### Video Transcript

A triangle π΄π΅πΆ, where π΄π΅
equals 33 centimeters, π΅πΆ equals 44 centimeters, πΆπ΄ equals 55 centimeters, and
π· and πΈ are on the midpoints of line segment π΄π΅ and line segment π΄πΆ,
respectively, is located in the first quadrant of a Cartesian plane such that π΅ is
at the origin and point πΆ is on the π₯-axis. Three equal masses are placed at
points π΅, π·, and πΈ. Determine the coordinates of the
center of gravity of the system given that the scales of the axes are defined such
that each unit represents one centimeter of distance.

In the triangle π΄π΅πΆ below we can
see that there are three masses labeled with the letter π. In order to work out the center of
gravity or center of mass of this system, weβll need to find the coordinates or
position vectors of each of these three masses. We can use the given lengths of the
line segments to help us along with the fact that point π΅ is at the origin. It will have the coordinates zero,
zero. And we can also give this as a
position vector. The position vector of point π΅ is
zero π’ plus zero π£.

We are also given that π΄π΅ is
equal to 33 centimeters. And the point π·, we are told, is
at the midpoint Of line segment π΄π΅. Point π· will therefore be at a
distance of 33 over two centimeters away from π΅. We could write the coordinates of
point π· as zero, 33 over two, which as a position vector is zero π’ plus 33 over
two π£.

We also need to find the position
vector of this mass at point πΈ. This vector can be written as
vector π©π, and it will be equal to vector π©π plus a half of vector ππ¨. The vector ππ¨ is also equivalent
to vector ππ© plus vector π©π¨. We can then substitute this into
the equation. Then since vector ππ© is equal to
negative vector π©π, the right-hand side simplifies to vector π©π plus one-half
times negative vector π©π plus vector π©π¨.

Distributing one-half across the
parentheses, the right-hand side of this equation becomes vector π©π minus one-half
vector π©π plus a half vector π©π¨. This gives us that vector π©π is
equal to one-half of vector π©π plus vector π©π¨. We can then work out both of these
vectors π©π and π©π¨ in terms of their horizontal and vertical components.

Remember, we were told that π¨π© is
33 centimeters, and so point π΄ is at a position vector of zero π’ plus 33π£. This will be the vector π©π¨. In the same way, π΅πΆ is 44
centimeters, so vector π©π can be given as 44π’ plus zero π£. Now when we add 44π’ plus zero π£
and zero π’ plus 33π£, we get 44π’ plus 33π£. And so we have done all this
working to find that this vector π©π is equal to 44 over two π’ plus 33 over two
π£. We can fill that onto the diagram
and clear some space for the next part.

In order to find the center of
gravity or center of mass of this system, we can apply this formula that 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, where π is the total mass. π sub π is the mass of object
π. And vector π« sub π is the
position vector of object π.

Notice that this is different than
simply finding the geometric center of the masses at points π΅, π·, and πΈ. This is because we need to take
account of the mass of each object. And so although this formula can
look quite complicated, what weβre really doing is multiplying each mass by its
position vector, adding up those products, and then multiplying by one over the
total mass.

So letβs see how we can apply this
formula in this context. The first thing we can calculate
here is π, the total mass. And since we have three equal
weights of πβ, then the total mass can be written as three π. The first mass times position
vector we can take is that at point π΅, which will have a mass of π multiplied by a
position vector of zero π’ plus zero π£. We can then add a second mass of
mass π times zero π’ plus 33 over two π£. Finally, we add on the third mass
times the position vector, which is π times 44 over two π’ plus 33 over two π£.

Notice how when we simplify this by
expanding the parentheses, the terms that have zero π’ or zero π£ will simply be
equal to zero. Before we carry out the next stage
of simplifying, which is distributing one over three π across the parentheses, we
might notice that each of the terms within the parentheses also has a variable
π. So these will cancel.

And so one-third multiplied by 33
over two π£ is 33 over six π£ or 11 over two π£. We then add one-third times 44 over
two π’, which is 44 over six π’ or 22 over three π’. Finally, we have one-third times 33
over two π£, which is 33 over six simplifying to 11 over two π£. We can write this as 22 over three
π’ plus 11π£. And so we have find the position
vector of the center of gravity, but we need to give this as a set of
coordinates. And so we can give the answer that
the center of gravity of this system of masses is 22 over three, 11.

Finally, itβs always worth checking
if the answer seems sensible. And if weβve already drawn a
diagram, this can be an easy check to make. If we draw on approximately where
the coordinates 22 over three, 11 would be, they would be here. And this does seem like a sensible
position for the center of mass. A point far outside the system of
masses, particularly when all the masses are equal, is unlikely to be correct. So here we can simply give the
answer as the coordinates 22 over three, 11.