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Stuck on Equilibrium problem

  1. Oct 6, 2005 #1
    The problem:

    A force is applied to the end of a 2ft. long, uniform board weighing 50 lbs, in order to keep it horizontal and stationary while it pushes against a wall at the left. If the angle the force makes is 30 degrees facing the wall, the magnitude of the applied force is?
    (hint: the board is stationary and thus is in translational and rotational equilibrium.)
    Answer needs to be in lbs.


    Please help me, I have no idea what formula to use. My book is so crappy.
     
  2. jcsd
  3. Oct 6, 2005 #2

    Pyrrhus

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    Well, in static equilibrium problems, the resultant of force and the resultant moment must be zero, therefore you have these equations to work with, in vectorial form:

    [tex] \sum \vec{F} = 0 [/tex]

    [tex] \sum \vec{M}{o} = 0 [/tex]

    where o is any point.
     
  4. Oct 6, 2005 #3
    The answer is between 20 and 100 lbs. I'm still not sure how to apply that equation to my problem?
     
  5. Oct 6, 2005 #4

    Pyrrhus

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    I suppose the problem assumes only a reaction force without the reaction moment. Remember the weight of the board is concentrated in its center of gravity which coincides which its centroid or center of mass in this case (In the case of a homogenous rod is Length/2). Now, you need to apply the moment equation with respect to the point where the board meets the Wall (where the reactions are acting at). This will cancel the x component of the applied force, because its lever arm will be 0 with respect to the point we chose.
     
  6. Oct 6, 2005 #5

    mukundpa

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    What about the friction between the wall and the board. Or what is the thickness of the Board. If both are not given please give the drawing.
     
  7. Oct 6, 2005 #6
    |
    | F \
    | \
    | 30 degrees--> \
    |_________________________\
    |___________=______________|
    | |
    | |
    | |
    | \|/
    W

    the picture dosent post right for some reason the weight is actully in the middle and I drew the 30 degree angle at the end of the board. Thanks for all the help
     
  8. Oct 6, 2005 #7

    Pyrrhus

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    Basicly both the friction and Normal force will be the components of the reaction force. Solve this problem by taking Moment about the wall.

    [tex] \sum M_{o} = F\sin 30^{o} (L) - W(\frac{L}{2}) = 0 [/tex]
     
  9. Oct 15, 2005 #8
    I dont know what F is so how can I use that equation?
     
  10. Oct 15, 2005 #9

    hotvette

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    Cyclovenom basically gave you the answer. The original question was to find F. You know W and L, so just solve the equation for F. Algebra.
     
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