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Statics - Crane boom

  1. Mar 6, 2016 #1
    rbd6X51.jpg
    1. The problem statement, all variables and given/known data

    W engine = 500 kg * 9.81 = 4905 N
    sin t = 3/5

    2. Relevant equations


    3. The attempt at a solution
    1. use section AE
    ΣMa = 0

    [itex] 3/5 * Fcd *2 - 4905 * 3 = 0 [/itex]

    [itex] F = 12262.5 N [/itex]

    2. use section EB

    ΣMe = 0

    [itex] 3/5 * Fcd *1 - 4905 * 2 = 0 [/itex]

    [itex] F = 16350 N [/itex]

    why 1 and 2 doesn't result same answer ?
     
  2. jcsd
  3. Mar 6, 2016 #1

    SteamKing

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    Beats me.

    You should write the equations of static equilibrium for this crane rather than try to jump into the middle of a calculation.

    Try drawing a free body diagram for the beam AEB and putting that member in equilibrium.
     
  4. jcsd
  5. Mar 6, 2016 #2
    There is only 3 equations right ? qDWn1Vk.png
    ΣMoment =0
    [itex] Fy * distance - weight * distance = 0 [/itex]

    Σshear = 0
    Σnormal = 0
     
  6. Mar 6, 2016 #3

    SteamKing

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    There are two equations of static equilibrium: ∑F = 0 and ∑M = 0.

    The force equation can be applied in the horizontal or the vertical direction, but generally only one moment equation can be written.
     
  7. Mar 6, 2016 #4
    ok, so moment is caused by force in the vertical direction so formula for ΣM = Fy * d - W * d

    For moment at point A
    ΣMa = 0
    Fy *d - W * d = 0
    [itex] sin( t) * Fcd - W *d [/itex]
    [itex](3/5 * Fcd *2) - (4905 * 3) = 0 [/itex]

    F=12262.5N
    is this correct ?

    If i cut to segment EB
    rxZcgoL.png
    sin t will be the same since angle t is not change

    ΣM = 0
    is my formula correct ?
     
  8. Mar 6, 2016 #5

    SteamKing

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    Your calculation for F is correct. However, you need to calculate the components of F in order to find the internal forces acting at point E.

    I would not use the same letter "d" for both moment arms as this suggests each arm is the same distance.
     
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