The question states:
Determine the resultant couple moment by (a) summing moments about point $O$ and (b) summing the moments about point $A$.
I used scalar analysis for solving this question, wherein
$$M=\sum{F_xd_y}+\sum{F_yd_x}$$
When completing part (a), I found the following answer:
$$M_O=\sum{F_xd_y}+\sum{F_yd_x}=9.686\mathrm{ kNM}$$
for $$\sum{F_xd_y}=(-8\sin(45)+2\sin(30)) \mathrm{kN}(0 \mathrm{m})+(-2\sin(30)+8\sin(45)) \mathrm{kN}(-0.3 \mathrm{m})=-1.3971 \mathrm{kNm}$$
and $$\sum{F_yd_x}=(-8\sin(45)-2\sin(30)) \mathrm{kN}(-3.3 \mathrm{m})+(2\sin(30)+8\sin(45)) \mathrm{kN}(-1.8 \mathrm{m})=11.083 \mathrm{kNm}$$
I solved part (b) similarly, and got the same result for $M_A$, with the caveat that both $\sum{F_xd_y}$ and $\sum{F_yd_x}$ have opposite polarity (negatives, i.e. $\sum{F_xd_y}=1.3971 \mathrm{kNm}$).
Where I'm having trouble is two-fold: I'm uncertain that I'm solving for the moments correctly, and I'm not sure what how to calculate the final resultant moment.
If I did solve for the moments correctly, would the the resultant couple moment be $\sum{M}$, which would result in a resultant moment of zero?
Thank you in advance for any clarifications or explanations you can provide.
Best Answer
The net force acting on the beam is zero but there is a couple acting on the system. A couple has the nice property that the moment about any point is the same.
So you should have found the same answer for both parts.
So find the vertical component of the two forces on the left which will equal the magnitude of the vertical component of the two forces on the left.
The moment of a couple is force $\times$ perpendicular distance between the two forces.
Later
Moment about $A$ is $Fa$ clockwise
Mpment about $O$ is $F(a+b)$ clockwise $+ Fb$ anticlockwise
$\Rightarrow F(a+b)$ clockwise $- Fb$ clockwise = $Fa$ clockwise the same as about $A$