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How do I find the line of action of this resultant force?

  1. Jun 3, 2009 #1
    Three forces are applied to a body. They are :

    F1 = (4,5) applied at (1,2)

    F2 = (2,-1) applied at (3,-1)

    F3 = (-3, 2) applied at (-2,1)


    i) Find the resultant force

    Answer : R = (3,6)

    ii) Find the total moment about the origin

    (This i know how to do)

    iii) The line of action of R cuts the Y-axis at (0,d). Find d

    iv) Find the equation of this line of action



    How do I solve iii) and iv)
     
  2. jcsd
  3. Jun 3, 2009 #2

    PhanthomJay

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    Did you try calculating the moments about the origin from the the x and y components of the forces separately? Then you can find a set of coordinates through which the resultant passes ([tex] X = \Sigma{(F_y(x))}/R_y[/tex], etc.), the slope of which resultant is R_y/R_x.
     
  4. Jun 4, 2009 #3
    Is this assumption correct? (force vectors and moment)

    Three forces are applied to a body. They are :

    F1 = (4,5) applied at (1,2)

    F2 = (2,-1) applied at (3,-1)

    F3 = (-3, 2) applied at (-2,1)


    i) Find the resultant force

    Answer : R = (3,6)

    ii) Find the total moment about the origin

    Answer: I calculated this and it is -5 Nm

    iii) The line of action of R cuts the Y-axis at (0,d). Find d

    Question: Can I assume that R = (3,6) is applied at (0,d) and its moment about the origin is - 5Nm? In order to find what d is?


    If that assumption is wrong then how do i find d?
     
  5. Jun 4, 2009 #4

    LowlyPion

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    Re: Is this assumption correct? (force vectors and moment)



    Can you show your work on determining the moment in part ii)?

    Isn't it the cross product of the F X r ?

    Doesn't that yield

    (4*2 + 5*1) + (2*(-1) + (-1)*3) + ((-3)*1 + 2*(-2)) = ... but ≠ -5
     
  6. Jun 4, 2009 #5

    berkeman

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    I merged two duplicate threads.
     
  7. Jun 5, 2009 #6

    PhanthomJay

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    Re: Is this assumption correct? (force vectors and moment)

    Yeah, as long as you are consistent with your plus and minus signs, that is to say, is d on the positive or negative y axis?
     
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