Force along $CA = 100 N,$ along $OD = 250 N$, along $ED = 150N$, along $OE = 100N.$ A clockwise moment of 5000 N-cm is also acting at the point O.

To locate the single force, Varignon’s theorem is required.

To find a single force (Resultant), we need to first find its horizontal and vertical components ($∑Fx$ and $∑Fy$)

To find the components, we need to resolve all the forces and point loads first.

And for that we need FBD.

Going in the reverse order now,

Step 1:

FBD

Step 2: Finding angles

$α = \tan^{-1}(20/30) = 33.82 \\ β = \tan^{-1}(50/30) = 59 $

Finding R

$∑Fx $ and $ ∑Fy$

$∑Fx$ (right positive) $= 100 + 250\cos β – 100\cos α = 145.68→ $

$∑Fy $(up positive) $= 150 + 250\sin β – 100\sin α = 308.63↑\\ R^2= ∑Fx^2 + ∑Fy^2 = 145.68^2 + 308.63^2 \\ R = 341.3 N↗ . θ = \tan^{-1}(∑Fy/∑Fx) = 65 $

Step 4: Position from O

By varignons theorem,

$R \times D = ∑M$ (anticlockwise positive)

$= 150 \times 30 + 100 \cos α \times 90 - 100\sin α \times 30 + 5000 \\ R \times d = 15307 \\ . d = 15307/341.3 = 44.85 cm $

Since, we want its X-intercept from O, we want X (as shown in diagram)

$X= d/\sin θ = 44.85 / \sin 65 = 49.49 cm$

Hence, resultant will be acting at a point $49.49$ cm on the right side of O.