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General equivalent couple - force system

  1. Jan 18, 2017 #1
    1. The problem statement, all variables and given/known data

    Hello gentlemen!

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

    As it is a little more difficult problem, the author provided a solution:
    [/tex] and

    2. Relevant equations

    My first attempt was to use the equivalence of the couples:
    and the BAC - CAB rule for double cross product: [itex] \vec A \times \vec B \times \vec C = \vec B (\vec A \ldotp \vec C) - \vec C (\vec A \ldotp \vec B) [/itex]

    3. The attempt at a solution
    That [itex]\vec{F}_1 = \vec{F}_2[/itex], 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}}.

    So my point is: if there is freedom to choose both [itex]\vec{r}_{2}[/itex] and [itex]\vec{F}_{2}[/itex] , it's not possible to assume that these vectors should be related to the parameters of the problem, [itex]\vec{r}_{1}[/itex], [itex]\vec{F}_{1}[/itex] and [itex]M_{1}\hat{k}[/itex] as the author requires in his solution. Am I right or am I missing something really important on this issue?
  2. jcsd
  3. Jan 18, 2017 #2
    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?
  4. Jan 18, 2017 #3
    Well, basically, if [itex]\vec{F}_1 = \vec{F}_2[/itex], then it will be mandatory that [itex]\vec{r}_1 = \vec{r}_2[/itex], 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}}
    Last edited: Jan 18, 2017
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