# Center of mass of an equilateral triangle (Kleppner)

geoffrey159

## Homework Statement

Find the center of mass of an equilateral triangle with side ##a##

## Homework Equations

## \vec R = \frac{1}{M} \int \vec r \ dm ##

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

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

## 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?

Homework Helper
Gold Member
2022 Award
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.

geoffrey159
I did not think about that! Thank you!

duarthiago
##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}}##

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 $\int_0^{\sqrt{3}(a-x)} x \ dy$?

Staff Emeritus
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 $\int_0^{\sqrt{3}(a-x)} x \ dy$?
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.