# Homework Help: Center of Mass of a Hollow 2D Triangle

1. Apr 3, 2009

### datdo

1. The problem statement, all variables and given/known data

What is the center of mass a hollow 2D right triangle having legs of length A and B. The sides of the triangle are also extremely thin. Assume uniform density.

2. Relevant equations

$$\overline{x}$$=∫xmdx/∫mdx

3. The attempt at a solution

$$\overline{x}=\int\frac{ax^2dx}{b}/\int\frac{axdx}{b}$$<-something like that
my final answer was ($$\frac{a^2}{b}$$,$$\frac{b^2}{a}$$)

File size:
143.6 KB
Views:
136
2. Apr 3, 2009

### tiny-tim

Welcome to PF!

Hi datdo! Welcome to PF!

(i can't see your picture yet, but …)

Wouldn't it be simpler to replace each side by an equal mass at its centre?

3. Apr 4, 2009

### datdo

hmm.... i didn't think of that...
A:(.5A,0)
B:(0,.5B)
hypotenuse:(.5A,.5B)
$$\overline{x}=\frac{A*.5A+B*0+\sqrt{A^2+B^2}*.5A}{A+B+\sqrt{A^2+B^2}}=\frac{.5A^2+\sqrt{A^2+B^2}*.5A}{A+B+\sqrt{A^2+B^2}}$$
$$\overline{y}=\frac{A*0+B*.5B+\sqrt{A^2+B^2}*.5B}{A+B+\sqrt{A^2+B^2}}=\frac{.5B^2+\sqrt{A^2+B^2}*.5B}{A+B+\sqrt{A^2+B^2}}$$

Oh but I had to use integrals. I should have put in the problem.

Last edited: Apr 4, 2009
4. Apr 4, 2009

### tiny-tim

Hi datdo!
yeeees

ok, then for the x-coordinate divide the triangle into "vertical" slices of thickness dx …

then for the bottom side just integrate xdx, and for the slanty side integrate … ?