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Static equilibrium, trusses/section method

  1. Jul 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the forces in members BC, BE, and EF. Solve for each force from an equilibrium equation which contains that force as the only unknown. The forces are positive if in tension, negative if in compression.
    20150718_153115_zps3gcjaafw.jpg
    2. Relevant equations
    Sum of forces and moments equal zero.

    3. The attempt at a solution
    20150718_153056_zpsr4lvcjn7.jpg

    I'm on force BE, i cannot see what is wrong with my moment equation. what am i missing here? thanks.
     
  2. jcsd
  3. Jul 18, 2015 #2
    How many torques are acting on B?
     
  4. Jul 18, 2015 #3
    B has two torques acting on it.
     
  5. Jul 18, 2015 #4
    Are you sure? Why only two?
     
  6. Jul 18, 2015 #5
    because i made a cut between B and C. and the only torques are: [tex] Dy(5) and 2.5EFcos∅ [/tex]
    im sure i think its only two. but im often wrong
     
    Last edited: Jul 18, 2015
  7. Jul 18, 2015 #6
    Should i include G(2.5)? if yes then why because i cut it out?
     
  8. Jul 18, 2015 #7

    TSny

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    Your distance between D and P doesn't look correct.
     
  9. Jul 18, 2015 #8
    I created point P to be 2.5m from point D. was i wrong to do that?
    edited
     
  10. Jul 18, 2015 #9

    TSny

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    I thought you chose P so that P lies on the extension of segment FE. Then force FEF would not produce any moment about P.

    If P is not on the extension of EF then you will need to include the moment due to FEF when summing moments about P.
     
  11. Jul 18, 2015 #10
    Oh ok i get the trick now. that makes much more sense. let me try again. thanks
     
  12. Jul 18, 2015 #11
    so how do i find the length of DP if i only know the angles and one side of a similar triangle CED?
    Is using similar triangles the correct way to solve for DP?
     
    Last edited: Jul 18, 2015
  13. Jul 18, 2015 #12

    TSny

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    Consider triangle CEP. What is the angle at vertex P? What is the length of side CP?
     
  14. Jul 18, 2015 #13
    the angle at P is 17.74 degrees. Im still thinking about how to find length CP
     
  15. Jul 18, 2015 #14
    CP is 5.8 using the law of sines.
     
  16. Jul 18, 2015 #15

    TSny

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    OK
     
  17. Jul 18, 2015 #16

    TSny

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    I don't think this is correct. I don't see how you are using the law of sines.

    If you know the angle at P and the length of the side CE, you should be able to get side CP.
     
  18. Jul 18, 2015 #17
    Im struggling to see how. I used the law of sines knowing two angles of CEP and one side.
    [tex] sin(72.26)/CP = sin(17.74)/1.7 [/tex]
     
  19. Jul 18, 2015 #18

    TSny

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    OK. That will work. But I don't get 5.8 for CP using this formula.

    I was thinking of just using 1.7/CP = tan(17.74) to find CP.
     
  20. Jul 18, 2015 #19
    Yea i made an algebra mistake...i should of taken an algebra class this summer instead.
     
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