- #1
Portuga
- 56
- 6
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:
<br /> \vec{r}_{2}=\vec{r}_{1}+\frac{\vec{F}_{1}\times\hat{k}M_{1}}{F_{1}^{2}},<br /> and
<br /> \vec{F}_{2}=\vec{F}_{1}.<br />
Homework Equations
My first attempt was to use the equivalence of the couples:
<br /> \vec{r}_{1}\times\vec{F}_{1}=\vec{r}_{2}\times\vec{F}_{2}=M_{1}\hat{k},<br />
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:
<br /> \begin{aligned} & \vec{F}_{1}\times\vec{r}_{2}\times\vec{F}_{2}=\vec{F}_{1}\times M_{1}\hat{k}\\<br /> \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}\\<br /> \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}}.<br /> \end{aligned}<br />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?