Question

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.

Short answer

a. $\mathrm{X}$coordinate of center of mass of the box is $20\mathrm{cm}$.

b. $\mathrm{Y}$coordinate of center of mass of the box is 20 cm.

c. $Z$coordinate of center of mass of the box is 16 cm.

Step-by-step solution

Listing the given quantities

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

Understanding the concept of center of mass

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.

Determination of the coordinates of different planes

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.

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

(a) Determination of the x coordinate of the center of mass of the box

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.

(b) Determination of the y coordinate of center of mass of box

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.

(c) Determination of the z coordinate of center of mass of box

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.