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Engineering Statics Question

  1. Apr 20, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img512.imageshack.us/img512/2617/staticsquestion4.jpg [Broken]

    2. Relevant equations
    - Ability to resolve forces into components
    - M = Fd

    3. The attempt at a solution
    Resolve the force acting at an angle into components...
    10kN force = -10cos50i - 10 sin 50 j kN

    Therefore, resultant R becomes...

    = (-10cos50)i + (15 - 3 - 10 sin 50)j kN
    = -6.428i + 4.3396j kN

    (with a magnitude of 7.756kN)

    (b) This is where I get stuck.

    I basically calculated the moments created by the individual elements of R,

    F1 = 15kn force
    F2 = 3kn force
    F3 = vertical component 10kn force
    and, because I am using the joint with the wall as an origin, the line of action of the horizontal component passes through my origin thus no moment.

    M1 = 15*3 = 45kNm
    M2 = -3*1.5 = -4.5kNm
    M3 = (-10 sin 50)*9 = -68.944kNm

    Sum of moments = 45 - 4.5 - 68.944 = -28.444kNm

    and then I get stuck. Do you add the moment you've just found to the moment provide in the couple, then sub. back into M=Fd where F=R?

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 20, 2009 #2


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    Homework Helper

    andrew.c: Nice work. The answer to the question in your last paragraph is yes. Also, notice a mistake in your M2 equation.
  4. Apr 21, 2009 #3
    Yes, should use 4.5, not 1.5.
    Thanks, I got the right answer!
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