General equivalent couple - force system

[Portuga]

Homework Statement

Hello gentlemen!

I came across a real challeging problem about static, which states
the following: assume a force system is equivalent to a force $\vec{F}_{1}$ and a couple $M_{1}\vec{k}$ acting at a point $\vec{r}_{1}$. Find some point $\vec{r}_{2}$ and a force $\vec{F}_{2}$ so that $\vec{F}_{2}$ acting at $\vec{r}_{2}$ is equivalent to $\vec{M}_{1}$ acting at $\vec{r}_{1}$.

As it is a little more difficult problem, the author provided a solution:
$$\vec{r}_{2}=\vec{r}_{1}+\frac{\vec{F}_{1}\times\hat{k}M_{1}}{F_{1}^{2}},$$ and
$$\vec{F}_{2}=\vec{F}_{1}.$$

Homework Equations

My first attempt was to use the equivalence of the couples:
$$\vec{r}_{1}\times\vec{F}_{1}=\vec{r}_{2}\times\vec{F}_{2}=M_{1}\hat{k},$$
and the BAC - CAB rule for double cross product: $\vec A \times \vec B \times \vec C = \vec B (\vec A \ldotp \vec C) - \vec C (\vec A \ldotp \vec B)$

The Attempt at a Solution

That $\vec{F}_1 = \vec{F}_2$, it's obvious, because of the equivalence of the forces systems. So I focused on the equivalence of the couples.
I realized that only the last two members of the vectorial equation were interesting for this:
$$\begin{aligned} & \vec{F}_{1}\times\vec{r}_{2}\times\vec{F}_{2}=\vec{F}_{1}\times M_{1}\hat{k}\\ \Rightarrow & \vec{r}_{2}\left(\vec{F}_{1}\ldotp\vec{F}_{2}\right)-\left(\vec{F}_{1}\ldotp\vec{r}_{2}\right)\vec{F}_{2}=\vec{F}_{1}\times M_{1}\hat{k}\\ \Rightarrow & \vec{r}_{2}=\frac{\left(\vec{F}_{1}\ldotp\vec{r}_{2}\right)\vec{F}_{2}+\vec{F}_{1}\times M_{1}\hat{k}}{\vec{F}_{1}\ldotp\vec{F}_{2}}. \end{aligned}$$So my point is: if there is freedom to choose both $\vec{r}_{2}$ and $\vec{F}_{2}$ , it's not possible to assume that these vectors should be related to the parameters of the problem, $\vec{r}_{1}$, $\vec{F}_{1}$ and $M_{1}\hat{k}$ as the author requires in his solution. Am I right or am I missing something really important on this issue?

 

[Dr.D]

Looks to me like you missed the easy part. If the net force on the system is originally F1, and the net force is to be held constant, then F2 = F1. Doesn't that enable you to finish your development of r2?

 

[Portuga]

Well, basically, if $\vec{F}_1 = \vec{F}_2$, then it will be mandatory that $\vec{r}_1 = \vec{r}_2$, don't you think? Even if I put this reasoning forward, the result would be strange:
$$\begin{aligned}\vec{r}_{2} & =\frac{\left(\vec{F}_{1}\ldotp\vec{r}_{2}\right)\vec{F}_{1}+\vec{F}_{1}\times M_{1}\hat{k}}{F_{1}^{2}}\\ & =\frac{\left(\vec{F}_{1}\ldotp\vec{r}_{1}\right)\vec{F}_{1}+\vec{F}_{1}\times M_{1}\hat{k}}{F_{1}^{2}}\\ & =r_{1}\cos\theta\hat{F}_{1}+\frac{\vec{F}_{1}\times M_{1}\hat{k}}{F_{1}^{2}} \end{aligned}$$