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FireKat97
 one year ago
I really need help with this question...
FireKat97
 one year ago
I really need help with this question...

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FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0this is how I labelled the reaction forces

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0So 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?

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0And b has me totally stumped, would I have to use Sections method?

surry99
 one year ago
Best ResponseYou've already chosen the best response.1Yes method of sections will work for b)

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0where would I cut the section @surry99?

surry99
 one year ago
Best ResponseYou've already chosen the best response.1Suppose 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?

surry99
 one year ago
Best ResponseYou've already chosen the best response.1Post your attempt and I would be happy to check it later.

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0so 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! :)

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0And also Im unsure on how to determine if forces are under compression or tension..

surry99
 one year ago
Best ResponseYou've already chosen the best response.1If 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.

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0So 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)

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0I would post a pic of my working out but my phone isn't letting me :/

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0@firekat97 have you solved the problem yet?

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0Not entirely... Still confused hahaha

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0@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.

FireKat97
 one year ago
Best ResponseYou've already chosen the best response.0@mathmate okay 👍 thanks
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