Three cubes, of side \({l_ \circ }\), \(2{l_ \circ }\), and \(3{l_ \circ }\), are placed next to one another (in contact) with their centers along a straight line as shown in Fig. 7–38. What is the position, along this line, of the CM of this system? Assume the cubes are made of the same uniform material.

FIGURE 7-38

Problem 52.

For symmetrical shape, the center of mass is located at the center.The cube is symmetrical in shape; so the CM of an individual cube is at its center.

The sides of the cubes are \({l_ \circ }\), \(2{l_ \circ }\), and \(3{l_ \circ }\).

Let \(\rho \)be the density of the material by which the cubes are made.

Now, the mass of the small cube is \({m_1} = \rho {\left( {{l_ \circ }} \right)^3}\).

The mass of the medium cube is:

\(\begin{array}{c}{m_2} = \rho {\left( {2{l_ \circ }} \right)^3}\\ = 8\rho {\left( {{l_ \circ }} \right)^3}\\ = 8{m_1}\end{array}\).

The mass of the large cube is:

\(\begin{array}{c}{m_3} = \rho {\left( {3{l_ \circ }} \right)^3}\\ = 27\rho {\left( {{l_ \circ }} \right)^3}\\ = 27{m_1}\end{array}\).

The cube is symmetrical in shape; therefore, the CM of an individual cube is in the middle of the cube.

So, the distance of the CM of the small cube is at \({x_1} = \frac{{{l_ \circ }}}{2}\).

The distance of the CM of the medium cube is at:

\(\begin{array}{c}{x_2} = {l_ \circ } + \frac{{2{l_ \circ }}}{2}\\ = 2{l_ \circ }\end{array}\)

The distance of the CM of the large cube is at:

\(\begin{array}{c}{x_3} = {l_ \circ } + 2{l_ \circ } + \frac{{3{l_ \circ }}}{2}\\ = \frac{9}{2}{l_ \circ }\end{array}\)

Now, the center of mass of the cube system is:

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{\left( {{m_1} \times {x_1}} \right) + \left( {{m_2} \times {x_2}} \right) + \left( {{m_2} \times {x_2}} \right)}}{{{m_1} + {m_2} + {m_3}}}\\ = \frac{{\left( {{m_1} \times \frac{{{l_ \circ }}}{2}} \right) + \left( {8{m_1} \times 2{l_ \circ }} \right) + \left( {27{m_1} \times \frac{9}{2}{l_ \circ }} \right)}}{{{m_1} + 8{m_1} + 27{m_1}}}\\ = \frac{{\frac{{{m_1}{l_ \circ }}}{2}\left( {1 + 32 + 243} \right)}}{{36{m_1}}}\\ = 3.83{l_ \circ }\end{array}\)

Hence, the center of mass of the cube system is at \(3.83{l_ \circ }\).