Tension

[MedPR]

Confused about tension..

In a game of tug of war, each group pulls on the rope with a force of 2000N. So, the tension on the rope is 2000N.

What is the tension on the rope if one group pulls with 150N and the other pulls with 100N? I it then the difference between the two Forces?

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[chiddler]

I think of it by comparing to pulling on a rope attached to a wall. If I put 100N of force backwards, then tension is simply 100 N. If somebody holds the opposite side and starts pulling 100 N opposite, then there's twice as much tension in the rope because i'm pulling and they're pulling. More tension.

So i'd expect it to be additive. I'm not the best at physics though so...am I correct? Do you have the answer?

 

[MedPR]

I think of it by comparing to pulling on a rope attached to a wall. If I put 100N of force backwards, then tension is simply 100 N. If somebody holds the opposite side and starts pulling 100 N opposite, then there's twice as much tension in the rope because i'm pulling and they're pulling. More tension.

So i'd expect it to be additive. I'm not the best at physics though so...am I correct? Do you have the answer?

No, I don't have the answer. However, in your example, if you pull with 100N, the wall is going to pull back with 100N. So your two situations (pulling against a wall vs pulling against another person) are analogous.

 

[ljc]

I think of it by comparing to pulling on a rope attached to a wall. If I put 100N of force backwards, then tension is simply 100 N. If somebody holds the opposite side and starts pulling 100 N opposite, then there's twice as much tension in the rope because i'm pulling and they're pulling. More tension.

So i'd expect it to be additive. I'm not the best at physics though so...am I correct? Do you have the answer?

This is wrong. T is still 100N.

For your system MedPR, draw a diagram. Assume the 100N man is on the left and 150N on the right.

So your diagram should look like this:

100N <-----[ Man 1 ]-----> T T <------[ Man 2 ]-------> 150N

You know the system will accelerate to the right, so write out your force equations accordingly:

T-100N=ma (T will be more than 100N, so we opt for T-100N instead of 100N-T)
and 150N-T=ma

Assuming they are the same mass, and knowing that accelerating for one will be equal to the acceleration of the other, we set them equal:
T-100N=150N-T
2T=250N
T=125N

So tension is 125N, which should make sense, because it can't possibly be more than 150N, given what we know about tension against a stationary wall, and it couldn't possibly be less than 100N, given that the system is accelerating and there are two opposing forces, both of which are greater than 100N.

 

[chiddler]

No, I don't have the answer. However, in your example, if you pull with 100N, the wall is going to pull back with 100N. So your two situations (pulling against a wall vs pulling against another person) are analogous.

huh. why?

so if we both hold a rope and you pull, the tension will be the same if i start pulling on the rope as well? why?

because it can't possibly be more than 150N

why not?

 

[ljc]

huh. why?

so if we both hold a rope and you pull, the tension will be the same if i start pulling on the rope as well? why?

If the rope is attached to the wall, yes. If you pull on a rope that is not attached to anything, there is no tension. You are simply dragging a rope. But when you pull at one end with tension T, and the rope remains stationary attached to a wall, tension throughout the rope will simply be T. Similarly, if you pull with T, and another person pulls the other way with T, tension is still T. In the original scenario, if the wall was NOT pulling with T, the rope would accelerate.


This also answers your other question. If the force on one end of the rope is X, and the force on the other end of the rope is less than X, tension will never be more than X. When you pull with a force and so does another person in the opposite direction, the tension is not cumulative. It is a factor of the difference of the two forces, and will always be somewhere in the middle.

 

[chiddler]

If the rope is attached to the wall, yes. If you pull on a rope that is not attached to anything, there is no tension. You are simply dragging a rope. But when you pull at one end with tension T, and the rope remains stationary attached to a wall, tension throughout the rope will simply be T. Similarly, if you pull with T, and another person pulls the other way with T, tension is still T. In the original scenario, if the wall was NOT pulling with T, the rope would accelerate.


This also answers your other question. If the force on one end of the rope is X, and the force on the other end of the rope is less than X, tension will never be more than X. When you pull with a force and so does another person in the opposite direction, the tension is not cumulative. It is a factor of the difference of the two forces, and will always be somewhere in the middle.

So in either case tension is the same, but the difference is the acceleration of one body because it's not pulling back. Right?

 

[ljc]

Essentially, yes. But be careful. You can't just say "well between 150N and 100N, 150N is greater so that must be the tension!" or vis versa. You have to draw the force body diagrams for each case. It requires some level of intuition to understand which direction the system will accelerate, but at that point, there is really nothing more to be done in your head (unless you're VERY comfortable with these). Once you know which way the system will accelerate (towards the side with greater force), draw out each half of the system, accounting for this acceleration.

