A--->------<---B--->20N
Let us assume mass of A and B = 1kg. Let us denote tension on rope = Ft. Let us denote acceleration of A&B=a

freebody diagram of A:
10N<---A--->Ft
Ft-10N=1*a

freebody diagram of B:
Ft<---B--->20N
Ft-20N=-1*a
-1*a because Ft and a are in opposite directions for B

Equation1: Ft-10N=a
Equation2: Ft-20N=-a
Equation1-Equation2: 10N=2a
a=5N

Ft=15N

 

[PeroK]

You should open a new thread. Your answer makes little sense to me.

 

[kuruman]

restating the question. A is pulling a massless rope with 10N, B is pulling in opposite direction of A with 20N. What is tension on string? It is 15N.

You realize that thread is 2 years old. That said, your free body diagram is incorrect. If the rope is the only entity that exerts a force on A and B, the direction of the force on A is from A to B and the direction of the force on B is from B to A. These two are equal and opposite according to Newton's third law.

 

[haruspex]

Ft-10N=1*a

No. That equation says -10N is the force A is exerting on the ground. Your equations lead to A exerting a force -15N on the rope and B exerting a force +15N on it.

 

[kkps]

You realize that thread is 2 years old. That said, your free body diagram is incorrect. If the rope is the only entity that exerts a force on A and B, the direction of the force on A is from A to B and the direction of the force on B is from B to A. These two are equal and opposite according to Newton's third law.

10N<-A---ROPE---B->20N
is not different from
<-A(say 1Kg)hanging on a rope on left of pulley---ROPE---B(say 2Kg)hanging on a rope on right of pulley
this is same as Atwood machine problem
of course, we can add force on ground etc. to problem and complicate, but it is not needed to solve the problem IMO
I think it can be solved by my simple free body diagram which are drawn similar to Atwood machine solution
and, of course, i am assuming massless rope to simplify

>You realize that thread is 2 years old
yes, but i think someone will still search for this, and i think the question was not answered well

 

[kkps]

No. That equation says -10N is the force A is exerting on the ground. Your equations lead to A exerting a force -15N on the rope and B exerting a force +15N on it.

please see my answer to kuruman. A is pulling on rope with 10N to left, B is pulling on rope with 20N to right. we do not have to add Ground in picture, A is somehow generating 10N, B 20N. Then we can translate the question to an Atwood machine problem

I have still not answered the "rope with mass" part of the original question. I am working on it

 

[kkps]

You should open a new thread. Your answer makes little sense to me.

My answer is treating the problem like an Atwood machine problem. I think I forgot to mention that I am treating the string as massless. I am working on a simple explanation for the "string with mass" case. I think this thread is a good interaction, though there is lot of unnecessary confusion in the discussion. But the original question is a doubt anyone could have

 

[PeroK]

there is lot of unnecessary confusion in the discussion.

Which you've just added to! Your solution is conceptually wrong, as it violates Newton's third law, which I think has already been pointed out.

 

[kkps]

Which you've just added to! Your solution is conceptually wrong, as it violates Newton's third law, which I think has already been pointed out.

can you explain why this question cannot be simplified to an Atwood machine question as I pointed out in reply to kuruman? After that, can you explain why my solution is conceptually wrong and how it violates Netwon's third law?

FBD of A
10N<-A->15N A is moving to right with a=5, if m of A=1kg

FBD of B
15N<-B->20N B is moving to right with a=5, if m of B=1kg

FBD of rope
15N<-Rope(massless)->15N

what force is not balanced violating third law?

 

[PeroK]

what force is not balanced violating third law?

B cannot be pulling on the string with $20N$ and the string pulling back with $15N$. That violates Newton's third law.

 

[haruspex]

A is pulling on rope with 10N to left, B is pulling on rope with 20N to right

That is the proposition in post #1. It appears to be a situation invented by the student, not a problem he or she was given.
If the rope has zero mass then it is not physically possible, and if you start with an impossible combination of facts you can deduce anything.

 

[haruspex]

we do not have to add Ground in picture

In post #68 you described your equations as based on free body considerations for A and B. Necessarily that means your are considering all horizontal forces on those bodies, namely, the force from the ground and the force from the rope.

If your FBD has the force of tension and the force of a person pulling, what is the body those two forces are acting on?

You seem not to understand the concept of an FBD.

 

[kkps]

B cannot be pulling on the string with $20N$ and the string pulling back with $15N$. That violates Newton's third law.

to make it simple, translate the question to this:
A(1Kg)hanging on a rope on left of pulley---massless ROPE---B(2Kg)hanging on a rope on right of pulley
in the above (pulley) case, the entire system (A-rope-B) will fall to right

This is an Atwood machine problem
The original question is exactly same as I see it
solution to above Atwood problem is: a=5, Ft=15

>B cannot be pulling on the string with $20N$ and the string pulling back with $15N$.
it is very much possible. that is what causes motion!

 

[kkps]

That is the proposition in post #1. It appears to be a situation invented by the student, not a problem he or she was given.
If the rope has zero mass then it is not physically possible, and if you start with an impossible combination of facts you can deduce anything.

you are trying to prove that the question makes no sense, but the question makes a lot of sense to the original poster. I am trying to understand what the original poster meant, and answer the intent of the original poster.

Most textbooks have similar questions where, say, A pulls something to left. All that the student has to consider to answer the question is that there is a force to left. There is no need to consider where the force is coming from, the type of shoes he is wearing, the surface he is standing on (ice vs slush), etc.

