$1.$ Force F$1=250$NF$\_1=250\backslash ,\backslash$mathrm$\{$N$\}$F$1=250$N

Acts at a distance of $2$m$2\backslash ,\backslash$mathrm$\{$m$\}2$m from AAA.

Angle with the beam: $3030^\backslash$circ$30.$

$2.$ Force F$2=300$NF$\_2=300\backslash ,\backslash$mathrm$\{$N$\}$F$2=300$N

Acts at a horizontal distance of $5$m$5\backslash ,\backslash$mathrm$\{$m$\}5$m from AAA $($total beam length$).$

Angle with the vertical: $6060^\backslash$circ$60$

To calculate the resultant moment about point AAA, we need to evaluate the contribution of the moments generated by the forces F$1$F$\_1$F$1,$ F$2$F$\_2$F$2,$ and F$3$F$\_3$F$3.$

Key Steps:

Resolve forces into components $($if necessary$).$

Determine perpendicular distances from point AAA to the line of action of each force.

Calculate moments about point AAA using M$=$F$$dM $=$ F $\backslash$cdot dM$=$F$$d$,$ where ddd is the perpendicular distance.

Assign signs: Use the right$-$hand rule to assign positive or negative signs to moments $($typically counterclockwise moments are positive$).$

$1.$ Force F$1=250$NF$\_1=250\backslash ,\backslash$mathrm$\{$N$\}$F$1=250$N

Acts at a distance of $2$m$2\backslash ,\backslash$mathrm$\{$m$\}2$m from AAA.

Angle with the beam: $3030^\backslash$circ$30.$

Moment arm $($perpendicular distance$)$: d$1=2$sin$(30)=1$md$\_1=2\backslash$sin$(30^\backslash$circ$)=1\backslash ,\backslash$mathrm$\{$m$\}$d$1=2$sin$(30)=1$m$.$

Moment about AAA: M$1=$F$1$d$1=2501=250$Nm$.$M$\_1=$ F$\_1\backslash$cdot d$\_1=250\backslash$cdot $1=250\backslash ,\backslash$mathrm$\{$Nm$\}.$M$1=$F$1$d$1=2501=250$Nm$.$ Direction: Counterclockwise $(+)(+)(+).$

$2.$ Force F$2=300$NF$\_2=300\backslash ,\backslash$mathrm$\{$N$\}$F$2=300$N

Acts at a horizontal distance of $5$m$5\backslash ,\backslash$mathrm$\{$m$\}5$m from AAA $($total beam length$).$

Angle with the vertical: $6060^\backslash$circ$60.$

Moment arm: d$2=5$cos$(60)=2\mathrm{.}5$md$\_2=5\backslash$cos$(60^\backslash$circ$)=2\mathrm{.}5\backslash ,\backslash$mathrm$\{$m$\}$d$2=5$cos$(60)=2\mathrm{.}5$m$.$

Moment about AAA: M$2=$F$2$d$2=3002\mathrm{.}5=750$Nm$.$M$\_2=$ F$\_2\backslash$cdot d$\_2=300\backslash$cdot $2\mathrm{.}5=750\backslash ,\backslash$mathrm$\{$Nm$\}.$M$2=$F$2$d$2=3002\mathrm{.}5=750$Nm$.$ Direction: Counterclockwise $(+)(+)(+).$

$3.$ Force F$3=500$NF$\_3=500\backslash ,\backslash$mathrm$\{$N$\}$F$3=500$N

Acts at an inclined distance. Resolve into horizontal and vertical components:

Vertical component: F$3$y$=50045=400$NF$\_\{3$y$\}=500\backslash$cdot $\backslash$frac$\{4\}\{5\}=400\backslash ,\backslash$mathrm$\{$N$\}$F$3$y$=50054=400$N$.$Horizontal component: F$3$x$=50035=300$NF$\_\{3$x$\}=500\backslash$cdot $\backslash$frac$\{3\}\{5\}=300\backslash ,\backslash$mathrm$\{$N$\}$F$3$x$=50053=300$N$.$

Perpendicular distances from AAA:

For F$3$yF$\_\{3$y$\}$F$3$y$$: Horizontal distance d$3$y$=5$md$\_\{3$y$\}=5\backslash ,\backslash$mathrm$\{$m$\}$d$3$y$=5$m$.$For F$3$xF$\_\{3$x$\}$F$3$x$$: Vertical distance d$3$x$=4$md$\_\{3$x$\}=4\backslash ,\backslash$mathrm$\{$m$\}$d$3$x$=4$m$.$

determine the resultant moment about point A