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owlet

  • one year ago

My answer is wrong. Can anyone check my solution? I got approx, 490 N but the correct answer is 550N. Why is that? I'll attached both the question and my solution.

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  1. owlet
    • one year ago
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  2. owlet
    • one year ago
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    @chris00 physics?

  3. anonymous
    • one year ago
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    is the answer highlighted in yellow?

  4. owlet
    • one year ago
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    yeah

  5. anonymous
    • one year ago
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    okay lets do this

  6. anonymous
    • one year ago
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    lets get forces in terms of their x and y components

  7. owlet
    • one year ago
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    F1 first... F1x= 850(4/5)=680 F1y= 850(3/5)=510 right?

  8. anonymous
    • one year ago
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    yep or we can write it as\[F _{1}=\left[ 680i-510j \right] {\space}kN\]

  9. anonymous
    • one year ago
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    now lets do F(2)

  10. owlet
    • one year ago
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    oh okay.. next f2: F2x= 625sin(30)=312.5N F2y=625cos(30)=541.3N oh i think this is the part i got wrong

  11. anonymous
    • one year ago
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    \[F _{2}=\left[ (-625\sin30)i-(624\cos30)j \right] kN\]

  12. anonymous
    • one year ago
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    remember unit circle

  13. anonymous
    • one year ago
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    |dw:1444531507696:dw|

  14. anonymous
    • one year ago
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    so in the third quadrant, only tan is positive, hence since and cos are negative

  15. anonymous
    • one year ago
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    \[F _{2}=\left[ -312.5i-541.3j \right]{\space} \]

  16. anonymous
    • one year ago
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    now do F3

  17. owlet
    • one year ago
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    okay.. F3: F3x=750 sin(45)=530.33 N (-x direction) F3y=750cos(45)=530.33 N (+y direction) so if i'll put it like your way it would be like (-530.33i, 530.33j) ?

  18. anonymous
    • one year ago
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    yep!

  19. anonymous
    • one year ago
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    so lets write out full equations out

  20. anonymous
    • one year ago
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    \[F _{1}=\left[ 880i-510j \right]{\space} kN\] \[F _{2}=\left[ -312.5i-541.3j \right]{\space} kN\] \[F _{3}=\left[ -530.3i+530.3j \right]{\space} kN\]

  21. anonymous
    • one year ago
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    Now, \[F _{R}=\sum_{}^{}F _{i}+\sum_{}^{}F _{j}\]

  22. owlet
    • one year ago
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    add all the x and all the y components Fnet= (-162.8i, 520.6j)

  23. owlet
    • one year ago
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    then get the magnitude of Fnet: sqrt( (162.8)^2 + (520.6)^2)=545.5 N or 550 N :D thanks!

  24. anonymous
    • one year ago
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    \[F _{R}=(680-312.5-530.3)i+(-510-541.3+530.3)j\] \[F _{R}=-162.8i+520.6j\]

  25. anonymous
    • one year ago
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    yep, spot on!

  26. anonymous
    • one year ago
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    haha why did you choose me to do this question? haha

  27. owlet
    • one year ago
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    oops sorry late response. I choose you because you're the only one online that I fanned and you're nice so yeah xD thanks again!

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