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Zale101

  • one year ago

Find the magnitude of the resultant force and the angle that makes with the positive x-axis

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  1. Zale101
    • one year ago
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    |dw:1440990161648:dw|

  2. Zale101
    • one year ago
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    \(F1=75 N; θ=0\) \(F2=50 N; θ=30\) \(F3=40 N; θ=150\) F1= \(f_1x=75cos(0)=75 N\) \(f_1y=75sin(0)=0 N\) F2= \(f_2x=50cos(30)=25\sqrt{3} N\) \(f_2y=50sin(30)=25 N\) F3= \(f_3x=40cos(150)=−20\sqrt{2}N\) \(f_3y=40sin(150)=20 N\)

  3. Zale101
    • one year ago
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    Is this good so far?

  4. jim_thompson5910
    • one year ago
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    I don't agree with f3x

  5. jim_thompson5910
    • one year ago
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    everything else looks good

  6. ganeshie8
    • one year ago
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    Looks good! next simply add them component by component

  7. ganeshie8
    • one year ago
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    Ahh I didn't check the arithmetic..

  8. ganeshie8
    • one year ago
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    this should fix it : \(F1=75 N; θ=0\) \(F2=50 N; θ=30\) \(F3=40 N; θ=150\) F1= \(f_1x=75cos(0)=75 N\) \(f_1y=75sin(0)=0 N\) F2= \(f_2x=50cos(30)=25\sqrt{3} N\) \(f_2y=50sin(30)=25 N\) F3= \(f_3x=40cos(150)=−20\sqrt{\color{red}{3}}N\) \(f_3y=40sin(150)=20 N\)

  9. jim_thompson5910
    • one year ago
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    much better

  10. Zale101
    • one year ago
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    Okay. Thanks! @jim_thompson5910 and @ganeshie8 !!! I'll deal with the arithmetic, it's not the problem for me now. But the next step is what i'm having problems with.

  11. jim_thompson5910
    • one year ago
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    you have the vector components, add up the corresponding components

  12. jim_thompson5910
    • one year ago
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    u = <a,b> v = <c,d> w = <e,f> u+v+w = <a+c+e, b+d+f>

  13. ganeshie8
    • one year ago
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    |dw:1440991132300:dw|

  14. jim_thompson5910
    • one year ago
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    |dw:1440991352564:dw|

  15. Zale101
    • one year ago
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    Okay, thanks. makes sense !

  16. jim_thompson5910
    • one year ago
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    you'll end up with some vector z = <p,q> p = sum of the Fx q = sum of the Fy magnitude of z = |z| = sqrt(p^2 + q^2)

  17. Zale101
    • one year ago
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    what about theta?

  18. jim_thompson5910
    • one year ago
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    theta = arctan(q/p)

  19. Zale101
    • one year ago
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    Oh yes. I forgot about that xD

  20. jim_thompson5910
    • one year ago
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    theta = arctan(q/p) if theta is in Q1 or Q4 theta = arctan(q/p)+180 if theta is in Q2 or Q3

  21. jim_thompson5910
    • one year ago
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    to figure out which quadrant theta is in, simply look at the vector <p,q>

  22. Zale101
    • one year ago
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    Absolutely. Alright thanks for all the help.

  23. jim_thompson5910
    • one year ago
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    sure thing

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