# Finding the center of mass of a composite lamina

The following lamina is given and I have to find the distance from center of mass from AB and AC

I started by splitting the lamina into two shapes, one rectangle of width $$4a$$ height $$5a$$ and one of height $$3a$$ and width $$3a$$ which is a square.

$$\because \bar{x}=\frac{\sum_{i=0}^n \widehat{x}_i A_i}{\sum_{i=0}^n A_i}$$
$$\therefore \bar{x}=\frac{(\frac{4a}{2}((4a)(5a))+(\frac{3a}{2}(3a)^2)}{(3a)^2+(4a)(5a)}=\frac{107a}{58}$$

And similarly

$$\bar{y}=\frac{(\frac{5a}{2}((4a)(5a))+(\frac{3a}{2}(3a)^2)}{(3a)^2+(4a)(5a)} =\frac{127a}{58}$$ Hence the coordinates of the center of mass are $$<\frac{107a}{58}, \frac{127a}{58}>$$

$$\frac{179a}{58}$$ from AB and $$\frac{127a}{58}$$ from AC

I am entirely baffled on how to get this answer and have little to no clue what Ive done wrong in solving this.

• I think some data is missing, we know the overall height but not the height of the 2 pieces Jan 20, 2021 at 18:54
• I have added the missing data Jan 20, 2021 at 18:59

$$\therefore \bar{x}=\frac{(\frac{4a}{2}((4a)(5a))+((\frac{3a}{2}+4a)(3a)^2)}{(3a)^2+(4a)(5a)}= ?$$
Where $$4a$$ is a shift from the left side. The right rectangle does not start at the origin so you must take this shift into account
Your mistake is the $$x$$ cordinate of the square that you have written. If you are taking the origin at A, so the $$x$$ coordinate would be $$4a+\frac{3a}{2}=\frac{11a}{2}$$. Now if you do the calculation, you will get the right answer.