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.

 

[rwisz]

Hello,

Your approach of drawing an imaginary horizontal line to find the center of mass is a valid method. However, there are a few things to consider in order to ensure the accuracy of your answer.

Firstly, it is important to note that the center of mass of any object is the point at which the mass is evenly distributed, meaning that the sum of the moments of all the masses on one side of the center of mass is equal to the sum of the moments of all the masses on the other side. In the case of a triangle, the center of mass is indeed 2/3 of the way down its height, but this is only true for a triangle with uniform mass distribution. In your given scenario, the mass distribution is not uniform, so the center of mass will be at a different distance from the base.

Secondly, when finding the center of mass, it is important to take into account the position of each mass relative to the chosen reference point. In this case, your reference point is Q, but you have not specified the positions of the masses relative to Q. It is possible that the two 50g masses are at different distances from Q, which would affect the calculation of the center of mass.

Lastly, your calculation of the center of mass distance from Q seems to be missing a few steps. It would be helpful to see your working and calculations in order to determine the accuracy of your answer. Additionally, it would be beneficial to label your diagram with the distances and masses in order to clearly understand your approach.

In summary, your idea of drawing an imaginary horizontal line to find the center of mass is a valid method, but it is important to consider the non-uniform mass distribution and the positions of the masses relative to the reference point. It would also be helpful to provide more information and show your calculations in order to confirm the accuracy of your answer.