Calculating center of gravity asymmetrical objects
find the position of the center of gravity of relative to the edges AB and AC
The answers are 32.5mm from AB and 7.5 from BC
I can't seem to solve this problem for the side where the answer 7.5 mm is obtained
I know that the object consists of three smaller squares so I found the area of each square and made my reference lines and solved each axis for the perpendicular midpoints of each rectangle
I found each area : A1=400 ,A2=500,A3=300 , total area=1200mm^2
then I input 20x400+35x500+45x300/1200=32,5mm
when I input the formula for the other side I obtain 5x400+35x500+65x300/1200 i obtain 32.5 ,which is the same answer as before
I would appreciate any help solving this problem and some advice with calculating the center of gravity as I have spent several hours unsuccessfully solving it.I am sure I am making some mistake somewhere. Thank you
Draw a vertical line $\ell$ through $A$. We use the same decomposition as you did, and find that the distance of the centre of mass from this line is $$\frac{(20)(400)+(35)(500)+(45)(300)}{1200}.$$ This is $32.5$, so our centre of mass is $32.5$ to the right of $\ell$. It is therefore $7.5$ to the left of $BC$.
Or else we could take distances to the left of $BC$ as negative, and distances to the right as positive. So we are taking the line through $B$ and $C$ as the $y$-axis. Then the centre of mass has $x$-coordinate $$\frac{(-20)(400)+(-5)(500)+(5)(300)}{1200}.$$ This is $-7.5$.
Remark: Your calculation was exactly right, except that the question asked one to describe the location of the centre of mass in terms of distance from $BC$, not in terms of distance from $\ell$.