Answer
$$
\begin{aligned}
& \left(M_{F_1}\right)_A=433 \mathrm{~N} \cdot \mathrm{m}↻ \\
& \left(M_{F_2}\right)_A=1.30 \mathrm{kN} \cdot \mathrm{m}↻ \\
& \left(M_{F_3}\right)_A=800 \mathrm{~N} \cdot \mathrm{m} ↻
\end{aligned}
$$
Work Step by Step
The moment arm measured perpendicular to each force from point $A$ is
$$
\begin{aligned}
& d_1=2 \sin 60^{\circ}=1.732 \mathrm{~m} \\
& d_2=5 \sin 60^{\circ}=4.330 \mathrm{~m} \\
& d_3=2 \sin 53.13^{\circ}=1.60 \mathrm{~m}
\end{aligned}
$$
Using each force where $M_A=F d$, we have
$$
\begin{aligned}
↺+\left(M_{F_1}\right)_A & =-250(1.732) \\
& =-433 \mathrm{~N} \cdot \mathrm{m}=433 \mathrm{~N} \cdot \mathrm{m}(\text { Clockwise }) \\
↺+\left(M_{F_2}\right)_A & =-300(4.330) \\
& =-1299 \mathrm{~N} \cdot \mathrm{m}=1.30 \mathrm{kN} \cdot \mathrm{m}(\text { Clockwise }) \\
↺+\left(M_{F_3}\right)_A & =-500(1.60) \\
& =-800 \mathrm{~N} \cdot \mathrm{m}=800 \mathrm{~N} \cdot \mathrm{m}(\text { Clockwise })
\end{aligned}
$$
