Tension is causing me to be tensed

[EddiePhys]

There's two questions about tension(the force), that no matter how much I try to find answers to and understand since the past few days, I'm simply not able to. Any help in this matter would be highly appreciated.

1) If person A and person B are holding a rope, and person A applies 100N of force, he would get a reaction of 100N and the tension in the rope would be 100N. Fair enough. Now, if person B pulls with 100N too, person B would be pulled not only by the reaction force of 100N, but also by the force exerted by person B. In this case, shouldn't the tension be = 200N?

Related sub-question: If a block of 10 kg is hanging from the ceiling, It would be exerting 100N downwards and get a reaction from the rope of 100N, and the ceiling would pull it with 100N too. So shouldn't the tension be 200N?

2) If person A is pulling the string with 100N and person B with 200N, why is the tension in the rope 100N?
I've read somewhere that in a string with negligible mass, this would result in the string slipping off the hands of person A, which makes sense. But why wouldn't that be the case with a string with mass too?

 

[CWatters]

1) If person A and person B are holding a rope, and person A applies 100N of force, he would get a reaction of 100N and the tension in the rope would be 100N.

Fair enough. Now, if person B pulls with 100N too...

Hold it right there. In the first situation how does A "get a reaction force" if B doesn't also apply 100N to the rope?

 

[OldYat47]

1. It's the same force, not additional forces. Think about the rope. There's 100N of tension in the rope. Person A is pulling with 100N force. If person B didn't pull with an equal force the rope would move. If you imagine cutting the rope at any point and putting scales on the cut ends they would both read 100N. If you put one scale in the rope at any point it would read 100N.
2. If person A is pulling with 100N force and person B pulls with 200N force then the tension in the rope would not be 100N. Person A would accelerate in the direction of person B, the magnitude of the acceleration would be (200N - 100N)/(mass of person A). The force would be (acceleration) X (mass of person A).

 

[jbriggs444]

1. It's the same force, not additional forces. Think about the rope. There's 100N of tension in the rope. Person A is pulling with 100N force. If person B didn't pull with an equal force the rope would move. If you imagine cutting the rope at any point and putting scales on the cut ends they would both read 100N. If you put one scale in the rope at any point it would read 100N.

Agreed, completely.

2. If person A is pulling with 100N force and person B pulls with 200N force then the tension in the rope would not be 100N. Person A would accelerate in the direction of person B, the magnitude of the acceleration would be (200N - 100N)/(mass of person A). The force would be (acceleration) X (mass of person A).

We cannot know the acceleration of person A because we do not know all the forces to which person A is subject. In any case, the acceleration of person A will be the net force on person A divided by the mass of person A. That net force will not, in general, happen to match the discrepancy in tension between the two ends of the rope.

We do know the all the relevant horizontal forces on the rope. If there is a 200N force on one end and a 100N force oppositely directed on the other end then the net force on the rope is 100 N. The acceleration of the rope is then given by 100N divided by the mass of the rope.

A rope with a small mass subject to a large net force will accelerate quickly. That explains why trying to pull on one end of a rope can result in a collision between the ground and one's posterior. The rope will move.

 

[OldYat47]

We know that person A is in equilibrium pulling 100N against person B pulling 100N. If person B starts pulling 200N (all other things remaining the same) then my statements work just fine for the purposes of this thought experiment.

 

[jbriggs444]

We know that person A is in equilibrium pulling 100N against person B pulling 100N. If person B starts pulling 200N (all other things remaining the same) then my statements work just fine for the purposes of this thought experiment.

No. They do not. We have no information about the stability of the supposed equilibrium in the face of disturbances.

Edit: Nor do we have an assurance that an equilibrium exists. We know only the force between A and the rope, not whether A is already accelerating under the unknown net force upon him.

 

[EddiePhys]

Hold it right there. In the first situation how does A "get a reaction force" if B doesn't also apply 100N to the rope?

If I'm in space, and I pull a rope with 100N that is just freely floating or even attached to another person, the rope will exert a 100N force on me

 

[CWatters]

If I'm in space, and I pull a rope with 100N that is just freely floating or even attached to another person, the rope will exert a 100N force on me

Correct but only if B accelerates at the same rate as the rope so he doesn't also apply a force to it. Did you mean to make it that complicated?

Now, if person B pulls with 100N too

How does that differ from a situation where A pulls one end with 100N and the other end is fixed to a pole? The pole also exerts 100N on the rope.

In this situation the rope isn't accelerating so does the rope still provide the reaction force you mention? or is it provided by B or the pole?

