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Resolution of Forces Help Please

  1. Feb 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle is in equilbrium with the forces of 4N north, 8N west, 5√2N south east and P.

    Find the magnitude and the direction of P


    2. Relevant equations
    ?


    3. The attempt at a solution
    A am a bit lost on this question, I can make an educated guess on the magnitude and direction having done what I have done so far...
    [itex]
    y=4-5\sqrt{2}cos45 \\
    y=-1 \\
    x=-8-5\sqrt{2}sin45 \\
    x=-13
    [/itex]
    This would put the magnitude at near 12N and the direction at about 80° from north. However I am a bit lost on how to actually work it out (and even if what I have done so far is any good).

    Any help is appreciated.
     
  2. jcsd
  3. Feb 12, 2013 #2

    tiny-tim

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    Hi FaraDazed! :smile:

    Yes, that looks ok (except that you've got one of the "south-east" signs wrong) …

    you're adding the three known vectors, and then your answer will be minus that.
     
  4. Feb 12, 2013 #3
    Hi, thanks for your reply :)

    OK, should it be this?
    [itex]
    y=4-5\sqrt{2}cos45 \\
    y=-1 \\
    x=-8+5\sqrt{2}sin45 \\
    x=-3.01
    [/itex]

    Or this?
    [itex]
    y=4+5\sqrt{2}cos45 \\
    y=8.99 \\
    x=-8-5\sqrt{2}sin45 \\
    x=-13
    [/itex]
     
  5. Feb 12, 2013 #4

    tiny-tim

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    yes! :smile:

    (but where did the .01 come from?

    did you use a calculator for that? :confused:)
     
  6. Feb 12, 2013 #5
    probably a rounding error, I did 5√2 = 7.07, then did 7.07sin45 = 4.99 (just realised this was wrong as rounding it makes it 5).

    Thanks for point that out.

    Not sure where to go from there though. Its doing my head in as we have a problem sheet to do and this is one of the easy ones, and I am ok with the harder ones about rough inclined planes with co-efficient of friction etc yet its the 'easy' ones that always get me!

    OK well I think I got the magnitude to be 10N as 1^2+3^2=10 , and the direction involves 18.34° because arctan(1/3)=18.34, but not sure if is 90-18.34 = 71.66° from north?
     
  7. Feb 12, 2013 #6

    tiny-tim

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    Hi FaraDazed! :smile:
    hmm … you do know that sin45° is exactly 1/√2, don't you? :redface:
    That gives you the sum of the given forces.

    So the balancing force must be minus that sum. :wink:
     
  8. Feb 12, 2013 #7
    No, im not all that up on my surd notation just yet. I am getting better, slowly, though :smile:

    OK cheers, could you check my edit to my last post please?
     
  9. Feb 12, 2013 #8

    tiny-tim

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    you mean, is it x = -8 - 5 or -8 + 5 ?

    what do you think, and why? :smile:
     
  10. Feb 12, 2013 #9
    Its -8+5 because its south east and east on an x-y plane is positive. When doing my original diagram I put the 5√2 as south-west :redface: .

    I meant could you check the last bit of it actually, Ive just c+p it below (and made changes after realsing it is SE not SW) anyway :)

    OK well I think I got the magnitude to be 10N as 1^2+3^2=10 , and the direction involves 71.56° because arctan(3/1)=71.56, but as it would be in a south(ish)-west direction it would be 180+71.56=251.56° from north?
     
  11. Feb 12, 2013 #10

    tiny-tim

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    no, it's √10 !! :rolleyes:
    let's see …

    P = (3,1), so that's a bit north of east (P is the opposite of the sum of the other forces)

    but yes the angle from north is arctan3 :smile:
     
  12. Feb 12, 2013 #11
    Of course it is, :redface:.
    Right OK, of course as it needs to balance out, so it would simply be 71.56° from north?
     
  13. Feb 12, 2013 #12

    tiny-tim

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    yup! :smile:

    (though i suspect from the "south-east" in the question that they want you to say "18.44° north of east" :wink:)
     
  14. Feb 12, 2013 #13
    Thanks for your help :)
     
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