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

  • Thread starter Mrchilko
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In summary: I did it wrongFnet = 1.25 x 1.4 = 1.75 net force hahahah I get it......I didYes, this is the net force on the rope. This force acts on both ends of the rope, so half of that force acts on one end of the rope. You know that Rhys exerts an additional force of 24 N on the rope. So the force that Gareth exerts on the rope is ...?In summary, Rhys and Gareth are having a tug of war with a 1.25kg climbing rope. Rhys pulls with a force of 24N and the rope accelerates away from him at 1.4m
  • #1
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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  • #2
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?
 
  • #3
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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  • #4
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.
 
  • #5
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
 
  • #6
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.
 
  • #7
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?
 
  • #8
I just Started physics 3 months ago... II am only assuming it takes very much practice to move on from free body digrams
 
  • #9
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
 

1. How can you determine the opposite person's force in a game of tug of war?

The opposite person's force in a game of tug of war can be determined by using a force meter or a dynamometer. This device measures the amount of force being applied by each team and displays the results in units of weight, such as pounds or kilograms.

2. What factors can affect the opposite person's force in a game of tug of war?

The opposite person's force in a game of tug of war can be affected by several factors, including their body weight, muscle strength, grip strength, and technique. Additionally, external factors such as friction between the ground and their feet, the type of rope used, and the surface on which the game is being played can also impact their force.

3. Is there a specific technique to use in order to increase the opposite person's force in a game of tug of war?

Yes, there are several techniques that can be used to increase the opposite person's force in a game of tug of war. These include maintaining a low center of gravity, keeping a tight grip on the rope, and using leg muscles to push against the ground instead of relying solely on arm strength. Proper coordination and communication with teammates can also improve overall force.

4. How can you improve your own force in a game of tug of war?

To improve your own force in a game of tug of war, you can focus on building overall muscle strength, particularly in the arms, legs, and core. It is also important to maintain good technique and form throughout the game, as well as work on coordination and communication with your teammates. Regular practice and training can also help improve your force in tug of war.

5. Is there a limit to the opposite person's force in a game of tug of war?

There is no specific limit to the opposite person's force in a game of tug of war. However, the maximum force that can be generated is limited by the strength and physical abilities of the individuals participating in the game. Additionally, external factors such as the type of rope and surface being used can also impact the maximum force that can be exerted by the opposite person.

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