Tug of war.... Finding oppostite person's force

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Homework Help Overview

The discussion revolves around a physics problem related to Newton's laws, specifically focusing on a tug of war scenario involving two individuals, Rhys and Gareth, and a rope. The problem presents a situation where Rhys pulls with a force of 24N, and the rope accelerates at 1.4m/s². Participants are tasked with determining the force exerted by Gareth on the rope.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between force, mass, and acceleration, questioning how to apply Newton's second law. Some express confusion about the relevance of the rope's weight and the implications of the forces involved in the tug of war.

Discussion Status

There is an ongoing exploration of different interpretations of the problem. Some participants have attempted calculations based on their understanding of the forces, while others have raised questions about the assumptions made regarding the mass of the individuals involved and the impact of gravity on the rope. Guidance has been offered regarding the correct application of equations, but no consensus has been reached on the final solution.

Contextual Notes

Participants note the absence of specific mass values for Rhys and Gareth, which complicates the calculations. There is also a discussion about the importance of free body diagrams in visualizing the forces at play, with some participants expressing a need for further practice in this area.

Mrchilko
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Homework Statement


Don't know where to start cause this is part of Newtons laws apparently but seems to be more varied around his second law... Just can't figure out what equation to use.. Question : Rhys and Gareth are having a tug of war with 1.25kg of climbing rope. If Rhys pulls with a force of 24N and the rope accelerates away from him at 1.4m/s2, with what force is Gareth pulling the rope ?

Homework Equations


ΣNet = m*a
Fnet= FA + FF

The Attempt at a Solution


I'm only assuming that you have to take the mass of Rhys and use that plus the weight of the rope to your advantage..
 
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Mrchilko said:

Homework Statement


Don't know where to start cause this is part of Newtons laws apparently but seems to be more varied around his second law... Just can't figure out what equation to use.. Question : Rhys and Gareth are having a tug of war with 1.25kg of climbing rope. If Rhys pulls with a force of 24N and the rope accelerates away from him at 1.4m/s2, with what force is Gareth pulling the rope ?

Homework Equations


ΣNet = m*a
Fnet= FA + FF

The Attempt at a Solution


I'm only assuming that you have to take the mass of Rhys and use that plus the weight of the rope to your advantage..
Hmm ... It's too bad that you don't have the mass of Rhys (or Gareth) to plug into your equation. :frown:

However, if you read the problem statement carefully, you will find that neither Rhys nor Gareth are doing any movement.

When Rhys pulls with 24 N, what accelerates away from him? Is Rhys accelerating away from himself? Is that possible?
 
I figured it out.. Don't know if its the right way.. But I did: 24N / 1.4m/s2 to find mass... Because m= Fnet/a... And seeing that this seems to be the only force mentioned.. And since it never mentioned them moving. That equaled to 17kg... 17kg+ 1.25kg( weight of rope) seeing that he is pulling the rope and Rhys. Then I got 18kg... 18kg * 1.4m/s2 = 26N... That was the answer in the book too... Haha now I'm dealing with a much harder question
 

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Mrchilko said:
I figured it out.. Don't know if its the right way.. But I did: 24N / 1.4m/s2 to find mass...

You were already given the mass of the rope (1.25 kg). You were asked to find out the force with which Gareth is pulling on the rope.

Because m= Fnet/a... And seeing that this seems to be the only force mentioned.. And since it never mentioned them moving. That equaled to 17kg... 17kg+ 1.25kg( weight of rope) seeing that he is pulling the rope and Rhys. Then I got 18kg... 18kg * 1.4m/s2 = 26N... That was the answer in the book too... Haha

What, it was too complicated to write a net force equation involving Rhys and Gareth?

A stopped clock is right twice a day, too. That doesn't mean it's right all the time.
now I'm dealing with a much harder question

Then you should start a new thread for this question rather than hijacking your own thread for the first problem.
 
Ahhh... Too many questions and a important test tmr... I don't have time to post now... I have to figure it out on my own
 
Your relevant equations are correct. You are given the mass and acceleration so you can work out fnet. You are given one of two forces so can work out the other.
 
Have you drawn a free body diagram of the rope, or do you feel like you've advanced in physics beyond the need to draw free body diagrams?
 
I just Started physics 3 months ago... II am only assuming it takes very much practice to move on from free body digrams
 
Mrchilko said:
I just Started physics 3 months ago... II am only assuming it takes very much practice to move on from free body digrams
Excellent. So, based on your free body diagram, please write out your Newton's 2nd law force balance for the rope.

Chet
 
  • #10
I'm just confused with the weight of the rope... Cause isn't it possible that there is gravity beeing exerted onto it... And wouldn't that have an impact on the result
 

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  • #11
Mrchilko said:
I'm just confused with the weight of the rope... Cause isn't it possible that there is gravity beeing exerted onto it... And wouldn't that have an impact on the result

Haven't you ever been in a tug of war contest?

The weight of the rope is irrelevant to the force of the pull being applied by the people on each end of the rope.
 
  • #12
Mrchilko said:
I'm just confused with the weight of the rope... Cause isn't it possible that there is gravity beeing exerted onto it... And wouldn't that have an impact on the result
No. Neglect the weight. That's in the vertical direction, and we're doing a force balance in the horizontal direction. So...
 
  • #13
By the way, your free body diagram is correct.

Chet
 
  • #14
Ye so I will figure out when I can. I'm guessing it's just going to be Netforce - Rhys = gareth in some form or not
 
  • #15
Mrchilko said:
Ye so I will figure out when I can. I'm guessing it's just going to be Netforce - Rhys = gareth in some form or not
I have no idea what Rhys and gareth are.

Chet
 
  • #16
Ye woops lol that wasn't really meaning that I meant more Fnet = F( of Gareth) - 24N
 
  • #17
Mrchilko said:
Ye woops lol that wasn't really meaning that I meant more Fnet = F( of Gareth) - 24N
Excellent. And, in terms of the mass of the rope m and the acceleration of the rope a, the net force is equal to ?
 
  • #18
Well Fnet = m*a
a= Fnet/ m
M= Fnet/a

So Fnet = 0.400* 1.4 = 0.56
 
  • #19
Mrchilko said:
Well Fnet = m*a
a= Fnet/ m
M= Fnet/a

So Fnet = 0.400* 1.4 = 0.56
Wrong problem, Mrchilko.
 
  • #20
The problem statement tells you that the mass of the rope is 1.25 kg and its acceleration is 1.4 m/s2. So what does this give you for the net force?

Chet
 
Last edited:
  • #21
Fnet = 1.25 x 1.4 = 1.75 net force hahahah I get it... Is it not 24N + 1.75N = 25.75= only 2 SIG figs so 2.6 x 101N = the force Gareth exerts on the rope
 

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