Figure 9-39 shows a cubical box that has been constructed from uniform metal plate of negligible thickness. The box is open at the top and has edge length$\mathbf{L}\mathbf{=}\mathbf{40}\mathbf{cm}$Find (a) The xcoordinates, (b) The ycoordinate, and (c) The zcoordinates of the center of mass of the box.

Edge length of the box,$\mathrm{L}=40\mathrm{cm}$

For a system of particles, the whole mass of the system is concentrated at the center of mass of the system.

The expression for the coordinates of the center of mass are given as:

${\overrightarrow{r}}_{com}=\frac{1}{M}\sum _{i=1}^n{m}_i$ (i)

Here, $\mathrm{M}$is the total mass, ${\mathrm{m}}_{\mathrm{i}}$ is the individual mass of ith particle and ${\mathrm{r}}_{\mathrm{i}}$ is the coordinates of ithparticle.

We can determine the center of mass of each side of the box using symmetry. We can use these coordinates in the formula of the center of mass. Since the sides are identical, we can eliminate the mass term from the formula.

For our convenience, we separate the center of mass for each side of the cube.

$$\begin{gathered} \mathrm{For}\mathrm{the}\mathrm{side}\mathrm{in}\mathrm{yz}\mathrm{plane},({\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1)=(0,20\mathrm{cm},20\mathrm{cm}) \\ \mathrm{For}\mathrm{the}\mathrm{side}\mathrm{in}\mathrm{xz}\mathrm{plane},({\mathrm{x}}_2,{\mathrm{y}}_2,{\mathrm{z}}_2)=(20\mathrm{cm},0,20\mathrm{cm}) \\ \mathrm{For}\mathrm{the}\mathrm{side}\mathrm{in}\mathrm{xy}\mathrm{plane},({\mathrm{x}}_3,{\mathrm{y}}_3,{\mathrm{z}}_3)=(20\mathrm{cm},20\mathrm{cm},0) \\ \mathrm{For}\mathrm{the}\mathrm{side}\mathrm{parallel}\mathrm{to}\mathrm{yz}\mathrm{plane},({\mathrm{x}}_4,{\mathrm{y}}_4,{\mathrm{z}}_4)=(40\mathrm{cm},20\mathrm{cm},20\mathrm{cm}) \\ \mathrm{For}\mathrm{the}\mathrm{side}\mathrm{parallel}\mathrm{to}\mathrm{xz}\mathrm{plane}({\mathrm{x}}_5,{\mathrm{y}}_5,{\mathrm{z}}_5)=(20\mathrm{cm},40\mathrm{cm},20\mathrm{cm}) \end{gathered}$$

Using equation (i), the x coordinate of center of mass is,

${\mathrm{x}}_{\mathrm{com}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2+{\mathrm{m}}_3{\mathrm{x}}_3+....}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+....}$

Since mass is the same for each side, so we can eliminate it easily.

$$\begin{gathered} {\mathrm{x}}_{\mathrm{com}}=\frac{0+20\mathrm{cm}+20\mathrm{cm}+40\mathrm{cm}+20\mathrm{cm}}{5} \\ =20\mathrm{cm} \end{gathered}$$

Thus, the x coordinate of the center of mass is 20 cm.

Using equation (i), the y coordinate of center of mass is,

$$\begin{gathered} y_{\mathrm{com}}=\frac{{\mathrm{m}}_1{\mathrm{y}}_1+{\mathrm{m}}_2{\mathrm{y}}_2+{\mathrm{m}}_3{\mathrm{y}}_3+....}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+....} \\ =\frac{20cm+0+20cm+20cm+40cm}{5} \\ =20cm \end{gathered}$$

Thus, the y coordinate of the center of mass is 20 cm.

Using equation (i), the z coordinate of center of mass is,

$$\begin{gathered} z_{\mathrm{com}}=\frac{{\mathrm{m}}_1{\mathrm{z}}_1+{\mathrm{m}}_2{\mathrm{z}}_2+{\mathrm{m}}_3{\mathrm{z}}_3+....}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+....} \\ =\frac{20cm+20cm+0+20cm+20cm}{5} \\ =16cm \end{gathered}$$

Thus, the z coordinate of center of mass is 16 cm.