Calculate the position of the centroid from MO of the lamina shown in FIGURE 2 below.

• Shape 1 (Leftmost rectangle):

  • Width ($w_1$) = $200 \text{ mm}$
  • Height ($h_1$) = $400 \text{ mm}$
  • Area $A_1 = w_1 \times h_1 = 200 \text{ mm} \times 400 \text{ mm} = 80000 \text{ mm}^2$
  • Centroid x-coordinate $x_1 = \frac{w_1}{2} = \frac{200 \text{ mm}}{2} = 100 \text{ mm}$

• Shape 2 (Middle rectangle):

  • Width ($w_2$) = $250 \text{ mm}$
  • Height ($h_2$) = $220 \text{ mm}$
  • Area $A_2 = w_2 \times h_2 = 250 \text{ mm} \times 220 \text{ mm} = 55000 \text{ mm}^2$
  • Centroid x-coordinate $x_2 = 200 \text{ mm} + \frac{w_2}{2} = 200 \text{ mm} + \frac{250 \text{ mm}}{2} = 200 \text{ mm} + 125 \text{ mm} = 325 \text{ mm}$

• Shape 3 (Rightmost rectangle):

  • Width ($w_3$) = $300 \text{ mm}$
  • Height ($h_3$) = $90 \text{ mm}$
  • Area $A_3 = w_3 \times h_3 = 300 \text{ mm} \times 90 \text{ mm} = 27000 \text{ mm}^2$
  • Centroid x-coordinate $x_3 = 200 \text{ mm} + 250 \text{ mm} + \frac{w_3}{2} = 450 \text{ mm} + \frac{300 \text{ mm}}{2} = 450 \text{ mm} + 150 \text{ mm} = 600 \text{ mm}$

The position of the centroid from MO is $\text{259.72 mm}$.