Is the Slipping Ladder Problem a Torque or Static Equilibrium Problem?

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SUMMARY

The discussion clarifies that the Slipping Ladder Problem is fundamentally a static equilibrium problem, which inherently involves torque considerations. Participants agree that while the ladder does not rotate, the analysis of forces and torques at equilibrium is essential for understanding the conditions leading to slipping. The key takeaway is that static equilibrium encompasses both the sum of forces and the sum of torques being zero, making torque analysis integral to solving these problems.

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  • Understanding of static equilibrium principles
  • Basic knowledge of torque and its calculations
  • Familiarity with force interactions in physics
  • Concept of angles and their impact on force distribution
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  • Study the principles of static equilibrium in physics
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  • Explore real-world applications of torque in ladder stability
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Physics students, educators, and engineers interested in understanding the dynamics of static equilibrium and torque in practical applications, particularly in scenarios involving ladders and similar structures.

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I'm a little confused as to the conceptual premise behind torque-ladder problems? Like "at what height up the ladder is the man and ladder most likely to slip?"

I understand how to solve such problems, but i have trouble understanding how it is a torque problem.

Is it a torque problem or rather a static equilibrium problem?
I know they're kind of the same thing, but as a torque problem, I can't imagine the object actually rotating. Rather i imagine it slipping but once it starts slipping the torque conditions constantly change as the angle theta of the ladder against the ground changes.

As an equilibrium problem, it makes sense because in static equilibrium using torue allows us to use more information to understand the force interactions.

Am I "right" in being confused about this as a traditional torque problem?
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Hereformore said:
Is it a torque problem or rather a static equilibrium problem? I know they're kind of the same thing,
As you mention, they are essentially the same thing. Specifically, static equilibrium means that the sum of the forces are zero and also that the sum of the torques are 0. So any static equilibrium problem has a torque problem buried inside.
 
DaleSpam said:
As you mention, they are essentially the same thing. Specifically, static equilibrium means that the sum of the forces are zero and also that the sum of the torques are 0. So any static equilibrium problem has a torque problem buried inside.
I see. So in considering such scenarios, it has to be in static equilibrium right? I guess it just seems odd that the question asks about a scenario outside of equilibrium, when you can really only accurately determine what's going on when it is in equilibrium.

It threw me off because i was trying to picture the ladder rotating and it didnt make sense to me. But it's more exploiting the torque at equilibrium right?
 
Hereformore said:
But it's more exploiting the torque at equilibrium right?
Right. The forces still exert torques at equilibrium, even though at equilibrium it is not rotating and all of the torques sum to 0.
 

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