Atwood Machine Help: Magnitude of T When M1 Goes to Infinity

Click For Summary

Homework Help Overview

The discussion revolves around an Atwood machine problem involving two masses, m1 and m2, where m1 approaches infinity while m2 remains constant. Participants are exploring the implications of this scenario on the tension in the rope connecting the masses.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including free body diagrams and the forces acting on each mass. Questions arise regarding the behavior of tension as m1 increases and the acceleration of m2. There is confusion about the relationship between tension and the forces acting on m2.

Discussion Status

Some participants have provided guidance on analyzing the free body diagram and understanding the physical implications of the problem. Multiple interpretations of the tension and acceleration are being explored, indicating a productive discussion without explicit consensus.

Contextual Notes

There is an ongoing examination of the assumptions regarding the masses and the effects of m1's increasing size on the system's dynamics. Participants are encouraged to reconsider their understanding of tension and acceleration in this context.

Meteo
Messages
12
Reaction score
0
An atwood machine holds two masses at its ends by a massless rope. m1 and m2. Assume the pully is frictionless.
What will the magnitude of the tension(T) go to as m1 goes to infinity but m2 stays the same?
Im very lost on this problem. I have drawn a free body diagram showing m1g and m2g pointing downwards and T pointing upwards.

I have derived two equations

m_1g-T=-m_1a where m1 is going down. I subtracted T from m1g because m1g is much bigger.
T-m_2g=m_2a where m2 is going up. I did this because T should be greater than m2g. i think.
but apparently that's not right, and I don't know why.
 
Last edited:
Physics news on Phys.org
You need to analyze your free body diagram a little more carefully!

Alternatively, you should realize that in the limit of large m_1, m_2 will be accelerated upward at a rate g, i.e. it pulls on the string with a force equal to m_2 g due to the acceleration. That, plus its weight, gives you the tension.
 
why is m2 accelerated upwards at g?
its weight will be m2g... so T=2m_2g?

I think I am comfused on what tension is...so the force of it pulling up and the force of the weight pulling down = tension?
 
Tide was talking about the limiting case as m1 gets very large. Suppose m1 were 30,000 tons and m2 were 2 ounces. Can you see that m2 would make very little difference in that case? So, for all practical purposes, m2 would accelerate upwards at g because m1 would accelerate downwards at g.

As for what tension is - well, you can consider it a force exerted by the rope. It'll pull both m1 and m2 upwards - exactly how much depends on how m1 and m2 pull downward on their ends.

Go back and do as Tide suggested - analyze the free body diagram. But when you do it, try to picture it as a physical system rather than just a system of equations. If the math confuses you, the physics will always pull you out. Just ask yourself what should happen.
 
m_2 is accelerated upward at g because it offers negligible resistance to m_1 falling.

The tension is the force exerted by the string on the mass. The forces acting on the mass are the tension and force of gravity so T - m_2 g = m_2 a. But a = g, therefore ...!
 
Meteo,

Listen to Diane - she's good! :)
 
How does one blush online? :)
 
Like this? :blushing:

I didn't mean to make you blush - just offering some sound advice! :smile:
 

Similar threads

Replies
25
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
Replies
1
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 7 ·
Replies
7
Views
4K