I'm new here and this is my first post so please forgive me if I've not posted it in the right place!
It's an Engineering related question but as maths is involved I thought it may be appropriate to ask here...
I'm given the co-ordinates of a massless quadrilateral (z) with 4 mass particles attached to the vertexes; I need to find the co-ordinates of the centre of gravity (or mass?).
Now, I know that if (z) were a uniform mass it would be a case of finding the sum of the x and y co-ordinates and dividing by the number of vertexes. The question also defines (z) as being a triangular system, so I'm wondering if that is a hint in how to find the COG.
Any suggestions or advice on how to go about solving it would be really appreciated.
Thanks,
Jack
You can solve it from first principles. Suppose that masses $m_1,m_2,m_3$, and $m_4$ are located at $\langle 0,0\rangle,\langle a,0\rangle,\langle 0,b\rangle$, and $\langle a,b\rangle$, respectively. (You can always impose such a coordinate system.) Suppose that the centre of mass is at $\langle\bar x,\bar y\rangle$; then the moments of the system about the lines $x=\bar x$ and $y=\bar y$ must be $0$. The moment about $x=\bar x$ is
$$m_1(0-\bar x)+m_2(a-\bar x)+m_3(0-\bar x)+m_4(a-\bar x)=a(m_2+m_4)-m\bar x\;,$$
where $m=m_1+m_2+m_3+m_4$, so $$\bar x=\frac{a(m_2+m_4)}m\;.$$
I’ll leave $\bar y$ to you; the computation is very similar.