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Slipping Ladder and Torque?

  1. Oct 17, 2014 #1
    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?
    upload_2014-10-17_16-5-18.png
     

    Attached Files:

  2. jcsd
  3. Oct 17, 2014 #2

    Dale

    Staff: Mentor

    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.
     
  4. Oct 17, 2014 #3
    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?
     
  5. Oct 17, 2014 #4

    Dale

    Staff: Mentor

    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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