# Moment of a bent rod

In my mechanics textbook there is an example where they take the moment of the external forces in a system. The example states:

Two uniform rods $OX$ and $XY$, pin jointed together at $X$, hang from a fixed hinge at $O$. The rods have length $a$ and $b$, and weight $ka$ and $kb$ respectively. The lower end $Y$ is now pulled aside with horizontal force $F$. Find the angles $\alpha$ and $\beta$ which the rods make with the vertical in equilibrium.

They then go on to take the moment for the rod $XY$ giving: $F(b\cos{\beta}) = kb(\frac{1}{2}b\sin{\beta})$ , and I understand where this comes from. However, they then go on to take the moments of the whole system about $O$ to give: $$F(a\cos{\alpha}+b\cos{\beta}) = ka(\frac{1}{2}a\sin{\alpha})+ kb(a\sin{\alpha} +\frac{1}{2}b\sin{\beta})$$

I understand where the $ka(\frac{1}{2}a\sin{\alpha})$ came from but not how to get the other two parts. For the left side, it looks to me like they have taken the moments caused by $F$ separately for each rod then added them together, and that they have done the same for the right side. I haven't come across this before, so why are they able to do this? Is there some general rule that covers this? Here is a diagram they have given to help illustrate the problem Since the system is in equilibrium, the net moment acting about $O$ must be $0$. Now, we see that F has no vertical component, hence the moment produced by F about O is the magnitude of F multiplied by the vertical distance of the point of application of F from O (keep in mind that the torque vector is $\tau=r \times F$, that is the product of the force component that is perpendicular to the position vector of the force from the origin). Now the vertical distance can be found from some trigonometry. The same idea applies for B, where as it has only a vertical component, the moment is the product of the horizontal distance between the point of applied force and the origin, which is again found through some trigonometry.