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Calculating forces in an object that is leaning

  1. Nov 9, 2006 #1
    A question came up in physics class that I'm curious about.

    The question involved a ladder leaning against a frictionless horrizontal surface with a man climbing up it. The ground the ladder was resting on had a friction coefficient and the angle of the lean of the ladder was given. The question asked which point on the ladder was the highest the man could climb before the ladder would slide out from under him. Now, my question to you is this:

    How does one determind the the components of the forces caused by the weight of a leaning object? The weight vector of an oject always points downard, but an object that is leaning or falling has forces acting upon it in both the X and Y direction, how do I determine these?
     
    Last edited: Nov 10, 2006
  2. jcsd
  3. Nov 10, 2006 #2
    Draw a free body diagram.
     
  4. Nov 10, 2006 #3
    I did so, but I am still a little confused.

    The Weight vectors point straight down, and any normal force from the ground must point straight up because the ground is assumed to be perfectly flat. Eventhough the ladder is leaning, the weight vectors and the normal force completely cancel each other?
    I am skeptical that this is the case because some force must be canceled by the force that the horrizontal wall is exerting on the ladder. Is the the firction betweent eh base of the ladder and the ground?
     
  5. Nov 10, 2006 #4
    You should be able to determine that from your free body diagram. You know the answer to this yourself.

    Please draw one and put it on image shack.

    Draw an image, and post some force balance equations. I'm not going to 'give' you the answer, you're going to give it to me.
     
  6. Nov 10, 2006 #5
    I would love to, but I am lacking a scanner at the moment. I'm just going to treat the points of contact between the ladder and the wall/floor as pivots and mess with torques and see what happens. Even if I'm dead wrong, it s fun to screw around with physics.
     
  7. Nov 10, 2006 #6
    No, you can use MS paint and upload it on www.imageshack.com and put the link to the picture here.

    And start writing down equations.

    If you know how to use word, you can make the picture very nice very fast using the autoshapes, it will take you 5 mins.
     
  8. Nov 10, 2006 #7
    Alright, I will try to keep it neat
     
  9. Nov 10, 2006 #8
  10. Nov 10, 2006 #9
    OOps, I left friction out, hold on
     
  11. Nov 10, 2006 #10
    You should assume forces in both the x and y directions at both edges of the ladder. You need to fix your picture.
     
  12. Nov 10, 2006 #11
    Alright, it has been fixed
     
  13. Nov 10, 2006 #12
    Ok, will that suffice?
     
  14. Nov 10, 2006 #13
    What about the other wall?
     
  15. Nov 10, 2006 #14
    If I draw cooridnate axis with the X axis parallel to the ladder, cna we assume that it holds true for both ends of the ladder?

    Just to make sure I'm not confused as to what you are asking of me. You want to orient the coordinate axis in such a was so that the components of the forces acting on the ladder act in the X and Y directions, correct?
     
  16. Nov 10, 2006 #15
    All I said was your are missing a force on the other end of the ladder. Where is the y component of force?
     
  17. Nov 10, 2006 #16
    But I do not understand how I am missing a force. The only force the horizontal wall cna exert is the normal force because that horrizontal wall is frictionless, isn't that right? I appologize if I am missing something obvious.
     
  18. Nov 10, 2006 #17
    Yep, I was making sure you were paying attention. Now write some force balances.
     
  19. Nov 10, 2006 #18
    Ok, I'm going to bed. My advice for now. Do a force balance and start taking moments and write down your results.
     
  20. Nov 10, 2006 #19
    I made a mistake, the angle was nto given so I'm going to go with the traditional orientation of the axes to simplify things.

    sumF(x) = W(man)-W(ladder)+N=0
    sumF(y) =F(wall)-f=0
     
  21. Nov 10, 2006 #20
    Alright, thank you
     
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