Tension in a String/Rope (tug-of-war)

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Homework Statement:

Two students are having a tug of war. A 60kg student pulls on a rope with a force of 40N to the left. A 70kg student pulls on the other end of the rope with a force of 50N to the right. Find the
a) acceleration of the two students
b) force of tension in the rope

Relevant Equations:

Newton's Third Law
a) I think you can just use Fnet = m*a, so for
student 1: a = 40N/60kg a = 0.667 m/s^2
student 2: a = 50N/70kg → a = 0.714 m/s^2


b) Fnet = F - T, rearrange to solve for tension, → T = F - ma
Student 1, T = 40N - (60kg*0.667m/s^2) T = -0.02N
Student 2, T = 50N - (70kg*0.71m/s^2) T = 0.02N

The answer seems wrong since its too small. From my understanding the force of tension of the rope needs to be equal on both sides of the rope. But I'm not sure if that also applies to different applied forces on both ends. The answers to the question aren't given.


 

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  • #2
PeroK
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Where did you get this question?

What happens if you apply Newton's second law to the rope?
 
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Where did you get this question?
The teacher gave it for practice.

What happens if you apply Newton's second law to the rope?
I think I've used the equations from the second law throughout the question. The only thing applied from Newton's third law could be that since every action force has an equal reaction force, the tension would be equal on both ends of the rope.
 
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PeroK
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I think I've used the equations from the second law throughout the question. The only thing applied from Newton's third law could be that since every action force has an equal reaction force, the tension would be equal on both ends of the rope.
What did you use for the mass of the rope?
 
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What did you use for the mass of the rope?
The mass of the rope isn't given. So i assume you can only find the tension after you've calculated the acceleration of the students.
 
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PeroK
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The mass of the rope isn't given. So i assume you can only find the tension after you've calculated the acceleration of the students.
Okay, what about the forces on the two students from the ground, where they are no doubt pushing with their feet?

I should say that I think the question is nonsense. Which is why I asked where you found it.
 
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Okay, what about the forces on the two students from the ground, where they are no doubt pushing with their feet?

I should say that I think the question is nonsense. Which is why I asked where you found it.
If they're pushing on the ground, that means the ground may be creating friction helping them each to move to the left or right, but I believe the student with the greater acceleration wins? I feel like there's multiple ways to do this problem that I can't wrap my head around.

Lol i guess having a phd doesn't make my teacher any smarter XD
 
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PeroK
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If they're pushing on the ground, that means the ground may be creating friction helping them each to move to the left or right, but I believe the student with the greater acceleration wins? I feel like there's multiple ways to do this problem that I can't wrap my head around.

Lol i guess having a phd doesn't make my teacher any smarter XD
If you apply Newton's second law to the rope, then you have an unbalanced force of ##10N## on the rope, so there will be acceleration of the rope. That's the only way there can be an unbalanced force on the rope.

The students don't accelerate from the force on the rope alone: they also have forces on them from the ground.

I'm not sure whether whoever set the problem had no idea about any of this.
 
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If you apply Newton's second law to the rope, then you have an unbalanced force of ##10N## on the rope, so there will be acceleration of the rope. That's the only way there can be an unbalanced force on the rope.

The students don't accelerate from the force on the rope alone: they also have forces on them from the ground.

I'm not sure whether whoever set the problem had no idea about any of this.
I think I understand now, thanks. But how will we solve that with the equations? Can I still use Fnet = m*a? Or do I just figure it out by recognizing the forces are unbalanced, so the rope would move in the direction of the unbalanced force?
 
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PeroK
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I think I understand now, thanks. But how will we solve that with the equations? Can I still use Fnet = m*a? Or do I just figure it out by recognizing the forces are unbalanced, so the rope would move in the direction of the unbalanced force?
You could do a general free-body diagram for a tug-o-war. That's worth doing.
 
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Thank you for you help!
 
  • #12
haruspex
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Can I still use Fnet = m*a?
Only if you use the right m. As @PeroK has indicated, it would be the mass of the section of rope between them, which is unknown. If it has mass 1kg then it will be accelerating at roughly g. Even then, you wouid not be able to state "the tension" since it will vary along the rope, from 40N at one end to 50N at the other.

Either this is a trick question (to which the correct answer is "not enough info") or the setter has blundered. Consider a slightly different question:
One end attaches to a 56kg mass lying on a smooth horizontal surface, and on the other side of that mass a second rope goes over a pulley to a suspended 4kg mass: 56+4=60.
The other end attaches to a 65kg mass, thence by another rope to a suspended 5kg mass: 65+5=70.
While something in the middle holds everything static, there is a tension of 40N at one end and 50N at the other. But as soon as the system is released it will accelerate and the tension in the central rope will become some value between the two.

Sounds like the intended question, except that the two systems are now not pulling on the rope with different forces; each pulls with the same force, namely, the tension.
 
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Is it possible for there to be two different forces of tension on opposite ends if the rope is being pulled by forces with different magnitudes?
 
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PeroK
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Is it possible for there to be two different forces of tension on opposite ends if the rope is being pulled by forces with different magnitudes?
We had a long thread about this recently. For a static scenario, no - because of Newton's second law for the rope. The rope must accelerate if there is an unbalanced force on it. And, if you consider the rope to be of negligible mass, then the tension must be equal throughout the rope.
 
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  • #15
haruspex
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We had a long thread about this recently. For a static scenario, no - because of Newton's second law for the rope. The rope must accelerate if there is an unbalanced force on it. And, if you consider the rope to be of negligible mass, then the tension must be equal throughout the rope.
And, to be absolutely clear, the forces with which the rope is being pulled are equal to this tension, so are themselves equal. That is the flaw in the question in this thread.
 

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