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Normal force in man-on-ladder torque questions?

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data

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    This is just a more generalized question, and not really about a specific value. The question was regarding torque, and had a man standing on a ladder like so. The only force (exerted by the man on the ladder) that provided torque was indicated to be mg, but I don’t really understand—why isn’t Fn included? Obviously there is a downwards Fn exerted by the man on the ladder, so why doesn’t the diagram account for that? (I’m a bit confused since in loop-the-loop roller coaster problems, we always have to account for downwards Fn.)

    Thank you in advance!
     
  2. jcsd
  3. Nov 5, 2012 #2

    ehild

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    Good question! The man stands on a step of the ladder which is (hopefully) horizontal. He exerts some normal force FN on that step. The step acts with the same force, but in opposite direction on the man. Gravity also acts on the man. The man is steady, so the resultant force on him is zero. That means FN-mg=0, FN=mg. At the end, the man exerts a force, equal to its weight, on the ladder.

    ehild
     
  4. Nov 6, 2012 #3

    haruspex

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    Here's another way to think about it. It needn't be a ladder; it could just be a plank, provided there is enough static friction that the man does not slide. In this case there would be a normal force, and a frictional force up the plank. But since these have to be balanced exactly by the vertical weight of the man, their resultant is vertical.
     
  5. Nov 7, 2012 #4
    I get it now! Thanks so much to both of you. Whew, thinking about these things can be a real doozy sometimes :tongue:
     
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