Answer
$523.33$ $N m$ in clockwise direction
Work Step by Step
First, we resolve both the forces $F_{1}$ & $F_{2}$ to their components parallel (x-axis) and perpendicular (y-axis) to the ground.
[Sign convention: x-axis: +ve when force is directed towards right, y-axis: +ve when force is directed towards up.]
$F_{1,x}= 500 \times \frac{4}{5} = (+) 400 $ $N$ (because, cosine of angle between $F_{1}$ and x-axis is determined through the given triangle in the diagram)
$F_{1,y}= 500 \times \frac{3}{5} = (+) 300 $ $N$
$F_{2,x} = 600cos(60) = (+) 300$ $N$
$F_{2,y} = 600sin(60) = (-) 519.61 $ $N$
$ΣF_{x} = F_{1,x} + F_{2,x} = 400 + 300 = 700 $ $N$
$ΣF_{y} = F_{1,y} + F_{2,y} = 300 + - 519.61 = -219.61 $ $N$
Now, we determine resultant moment about point O
[Sign convention: Moment is +ve when it is in clockwise direction]
$= F_{x}\times y + F_{y} \times x = 700 \times 0.25 + 219.61 \times (0.125+0.3) =(+) 523.33$ $N m$
