A (lightweight) pallet has a load of ten identical cases of tomato paste (see Fig. 7–39), each of which is a cube of length \(l\). Find the center of gravity in the horizontal plane, so that the crane operator can pick up the load without tipping it.

FIGURE 7-39

Problem 53.

To find the position of the CM in two dimensions, you have to find both the x and y positions of the CM.To find the center of mass, you have to find the x and y coordinates of the CM for individual cases.

The sides of each cube are l.

You can assume the mass of the cubes to be m as they are the same in size.

Now, the top view of the pallet is shown below:

Here, you can also consider the upper left corner as the origin.

As the cases are symmetrical in shape, the CM of an individual box is at the center of the box.

Therefore, the x coordinate of the first column masses \(\left( {3m,m,m} \right)\) is \(\frac{l}{2}\).

The x coordinate for the second column masses \(\left( {2m,m} \right)\) is \(\frac{{3l}}{2}\).

The x coordinate for the last column mass \(2m\) is \(\frac{{5l}}{2}\).

The y coordinate for the first row masses \(\left( {3m,2m,2m} \right)\) is \(\frac{l}{2}\).

The y coordinate for the second row masses \(\left( {m,m} \right)\) is \(\frac{{3l}}{2}\).

The y coordinate for the third row mass \(\left( m \right)\) is \(\frac{{5l}}{2}\).

Now, the x coordinate of the center of mass of the cases is:

\(\begin{array}{c}{x_{{\rm{CM}}}} = \frac{{\left\{ {\left( {3m + m + m} \right) \times \frac{l}{2}} \right\} + \left\{ {\left( {2m + m} \right) \times \frac{{3l}}{2}} \right\} + \left( {2m \times \frac{{5l}}{2}} \right)}}{{\left( {3m + m + m} \right) + \left( {2m + m} \right) + 2m}}\\ = \frac{{\frac{{5ml}}{2} + \frac{{9ml}}{2} + \frac{{10ml}}{2}}}{{10m}}\\ = 1.2l\end{array}\)

Now, the y coordinate of the center of mass of the cases is:

\(\begin{array}{c}{y_{{\rm{CM}}}} = \frac{{\left\{ {\left( {3m + 2m + 2m} \right) \times \frac{l}{2}} \right\} + \left\{ {\left( {m + m} \right) \times \frac{{3l}}{2}} \right\} + \left( {m \times \frac{{5l}}{2}} \right)}}{{\left( {3m + m + m} \right) + \left( {2m + m} \right) + 2m}}\\ = \frac{{\frac{{7ml}}{2} + \frac{{6ml}}{2} + \frac{{5ml}}{2}}}{{10m}}\\ = 0.9l\end{array}\)

Hence, the center of mass of the mass system is at \(\left( {1.2l,0.9l} \right)\) point relative to the origin at the upper left corner.