[Please also refer to the EDIT appended below.]
Here, $P>Q$. $O$ is the center of mass of the rigid and uniform bar/stick.
As $P>Q$, the resultant is situated to the right of $\vec{P}$ and is parallel to $\vec{P}$. The magnitude of the resultant is $P-Q$.
To convince you that the figure is correct, I'll do some math to prove it.
Let us obtain the sum of torques about the center of mass,
$$(P-Q)b=Pa+Qa$$
$$b=\frac{P+Q}{P-Q}a$$
$$b=fa\ \left[\text{Let $f=\frac{P+Q}{P-Q}$}\right]$$
As $P>Q$, $f>1$, and $b>a$ [see edit below]. So, the correct figure will be,
I hope you're satisfied that the figure is correct.
My comments:
Is it possible to replace $\vec{P}$ and $\vec{Q}$ with a single force? I mean practically, not theoretically. From the figure, we can see that the resultant force is outside the bar. In other words, $\vec{P}$ and $\vec{Q}$ can be replaced by a force of magnitude $P-Q$, which will act outside the bar. This may be possible theoretically; however, this is not possible practically as the resultant force will be acting on literally nothing as it is outside the bar. Therefore, I conclude that it is impossible to replace $\vec{P}$ and $\vec{Q}$ with a single force practically. Theoretically, it is possible, but practically, no.
My question:
- Can $\vec{P}$ and $\vec{Q}$ be replaced by a single force? Is my conclusion correct?
These may help you to answer this question:
This question was posted with the help of @Eli.
EDIT (from a now-deleted answer)
Clarifications:
- Point $O$ is not affixed.
- $\vec Q\ne\mathbf0.$
Correction:
- Regarding the mathematics in the second block, $f>1$ not merely because $P>Q,$ but because $P>Q>0.$