2
$\begingroup$

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

Lamina

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 $$<\frac{107a}{58}, \frac{127a}{58}>$$

However the actual answer is

$\frac{179a}{58}$ 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.

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

2 Answers 2

2
$\begingroup$

Try

$$\therefore \bar{x}=\frac{(\frac{4a}{2}((4a)(5a))+((\frac{3a}{2}+4a)(3a)^2)}{(3a)^2+(4a)(5a)}= ? $$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.