 

[OldYat47]

I was trying to stay as basic as possible without adding any unknown scenarios, taking the question as stated.
Quoting from above:
"2) If person A is pulling the string with 100N and person B with 200N, why is the tension in the rope 100N?
I've read somewhere that in a string with negligible mass, this would result in the string slipping off the hands of person A, which makes sense. But why wouldn't that be the case with a string with mass too?[/QUOTE]"

Note that this is physically impossible. You can't have two different tension forces on a simple straight piece of rope. If the end points aren't fixed then one will move, and my equations are correct in this context. If the end points are fixed then the tension throughout the rope must be the same.

Edit: Nor do we have an assurance that an equilibrium exists. We know only the force between A and the rope, not whether A is already accelerating under the unknown net force upon him.

No, we don't. But let's assume for the purposes of this experiment that the facts are as stated, and not conditions at some undefined instantaneous moment in time. I am sure that there are many scenarios that are not part of the stated problem that could be imagined. .

 

[EddiePhys]

Correct but only if B accelerates at the same rate as the rope so he doesn't also apply a force to it. Did you mean to make it that complicated?
How does that differ from a situation where A pulls one end with 100N and the other end is fixed to a pole? The pole also exerts 100N on the rope.

In this situation the rope isn't accelerating so does the rope still provide the reaction force you mention? or is it provided by B or the pole?

So is the reaction provided by the pole or the rope? It could be the rope since in space when it is unattached to something, that's what it does.
Also, what would be the tension in the case that the rope is fixed to another person while being pulled?

 

[OldYat47]

The reaction is due to the pole. If the rope were unattached there would be no tension, unless you were accelerating hard enough while holding the rope that (mass of rope) X (rate of acceleration) = 100N.

 

[CWatters]

+1

So is the reaction provided by the pole or the rope?

If the rope is tied to a fixed pole that means the rope isn't accelerating. Therefore the inertia of the rope can't be providing a reaction force. Therefore the reaction force must be provided by the pole.

 

[nasu]

If you pull on the rope with your hand, the reaction, if any, on the hand, can be provided only by the rope. Your hand is not interacting with anything else but the rope.
The Newton third law pair involving your hand regards the rope and the hand.
The interaction between the rope and the pole is a different interaction, characterized by a different pair of forces. Even if the magnitudes of the forces in the two pair may be equal, these are different forces.
If the rope is not attached, both your force and the reaction of the rope may be very small (zero if the rope is considered massless). But you should realize that not only the "reaction" is zero, your force is zero as well. Both forces in the pair are zero. Or actually small.

Attaching the other end of the rope to a pole you introduce another pair of forces, at the pole end.
This time you can pull the rope with a larger force and the rope will pull you with the same larger force.
So the effect of the pole is to allow you to increase the forces at your end. However this is not the same as saying that the pole produces the reaction force at your hand.

 

[EddiePhys]

+1
If the rope is tied to a fixed pole that means the rope isn't accelerating. Therefore the inertia of the rope can't be providing a reaction force. Therefore the reaction force must be provided by the pole.

Oh, so the rope in this case only serves the purpose of "transfering" the force to the pole, and the pole provides a reaction which is "transfered" back to the person holding the rope. Is this correct?

 

[EddiePhys]

Attaching the other end of the rope to a pole you introduce another pair of forces, at the pole end.
This time you can pull the rope with a larger force and the rope will pull you with the same larger force.
So the effect of the pole is to allow you to increase the forces at your end. However this is not the same as saying that the pole produces the reaction force at your hand.

This is the part that confused me. Introduce a new pair of forces at the pole end?

Could you make a free body diagram on this including these new set of forces?

 

[jbriggs444]

What body do you want to focus your attention on? The person, the rope, the pole or the Earth?

 

[EddiePhys]

What body do you want to focus your attention on? The person, the rope, the pole or the Earth?

The person and the pole

 

[jbriggs444]

That's two bodies.

 

[EddiePhys]

That's two bodies.

There's enough space. Or maybe make two separate fbd's

 

[Dale]

The person and the pole

What forces act on the person?

What forces act on the pole?

 

[EddiePhys]

What forces act on the person?

What forces act on the pole?

That's two bodies.

Alright, so far:
1. I've understood that if I hold a rope which is tied around a person's waist and I pull with 100N, I'm going to get a reaction of 100N. What I don't understand is what will happen when the same rope is tied to that person's hand and he exerts 100N using his muscles while I'm pulling with 100N
2. The two forces at the ends of the rope(in a case of a massless rope) have to be equal. I even saw a mathematical proof for this. Intuitively I imagine that if the forces are unequal, the rope would slip off the hands of the person pulling with the lower force (not sure if this is correct)

 

[Dale]

The two forces at the ends of the rope(in a case of a massless rope) have to be equal

Yes. The net force on a massless object must always be 0.

Intuitively I imagine that if the forces are unequal, the rope would slip off the hands of the person pulling with the lower force (not sure if this is correct)

If the forces were unequal then the rope would accelerate with an infinite acceleration. Or rather, the acceleration would be undefined by Newton's 2nd law.

