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Statics: Determine the magnitude of the resultant force

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data:

    Untitled-2.png



    3. The attempt at a solution
    I keep attempting this problem and so far I have got this:

    F1x: 150
    F1y:260
    F2x:424
    F2y:424
    F3x:600
    F3y:250

    Then I add the x's together and y's together, square both and take the square root for magnitude, but can't seem to get the right answer
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 2, 2012 #2

    rock.freak667

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    Homework Helper

    F1= 800 N at 60 degrees, so F1x = 800cos(60) = 400 N. You have it as 150.
     
  4. Sep 2, 2012 #3
    alright so I get 1697 in the end...still didn't work, and I noticed it says to 2 significant digits...I am completely stuck...I have no idea what else would be wrong
     
  5. Sep 2, 2012 #4
    Have you gotten any more progress? I have gotten an answer but nowhere near 1697 or 1526. I would recommend splitting up the equation into the x and y components. As you probably have already done. Also when doing it make sure that you are using the correct angle.
     
  6. Sep 2, 2012 #5
    No, I kinda gave up for now...as seen in my post I already split up the components into x and y, so not sure what else to do..its such a simple problem I am not sure why I am having trouble with it
     
  7. Sep 2, 2012 #6

    PhanthomJay

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    Science Advisor
    Homework Helper
    Gold Member

    Please show your workings so we can better assist in finding your errors. Might be due to an error in detremining force components ( as has been pointed out), or might be an error in the pesty plus or minus sign,...or both.
     
  8. Sep 2, 2012 #7
    Well, if you decide to try the problem again we will be here to help you if you still cannot figure it out.
     
  9. Sep 3, 2012 #8
    It looks like I have made some small mistakes, I am going to attempt it again this afternoon and hopefully it should work now.
     
  10. Sep 3, 2012 #9
    Well I tried again and I am running out of time..but here is my work:

    F1x: 400
    F1y:693
    F2x:424
    F2y:424
    F3x:600
    F3y:250

    Fx's:1424
    Fy's:1367

    (1424^2+1367^2)^(1/2)

    As far as how I found the x and y components I simply used the magnitude of the vector times cosine of the angle for x's and sine for y's...

    I am very frustrated because this is such a simple problem and I don't see what else I would be doing wrong
     
  11. Sep 3, 2012 #10
    What angles are you using for each force? Primarily F3?

    Check the angle that you are using for F2. Are you sure you want to use 45 degrees?
     
    Last edited: Sep 3, 2012
  12. Sep 3, 2012 #11
    Figured it out...and of course it was something stupid...I forgot to add the negatives to the numbers that needed it...I just figured I didn't need them since I was finding the magnitude anyways, but of course that affects the x and y resultant that you square
     
  13. Sep 3, 2012 #12
    Exactly! Force is a vector so its direction matters ergo the sign also matters. I'm glad that you figured it out! :smile:
     
  14. Sep 4, 2012 #13
    I find it quite easy to always use force triangles when given problems like this since its always easy to make mistakes when resolving( you have more equations obviously high chances of sign mistakes)... wherez you could pplug all the values in one geometric equation and instantly find the magnitude and direction! :)
     
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