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:

Converting udl to point load (acting always at the centre of the original udl)

$W1= w1 \times L1 = 10 \times 4 = 40 kN $

Converting uvl to point load (acting always 1/3rd of the base away from right angle)

$W2 = 0.5 \times w2 \times L2 = 0.5 \times 12 \times 2 = 12kN$

Step 3: Finding R

$∑Fx$ and $∑Fy$

$∑Fx$ (right positive)= $30 + 50\cos 30 = +73.3 kN = 73.3 kN →$

$∑Fy$ (up positive) $= 50\sin 30 – 40 – 12 = -27 = 27 kN ↓ \\ R^2= ∑Fx^2 + ∑Fy^2 = 73.3^2 + 27^2 \\ \\ R = 78.11 kN↘ \\ . α = \tan^{-1}(∑Fy/∑Fx) = 20.22 $

Step 4: Position from C

By varignons theorem,

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

$-78.11 \times d = -30 \times 1 – 50\cos 30 \times 2 – 40 \times 2 + 10 – 12 \times 8.67 \\ = -290.64 \\ . d = 290.64/78.11 = 3.72 m $

Since, we want its position from C, we want X (as shown in diagram)

$X= d/\sin α = 3.72 / \sin 20.22 = 10.77 m $

Hence, resultant will be acting at a point $10.77m$ on the right side of C.