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Scenario 1:

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Here, $P>Q$. $O$ is the center of mass of the rigid and uniform bar/stick. The resultant acts to the right of $\vec{P}$ as $P>Q$.

Scenario 2:

enter image description here

Here, $P>Q$ also. $O$ is the center of mass of the rigid and uniform bar/stick. Now, the problem here is that as $\vec{P}$ and $\vec{Q}$ act at the two ends of the bar/stick, there is no place left to the right of $\vec{P}$. So, where will the resultant of magnitude $(P-Q)$ act?

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The $P-Q$ resultant is in the incorrect position.

Because the rod is subjected to both a net force and a net couple a way of considering the situation is as follows.

enter image description here

Add forces $Q'$ and $Q''$ acting at the centre of mass $O$ of the same magnitude as force $Q$ as shown in the diagram and repeat by adding forces $P'$ and $P''$ acting at the centre of mass $O$ of the same magnitude as force $P$ as shown in the diagram.

Forces $Q$ and $Q'$ constitute a couple magnitude $Qq$ in an anticlockwise direction and forces $P$ and $P''$ constitute a couple magnitude $Pp$ also in an anticlockwise direction, so the net torque on the rod is $Qq+Pp$ anticlockwise.

The net force acting at the centre of mass of the rod is $P-Q$ and this is just as true for your second diagram.

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  • $\begingroup$ Wow! Your solution is very elegant! However, @Eli's solution is also correct. There he showed that $P−Q$ is acting to the left of $Q$, not at the center of mass, but to the left of $Q$, there is no bar left, so $P−Q$ is acting on nothing?! How does $P−Q$ act at the center of mass of the bar according to @Eli's figure? $\endgroup$
    – tryingtobeastoic
    Commented Mar 21, 2022 at 13:04
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    $\begingroup$ All I can add is that the centre of mass of the rod will undergo a linear acceleration under the influence of the resultant force acting on the rod and also an angular acceleration as a result of the net torque acting on the rod. I have assumed that you are suggesting that the $P-Q$ force in your diagram is the resultant force acting on the rod. $\endgroup$
    – Farcher
    Commented Mar 21, 2022 at 17:02
  • $\begingroup$ I see. I understood that from your answer. According to @Eli's answer, the resultant will actually act outside the bar/stick. How is that possible?! Could you please elaborate on that? $\endgroup$
    – tryingtobeastoic
    Commented Mar 21, 2022 at 17:50
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    $\begingroup$ I am afraid that I cannot elaborate on what @Eli has done. Perhaps you should ask the question of the author of that answer? $\endgroup$
    – Farcher
    Commented Mar 21, 2022 at 18:05
  • $\begingroup$ "The $P−Q$ resultant is in the incorrect position". Sir, I think that the aforementioned sentence is incorrect. The $P-Q$ resultant is a force that produces the same net force and the same torque on the bar as $P$ and $Q$ combined. In that sense, the $P-Q$ force is in the correct position (in both scenarios 1 and 2). Even though it is in the correct position, it is true, however, that the $P-Q$ resultant in scenario 2 can't be physically implemented. (...) $\endgroup$
    – tryingtobeastoic
    Commented Apr 2, 2022 at 3:54

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