How to replace these two forces with one force?

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 $$b>a$$. So, the correct figure will be,

I hope you're satisfied that the figure is correct.

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:

1. Can $$\vec{P}$$ and $$\vec{Q}$$ be replaced by a single force? Is my conclusion correct?

This question was posted with the help of @Eli.

• Crossposted from PSE and MSE Mar 23, 2022 at 5:12
• The question has an answer that is linked to in the question. Mar 23, 2022 at 6:57

As you have calculated the torque on the bar is

$$\tau= (P+q)A$$

and a net force

$$F=P-Q$$

This will cause the bar to turn with an angular acceleration,

$$\alpha=\frac{\tau}{I}$$

and also accelerate with,

$$a=\frac{P-Q}{m}$$

Any substitute pair of forces acting within the length of the bar can be scaled by the factor of $$A/D$$ to impart the same torque. But the new net force will not be the same.

$$P_N-Q_N\neq P-Q$$.

• A= half-length of bar
• m= mass
• a= linear acceleration
• D = distance of new pair of force Pn, Qn, from the center of the bar
• $$\alpha$$= angular acceleration
• I= bar's moment of inertia
• $$\tau$$= torque

So depending on what you demand the answer varies, if you require just the same torque, yes. If you require the same torque and linear acceleration no!