Center of Mass — Question 5

Question 5 – Solution

Step 1: Treat the removed square as negative area. The full $2\times2$ square has $$A_1=4,\qquad (x_1,y_1)=(1,1).$$ The removed $1\times1$ square has $$A_2=1,\qquad (x_2,y_2)=\left(\frac32,\frac32\right).$$

Step 2: Find the remaining area. $$A=A_1-A_2=4-1=3.$$

Step 3: Subtract the moments about the $y$-axis. $$M_y=A_1x_1-A_2x_2 =4(1)-1\left(\frac32\right)=\frac52.$$

Step 4: Subtract the moments about the $x$-axis. $$M_x=A_1y_1-A_2y_2 =4(1)-1\left(\frac32\right)=\frac52.$$

Step 5: Divide by the remaining area. $$\bar x=\frac{M_y}{A}=\frac56,\qquad \bar y=\frac{M_x}{A}=\frac56.$$ $$\boxed{(\bar x,\bar y)=\left(\frac56,\frac56\right)}$$