# Center of mass of a triangle

## Homework Statement

Find the center of mass of a triangle with two equal sides of length a. The triangle's third side is length b and it has a uniform mass of M.

## Homework Equations

$R = \frac{1}{M} \int dm \vec{r}$
$dm = \frac{M}{A}$
$A = \frac{1}{2}base*height$

## The Attempt at a Solution

Right now I have my triangle set up with one of the "a" length sides on the x-axis. I'm having trouble defining the y limit in the integral:
$X = \frac{1}{M} \rho \int_{0}^{a} x dx \int dy$​
I know that it has to have length "b" for the triangle I have drawn. I also found that it was equal to tan$\theta_{1}$, but I need it in terms of x to complete the integral. It's that or change variables. I've also been throwing around the idea of doing this in polar coordinates.

I don't see why you are going through all this trouble when you can transform this into a one-dimensional problem.

The triangle's center of mass is obviously down the perpendicular bisector of side b. Therefore, you can just calculate the problem as though all the mass is concentrated on that bisector.

To do that, put the triangle such that b is on the y axis and its perpendicular bisector is on the x axis. You'll end up with two lines beginning at y=b/2 and converging at x=a. These lines can be found in slope-intercept form as y=∓b/(2a)x±b/2. Considering that this simplification requires dictates that the mass is completely concentrated along the x axis, you can just do a single integral in which y=-(b/a)x+b from x=0 to a. (This gives you the mass, if you really wanted to do it by calculus.)

Go back and find the point x=c at which x ∫0c y dx = ∫ca y dx = 1/2 ∫0a y dx.

HallsofIvy