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Equilibrium Ladder Problem

  1. Feb 24, 2014 #1
    1. The problem statement, all variables and given/known data
    A uniform ladder is 10 m long and weighs 180 N. In the figure below, the ladder leans against a vertical, frictionless wall at height h = 8.0 m above the ground. A horizontal force is applied to the ladder at a distance 1.0 m from its base (measured along the ladder).

    (a) If F = 50 N, what is the force of the ground on the ladder, in unit-vector notation?

    (b) If F = 150 N, what is the force of the ground on the ladder, in unit-vector notation?

    (c) Suppose the coefficient of static friction between the ladder and the ground is 0.38; for what minimum value of F will the base of the ladder just start to move toward the wall?

    2. Relevant equations
    ∑[itex]\tau[/itex] = r x F and ∑F = ma


    3. The attempt at a solution
    I got both parts (a) and (b) right so I don't need those.

    For part (c), I summed the horizontal forces: ∑F[itex]_{x}[/itex] = 0 = -F[itex]_{f}[/itex] + F[itex]_{A}[/itex] - N[itex]_{W}[/itex]
    where F[itex]_{A}[/itex] is the applied force, F[itex]_{f}[/itex] is the frictional force, and N[itex]_{W}[/itex] is the normal force from the wall.

    Then I summed the vertical forces: ∑F[itex]_{y}[/itex] = 0 = N - W

    Then I summed the torques using the base of the ladder as the axis of rotation: ∑[itex]\tau[/itex] = 0 = Fsin(53) + 5(180)sin(37) - 10N[itex]_{W}[/itex]sin(53)

    Then I used N[itex]_{W}[/itex] = μmg - F[itex]_{A}[/itex] and substituted it into the torque equation and then plugged in all my numbers to solve for F.

    I got F = 133 N, which is wrong and I'm not sure what I'm doing wrong.
     

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    Last edited: Feb 24, 2014
  2. jcsd
  3. Feb 24, 2014 #2
    This equation NW = μmg - Ff is not consistent with your earlier relationship. It should read:
    NW = FA-μmg
     
  4. Feb 24, 2014 #3
    Okay so after fixing that I have ∑[itex]\tau[/itex] = 0 = F[itex]_{A}[/itex]sin(53) +5(180)sin(37) - 10(F[itex]_{A}[/itex] - (.38)(180))

    Solving for F[itex]_{A}[/itex], I got 118.23 N which is still wrong. Did I make a mistake in summing the torques?
     
  5. Feb 25, 2014 #4

    haruspex

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    There's no value in determining the angle. Just work with its trig ratios. These are handily rational here.
    Think again about that 10.
     
  6. Feb 25, 2014 #5

    ehild

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    You left out sin(53) from the last term.
    And better to use the ratios as haruspex said. Taking the angle 53° means a big rounding error.

    ehild
     
  7. Feb 25, 2014 #6
    Should I use 6 (the horizontal distance from the base of the ladder to the wall) instead?
     
  8. Feb 25, 2014 #7

    ehild

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    No. What is the leve[STRIKE]l[/STRIKE]r arm for NW?

    ehild
     
    Last edited: Feb 26, 2014
  9. Feb 26, 2014 #8

    haruspex

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    "Level arm"? I'm unfamiliar with the term, so bnoone might be too.
    bnoone, NW acts horizontally. What is its distance from the point you're taking moments about?
     
  10. Feb 26, 2014 #9

    ehild

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  11. Feb 26, 2014 #10

    haruspex

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  12. Feb 26, 2014 #11

    ehild

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    Yes, I mistyped it...

    ehild
     
  13. Feb 26, 2014 #12
    Isn't it just the length of the ladder?
     
  14. Feb 26, 2014 #13

    haruspex

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    Torque = force * perpendicular distance. Is the ladder perpendicular to NW?
     
  15. Feb 27, 2014 #14
    Ohhh okay that makes sense. I forgot to include the sine of the angle between them. Thank you!
     
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