# Center of mass of an equilateral triangle (Kleppner)

1. Dec 8, 2014

### geoffrey159

1. The problem statement, all variables and given/known data
Find the center of mass of an equilateral triangle with side $a$

2. Relevant equations
$\vec R = \frac{1}{M} \int \vec r \ dm$

$dm = \frac{M}{A} dx dy$

$A = \frac{\sqrt{3}}{4}a^2$

3. The attempt at a solution

I set a pair of orthogonal axis $(\vec x,\vec y)$ so that one side of the triangle lies on the $x$ axis, and one vertex is at the origin.
I find the following position for the center of mass:
$R_x = \frac{1}{A} (\int_0^{\frac{a}{2}}\int_0^{\sqrt{3}x} x \ dy\ dx + \int_\frac{a}{2}^{a}\int_0^{\sqrt{3}(a-x)} x \ dy\ dx ) = \frac{a}{2}$
$R_y = \frac{1}{A} (\int_0^{\frac{a}{2}}\int_0^{\sqrt{3}x} y \ dy\ dx + \int_\frac{a}{2}^{a}\int_0^{\sqrt{3}(a-x)} y \ dy\ dx ) = \frac{a}{2\sqrt{3}}$

Do you think it is correct?

2. Dec 8, 2014

### PeroK

That's the correct answer. It would have been easier to put the apex of the triangle on the y-axis: then, by symmetry, R_x = 0.

3. Dec 8, 2014

### geoffrey159

I did not think about that! Thank you!

4. Mar 19, 2015

### duarthiago

Please let me ask: I think I understood why the upper term from integration interval is equal to $x\sqrt{3}$, is the height in function of x, but what is the idea behind the upper term from $\int_0^{\sqrt{3}(a-x)} x \ dy$?

5. Mar 19, 2015

### SteamKing

Staff Emeritus
What are the equations of the two lines, which intersect at the apex, forming the sides of this triangle?

6. Mar 20, 2015

### duarthiago

Oh, of course, it would be something like a piecewise function where $x \sqrt{3}$ if $0 \leq x \leq \frac{a}{2}$ and $a \sqrt{3} - x \sqrt{3}$ if $\frac{a}{2} < x \leq a$. Thank you for your answer.