How will the force act outside the stick?

Question

Scenario 1:

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:

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?

Answer 1

Here, $P>Q$. $O$ is the center of mass of the rigid and uniform bar/stick.

It's more useful and natural, and less confusing/misleading, to label vectors with their names $(\mathbf Q,\mathbf P,\mathbf P+\mathbf Q)$ rather than their lengths $(|\mathbf Q|,|\mathbf P|,|\mathbf P|-|\mathbf Q|).$

Likewise, lengths $(|\mathbf Q|,|\mathbf P|)$ ought to be notationally distinguished from points $(Q,P,O).$

Scenario 2:

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?

Assume that the rotation axis passes through $O,$ that all forces are vertical, and that $\mathbf Q\ne\mathbf0.$ Then the resultant force-torque $(\mathbf P+\mathbf Q),$ which has the combined effect of $\mathbf P$ and $\mathbf Q,$ indeed has a vertical line of action on $\mathbf P$'s right. Obviously, this resultant force cannot be physically implemented.

(Circling back to my opening point: “resultant of magnitude $(P−Q)$” is confusing to parse, unlike “resultant force with magnitude $(P-Q)$” or simply “resultant force $\mathbf P+\mathbf Q$”. Still, since we're already using unboldfaced capital letters to denote points, $|\mathbf P|$ is better than $P$ to denote $\mathbf P$'s length.)

Answer 2

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.

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.