I'm trying find the center of mass ($R$) of a uniform density equilateral triangle without using symmetries to find the x coordinate of the vector $R$, but I can't get the expected $R_x = \frac{a}{2}$, do not matter what I try. Here is my attempt:

With the base of triangle (=$a$) coinciding with the x axis and the origin coinciding with the origin of my coordinate system I know that $h = \frac{a \sqrt{3}}{2}$. My approach is split the triangle in two right triangles and then sum the $x$ coordinate of the center of mass of both of them.

By definition $R_x=\frac{1}{M}\sum_{i}^{N}m_i r_{x_i}$. Then when $\lim_{N \to \infty}$$R_x$ becomes $\frac{1}{M}\int r_{x}dm$.

Let $\frac{M}{A}$ be the mass per unity of area, then $dm=\frac{Mdydx}{A}$. Writing the height $y$ (of the first half of the triangle) in function of $x$ results in $y=\frac{2xh}{a}$, hence $ y=x\sqrt{3}$.

Then, intending add the center of mass of both right triangles, I can write $\frac{1}{M}\int r_{x}dm$ as:

$$\frac{1}{M}\int_{0}^{\frac{a}{2}} \int_{0}^{x\sqrt{3}} x \frac{M}{A}dydx + \frac{1}{M} \int_{\frac{a}{2}}^{a} \int_{x\sqrt{3}}^{0} x \frac{M}{A}dydx $$ $$=\frac{1}{A} {\left ( \int_{0}^{\frac{a}{2}} \int_{0}^{x\sqrt{3}} x dydx + \int_{\frac{a}{2}}^{a} \int_{x\sqrt{3}}^{0} x dydx\right )}.$$

From that point (which I suppose/hope not be wrong, except for the integration intervals) I've tried a plenty of changes and I got anything but $R_x = \frac{a}{2}$. Could someone please explain what is wrong and how to fix it?


1 Answer 1


Check the bounds in the second integral, and plug in values at the extremes. When $x=a$, the far corner of the triangle, you should be integrating $y$ from $0$ to $0$, but the integral runs from $0$ to $a \sqrt{3}$. This is because you used the equation $y = \sqrt 3 x$ for the upper bound of the second half of the triangle, when it only applies to the first half. The second half of the triangle is bounded by a line with negative slope and an $x$ intercept at $a$, namely $y = -\sqrt 3 (x - a)$. Using that as your bound, your second integral would be $$ \int_{\frac a2}^{a}\int_0^{-\sqrt 3 (x - a)} x \, dy\,dx $$ which should provide the correct result.


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