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Free Body Diagram Finding Moments

  1. Mar 6, 2012 #1
    1. The problem statement, all variables and given/known data
    I have to find the forces FGB FCB AND FGH
    http://desmond.imageshack.us/Himg696/scaled.php?server=696&filename=img59wa.gif&res=medium [Broken]
    P1 = 5 kN, P2 = 10 kN, JY = 21.25 kN


    a1 = 2 m, a2 = 1 m, a3 = 0.5 m, a4 = 1 m, a5 = 2.5 m

    Angle FGH makes with the horizontal is: 26.57 degrees


    Angle FGB makes with the horizontal is –tan-1(1.5/2)

    2. Relevant equations




    3. The attempt at a solution

    Sum the forces up and down = 5+10 -21.25 +FGHsin26.57 - FGB sin38.87 = 0


    Then summing moments =
    2x10 - 2 x 21.25 +(a2+a3+a4)FGHcos26.57 + FGB cos 38.87= 0

    I don't get what i can do from here and im a bit stuck and i don't no if the force Fcb would have an effect??
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 7, 2012 #2

    tiny-tim

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    welcome to pf!

    hi steve2510! welcome to pf! :smile:
    You also have to sum the forces left and right (to equal zero). :wink:
     
  4. Mar 7, 2012 #3
    Okay So this is what I've got
    ƩFx = Fcb + Fghcos26.57 + Fgbcos-36.87 = 0

    ƩFy = 21.25 + Fghsin26.57 - Fgbsin-36.87 -5 - 10 = 0

    ƩFy = Fghsin26.57 - Fgbsin-36.87 =- 6.25

    ƩMb = (Fghcos26.57 x1.5 )+ (Fgbcos-36.87 x 1) +(2x5) -(2x21.25) = 0

    ƩMb = (1.5 x Fghcos26.57 ) + (Fgbcos36.87) = 32.25

    ƩMb = (Fghcos26.57 ) + (Fgbcos36.87) = 21.5 (divided by 1.5)

    ƩFx = Fcb + Fghcos26.57 + Fgbcos-36.87 = 0
    -((Fghcos26.57 ) + (Fgbcos36.87)) = 21.5
    = Fcb = -21.5

    Am i doing this right at the moment ? I'm not sure if i took moments about the correct point and whether the sin and cos are correct
     
    Last edited: Mar 7, 2012
  5. Mar 7, 2012 #4

    tiny-tim

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    shouldn't the moment of FGB about B be 0 ? :confused:

    and the multiplier of FGH needs to be the distance from B to GH
     
  6. Mar 7, 2012 #5
    Am i right in saying Fgb is the force that g exerts on b ? So there would be a moment force in the x direction but not in the y direction as the perpendicular distance would be 0. And i thought 1.5 was the mutliplier as the horizontal force acts along the top of the a4 line
     
  7. Mar 8, 2012 #6

    tiny-tim

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    hi steve2510! :smile:

    (just got up :zzz:)
    but isn't the perpendicular distance zero in any direction? :confused:
    i was referring to the cos
     
  8. Mar 8, 2012 #7
    Cos is the horizontal component for both forces , am I right ?
     
  9. Mar 8, 2012 #8

    tiny-tim

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    you seem to be mixing up the methods for components of force (in a direction) and moments of force (about an axis or point)

    cos is for components

    this is moments

    with moments, it's usually sin :wink:
     
  10. Mar 8, 2012 #9
    I'm confused as the sin of both forces don't produce moments around B ? Is there any chance you could write what the sum of the moments is meant to equal
     
  11. Mar 9, 2012 #10

    tiny-tim

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    The moment of a force F about a point B is F*d,

    where d is the perpendicular distance from B to the line of F.​

    If A and C are any points on that line, then Fd = F*AB*sinBAC.

    If you're not familiar with this, you need to go back to your book and study and practise it.
     
  12. Mar 9, 2012 #11
    okay think its back to the library then, thank you for your help!
     
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