We want to find the centroid, or center of mass,
of a triangle with the vertices given here.
Notice how this would be the triangle
on the coordinate plan.
If we think of this triangle as a thin plate
with uniform density,
the center of mass, or centroid,
would be the point where this triangle would balance.
There's an illustration,
if we take a look at this diagram here.
If this thin plate has uniform density,
this point would be the center of mass
if at this point the plate would balance at that point.
The formula to find the center of mass
is given here at the bottom,
where the center of mass, or centroid,
would be the point X bar comma Y bar,
where the X coordinate is the moment about the Y axis,
divided by M, the total mass of the system
comma the moment about X axis divided by M.
In general, to find M
we multiply the density and the area.
So if K is the density, M would be equal to K
times this integral here.
We should recognize this integral
that we can use to find the area,
on or by two functions.
Here's our formula for the moment about the X axis,
and here's the formula for the moment about the Y axis.
We won't have time to explain
where these formulas for the moments come from right now,
but if you want to pause the video, but if you want to pause
the video, you may want to review this slide
that explains a formula that we use
to find the moment about the X axis,
as well as this slide here
that explains the formula for the moment about the Y axis.
But going back to our example,
the first step will will be to find the mass of our system,
which again is equal to the density times the area,
and we're going to use K for the density
because we don't know the density, but it is uniform.
We can find this area using a geometric formula,
area equals 1/2 base times height.
But let's also set this up as an integral.
To do that though, we have to know the functions
F of X and G of X,
where F of X would be the top function
and G of X would be the bottom function.
Notice how this is a line
where the Y recept is positive five,
and the slope would be negative 5/3.
From this point to this point,
we go down five units and right three, and therefore,
F of X would be equal to negative 5/3 X plus five,
and G of X, this horizontal line here
which is the X axis,
would just be G of X equals zero.
And notice the X values are from zero to three.
And therefore, the mass, M, is equal to K the density
times the integral of F of X minus G of X.
Well, if G of X is zero,
then this difference would just be negative 5/3 X plus five,
integrated with respect to X from zero to three.
So now we'll find the anti-derivative.
So we'd have negative 5/3 times X squared divided by two
plus five X.
Now we'll substitute three for X and then zero.
So when X is three, we'd have negative 5/3
times three squared, that'll be nine over two
plus five times three.
Notice when X is zero these are both zero.
So notice this simplifies.
There is one, three, and three, and three threes and nine.
So we have K times this would be negative 15/2 plus 15.
Of course, that's minus zero.
Well negative 15/2 plus 15 would be 15/2,
so the mass of the system is 15/2 K.
Notice how we could have found this mass much faster
by determining the area of this triangle
and then multiplying by the density K.
Using the geometric formula,
we'd have the area equals 1/2,
the base, which is three,
times the height, which is five,
giving us an area of 15/2,
and therefore the mass, as we found, is 15/2 times K.
Next let's find the moment above the X axis.
We would have the density divided by two
times integral of here we have F of X squared,
so we'll have negative 5/3 X plus five squared
minus G of X squared, which is zero squared,
integrated with respect to X from zero to three.
Here we'll have to square this binomial.
So I have K over two times integral from zero to three of...
We'd have negative 5/3 X times negative 5/3 X,
that's 25 ninths X squared.
Then we'll have negative 25/3 X minus 25/3 X,
which is negative 50/3 X minus 50/3 X, and then plus 25.
Now we'll find the anti-derivative,
so we'll have 25 ninths times X to the third
divided by three, minus 50/3 times X squared
divided by two plus 25X.
Now we'll substitute three and then zero.
So an X is three, we'd have 25 ninths
times three cubed is 27.
So 27/3 minus 50/3 times three squared is nine, 9/2,
plus 25 times three,
and notice that when X is zero, each term is zero.
So we have K divided by two times, now we'll simplify,
there's one, nine and nine, and three nines and 27.
And we also have three over three that simplifies to one.
So we have 25, and notice here we have one, three and three,
and three threes and nine.
So we have -150 divided by two, or minus 75.
And then here we have plus 75.
So the moment about the X axis is equal to 25/2 K.
Next we'll find the moment about the Y axis,
and then we'll find the centroid, or center of mass.
So to find the moment about the Y axis,
we'll use this formula here.
So the moment about the Y axis is equal to K, the density,
times interval of X times, F of X minus G of X,
again, G of X is zero,
so we just have X times negative 5/3 X plus five.
Integrate from zero to three.
Let's go ahead and distribute.
So we have negative 5/3 X squared plus five X.
Now we'll find the anti-derivative.
So we have K times, we'd have negative 5/3
times X cubed over three plus five, or five over one,
times X squared over two.
And now we'll perform our substitutions.
So when X is three, we'd have negative 5/3 times
three cubed would be 27, 27 divided by three would be nine,
so we have nine over one,
plus five over one times three squared is nine, so 9/2,
and with X as zero, both terms are zero.
So we have K times, simplifying, we have negative 15
plus 45/2,
so negative 15 plus 45/2 would be 15/2,
so the moment about the Y axis is equal to 15/2 K.
And finally, we have all the information we need
in order to find the centroid, or center of mass.
The moment about the Y axis divided by M
is 15/2 K divided by 15/2 K,
notice how this just simplifies to one,
and then for the moment about the X axis divided by M,
that would be 25/2 K divided by 15/2 K,
the Ks simplify out.
Okay, multiply top and bottom by two,
then we'll simplify these twos out, which gives us 25/15,
with a common factor of five simplifies to 5/3.
So if we were to pop this point on the coordinate plan,
one 5/3 would be somewhere in here,
this is our centroid, or center of mass.
So if we think of this as a thin plate with uniform density,
the plate would balance at that point.
It's also true if we have a triangle,
the center of mass or centroid
would be the inner section of the three medians,
where the medians are the segments from each vertex
that connect to the midpoint of the opposite side.
So notice how here is one median,
here's another, and here's the third,
where these would be the midpoints of each side.
And notice how the point of intersection is the centroid,
or center of mass.
I hope you found this helpful.