Question

Two square blocks of size $a\times a$ is removed from a big square $4a\times 4a$ as shown in the figure. Find the COM of the system after removing the square blocks.

• A. $(\frac{3a}{28},0)$
• B. $(\frac{-3a}{28},0)$
• C. $(\frac{-3a}{14},0)$
• D. $(\frac{3a}{14},0)$

Answer 1

The correct option is C $(\frac{-3a}{14},0)$

	Area	COM
Parent square	$4a\times 4a=16a^2$	$(0,0)$
Removed square 1	$a\times a=a^2$	$(\frac{3a}{2},\frac{3a}{2})$
Removed square 2	$a\times a=a^2$	$(\frac{3a}{2},\frac{-3a}{2})$

From the symmetry along $x-$ axis we can say that
$x_{COM}=\frac{A_1{x}_1+A_2{x}_2+A_3{x}_3}{A_1+A_2+A_3}$
Here, $A_1=$ Area of parent square $=16a^2$
$A_2=A_3=$ Area of removed squares $=a^2$

COM of parent square, $x_1$$=(0,0)$
COM of removed square 1, $x_2$$=\frac{3a}{2}$
COM of the removed square 2, $x_3=$ $\frac{3a}{2}$
$x_{com}=\frac{16a^2\left(0\right)+(-a^2)\left(\frac{3a}{2}\right)+(-a{)}^2\left(\frac{3a}{2}\right)}{16a^2+(-a^2)+(-a^2)}$
$x_{com}=\frac{\frac{-3a^3}{2}-\frac{3a^3}{2}}{14a^2}=\frac{-3a}{14}$
$y_{com}=0$
So, new co-ordinates of COM is $(\frac{-3a}{14},0)$