How Do You Solve a Non-Equilibrium Pulley System with a 154 Degree Angle?

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Homework Help Overview

The discussion revolves around a non-equilibrium pulley system involving a specific angle of 154 degrees. Participants are exploring the dynamics of tension in the system and the relationships between the components involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the assumption of equal accelerations in the system and discussing the relationships between the positions of the masses and the geometry of the setup. Some are considering the implications of fixed lengths of the ropes and drawing parallels to related rates problems.

Discussion Status

The discussion is active, with participants offering insights into the relationships between the variables involved. There is a mix of interpretations regarding the nature of tension and its behavior as the angle changes, but no consensus has been reached on the specifics of the calculations or assumptions.

Contextual Notes

Participants note that the problem is framed by a teacher's prompt, emphasizing the non-equilibrium nature of the system. There are discussions about the minimum tension required and the implications of the system's geometry.

ashley2024
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Homework Statement
Find the tension in the following system (see image).
Relevant Equations
Fnet = ma, Ff = uFn,
Please note that the system is not in equilibrium, and that tension must be solved for the instant where the angle is 154 degrees.

inst V.png


My attempt (correct ans is Ft = 626N)
image (4).png
 
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##a_1=a_2##? Sure?
 
haruspex said:
##a_1=a_2##? Sure?
Nope, but I don't really have any other ideas.
 
ashley2024 said:
Nope, but I don't really have any other ideas.
There is a simple relationship between the two, just not quite that simple.
Start with positions. If m is x from the vertical through M, the knot is y below the horizontal through m, the string connecting them has length L, and it is at θ to the vertical, write expressions for x and y and differentiate twice.
 
haruspex said:
There is a simple relationship between the two, just not quite that simple.
Start with positions. If m is x from the vertical through M, the knot is y below the horizontal through m, the string connecting them has length L, and it is at θ to the vertical, write expressions for x and y and differentiate twice.
Would this be similar to the related rates problems? Since the ropes have fixed length, it seems similar to the ladder problem where you have to differentiate its rate as it falls...
 
ashley2024 said:
Please note that the system is not in equilibrium...
Welcome, Ashley! :smile:

What makes you state that?
 
Lnewqban said:
Welcome, Ashley! :smile:

What makes you state that?
The teacher who gave this problem.
 
ashley2024 said:
The teacher who gave this problem.
Then, the tension in the V=shaped wire should be the minimum to keep masses #1 sliding toward each other.
We could then, disregard the value of the force exerted by the falling mass #2, since we know that it is plenty to have induced and to keep the sliding movements of both masses #1.
Therefore, calculating T2 seems not to be necessary.
As the system is geometrically symmetrical, there is a unique value for T1.
 
Interesting! But I don't understand why it would be the minimum though? Since its asking for the tension when the angle is 154, I assumed that the tension could be still in the process of changing.
 
  • #10
ashley2024 said:
But I don't understand why it would be the minimum though
It isn't. Don’t be distracted by that.
 
  • #11
ashley2024 said:
Would this be similar to the related rates problems? Since the ropes have fixed length, it seems similar to the ladder problem where you have to differentiate its rate as it falls...
Yes.
 

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