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Leaning forces

  1. Jul 16, 2009 #1
    Lets say you take a ladder and put it on a scale and measure it's weight while balancing it vertical. If you then lean the ladder at varying angles against a wall, would the scale's reading remain unchanged? I take it that it would, but for some reason this seems odd to me.

    Lets say the wall was basically a giant scale too, it would vary but the one on the ground would not? Can someone please enlighten me? thanks!
     
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  3. Jul 16, 2009 #2

    Lok

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    If there is no friction between the wall and the ladder than there would be no force and all the weight would go to the scale. But as there's always a bit of friction, then that means that the scale will not show the full weight of the ladder.

    Sit on a scale and lean with one hand on a wall see what happens.
     
  4. Jul 16, 2009 #3

    Nabeshin

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    A quick derivation I'm not terribly confident in yields the following:
    [tex]F_{nf}=W*\left(\frac{\frac{1}{2}+\mu_f \tan{\theta}}{1+\mu_w \mu_f ^2 \tan{\theta}}\right)[/tex]

    Where W is the weight of the ladder, mu_f and mu_w are the coefficients of friction for the floor and wall respectively, and theta is the angle between the ladder and the floor.

    So, assuming this equation is right, there is a dependence on theta.

    (Can someone double check this? Doesn't quite look right to me, but I can't find much wrong with it.)
     
  5. Jul 16, 2009 #4

    rcgldr

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    If the friction forces are high enough to prevent slippage, then it seems that the vertical forces would be equally shared by wall and floor, regardless of the angle. An alternative scenario would be to imagine the ladder suspended at both ends by vertical ropes. Each rope bears 1/2 the weight of the ladder regardless of it's angle.
     
    Last edited: Jul 16, 2009
  6. Jul 17, 2009 #5
    What's [tex]F_{nf} [/tex] here?
     
  7. Jul 17, 2009 #6

    Nabeshin

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    Uhh, Force normal floor, the notation I used during the derivation to distinguish the different normal, frictional forces is F(type)(object).
     
  8. Jul 17, 2009 #7

    rcgldr

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    Assuming the ladder isn't sliding, then why do the coefficients of friction matter, since the actual forces are less than the maximum static friction forces?
     
  9. Jul 18, 2009 #8
    In that case your expression gives the value of the normal force exerted by the floor as 0.5W even when there is no friction between the ladder and the wall or floor. It should be W right?
     
  10. Jul 18, 2009 #9

    Nabeshin

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    Uhh, technically yeah, but since in my derivation I assumed friction from both the wall and the floor, I don't think the expression should be valid for the non-friction case.

    This is actually a really good point that I did overlook. However, without the ability to relate the frictional forces to the normal forces via the coefficient of friction, this looks like an under determined system to me. So I guess my posted solution is completely incorrect.

    Anyone have any ideas if this is actually solvable?
     
  11. Jul 18, 2009 #10

    rcgldr

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    I don't think it is. Replace the wall and floor contact points with hinges (infinite friction) and there is no limit to the magnitude or direction of force at either hinge, other than the sum of the vertical forces equals the weight. Since it is friction based, then the wall can't exert an inwards horizontal force and the ground can't exert a downwards veritcal force. Other than these limitations I don't see an answer. You could get similar results by hanging both ends of a ladder from two ropes.

    If the ropes were the ends of a single rope hanging from a pair of horizontally spaced pulleys, then their tension would be equal, and each would support 1/2 the weight of the ladder, regardless of the ladders angle.

    If the space between the pulleys >= length of ladder the ladder is horzontal.

    If the space between the pulleys < length of ladder, then the ladder rotates until the ropes are vertical, which is the ladders lowest position. If the initial state is a horizontal ladder, it's an unstable balance point.
     
  12. Jul 18, 2009 #11

    Nabeshin

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    Bummer, problems without definite solutions give me the heebie-jeebies.

    The question of whether or not the ladder would be supported at a given angle would be answerable given the coefficients of static friction of both surfaces though, right? I guess that's as close as we can get... Or, given the coefficients of friction, the maximum angle the ladder can remain supported at?
     
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