Answer
$M_{O} = $ { $1200k$ } $N$ $m$
Work Step by Step
Moment of force about a point $O$ is expressed in cartesian vector as
$M_{O} = r \times F = \begin{vmatrix}
i & j & k \\
r_{x} & r_{y} & r_{z} \\
F_{x} & F_{y} & F_{z}
\end{vmatrix}$
Where $r$ is the position vector from $O$ to the point of application of force and $F$ is the force vector
In the given problem, position vector $r = 4i$
To calculate force vector, we resolve the force F into its components in $x$ and $y$ direction.
From Pythagoras theorm $AB = \sqrt(3^2 + 4^2) = 5$
$F_{x} = 500cos(θ) = 500 \times \frac{4}{5} = 400$
[the force vector is in the same direction as AB, cosine & sine can be calculated from the given triangle]
$F_{y} = 500sin(θ) = 500 \times \frac{3}{5} = 300$
$F_{z} = 0$
Thus $F = $ {$400i + 300j$ } $N$
Now,
$M_{O} = r \times F = \begin{vmatrix}
i & j & k \\
4 & 0 & 0 \\
400 & 300 & 0
\end{vmatrix}
= 4(300k) = 1200k$
Thus, $M_{O} = $ { $1200k$ } $N$ $m$
