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The force of magnitude F acts along the edge of the triangular plate

  1. Aug 28, 2011 #1
    1. The problem statement, all variables and given/known data
    I attached the drawing.
    The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O.

    2. Relevant equations


    3. The attempt at a solution
    What I did was I just picked the top left point of the triangle and labeled that (0,h) which give me my r vector 0i+hj. After that I really don't know how to proceed except to find the angle by doing inverse tangent of h/b, but I just keep getting variable after variable. What other way is there to do this?

    Attached Files:

  2. jcsd
  3. Aug 28, 2011 #2
    Re: Moments

    F is

    b i - h j

    what matters is the y direction for it is perpendicular and x is forming no moment

    you just (bi) cross (-hj) to get ( -bh ) k, a vector of bh magnitude into the page.
  4. Aug 28, 2011 #3
    Re: Moments

    The answer is supposed to be Fbh/(h^2+b^2)^.5. Answers do vary tho according to method, but I don't think yours matches this
  5. Aug 28, 2011 #4
    Re: Moments

    YES sorry I made a very Fundamental mistake and will try to rectify it

    F= F *( b / ( h**2+b**2 ) ** 0.5 ) i + F*( h / ( h**2+b**2 ) ** 0.5 ) j

    Fx * h = is the answer F*b*h *(h**2+b**2 ) ** -0.5

    which is into the page
  6. Aug 28, 2011 #5
    Re: Moments

    I'm still a little confused. I would like to do all moment problems by using M=rxF. How would that work in this problem. i can get the F vector but it's going to look really messy because I gotta define the angle theta as tangent inverse of (b/h) then take the cosine of the tangent inverse of (b/h) to get the x component and same for the y.
  7. Aug 28, 2011 #6
    Re: Moments

    M=r X F is already defined to be M= ( rx i + ry j + rz k ) X ( Fx i + Fy j + Fz k )

    here simply r X ( Fx i + Fy j ) = r X Fx i + rXFy j by distribution property

    but here : r= h j what do you get when you cross j by j ? for the latter part, of course a zero

    here rXFyj drops and we are left with r X Fx i , here you cross jXi which points in the direction into the page.
  8. Aug 28, 2011 #7
    Re: Moments

    its so hard to write all this online w/o subscripts and stuff. can r=b i too?
  9. Aug 28, 2011 #8
    Re: Moments


    r is starting from ( 0, 0 ) ---> take origin


    draw a vector to the APPlication point of F.

    r begins at ( 0,0 ) ends at application point of F.
  10. Aug 28, 2011 #9
    Re: Moments

    what I asked was basically the ending point can be any point on the line of action right?
  11. Aug 28, 2011 #10
    Re: Moments

    No Sir, position vector starts at wherever you take the moment ( here (0,0) ) with respect to, and ends at starting point of F. ( which is h units above the (0,0)
  12. Aug 29, 2011 #11


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    Staff: Mentor

    Re: Moments

    You need to stop doing the student's homework for them. You may provide hints, ask questions, find mistakes, etc. But it is against teh PF rules to do the student's work for them.
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