# Thread: Finding center of mass

1. ## Finding center of mass

I've sat around for hours and cannot get the right answer for this past exam question and my exam is in a week!

The diagram shows a sheet ABCDEFG of uniform thickness. An open envelope is formed by folding the sheet along CF and by glueing CD to CB and FE to FG.

Show that the distance of the centre of mass of the envelope from BG is 28/(8 + 3^0.5). You should ignore the mass of the glue.

Thank you so much in advance!

2. Hello, meru!

The diagram shows a sheet $ABCD{E}FG$ of uniform thickness. An open envelope
is formed by folding along $CF$ and by gluing $CD\text{ to }CB$ and $FE\text{ to }FG.$

Show that the distance of the centre of mass from $BG$ is: $\frac{28}{8 + \sqrt{3}}$
Code:
C           B
* - - - - - *
|           |  *
|           |     *
- + - - o - - + -o- - -* - -
|     Q    O|  P  *    A
|           |  *
* - - - - - *
F           G

The area (mass) of $\Delta ABG$ is: . $\frac{\sqrt{3}}{4}(8^2) \:=\:16\sqrt{3}$

The distance from its center of mass $P$ to the origin $O$
. . is: . $\frac{1}{3}(OA) \:=\:\frac{1}{3}(4\sqrt{3}) \:=\:\frac{4\sqrt{3}}{3}$
Hence, its moment is: . $(16\sqrt{3})\left(\frac{4\sqrt{3}}{3}\right) \:=\:64$

The center of mass of the square is at $Q(-4,0).$
Its mass is: . $2 \times 8^2 \:=\:128$
Its moment is: . $(128)(-4) \:=\:-512$

The total moment is: . $64 -512 \:=\:-448$
The total mass is: . $128 + 16\sqrt{3}$
Hence, the center of mass is at: . $\frac{-448}{128 + 15\sqrt{3}} \;=\;\frac{-28}{8+\sqrt{3}}$

It is $\frac{28}{8+\sqrt{3}}$ units to the left of $BG$.