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

AI Thread Summary
The discussion focuses on solving a non-equilibrium pulley system at a 154-degree angle, emphasizing the need to calculate tension. Participants explore the relationship between the positions of the masses and the geometry of the system, suggesting that differentiating their positions could reveal insights. There is a debate about whether the tension is at a minimum or still changing, with some arguing that the system's symmetry allows for a unique tension value. The problem is likened to related rates problems, particularly the ladder problem, due to the fixed lengths of the ropes involved. Overall, the conversation highlights the complexities of analyzing non-equilibrium systems in physics.
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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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