Resultant Force on Eyebolt

Convention for unit vector
Upward (+y axis): j; downward (-ve y axis): -j; right side (=ve x axis): i; left side (-ve x axis): -i
Hence, foces in vector form in cartesian system

F1 = 800kN (-j)
F2 = 400 kN (-j)
F3 = 300kN (-i cos(50) - j sin(50)) = 300 (- 0.643 i - -.766 j) = (- 192.84 i -229.81j)
F4 = 20 kN (-i cos(45) + j sin(45)) = 20 (- 0.70 i + 0.707 j) = (-14.14 i + 14.14 j)

Note: Fi == Fi vector; |Fi| == magnitude of Fi vector

Resultant force,
R = F1 + F2 + F3 + F4
= - 800 j - 400 j + (- 192.84 i -229.81j) + (-14.14 i + 14.14 j)
= - 206.98 i - 1415.77 j

Magnitude of resultant force,
|R| = sqrt (206.98^2 + 1415.77^2) = 1430.82 kN --Answer

Direction:
theta (-ve x to -ve y direction) = tan-1(1415.77/206.98) = 81.685 degree --Answer

For graphical part, see attachment.
Note: Graphical part: Please measure the length of the resultant, and see result. Result should be more or less similar to what we are getting here. There may be a small error in graphical method, because of inaccuracy in drawing exact angle and length.

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