Video Transcript
A uniform rectangular lamina has a
length of 63 centimeters and a width of 59 centimeters. It is divided into three
equal-sized rectangles along its length. The last of these rectangles has
been folded over so that it lies flat on the middle rectangle as shown. Find the coordinates of the center
of gravity of the lamina in this form.
Remember, a lamina is a
two-dimensional shape of negligible thickness. The lamina in question is
uniform. This means essentially that its
mass is evenly distributed throughout the shape. In fact, for a uniform lamina, the
center of mass of the body is the geometric center of the body.
Now, we have a little problem here
though. This third rectangle is folded over
so that it rests on the second rectangle. This means that the mass of the
shape created is not evenly distributed in the ๐ฅ-direction. It is, however, still evenly
distributed in the ๐ฆ-direction. And so the ๐ฆ-coordinate of our
center of mass will lie exactly halfway along the line segment ๐ด๐ท. This means that the ๐ฆ-coordinate
of the center of mass or center of gravity will be half of 59; itโs 59 over two. But how do we find the
๐ฅ-coordinate of the center of gravity?
Well, if we think about the
rectangular lamina as being separated into three equally sized rectangles of masses
๐ sub one, ๐ sub two, and ๐ sub three, where ๐ฅ sub one, ๐ฅ sub two, and ๐ฅ sub
three are their center of gravities in the ๐ฅ-direction, then the ๐ฅ-coordinate of
our center of gravity will be the sum of ๐ sub one, ๐ฅ sub one; ๐ sub two, ๐ฅ sub
two; and ๐ sub three, ๐ฅ sub three divided by the total mass of the system.
Letโs deal with the first
rectangle, the one furthest to the left. This is a uniform rectangle. And so we know its center of mass
will be at a point exactly halfway along the shape. Dividing 63 by three gives us
21. So the width of each of our
rectangles is 21 centimeters. And so the ๐ฅ-coordinate of the
center of gravity of this first shape is 21 over two. The center of mass of the remaining
two rectangles will lie at exactly the same point. This will be 21 plus 21 over two
units from ๐ด, and thatโs 63 over two. So we have the ๐ฅ-coordinates of
the center of gravity of each of our three rectangles.
But what do we do about their
mass? Since the shape is uniform and
divided into three equally sized rectangles, then we can define the mass of each of
our rectangles as being equal. And in fact we can let that be
equal to one. So the ๐ฅ-coordinate of our center
of gravity will be one times 21 over two plus one times 63 over two plus one times
63 over two. And weโll divide this by the total
mass of our system, which is one plus one plus one; thatโs three.
Note at this point that we couldโve
dealt slightly differently with our second and third rectangles. We could instead have considered
this to be a rectangle double the mass of the first rectangle. This wouldโve still resulted in a
value of two times 63 over two and a total mass of three. So weโll get the same answers
either way. And this gives us a value of 49
over two. Since we measured this value from
point ๐ด, which lies at the origin, we have the coordinates of the center of
gravity. The coordinates of the center of
gravity are 49 over two, 59 over two.
