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Force System

  1. Dec 17, 2007 #1
    1. The problem statement, all variables and given/known data
    Determine the resultant of the three forces shown, given that P1= 50kN, P2=80kN, and P3=120kN and the dimension a is 4m.

    2. Relevant equations
    I think im supposed to use
    R=[tex]\Sigma[/tex]F=F1 + F2 + F3 + ....

    3. The attempt at a solution
    P1=50[tex]\lambda[/tex]=50(4i+aj / [tex]\sqrt{a^{2}+16}[/tex])
    P2=80[tex]\lambda[/tex]=120(2i+aj+3k / [tex]\sqrt{a^{2}+13}[/tex])
    P3=120[tex]\lambda[/tex]=80(4i+aj+3k / [tex]\sqrt{a^{2}+25}[/tex])

    R=(40,000i+2,500a^{2}j / a^{2}+16)
    + (57,600i+14,400a^{2}j+129,600k / a^{2}+13)
    +(102,400i+6,400a^{2}j+57,600k / a^{2}+25)

    I'm not too sure if I started the problem correctly, let me know what yall think, thanks in advance.
    Last edited: Dec 18, 2007
  2. jcsd
  3. Dec 17, 2007 #2


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    Homework Helper

    In the problem P2=80kN, and P3=120kN .But in the attempted solution the are different. And why you have taken j vector negative?
  4. Dec 18, 2007 #3
    Because I'm an idiot.
    Just kidding, when I redrew the graph on my paper I drew the start of the vectors on the origin, I really don't know why I did that, lol.
    I also made the corrections on the j vector.
    Thank you so much for pointing those out.

    If my work is correct im not sure what to do next.
    Any help would be appreciated.
  5. Dec 18, 2007 #4


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    Homework Helper

    Because I'm an idiot. No..No..You are right. You have calculated P1, P2 and P3 by using the position vectors. So keep j -ve. Put the value of a. Calculate (50/sqrt32) and so on. Then add the vectors. For the resultant you cannot square the vectors and add them.
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