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Normal and Friction Force of Leaning Ladder against a Corner

  1. Aug 30, 2009 #1
    How do we set up the normal and friction force at the place where the ladder (the red line) touches the corner?

    http://img56.imageshack.us/img56/6655/prob08021sketch01.gif [Broken]

    Worked Done So Far

    I used a normal force the headed off to the top left at 45 degrees and a perpendicular friction force but came up with the wrong answer.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 30, 2009 #2

    tiny-tim

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    Hi lizzyb! :wink:

    Show us your full calculations, and then we can see what went wrong, and we'll know how to help. :smile:
     
  4. Aug 30, 2009 #3
    We're to find the smallest mu_s given that the ladder is 6.5 m, m = 10 kg, and mu_s is at each surface.

    http://img408.imageshack.us/img408/4034/0821diagram1.png [Broken]

    Work completed thus Far

    http://img255.imageshack.us/img255/5127/08210001.th.jpg [Broken] http://img44.imageshack.us/img44/3205/08210002.th.jpg [Broken] http://img44.imageshack.us/img44/9206/08210003.th.jpg [Broken] http://img337.imageshack.us/img337/4198/08210004.th.jpg [Broken] http://img41.imageshack.us/img41/6672/08210005j.th.jpg [Broken] http://img188.imageshack.us/img188/6554/08210006.th.jpg [Broken]

    In an effort to solve it, I went on and on and finally came up with an incorrect answer.
     
    Last edited by a moderator: May 4, 2017
  5. Aug 30, 2009 #4

    tiny-tim

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    ah! :rolleyes:

    the normal force means normal to the ladder! :wink:

    Try again! :smile:
     
  6. Aug 30, 2009 #5
    oops there is a glaring error in the calculation of the distance between the ladder's center of gravity along the x axis - it should be 1.25.
     
  7. Aug 30, 2009 #6
    should the normal force at the bottom be slanted 15 degrees or should it, too, be at 67.38 degrees? I tried it with the bottom being 15 degrees and came slightly off but I had to use a computer program to come up with the answer - there's alot of computations.

    http://img408.imageshack.us/img408/5466/082120001.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  8. Aug 30, 2009 #7
    I came up with the right answer using 15 degrees on the bottom (there was an error in one of my equations). I used maxima to solve the simultaneous equations. Thanks for you help.
     
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