For instance, if two boxes hang from a string over a pulley, one weighing 1kg and the other 2kg, you should recognize the 2kg box will move downward. Now draw your force body diagrams, knowing that mg down on the 2kg will be greater than T up, and mg down on the 1kg will be less than T up. Solve for net force, and recognizing that the whole system will accelerate at the same rate, set the forces equal and solve for T. Without doing the math, I can see that the tension will be less than 20N but greater than 10N, because those are the two extremes provided by each box.


Edit: think I misread your question. Yes tension will be the same. The wall will not accelerate though, and if the force on both sides is equal, neither person will accelerate either. I'm not sure what you're asking?

 

[chiddler]

I am comfortable with solving these mathematically because it is pretty systematic. I was referring to the example I gave of two people pulling on a rope. Am I correct with this?

If person 1 does nothing and person 2 pulls 100 N, then tension is 100 N and person 1 accelerates towards person 2.

If person 1 pulls 100 N and person 2 pulls 100 N, then tension is 100 N and nobody moves.

i know the wall will not accelerate

 

[ljc]

I was referring to the example I gave of two people pulling on a rope. Am I correct with this?

If person 1 does nothing and person 2 pulls 100 N, then tension is 100 N and person 1 accelerates towards person 2.

If person 1 pulls 100 N and person 2 pulls 100 N, then tension is 100 N and nobody moves.

i know the wall will not accelerate

Edit: see below

 

[chiddler]

Correcto senor.

mucho thank yous

 

[ljc]

Err wait. I'm not super comfortable with your case #1. Give me a minute to work it out.

Edit: No, the tension would have to be less than 100N. If you think about it, person 2 is moving right with 100N of force. If tension were ALSO 100N, they would be stationary. So tension must be somewhat less than 100N in order to allow person 2 to accelerate.

 

[chiddler]

Err wait. I'm not super comfortable with your case #1. Give me a minute to work it out.

Edit: No, the tension would have to be less than 100N. If you think about it, person 2 is moving right with 100N of force. If tension were ALSO 100N, they would be stationary. So tension must be somewhat less than 100N in order to allow person 2 to accelerate.

that makes sense. thanks again.

 

[ljc]

Glad to help!

 

[MedPR]

If person 1 does nothing and person 2 pulls 100 N, then tension is 100 N and person 1 accelerates towards person 2.

This is only true if person 1 is massless, correct?

 

[ljc]

This is only true if person 1 is massless, correct?

If person 1 is massless, there is no tension. There must be some opposing force, or else by F=ma you would accelerate at an infinite rate, which is impossible.

 

[MedPR]

If person 1 is massless, there is no tension. There must be some opposing force, or else by F=ma you would accelerate at an infinite rate, which is impossible.

Yea that makes sense. Here's what I was thinking.

Say you wrap the rope around person 1's waist.

If person 1 weighs 100N, then person 2 will have to exert a tension greater than 100N to move person 1, right?

If person 1 weighs more than 100N, then person 1 will not move if person 2 exerts a tension of 100N on the rope. Just by weighing more than 100N, isn't person 1 creating more tension in the rope? So we would solve this by setting T1 = T2?

If person 1 weighs less than 100N, then person 1 will move towards person 2 when person 2 exerts a tension of 100N, but the tension on the rope again is not 100, right?

 

[ljc]

Yea that makes sense. Here's what I was thinking.

Say you wrap the rope around person 1's waist.

If person 1 weighs 100N, then person 2 will have to exert a tension greater than 100N to move person 1, right?

If person 1 weighs more than 100N, then person 1 will not move if person 2 exerts a tension of 100N on the rope. Just by weighing more than 100N, isn't person 1 creating more tension in the rope? So we would solve this by setting T1 = T2?

If person 1 weighs less than 100N, then person 1 will move towards person 2 when person 2 exerts a tension of 100N, but the tension on the rope again is not 100, right?

Not quite. Unless he's pulling him upwards, or coefficient of friction is 1.0. Any force will move any size object if there is no opposing force.

As stated, person 1 has no opposing force, so he would move in all of these cases. MG is pointing down, while tension is pulling left or right. They are perpendicular, and one will not affect the other (unless you include friction, or person 2 is trying to lift person 1 into the air)

 

[MedPR]

Not quite. Unless he's pulling him upwards, or coefficient of friction is 1.0. Any force will move any size object if there is no opposing force.

As stated, person 1 has no opposing force, so he would move in all of these cases. MG is pointing down, while tension is pulling left or right. They are perpendicular, and one will not affect the other (unless you include friction, or person 2 is trying to lift person 1 into the air)

Ah yea that makes sense. Thanks.

 

[SKation]

Confused about tension..

In a game of tug of war, each group pulls on the rope with a force of 2000N. So, the tension on the rope is 2000N.

What is the tension on the rope if one group pulls with 150N and the other pulls with 100N? I it then the difference between the two Forces?

Hey,

Could you explain how this is possible that if both of them pull then why isnt it 0N since the forces would cancel each other out.

Thanks!