1. I've understood that if I hold a rope which is tied around a person's waist and I pull with 100N, I'm going to get a reaction of 100N. What I don't understand is what will happen when the same rope is tied to that person's hand and he exerts 100N using his muscles while I'm pulling with 100N

If you are pulling to the left with 100 N and the person is pulling to the right with 100 N then the net force on the rope is 0, so everything is OK.

My general recommendation is for you to draw three free body diagrams. One for the rope and one for each of the objects touching the rope. As described above the sum of the forces on the (massless) rope must be 0. Also the two 3rd law pairs must be equal and opposite.

 

[EddiePhys]

If the forces were unequal then the rope would accelerate with an infinite acceleration. Or rather, the acceleration would be undefined by Newton's 2nd law.

Realistically, what would happen if two astronauts in space, one stronger and one weaker pull on a rope of negligible mass with all their strength?

 

[jbriggs444]

Realistically, what would happen if two astronauts in space, one stronger and one weaker pull on a rope of negligible mass with all their strength?

"All their strength" will turn out to result in identical forces as the one astronaut "climbs" the rope and the other astronaut pays out the rope. Both astronauts would move toward each other under the resulting identical force with different accelerations in [inverse] proportion to their respective masses.

Edit: Can you give up and accept Newton's third law instead of continuing to fight it?

 

[EddiePhys]

"All their strength" will turn out to result in identical forces as the one astronaut "climbs" the rope and the other astronaut pays out the rope. Both astronauts would move toward each other under the resulting identical force with different accelerations in [inverse] proportion to their respective masses.

Edit: Can you give up and accept Newton's third law instead of continuing to fight it?

How would "all their strength" result in identical forces and what would be the magnitude of this force?

 

[jbriggs444]

How would "all their strength" result in identical forces and what would be the magnitude of this force?

Have you ever tried to hurl a ping pong ball or a grain of sand with "all your strength". You cannot succeed in applying much force since it moves away so fast.

Edit: This is formalized to some extent when one discusses the concept of "work". The work done by a force is computed by the magnitude of the force applied to an object multiplied by the distance the object moves in the direction of the force while the force is being applied.

The work energy theorem shows that the work done (in this sense) to an object is equal to the kinetic energy gained by the object.

If a strong man attempts to move or throw an object with "all his strength", a lighter object will move more rapidly than a heavier object. If the man exerts a fixed force on the object, then he will have to expend energy more rapidly for lighter objects and less rapidly for heavier objects. For an infinitely light object he would expend energy at an infinite rate. There is a limit on how fast a strong man can expend energy and, hence, a limit on how much force he can exert on light objects.

 

[EddiePhys]

Have you ever tried to hurl a ping pong ball or a grain of sand with "all your strength". You cannot succeed in applying much force since it moves away so fast.

So you can only really use all your strength when you get a reaction of that same amount. Is this correct?

 

[jbriggs444]

So you can only really use all your strength when you get a reaction of that same amount. Is this correct?

I would say so, yes.

 

[EddiePhys]

I would say so, yes.

Alright. Thanks a lot for your help I really appreciate it.

 

[Dale]

Realistically, what would happen if two astronauts in space, one stronger and one weaker pull on a rope of negligible mass with all their strength?

They can't. The weaker astronaut will let go once the force exceeds his strength. So the stronger one will only be able to pull with a fraction of his strength.

 

[nasu]

Why would he let go? He can be pulled with any force, even 100 times larger than his "strength". And he doesn't have to do anything to be pulled.
He does not have to do anything for the "reaction" force on the first one to equal his "action" force.

When you are water-skying and are pulled by a motor boat, do you have to have the same strength as the engine of the boat?

 

[jbriggs444]

Why would he let go?

This sort of disagreement is what results when one tries to understand poorly defined terms like "all his strength".

On the one hand, Dale may have in mind the force limit at which an astronaut can pull in a rope hand over hand with some non-zero progress made and Nasu apparently has in mind the force limit at which a rope tied around the astronaut's body will cut him in twain.

 

[nasu]

You are right.
But my point was (tried to be) that what the second astronaut does (or can do) does not change the fact that the action-reaction pair between the first astronaut and the rope will always consist in forces with same magnitude. The forces at the other end of the rope are not "reaction" to the force of the first astronaut.

 

[Dale]

He does not have to do anything for the "reaction" force on the first one to equal his "action" force.

He has to hold on.

 

[Dale]

The forces at the other end of the rope are not "reaction" to the force of the first astronaut.

Correct. They are constrained to be equal and opposite by Newtons 2nd law (and the massless rope assumption). Newtons 3rd law is not relevant to the relationship between those two forces since both forces are acting in the same object.