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The purpose of this lab is to give you practice with using integrals
to determine the centroids of thin plates of uniform thickness and
density, but irregular shape.
In designing mechanisms or structures, one often has to deal with
distributed forces, that is, forces that do not act at a discrete,
finite set of points. The most common example of a distributed force
is the force of gravity, which acts on all parts of any body of
matter. Other examples are pressure in fluids and electrostatic
forces, though there are many others.
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
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 that we know the centers of mass are
. 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
. Assuming that the density is 1, we have the following
for the mass, , and center of mass,
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
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
, , for
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.
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
> 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 for
. Make sure
that you plot both functions on the same graph and check to see if
your center of mass coordinates make sense.
- Sometimes you have to find the center of mass of a composite
lamina, that is, a lamina made up of 2 or more parts. For
definiteness, suppose that you have a composite lamina made of two
parts. Part 1 is bounded by , , and , where . Part 2 is bounded by , , and , where . Show that the coordinates of the center of
mass of the composite lamina satisfy the following equations.
- Find the coordinates of the center of mass of a triangle whose
vertices are located at the points , , and
- Find the center of mass of a composite lamina bounded below by
and above the two curves
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