Center of mass of an inclined triangle

1. Apr 21, 2015

Karol

1. The problem statement, all variables and given/known data
Where's the COM

2. Relevant equations
The COM of a right triangle is a third of an edge apart of the right angle vertex

3. The attempt at a solution
Edge AC: $\frac{50}{\cos 20^0}=53.2$
Two thirds of edge AB: $\frac{53.2\cdot 30^0\cdot 2}{3}=30.7$
One third of edge BC: $\frac{53.2\cdot \sin 30^0}{3}=8.9$
Edge AO: $\sqrt {8.9^2+30.7^2}=32$
$$\cos \alpha=\frac{30.7}{32}\rightarrow \alpha=16.1^0$$
$$\beta=(30^0-\alpha)+20^0=33.9^0$$
Now i refer to drawing B:
$$x_{COM}=26.5 \surd,\ y_{COM}=17.8 \otimes$$

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2. Apr 21, 2015

paisiello2

Are you trying to say that the value for xCOM is correct but yCOM is wrong?

yCOM is suppose to be = 14.4?

Now why aren't you taking the intersection of the medians to determine the point O which is the COM?

Last edited: Apr 21, 2015
3. Apr 22, 2015

ehild

Your result is correct, but it would be easier to use the formula in the link, after having determined the coordinates of the vertexes.

http://www.mathopenref.com/coordcentroid.html

4. Apr 22, 2015

Thanks