One of the basic useful principles of analyzing distributed forces is
the idea of replacing them with a single, aggregate force
that acts at a single point and is somehow equivalent to the original
distributed force. This may not always be possible, but this technique
has found great use in engineering and science. As a simple example,
suppose we have gravity acting on a solid plate of uniform thickness and
density, but
irregular shape. Finding the equivalent force is really the problem of
finding the point where we could exactly balance the plate. This
balance point is often called the *center of mass* of the body.

For symmetric objects, the balance point is usually easy to find. For example, the balance point of an empty see-saw is the exact center. Similarly, the balance points for rectangles or circles are just the geometrical centers. For non-symmetric objects, the answer is not so clear, but it turns out that there is a fairly simple algorithm involving integrals for determining balance points.

We begin by restricting our attention to thin plates of uniform
density. In Engineering and Science, this type of object is called a
lamina. For mathematical purposes, we assume that the lamina is
bounded by , , , and , with
. Then the book gives the following formulas for the coordinates
of the center of mass.

To see where these formulas come from, we need the following result. Suppose we have a composite body that is made up of two masses and and that we know the centers of mass are and . Then according to the principles of mechanics, the center of mass of the composite body can be determined from the following equations.

This result can be easily generalized to the case where we have a composite body made up of masses to give the following result.

To go from these formulas to the integral formulas presented earlier,
we first partition the interval into subintervals

Then we approximate the lamina on each subinterval with a rectangle by letting be the midpoint of the subinterval and using as the top of the rectangle and as the bottom. This gives a rectangle of width and height . Assuming that the density is 1, we have the following for the mass, , and center of mass, of the rectangle.

Using these relations, we get the following equations.

The right hand sides of these equations should be easy to recognize as Riemann sums, so that when we take limits as goes to infinity, we get the following in our equations for the center of mass of our lamina.

These are easy to rearrange into the equations given in the text.

We end this section with an example, including Maple commands, for computing the center of mass. Suppose you have a lamina bounded by the curves , , for and you want to compute the center of mass. To do this in Maple, we first define the two functions.

> f := x -> x^3-3*x^2-x+3; > g := x -> 3-x;If you are not sure about the relative positions of the two curves, it is a good idea to plot them both.

> plot({g(x),f(x)},x=0..3);Notice that the graph of is above the graph of . This means that you must switch and in the formulas.

Now we are ready to compute the center of mass. Using labels, as shown below, can help you organize your calculations and avoid mistakes. Computing the mass separately also lets you check it. If you get a negative value for the mass, something is wrong and you have to check what you have done. A common mistake is reversing the order of the functions.

> mass := int(g(x)-f(x),x=0..3); > x_bar := int(x*(g(x)-f(x)),x=0..3)/mass; > y_bar := 1/2*int(g(x)^2-f(x)^2,x=0..3)/mass;

- Find the coordinates of the center of mass of the lamina bounded
by the curves
and . Plot both curves over the interval
along with the point
to see if your answer makes sense.
- Find the coordinates of the center of mass of the lamina bounded
by the curves
and . Plot both functions on the same graph over the interval
along with the point
to see if your answer makes sense.
- Find the coordinates of the center of mass of a triangle whose
vertices are located at the points , , and
.

2003-09-30