Tug Between Two People On Friction-less Ice.

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Discussion Overview

The discussion revolves around a physics problem involving two individuals of different masses (50 kg and 100 kg) on frictionless ice, connected by a rope. Participants explore the concept of center of mass, the implications of tension in the rope, and how the distances each person travels relate to their masses. The scope includes theoretical reasoning and mathematical relationships regarding motion and forces.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant questions why both methods of solving the problem work, specifically regarding the meeting point at the center of mass.
  • Another participant suggests that the center of mass can change if the two individuals approach each other, prompting further exploration of the conditions required for the center of mass to move.
  • There is a discussion about the necessity of an external force for the center of mass to accelerate, noting that the frictionless surface means no external forces are acting on the system.
  • Participants discuss how to apply the concept of center of mass to understand the reasoning behind the two methods of solving the problem.
  • One participant proposes that if the heavier person moves a certain distance, the lighter person must move a corresponding distance to keep the center of mass unchanged, leading to a mathematical relationship between their movements.
  • Newton's laws of motion are introduced to explain the relationship between the accelerations of the two individuals based on their masses.

Areas of Agreement / Disagreement

Participants generally agree on the principles of center of mass and the implications of frictionless motion, but there are varying interpretations of how to apply these concepts to the problem at hand. The discussion remains exploratory, with no consensus reached on the specific reasoning behind the methods proposed.

Contextual Notes

Participants express uncertainty about the conditions under which the center of mass remains constant and how to mathematically relate the movements of the two individuals. The discussion highlights the need for clarity on definitions and assumptions regarding motion and forces.

Bashyboy
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Okay, suppose we have two people, one being 50 kg and the other being 100 kg; they are separated by 600 m, with each person holding on the end of a rope of negligible mass. Why is it, that when one of the two people creates a tension in the string, they meet at their center of mass?

Also, how my friend solved a problem similar to this one, was that he said since the one person is half as heavy, that person will travel twice as far; so, 2x + x = 600 m.

Why do both of these methods work? Could someone fill in the reasoning behind each method, please? I'd like to understand why both work.
 
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Ask yourself: How can you calculate the location of the center of mass? Can the location of the center of mass change?
 
It would seem like if the two people approached it other, then the center of mass would change.
 
Bashyboy said:
It would seem like if the two people approached it other, then the center of mass would change.
Consider it more generally. Imagine a system at rest. For the center of mass to start moving, what is required?
 
Oooh, would it be that one of the people has to start moving, and only one?
 
Bashyboy said:
Oooh, would it be that one of the people has to start moving, and only one?
Well, that would do it. But it's not what I was going for. What is required for a system to accelerate?
 
It would be a force.
 
Bashyboy said:
It would be a force.
Right. In order for the center of mass of a system to accelerate, there needs to be a net force on the system. An external force. But here there are no such forces available, since the ice is frictionless.
 
Hmm, okay, I see. So, how can I apply that knowledge to the two different methods of solving a problem like this, to understand the reasoning behind them?
 
  • #10
Bashyboy said:
Hmm, okay, I see. So, how can I apply that knowledge to the two different methods of solving a problem like this, to understand the reasoning behind them?
Well, the first method is already done. Since there are no external forces acting, the only place they can meet is at the center of mass.

As far as the second method goes: If the 100 kg person moved 1 m to the left, how far must the 50 kg person move to keep the center of mass unchanged?
 
  • #11
I would guess two meters. But how can I justify that?
 
  • #12
Bashyboy said:
I would guess two meters. But how can I justify that?
What's the definition of the center of mass? How would you calculate it?
 
  • #13
For one dimensional motion: x_{cm}= \Large\frac{m_1x_1+m_2x_2+m_3x_3+ \cdots m_nx_n}{m_1+m_2+m_3+ \cdots m_n}. And the center of mass would be constant?
 
Last edited:
  • #14
Bashyboy said:
For one dimensional motion: x_{cm}= \Large\frac{m_1x_1+m_2x_2+m_3x_3+ \cdots m_nx_n}{m_1+m_2+m_3+ \cdots m_n}. And the center of mass would be constant?
Good. Now write it in terms of Δx for two masses. Then you can relate Δx1 to Δx2.
 
  • #15
Oh, so if I know the center of mass, and I know the change in position of one of the people, then I can find the corresponding change in position of the other person, so that the center of mass remains the same, is that right?
 
  • #16
Bashyboy said:
Oh, so if I know the center of mass, and I know the change in position of one of the people, then I can find the corresponding change in position of the other person, so that the center of mass remains the same, is that right?
Exactly.
 
  • #17
You can also think about it in the form of Newton's laws of motion:

a_1m_1=F=-a_2m_2

If m_1 is twice the mass of m_2, then a_2 must be twice the acceleration of a_1 (but in the opposite direction).
 
  • #18
Okay, I believe I understand. Thank you both for your insight.
 

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