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How to solve the moment questions, I tried couple of times but

  1. Nov 21, 2013 #1
    1. The problem statement, all variables and given/known data

    [URL=http://s1345.photobucket.com/user/Duk_Bato/media/Untitled_zps863917c8.png.html][PLAIN]http://i1345.photobucket.com/albums/p679/Duk_Bato/Untitled_zps863917c8.png[/URL][/PLAIN]

    2. Relevant equations

    Tan angle = opposite/adjacent Cos angle= adjacent/hypotenuse
    Sin angle= opposite/hypotenuse

    M=F*D

    3. The attempt at a solution

    40(2.83 sin45) -30(4.47 cos26.57) -20cos45 * 0.6 +10*0.6 sin15
    =-46.82

    which is not the correct answer
     
  2. jcsd
  3. Nov 21, 2013 #2

    adjacent

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    This is too vague.What the question?
    Moment is not just F*D
    Moment is force x perpendicular distance.
     
  4. Nov 21, 2013 #3
    sry, forget abt the question
     
  5. Nov 21, 2013 #4
    Find the moment about point hinge A due to all known forces
     
  6. Nov 21, 2013 #5

    SteamKing

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    ƩM[itex]_{A}[/itex] = Ʃ r x F

    where r is the position vector from the force to A and F is the force vector.
     
  7. Nov 21, 2013 #6
    is 20N moment = -20(6)?
     
  8. Nov 21, 2013 #7
    20N =20*6 =60NM
    30N= -97.53Nm
    40N= -40sin45 *2.83=80.04NM
    10N = 10sin15(6)=15.6NM

    Can anyone tell me any of the forces are wrong?
     
  9. Nov 21, 2013 #8

    haruspex

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    It's a little confusing to use equals signs like that... edited.
    I assume anticlockwise is positive in your scheme.
    Note the distances in the diagram are cm, not m.
    Check the arithmetic.
    Please show how you got to that number. Why negative?
    By plugging in numbers too soon you introduced a rounding error. It should be exactly 80. Can you see an easy way to obtain that? And you lost the minus sign.
    Sign?

    What about the two Fs?
     
  10. Nov 21, 2013 #9
    20N has a moment of 20*0.06 = +1.2Nm

    x component of force moment is -30cos(60° )*0.02

    y component of force moment is +30sin(60° )*0.04

    40N has a moment -40*0.02 = -80Nm

    10N has a moment of -10sin(15°)*6
     
  11. Nov 21, 2013 #10

    haruspex

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    That all looks right. I still don't understand where the two 'F' forces come in. Do they arise in a different part of the question?
     
  12. Nov 21, 2013 #11
    There is question B, which ask what magnitude of the F will keep it equilibrium.
     
  13. Nov 22, 2013 #12

    adjacent

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    If it is in equilibrium,Anticlockwise moment should be equal to clockwise moment.
     
  14. Nov 22, 2013 #13
    91.96= Fsin60 5.67-Fcos15
    F= 91.96/ (sin60 5.67 -cos15)
    =27.4N

    correct?
     
  15. Nov 22, 2013 #14

    haruspex

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    I get 27.48, but that's probably near enough.
     
  16. Nov 22, 2013 #15
    ty guys, you guys help me alot
     
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