# How will the force act outside the stick?

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?

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.

• 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? Commented Mar 21, 2022 at 13:04
• 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. Commented Mar 21, 2022 at 17:02
• 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? Commented Mar 21, 2022 at 17:50
• I am afraid that I cannot elaborate on what @Eli has done. Perhaps you should ask the question of the author of that answer? Commented Mar 21, 2022 at 18:05
• "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. (...) Commented Apr 2, 2022 at 3:54