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![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

$$(P-Q)b=Pa+Qa$$

$$b=\frac{P+Q}{P-Q}a$$

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

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

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.

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  • $\begingroup$ Crossposted to MSE. $\endgroup$ Commented Mar 22, 2022 at 13:59
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    $\begingroup$ there are some good answers here, but I wanted to mention that your use of the word "practically" is doing a huge amount of heavy lifting. It might absolutely trivial to replace forces P and Q with P-Q; say they are ropes, and the beam can be extended in some trivial way. Or, it might be neigh-impossible, if say the beam is a nuclear submarine and all you have is a crowbar. If you're working on paper, "practicality" is only limited by paper, without lots and lots of more context. $\endgroup$
    – levitopher
    Commented Mar 22, 2022 at 18:20
  • $\begingroup$ @levitopher "It might absolutely trivial to replace forces P and Q with P-Q; say they are ropes, and the beam can be extended in some trivial way." If we extend the beam, the center of mass will change. If the center of mass changes, the system will change completely. Then none of our old calculations will work. We will have to then find a new resultant for $P$ and $Q$, so I don't understand how we can extend the beam without changing the system entirely and negating all our previous calculations. Could you please give me some pointers, sir? $\endgroup$ Commented Mar 23, 2022 at 4:53
  • $\begingroup$ Crossposted to ESE $\endgroup$ Commented Mar 23, 2022 at 5:13
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    $\begingroup$ Extend the beam with a massless material (relative to the other materials in the system), which will have no effect on the CM. I can hear you now: "massless materials, what is this?!?" because of course, that's not literally possible - but without knowing much more about the system, we don't actually have any idea how "practical" it is (we have to know the mass of the other objects in the system). "Practical" is not a word that translates well to paper models, so I advocate against taking it too seriously here. $\endgroup$
    – levitopher
    Commented Mar 23, 2022 at 23:43

1 Answer 1

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To find the equipollent force and moment you just add up all the components together.

Say you have two forces $P$ and $Q$ (all values shown positive, like most FBD diagrams) and you want to find the equipollent force $F$ and its location $x$.

fig1

Put a coordinate system on the center of mass and add up all forces and torques

  • Sum of forces $$F = P -Q $$
  • Sum of torques about origin (COM) $$x\, F = a P +a Q$$ The reason for the positive sign on the torque is that both $P$ and $Q$ exert a counter-clockwise torque about the center of mass.

The solution of the two equations in terms of $F$ and $x$ is

$$ F = P -Q $$ $$ x = \frac{P+Q}{P-Q}\; a $$

Now you can examine the conditions that represent special cases

  • $x=0$ when $Q= -P$
  • $x=a$ when $Q = 0$
  • $x>a$ when $Q > 0$ (condition in op)

So theoretically yes you can use one force. In general, you need a force along the line of action and a parallel torque twisting the axis and it can represent any combined load system.

In practice, if by force you mean a contact force that requires a physical connection to the part, then when $x>a$ you can't create that. You will need two forces to create a force couple for the offset loading. Or material needs to extend where the force contact needs to be. But there are other ways of applying forces, that do not require physical contact, and so one cannot truly answer your question unless more information is provided on the system.


Now you extend this problem to 3D when multiple force vectors $\vec{P}_i$ are applied, each at a location $\vec{r}_i$.

$$ \begin{aligned}\vec{F} & =\sum_{i}\vec{P}_{i}\\ \vec{\tau} & =\sum_{i}\vec{r}_{i}\times\vec{P}_{i} \end{aligned} $$

The solution to $\vec{\tau} = \vec{r} \times \vec{F}$ in terms of $\vec{r}$ comes from the vector triple product

$$ \require{cancel} \begin{gathered}\vec{r}\times\vec{F}=\vec{\tau}\\ \vec{F}\times\left(\vec{r}\times\vec{F}\right)=\vec{F}\times\vec{\tau} \\ \vec{r}\left(\vec{F}\cdot\vec{F}\right)-\vec{F}\left(\cancel{\vec{F}\cdot\vec{r}}\right)=\vec{F}\times\vec{\tau}\\ \vec{r}=\frac{\vec{F}\times\vec{\tau}}{\vec{F}\cdot\vec{F}} \end{gathered} $$

This returns the location of the line of action of $\vec{F}$, ignoring any component parallel to the force. The $\vec{r}$ returned above is only the perpendicular distance to the line of action.

You can see the result is the same with $\vec{P}_1 = \pmatrix{0 \\ P \\0}$ and $\vec{P}_2 = \pmatrix{0 \\ -Q \\0}$ as well as $\vec{r}_1 = \pmatrix{a \\ 0 \\ 0}$ and $\vec{r}_2 = \pmatrix{-a \\ 0 \\ 0}$, yielding $\vec{F} = \pmatrix{0 \\ P-Q \\ 0}$ and $\vec{\tau} = \pmatrix{ 0 \\ 0 \\ a P + a Q}$

$$ \vec{r} = \frac{ \pmatrix{a (P-Q) (P +Q) \\ 0 \\ 0} }{ (P-Q)^2 } = \pmatrix{ a \frac{P+Q}{P-Q} \\ 0 \\ 0} $$

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    $\begingroup$ I think you made mistake. It should be $$ \vec{r} = \frac{ \pmatrix{a (P-Q) (P +Q) \\ 0 \\ 0} }{ (P-Q)^2 } = \pmatrix{ a \frac{P+Q}{P-Q} \\ 0 \\ 0} $$ $\endgroup$ Commented Mar 23, 2022 at 4:10
  • $\begingroup$ @tryingtobeastoic - yes, thank you. $\endgroup$ Commented Mar 23, 2022 at 11:52
  • $\begingroup$ I can't understand the gist of your answer. Do you agree with my conclusion that $\vec{P}$ and $\vec{Q}$ can't be replaced by a single resultant force? $\endgroup$ Commented Mar 23, 2022 at 12:10
  • $\begingroup$ @tryingtobeastoic - again sorry. I updated my answer. $\endgroup$ Commented Mar 23, 2022 at 12:24

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