# 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

### Karol

Thanks

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