![enter image description here

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,

enter image description here



$$b=fa\ \left[\text{Let $f=\frac{P+Q}{P-Q}$}\right]$$

As $P>Q$, $f>1$, and $b>a$. So, the correct figure will be,

enter image description here

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:

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

These may help you to answer this question:

  1. Comment by @Ivan
  2. Answer by @Farcher

This question was posted with the help of @Eli.

  • $\begingroup$ Crossposted from PSE and MSE $\endgroup$ Mar 23, 2022 at 5:12
  • 1
    $\begingroup$ The question has an answer that is linked to in the question. $\endgroup$
    – Solar Mike
    Mar 23, 2022 at 6:57

1 Answer 1


As you have calculated the torque on the bar is

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

and a net force


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


and also accelerate with,


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!


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