Center of Mass of a Hollow 2D Triangle

[datdo]

Homework Statement

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.

Homework Equations

\overline{x}=∫xmdx/∫mdx

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})

 

[tiny-tim]

Welcome to PF!

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.

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?

 

[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.

 

[tiny-tim]

Hi datdo!

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

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 … ?