How Does Mass Distribution Affect the Center of Mass in a Triangle?

[jono90one]

Homework Statement

I know traditionally the center of mass of a triangle is 2/3 the way down its height, but i believe it varies here due to the uneven mass distribution.
Here i have to find the center of mass distance from Q

http://img508.imageshack.us/f/scn0001p.jpg/

Here is an image of my working

Homework Equations

N/a

The Attempt at a Solution

See image.
Incase its unlegiable, i come to a conclusion that the that the center of mass is at h/2-(h/2 x 0.2). In other words if you drew a horizontal line through the two 50g's you'd get the center of mass just below that.

Im hopnig my idea of drawing an imaginary horizontal line is a way to approach it seeming the vertices are the same.
By doing this i get an answer (From Q!) of:
h - 2sqrt (3) = sqrt(75)-2sqrt(3) = 3sqrt(3)

Is this correct? Or is my "Horizontal line" idea not valid.

Thanks.

 

[vela]

Your answer isn't correct. For one thing, you're not taking into account the relative masses of the two triangles. Also, it appears you're trying to calculate the location of the center of mass of the second triangle and got 20% of its height. Think about if the bottom mass were 50 g instead of 20 g, so the second triangle is uniform. Does the formula you used give you the right answer?

 

[jono90one]

Ok I'm going to use a method i know works rather than a short cut (I don't know the actual answer):
Here's my working:
http://img600.imageshack.us/f/scn0002n.jpg/
I'm basically creating a coordinate system, then using;
M(x', y')= m1(x1,y1) + m2(x2,y2) + ...

This gives my the center of mass location which comes out as (0, -25/21) [This is cut off at the end]

Hence distance from Q is h/2 + magnitude of this answer (i.e. positive)
= sqrt(75)/2 + 25/21
= 5.52 (3sf)

So the above attempt was wrong as you said.

Is this correct?

 

[vela]

Your method sounds correct, but your final answer isn't. I got 5.36 cm.