### Video Transcript

Suppose three masses of six
kilograms, nine kilograms, and π kilograms are at points five, nine; zero, six; and
negative four, three, respectively. If the center of mass of the three
masses is one, π¦, what are the values of π and π¦?

If we wish, we can begin this
question with a coordinate grid representing the three different masses at the given
coordinates. With the circles representing the
masses, here we have a mass of six kilograms at the point five, nine; a mass of nine
kilograms at the point zero, six; and a mass of π kilograms at negative four,
three. We are also given that the
coordinates of the center of mass of these three masses is one, π¦. We need to work out both the
unknown mass of π kilograms and the value of π¦.

The center of mass of a set of
masses isnβt simply just the geometric center. The center of mass needs to take
into account the mass of the system. And there is a formula which we can
use to help us. This formula tells us that the
position vector of the center of mass, represented as vector π, is equal to one
over π times the sum from π equals one to π of π sub π times vector π« sub
π. π represents the total mass of the
system of masses. π sub π represents the mass of
object π. And vector π« sub π is the
position vector of object π.

Although this formula can look
complicated, what weβre really doing is multiplying every objectβs mass by its
position vector then adding up all these products and multiplying by one over the
total mass. Because this formula uses position
vectors and we have coordinates, then the first thing weβll do is work out these
four coordinates in terms of their position vectors.

Letβs take these in the order we
were given with the coordinate five, nine. It can be written in terms of π’
and π£ as five π’ plus nine π£. Then the second coordinate of zero,
six can be written as zero π’ plus six π£. Since zero π’ is simply zero, then
this position vector can be given as six π£. The third coordinate of negative
four, three can be written as negative four π’ plus three π£. We will also convert the
coordinates one, π¦ into a position vector as one π’ plus π¦π£. This is more simply written as just
π’ plus π¦π£.

So now we have enough information
to substitute these values into the given formula, remembering that we already have
the position vector of the center of mass. When we begin our substitution, π’
plus π¦π£ will be on the left-hand side of the equation. On the right-hand side, we have one
over six plus nine plus π, remembering that thatβs simply the total mass of the
system. This is then multiplied by six
times five π’ plus nine π£, which is the mass times the position vector of this
first object. We add this to the mass times the
position vector of the second mass, so we have nine times six π£. Finally, we add this to the mass of
the third object, which is π times its position vector of negative four π’ plus
three π£.

Once we have this, the next stage
is to simplify on the right-hand side. And so on the right-hand side, we
begin with the fraction of one over 15 plus π. We have then distributed six across
the parentheses, which gives us 30π’ plus 54π£ plus 54π£ minus four ππ’ plus three
ππ£. The next stage of simplifying will
really be like collecting like terms in terms of the horizontal and vertical
components π’ and π£.

The completed equation is then π’
plus π¦π£ equals one over 15 plus π times 30 minus four π π’ plus 108 plus three
π π£. Notice in this equation we still
have two unknowns; we have the unknown of π¦ and the unknown of π, which appears
twice. What we will do next is to
recognize that the horizontal and vertical components on each side of the equation
must be equal. In order to help us do this, letβs
distribute one over 15 plus π across the parentheses.

When we have done this, on the
right-hand side we have 30 minus four π over 15 plus π π’ plus 108 plus three π
over 15 plus π π£. Now we can equate the coefficients
of π’ and the coefficients of π£, remembering that on the left-hand side that π’
will simply be the same as one π’. We can clear some space above this
working to equate the coefficients. We now have two equations: one is
equal to 30 minus four π over 15 plus π and a second equation that π¦ is equal to
108 plus three π over 15 plus π.

Letβs see if we can solve equation
one first. Multiplying both sides by 15 plus
π would give us 15 plus π is equal to 30 minus four π. Then, in either one step or two
steps, we add four π and subtract 15 from both sides, leaving us with five π
equals 15. Dividing both sides by five, we get
that π is equal to three. This is the first answer that we
are required to find. And it means that the unknown mass
of π is three kilograms.

We can also use this to help us
solve the second equation by substituting π equals three into this equation. So we have π¦ is equal to 108 plus
three times three over 15 plus three. Simplifying the numerator, we have
108 plus nine, which is 117. And on the denominator, we have
18. As a decimal, this means that π¦ is
equal to 6.5. Letβs remember that the value of π¦
was the value in the coordinates for the center of mass. If we added this onto the diagram,
the center of mass at one, 6.5 would appear approximately here.

If we have already drawn a diagram,
then adding the center of mass to it can be a useful way to check the value is
sensible. Since we also worked out that the
unknown mass is three kilograms, then we might expect to see the center of mass
slightly to the right of this nine-kilogram mass. And so this is a good visual
confirmation that the center of mass is correct. And so we can give the answer that
π is equal to three and π¦ is equal to 6.5.