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Find the force of friction

  1. Nov 21, 2009 #1
    I need help on this problem. I thought I knew how to get it, but this is a multiple choice question, and my result wasn't any of the answers.


    An 80-kg man is one fourth of the way up a 10-m ladder that is resting against a smooth, frictionless wall. If the ladder has a mass of 20 kg and it makes an angle of 60 degrees with the ground, find the force of friction of the ground on the foot of the ladder.

    Multiple choice answers:
    *7.8 x 10^2 N
    *2.0 x 10^2 N
    *50 N
    *1.7 x 10^2 N
    *100 N


    Here is the equation I thought was relevant, and I used:
    Tnet = (distance from ground to top of ladder where it meet wall)n - (distance from foot of ladder to man)mm - (1/2 x 10m)ml = 0

    (ml = mass of ladder, mm = mass of man)


    Which for my attempt, came out like this:
    Tnet = 8.66n - 80 kg(2.5 m) - 20 kg(5 m) = 0

    (To get 8.66, I did 10sin60)

    Which, since I solved for n, gave me:
    n = (80(2.5) + 20(5))/8.66
    n = 34.642 N

    which was wrong. Any help showing me where I went wrong and what I need to do would be appreciated.
     
  2. jcsd
  3. Nov 21, 2009 #2
    We know that:

    [tex]\Sigma T = 0[/tex]

    [tex]\Sigma F = 0[/tex]

    What is the best choice for the axis of rotation?
     
  4. Nov 21, 2009 #3
    The bottom of the ladder (where the ladder touches the ground)?
     
  5. Nov 21, 2009 #4
    If we decided to take the bottom point of the ladder as our axis of rotation would the friction force produce any torque?
     
  6. Nov 21, 2009 #5
    No, at least I don't think it would.
     
  7. Nov 21, 2009 #6
    Correct, it wouldn't produce a torque because the angle between the lever arm and the friction force would be zero.

    [tex]T = L\times F = |F||L|\sin{\theta }=0[/tex]

    What is a more ideal point to place our axis? One in which the friction force does produce a torque.
     
  8. Nov 21, 2009 #7
    At the center of the ladder? Or the top, where it touches the wall?
     
  9. Nov 21, 2009 #8
    Where it touches the wall will help the most because it will save us from having to calculate the force the wall exerts on the ladder.

    So, which forces are acting on the ladder? And where along the ladder are these forces being applied? (What are their lever arm values?)
     
  10. Nov 21, 2009 #9
    There is a normal force on the ladder by the ground, a normal force on the ladder by the wall (which you said cancels out?), static friction between the ground and the ladder, the weight of the ladder, and the weight of the man.

    I do not understand lever arms, so I don't know how to explain your last request.
     
  11. Nov 21, 2009 #10
    Before we continue let's explain why the normal force on the ladder by the wall cancels.

    What is the angle between the force vector and the lever arm? (lever arm is just another term for "distance from axis of rotation to the force.")
     
  12. Nov 21, 2009 #11
    Oh, now that makes more sense to me. Thank you.

    I would say 150 degrees because the ladder makes a 30 degree with the wall. But I think I may be wrong because that doesn't explain why it cancels out.
     
  13. Nov 21, 2009 #12
    150 degrees is the angle between the ladder and the wall (outside the triangle, inside the triangle it is 30). What is the angle between the force normal to the wall and the lever arm?
     
  14. Nov 21, 2009 #13
    90 degrees?
     
  15. Nov 21, 2009 #14
    Normal to the wall goes straight through the axis of rotation. right?
     
  16. Nov 21, 2009 #15
    Yes, so it would be 90 degrees?
     
  17. Nov 22, 2009 #16
    Anyone want to help me finish this problem? I need more help, and I need it by tomorrow night.
     
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