Perpendicular Distance for Moments

Question

My question is: Is perpendicular distance in the equation $\tau=Fd$ the distance of the yellow or red line?

My belief was that it was the red line, however when I take moments about $O$ I get $$\begin{align*}\tau=4a\sin\theta\, mg&-2a\sin\theta\,4mg=0;\\ 4a\sin\theta\,mg&=8a\sin\theta mg;\\ 4&=8 \end{align*}$$

Which leads me to think my assumption was wrong. I understand the diagram is not accurate but I don't think I can assume $OC$ is perpendicular to the dashed line since the $\theta$ is not necessarily $45^{\mathrm{o}}$.

I have attempted to look elsewhere online but everything I have seen leads me to think the yellow line is the right distance.

edit: $\angle OBC=90^{\mathrm{o}}$

Answer 1

Consider the figure above. The torque about $O$ is given by

$$ \vec{T}=\vec{r}\times\vec{F}$$

Its magnitude is

$$||\vec{T}||=||\vec{r}||.||\vec{F}||.\sin{\theta}$$ As you can see in the figure, $||\vec{r}||.\sin{\theta}$ is equal to the length of the green line, which is the perpendicular drawn from $O$ to the line of action of $\vec{F}$ and is called the perpendicular distance.

Answer 2

It is the yellow and green lines. The perpendicular distance is always perpendicular to the direction of the force.

Answer 3

You have used the correct definition of the torque due to a force about a point as the force times the perpendicular distance from the point to the line of action of the force which are you green (length $2a \sin \theta$) and yellow (length $4a \sin \theta$) lines but not realised that there is a net torque about $O$ on the system of $4mg a \sin \theta$ anticlockwise.
That is why you got $4=8$.

Answer 4

The question is rather unclear: if the force under consideration is the weight $mg$ then the lever arm is of course the yellow red, since it must be perpendicular to the force.