91Ó°ÊÓ

The diagram is given for shape and size.

mass of plate = 800 g = 0.8kg

length and width is 20 and 30 cm (which is .2 m an d.3 m)

Use the hint and draw a small strip dx as shown in figure below

Area of the small strip is say dA, then

dA = I dx

assume the density of the plate is even

so mass of the small strip is dM

$dm=\frac{M}{A}dA$

where M is the mass and A is the area of the plate.

Area can be calculated using area of triangle

A = 1/2 x l x w =1/2 x 0.20 m x 0.30m = 0.03 m²

Substitute these values we get

$dm=\frac{(.8kg)}{(0.030m^2)}Idx=(26.67kg/m^2)ldx$

From the similar triangle property

$$\begin{gathered} \frac{l}{20cm}=\frac{x}{30cm} \\ l=\frac{20cm}{30cm}x \\ l=\frac{2}{3}x.............................\left(2\right) \end{gathered}$$

Substitute this in dm and then integrate

$dm=(26.67kg/m^2)\left(\frac{2}{3}x\right)dx$

$$\begin{gathered} x_{\mathrm{cm}}=\frac{1}{M}∫xdm \\ =\frac{1}{0.800\mathrm{kg}}∫x\left(26.67\frac{\mathrm{kg}}{{\mathrm{m}}^2}\right)\left(\frac{2}{3}x\right)dx \\ =\left(\frac{2}{3}\right)\left(\frac{26.67\frac{\mathrm{kg}}{{\mathrm{m}}^2}}{0.800\mathrm{kg}}\right){∫}_0^{0.300\mathrm{m}}{x}^{2}dx \\ =\left(\frac{2}{3}\right)\left(\frac{26.67\frac{\mathrm{kg}}{{\mathrm{m}}^2}}{0.800\mathrm{kg}}\right){\left[\frac{x^3}{3}\right]}_0^{0.300\mathrm{m}} \\ =\left(\frac{2}{9}\right)\left(\frac{26.67\frac{\mathrm{kg}}{{\mathrm{m}}^2}}{0.800\mathrm{kg}}\right)[0.300\mathrm{m}{]}^3 \\ =0.20\mathrm{m}=20.0\mathrm{cm} \end{gathered}$$

As it is clear from the diagram that it is an isosceles triangle , so y component is on the x=0