### Video Transcript

The figure shows a uniform square
lamina of side length 18 centimeters. It is divided into nine congruent
squares as shown. Given that the square πΆ was cut out
and stuck onto square π΄, determine the coordinates of the center of gravity of the
resulting lamina.

Letβs begin with a slightly
modified diagram of the situation. We have taken the square πΆ and
pasted it on top of the square π΄. We can imagine this system as three
separate square laminas. The first lamina is the original
large square with side length 18 centimeters, which has a mass of π one and a
center of mass at its geometric center. Note that we are assuming a uniform
gravitational field. So the center of gravity will be
the same as the center of mass in this case. The second lamina is the square πΆ,
which we pasted on top of the square π΄. It has a mass of π two and its
center of mass is located at its geometric center. And the third lamina is the missing
square lamina, which was occupied by the square πΆ. It has the same mass as the square
C, which we just pasted on top of square π΄. But we treat it as having negative
mass, so it has a mass of negative π two located at its geometric center.

We can model these three laminas as
three separate particles located at their centers of mass. And recall that the center of mass
of a system of particles, the position vector π«, is given by the sum of the
individual products of the particlesβ masses π π and their position vectors π« π
all divided by the total mass. The original square had a side
length of 18 centimeters. And the diagram indicates that its
bottom-left corner is at the origin zero, zero. Its geometric center and,
therefore, its center of mass is nine, nine.

Since the original square was
divided into nine congruent squares, each of these smaller squares must have a side
length of 18 over three, so six centimeters. We can deduce then that the
geometric center of the pasted lamina C is three, three and the geometric center of
the hole that it leaves is 15, three. We know that the total mass of the
system Ξ£ π π must be equal to the original mass of the large square π one. Since this large square was equally
divided into the nine smaller squares, each of the smaller squares must have a mass
of one-ninth of the original mass. So, π two equals π one over
nine.

We now have everything we need to
find the center of mass of this system of particles, which is equivalent to finding
the center of gravity of the resulting lamina. Adding together the products of the
masses of the laminas and the coordinates of their centers of mass, we get π one
times nine, nine plus π one over nine times three, three minus π one over nine
times 15, three all divided by the total mass π one. We have a common factor of π one
in all terms in the numerator and denominator. So, these will all cancel. This leaves us with the sum of the
vectors nine, nine plus one-ninth times three, three minus one-ninth times 15,
three. This gives us the position vector,
the center of gravity of the lamina, π« equals 23 over three, nine. This gives us our final answer the
coordinates of the center of gravity of the resulting lamina 23 over three,
nine.