October 24, 2002

Julia Howland
Franklin & Marshall College
Lancaster, PA 17604

F.A.R.
2468 Gowa Way
Shoe Chew City, PU 11248

Dear G. Olson Overby-Fitzpatrick,

I was very pleased to receive your letter requesting help with the packing of a train crate to create the lowest possible center of mass. After a fair amount of work, I determined that the lowest possible center of mass occurs when Mr. Moss's papier-mâché creations are stacked two feet high. This, in turn, makes the center of mass two feet. I hope to explain this as clearly as possible to both you and Mr. Moss.

The center of mass is defined by Dr. Crannell as the point at which all material above a certain height weights the same as all the material below this height. To determine this height, we must find a way to measure the weight of the crate and the weight of the papier-mâché and divide it evenly. To start, we must determine the density of the empty crate by dividing the weight of the empty crate by its surface area. The surface area can be calculated by adding the area of the six sides of the crate

Dividing the given weight of the empty crate, 736 pounds, by the surface area, 736 ft2 gives us the density of the train crate to be 1 lb/ ft2. To determine this, I had to make the large assumption that all sides were made of the same material. While this is probably not the case, you failed to provide sufficient information and I was left with no other choice.

In packing the papier-mâché creations, we have two choices. We can either pack the papier-mâché so that its height in feet (which I will now call H) is below the center of mass, also in feet because the center of mass is a height (now called C) or, we can pack them so that their height is above the center of mass. I will explore both these options in order to find the lowest possible center of mass.

We will first examine what happens when we pack the papier-mâché lower than the center of mass, or in terms of our variables, make H less than C. The diagrams below should help you visualize what this will look like. I have divided the train crate into two halves, the half above the center of mass, and the half below. Please keep in mind that these diagrams are not drawn to scale.

Since the center of mass is the point at which the weight of the crate and its contents is cut into two equal parts, we must now determine the weights of both the top half of the crate (now empty) and the bottom half of the crate (partially filled with papier-mâché). The weight in the top half of the crate is made up of the weight of the top of the crate added to the weight of the sides. Weight is determined by multiplying the area by the density of the material. So the weight of the top, plus the weight of the sides,

is the weight of the top half of the train crate. This reduces to

The weight of the bottom half is calculated in much the same manner, except that now we have the weight of the papier-mâché creations to consider. We know that the papier-mâché has a density of 0.65 lb/ ft3 and that its dimensions are 16 ft (length) by 10 ft (width) by H feet (height) as shown in the diagram. Therefore,

is the weight of the bottom half of the crate. This can also be written as

Because we need the weight of the top half to equal the weight of the bottom half, we need to set these two weights equal to each other by saying

This, when reduced, shows that

In other words, the center of mass equals four minus the height of the papier-mâché. This equation is only true when the height of the papier-mâché is less than or equal to the center of mass (H < C).

Now, we must examine what would happen if the papier-mâché is packed above the center of mass, or, in terms of our variables, H > C. The diagrams below, although not to scale, should help you visualize this.

The weight of the top half of the crate can be divided into three parts, the weight of the top of the crate, the weight of the sides of the crate, and the weight of the papier-mâché above the center of mass. The only difference between this situation and the top of the train crate described in the last example is the addition of the weight of the papier-mâché. This weight is

Therefore, the weight of the top half of the crate when H > C is

Similarly, the weight of the bottom half is exactly the same as the weight of the bottom half in the previous example except the weight of the papier-mâché is now

Therefore, the weight of the bottom half of the crate when H > C is

As with the last example, the center of mass is found when the weight of the top half of the crate is set equal to the weight of the bottom half of the crate. Therefore, the center of mass when H > C is

Or, when reduced and solved for C,

Often, the best way to determine the minimum value of a function, or group of functions, is to graph the functions on a set of axis and find the minimum point. Below is a graph of the functions describing both a crate where the papier-mâché is above the center of mass, and a crate where the height is below the center of mass.

While the graph clearly shows the minimum center of mass occurring when the two lines, C₁ and C₂ intersect, we still do not have an exact value for the lowest possible C. The graphic intersection of two lines occurs when the two lines are equal to each other. Thus, the lowest center of mass occurs when

Therefore,

As a result, if you pack the papier-mâché two feet high,

In conclusion, Mr. Overby-FitzPatrick, the lowest center of mass occurs at two feet when Mr. Moss's papier-mâché creations are packed only two feet high. You will have to make four times as many trips to transport the papier-mâché and not violate the terms of your contract. I wish you luck in your future travels.

Sincerely,

Julia Howland

Note to the reader: There are many people responsible for my completion of this paper. Among them is Dr. Crannell, for her definition of "center of mass" and her invaluable help with the solving of this problem. Also, Sam Krass for allowing me to read his paper. The second Buchanan math group, Jason and Ryan, were also very helpful. Thank you to the makers of the Microsoft Word clip art selection. It brings me endless joy. My love and gratitude to all of second Buchanan, but specifically Rachael Holland, for putting up with me for the past couple of weeks as I walk around mumbling about papier-mâché and center of mass. Last but not least, I would like to thank The Beatles for being wonderful.

Here comes the sun na na na na, here comes the sun

Go on to the first writing project second writing project the third writing project

Go back to Crannell's writing page Crannell's home page