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FireKat97

  • one year ago

I really need help with this question...

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  1. FireKat97
    • one year ago
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  2. FireKat97
    • one year ago
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    this is how I labelled the reaction forces

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  3. FireKat97
    • one year ago
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    So for part A using my FBD, I set up the following equations- ∑Fx = 0, -Lx = 0 therefore Lx = 0 ∑Fy = 0, Hy -P -Q + Ly = 0 ∑ML = 0, 8Hy - 4P - 2Q = 0 therefore Hy = P/2 + Q/4 so we sub Hy into ∑Fy equation, and get Ly = P/2 + 3Q/4 so those are the reactions I have calculated for the supports, and all my assumed directions are correct as all my "values" came out positive. Can someone please confirm if I have done this correctly?

  4. FireKat97
    • one year ago
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    And b has me totally stumped, would I have to use Sections method?

  5. surry99
    • one year ago
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    Yes method of sections will work for b)

  6. FireKat97
    • one year ago
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    where would I cut the section @surry99?

  7. surry99
    • one year ago
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    Suppose you want the force in BJ first. Then you must cut BJ, Now which other members would you cut at the same time? Try a few options and look at the resulting FBD's. Do the resulting FBD's allow you to determine the force in BJ?

  8. surry99
    • one year ago
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    Post your attempt and I would be happy to check it later.

  9. FireKat97
    • one year ago
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    so I had figured that DI = IE = KF = KG = 0 as they are all zero force members (I think) and if the above is correct, then since load at K is 2kN downward, from y equilibrium, KC should be 2kN upward and so CK would be 2 kN downward (action reaction forces) I then thought I'd make a cut down EC, JF and JK and use moments and equilibriums to solve for JC (JC = JF) and from doing this I found JF = √8 , JB = 1 (∑y equilibrium) and JK =2, but I am really uncertain of my answers @surry99 and thanks so much for your help! :)

  10. FireKat97
    • one year ago
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    And also Im unsure on how to determine if forces are under compression or tension..

  11. surry99
    • one year ago
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    If a member is being pulled on it is in tension. If a member is being pushed on, it is in compression. Please post your work and I will take a look.

  12. FireKat97
    • one year ago
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    So what I did was Mc = -(JK X 2) - (1 X 2) + (6 X 1) = 0 So -2JK -2 + 6 = 0 jK = 2 Then Mb = -(2 X JK) -( 2 X 1/sqrt(2) JC) + (4 X 1) = 0 So JC = 0 so JF = 0 and BJ = 1 (explained above) and CK = 2 (explained above) and for some reason I'm now getting JF = 0 instead of sqrt(8)

  13. FireKat97
    • one year ago
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    I would post a pic of my working out but my phone isn't letting me :/

  14. mathmate
    • one year ago
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    @firekat97 have you solved the problem yet?

  15. FireKat97
    • one year ago
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    Not entirely... Still confused hahaha

  16. mathmate
    • one year ago
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    @firekat97 If you have a chance to post your last version of your work (with your phone or otherwise), I'll be glad to look at it.

  17. FireKat97
    • one year ago
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    @mathmate okay 👍 thanks

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