Hello, meru!

The diagram shows a sheet $\displaystyle ABCD{E}FG$ of uniform thickness. An open envelope

is formed by folding along $\displaystyle CF$ and by gluing $\displaystyle CD\text{ to }CB$ and $\displaystyle FE\text{ to }FG.$

Show that the distance of the centre of mass from $\displaystyle BG$ is: $\displaystyle \frac{28}{8 + \sqrt{3}}$ Code:

C B
* - - - - - *
| | *
| | *
- + - - o - - + -o- - -* - -
| Q O| P * A
| | *
* - - - - - *
F G

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

The distance from its center of mass $\displaystyle P$ to the origin $\displaystyle O$

. . is: .$\displaystyle \frac{1}{3}(OA) \:=\:\frac{1}{3}(4\sqrt{3}) \:=\:\frac{4\sqrt{3}}{3}$

Hence, its moment is: .$\displaystyle (16\sqrt{3})\left(\frac{4\sqrt{3}}{3}\right) \:=\:64$

The center of mass of the square is at $\displaystyle Q(-4,0).$

Its mass is: .$\displaystyle 2 \times 8^2 \:=\:128$

Its moment is: .$\displaystyle (128)(-4) \:=\:-512$

The total moment is: .$\displaystyle 64 -512 \:=\:-448$

The total mass is: .$\displaystyle 128 + 16\sqrt{3}$

Hence, the center of mass is at: .$\displaystyle \frac{-448}{128 + 15\sqrt{3}} \;=\;\frac{-28}{8+\sqrt{3}}$

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