Answer:
<h2>
FA = 13 kN
</h2><h2>
FG = 15.3 kN</h2>
Explanation:
write each force in terms of magnitude and directions
Fx = F sin Ф
Fy = F cos Ф
where Ф is to be measured from x axis.
∑F at y = o
FAy + FBy + FCy + FDy + FEy + FGy = 0
∑F at x = o
FAx + FBx + FCx + FDx + FEx + FGx = 0
Let
FA = FA sin (110) + FA cos (110)
FB = 20 sin (270) + 20 cos (270)
FC = 16 sin (140) + 16 cos (140)
FD = 9 sin (40) + 9 cos (40)
FE = 20 sin (270) + 20 cos (270)
FG = FG sin (50) + FG cos (50)
add x and y forces:
FAx + FBx + FCx + FDx + FEx + FGx = 0
FAy + FBy + FCy + FDy + FEy + FGy = 0
FA sin (110) + 0 + 16 sin (140) + 9 sin (40) + 0 + FG sin (50) = 0
FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0
FA sin (110) + 0 + 10.285 + 5.785 + 0 + FG sin (50) = 0
FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0
FA sin (110) + 16.070 + FG sin (50) = 0
FA cos (110) - 45.363 + FG cos (50) = 0
solving for FA, and FG
FA = 13 kN
FG = 15.3 kN