Imagine you are in a classroom and the teacher has asked this question, and imagine the teacher walked away. You would answer the question making some assumptions yourself, rather than heckling the teacher with more questions. I am taking this approach

Also, zero mass of rope is used in all basic textbooks in all such problems. if you add mass, your answer is not going to be much different . I will post that answer soon

 

[kkps]

In post #68 you described your equations as based on free body considerations for A and B. Necessarily that means your are considering all horizontal forces on those bodies, namely, the force from the ground and the force from the rope.

If your FBD has the force of tension and the force of a person pulling, what is the body those two forces are acting on?

You seem not to understand the concept of an FBD.

please see post #80
IMO, post #1 is asking the same question as shown in the diagram of post #80
(as I mentioned in post #81, some simplification of original question can be applied to give a meaningful answer without further questioning - force from ground etc. are irrelevant to the original problem)

 

[jbriggs444]

you are trying to prove that the question makes no sense, but the question makes a lot of sense to the original poster. I am trying to understand what the original poster meant, and answer the intent of the original poster.

If the original poster intends a massless rope then the question posits an impossible scenario according to Newton's second law. $\sum F = ma$. $m = 0$ so $\sum F = 0$.

If the original poster intends a rope with mass then the scenario is possible.

Both possibilities have been contemplated in already. For instance, see post #6 which contemplates a massless rope and #8 which looks at both scenarios -- a rope with or without mass.

The possibility of A pulling on the rope with 20N and the rope pulling on A with 15N is a direct violation of Newton's third law and is nonsense.

 

[PeroK]

if you add mass, your answer is not going to be much different . I will post that answer soon

Please don't!

 

[kuruman]

Let's examine and interpret the equations in your drawing. You have

Equation 1: Ft - 10 = 1*a
Here Ft is the tension, 10 (N) is the weight of the mass on the left and 1*a is mass*acceleration

Equation 2: Ft - 20 = 1*(-a)

Here Ft is the tension, 20 (N) is the weight of the mass on the right and 1*(-a) $\dots~$ is what ? If the mass is 2 kg, the term should be 2*(-a) to be correct. Then one has
Ft - 10 (N) = a
Ft - 20 (N) = -2a
If the first equation is doubled and added it to the second, one gets
3 Ft - 40 (N) = 0 ⇒ Ft = 40/3 (N) = 13.3 (N) and that is the correct tension in the string connecting the masses. More generally, $$F_T=\frac{2m_1m_2}{m_1+m_2}g.

 

[haruspex]

B cannot be pulling on the string with 20N

Indeed, it is not. Gravity pulls on B with 20N, but if B is accelerating at 5m/s², B only pulls on the string with 15N, same as the tension.

What if we let A's mass dwindle to zero? B will be in free fall, so no tension, but according to your reasoning it is still pulling on the string with 20N.

 

[haruspex]

Please don't!

Well, it would be interesting to see that, and take the limit as the rope's mass tends to zero.

 

[kuruman]

Indeed, it is not. Gravity pulls on B with 20N, but B is accelerating at 5m/s², so B only pulls on the string with 15N, same as the tension.

For $m_1 = 1$ kg and $m_2 = 2$ kg, the acceleration is $$a=\frac{m_2-m_1}{m_2+m_1}g=\frac{1}{3}g$$and the tension is 13$\frac{1}{3}$ N as indicated in post #85.

 

[haruspex]

For $m_1 = 1$ kg and $m_2 = 2$ kg, the acceleration is $$a=\frac{m_2-m_1}{m_2+m_1}g=\frac{1}{3}g$$and the tension is 13$\frac{1}{3}$ N as indicated in post #85.

Thanks. I made the mistake of assuming @kkps's numbers were correct, just the interpretation that was wrong. I've edited it to avoid making that assumption.

 

[kuruman]

To @kkps :
Here is a multiple choice question to help you sort this out.
Two teams are engaged in a tug o' war as shown in the figure below. The team on the left is pulling on the rope with a force of 100 N and the team on the right with a force of 200 N. A force gauge is attached to the rope as shown. What does the gauge read?

(A) 100 N
(B) 150 N
(C) 200 N
(D) The people on the left read 100 N and the people on the right read 200 N.
(E) This question makes no physical sense.

Hint: How would your answer change if the right side of the gauge were attached to an immovable wall instead of being pulled by the team on the right? What about swapping the left team with the immovable wall?

 

[kkps]

Let's examine and interpret the equations in your drawing. You have

View attachment 316230
Equation 1: Ft - 10 = 1*a
Here Ft is the tension, 10 (N) is the weight of the mass on the left and 1*a is mass*acceleration

Equation 2: Ft - 20 = 1*(-a)

Here Ft is the tension, 20 (N) is the weight of the mass on the right and 1*(-a) $\dots~$ is what ? If the mass is 2 kg, the term should be 2*(-a) to be correct. Then one has
Ft - 10 (N) = a
Ft - 20 (N) = -2a
If the first equation is doubled and added it to the second, one gets
3 Ft - 40 (N) = 0 ⇒ Ft = 40/3 (N) = 13.3 (N) and that is the correct tension in the string connecting the masses. More generally, $$F_T=\frac{2m_1m_2}{m_1+m_2}g.

Thanks for pointing out the mistake
What I wanted to show by this example is that the following statement made by PeroK is wrong "B cannot be pulling on the string with 20N and the string pulling back with 15N. That violates Newton's third law."

In my FBD of B, there is 20N down and 13.3N up. This is similar to the question of the original poster - just that where the force comes from is different in the Atwood machine example vs the original poster's question

 

[berkeman]

Thread closed temporarily for Moderation and possible thread split.

Update -- thread will remain closed.