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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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.

Physics
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@chris00 physics?
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Other answers:

yeah
okay lets do this
lets get forces in terms of their x and y components
F1 first... F1x= 850(4/5)=680 F1y= 850(3/5)=510 right?
yep or we can write it as\[F _{1}=\left[ 680i-510j \right] {\space}kN\]
now lets do F(2)
oh okay.. next f2: F2x= 625sin(30)=312.5N F2y=625cos(30)=541.3N oh i think this is the part i got wrong
\[F _{2}=\left[ (-625\sin30)i-(624\cos30)j \right] kN\]
remember unit circle
|dw:1444531507696:dw|
so in the third quadrant, only tan is positive, hence since and cos are negative
\[F _{2}=\left[ -312.5i-541.3j \right]{\space} \]
now do F3
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) ?
yep!
so lets write out full equations out
\[F _{1}=\left[ 880i-510j \right]{\space} kN\] \[F _{2}=\left[ -312.5i-541.3j \right]{\space} kN\] \[F _{3}=\left[ -530.3i+530.3j \right]{\space} kN\]
Now, \[F _{R}=\sum_{}^{}F _{i}+\sum_{}^{}F _{j}\]
add all the x and all the y components Fnet= (-162.8i, 520.6j)
then get the magnitude of Fnet: sqrt( (162.8)^2 + (520.6)^2)=545.5 N or 550 N :D thanks!
\[F _{R}=(680-312.5-530.3)i+(-510-541.3+530.3)j\] \[F _{R}=-162.8i+520.6j\]
yep, spot on!
haha why did you choose me to do this question? haha
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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