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Determine the perpendicular distance for couple

  1. Dec 24, 2015 #1
    1. The problem statement, all variables and given/known data
    two 50N are applied to the corner B and D , determine the moment of couple formed by the two forces by resolving each force into the horizontal and vertical component and adding the two resulting couples . Use the result to determine the perpendicular distance between the line BE and DF

    2. Relevant equations


    3. The attempt at a solution
    i gt only Bx = 50sin50 = 38.3N (right)
    By = 50cos50 = 32.1N ( down)

    Dx= 50 sin50 = 38.2 (left)
    Dy = 50 cos50 = 32.1N (up)

    how to get the perpendicular distance ?
    btw , i have used another method which is not relevant to the question asked , whoch is tan 50 = O / 300
    O= 357.5mm
    hence , perpendicular deistance = 500-357.5mm = 142.5mm
     

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  2. jcsd
  3. Dec 24, 2015 #2

    SteamKing

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    This last calculation does not give the perpendicular distance between the lines of the two forces.

    The distance O = 357.5 mm is measured along the edge of the plate, and so is the difference 500 - 357.5 = 142.5 mm.

    You must find the distance between the line of each force measured perpendicular to each, similar to what is shown below:

    5.png
     
  4. Dec 24, 2015 #3
    CAN YOU GIVE SOME HINT HOW TO FIND ?
     
  5. Dec 24, 2015 #4

    SteamKing

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    Follow the original directions stated in the problem. Don't go straying off topic.
     
  6. Dec 24, 2015 #5
    how to find ? i have only
    Bx = 50sin50 = 38.3N (right)
    By = 50cos50 = 32.1N ( down)

    Dx= 50 sin50 = 38.2 (left)
    Dy = 50 cos50 = 32.1N (up)

    if i sum up them , i would get 0 .
     
  7. Dec 24, 2015 #6

    SteamKing

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    You're not supposed to sum them up.

    Read the problem statement again. Does it tell you to sum these forces up? Or does it ask you to do something else?
     
  8. Dec 25, 2015 #7
    well , force the force at B , i have moment = 50sin50(300) -50cos50(500) = -4560N
    for the force at D , i also have moment = 50sin50(300) -50cos50(500) = -4560N

    so i would get -9120Nmm / 50 = = 182.4mm , am i right ?
     
    Last edited: Dec 25, 2015
  9. Dec 25, 2015 #8
    You can calculate the stress by axes and the total couple force stress as ##F\times{D}##. Must be equals.
     
  10. Dec 25, 2015 #9

    SteamKing

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    You pick one point and calculate the moments of the force components about that point.

    If you choose either B or D as the point about which to calculate the total moment, your calculations are greatly simplified. :wink:
     
  11. Dec 25, 2015 #10
    so the ans = 500(32.1)-300(38.3)=4560Nmm
    4560Nmm / 50 = 91.2 mm ?
     
  12. Dec 25, 2015 #11

    SteamKing

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    Yes, that is correct.
     
  13. Dec 25, 2015 #12
    why not 91.2 x 2 = 184 mm ? this is beacuse when we calculate the sum of couple moment , we will need to consider both side .....
    so for the diagram below , the sum of couple moment is F1(d/2 ) +F1(d/2 )

    so for the question i asked , the 91.2 mm that we gt is only the d/2 only
     

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  14. Dec 25, 2015 #13

    SteamKing

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    Yes, but if you calculate the moments of each force about the midpoint of the distance separating them, then M = F * d/2 + F * d/2 = F * d, since both forces are of equal magnitude.

    If F = 500 N and M = 4560 N-mm, what must the distance d be equal to?
     
  15. Dec 25, 2015 #14
    the d = 91.2mm , but the d here that we gt is only half of the perpendicular distance of the question i asked originally , right ?
     
  16. Dec 25, 2015 #15

    SteamKing

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    No, it's not. You still have two forces creating the couple, as I showed you in Post #13.

    Take the small diagram you have in Post #12 and calculate moments about one end of d (it doesn't matter which end) and again about the center of d. Each moment you calculate should have the same magnitude.
     
  17. Dec 25, 2015 #16
    ok , i think i gt your point
     
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