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Center of mass of an inclined triangle

  1. Apr 21, 2015 #1
    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##
    $$\cos \alpha=\frac{30.7}{32}\rightarrow \alpha=16.1^0$$
    Now i refer to drawing B:
    $$x_{COM}=26.5 \surd,\ y_{COM}=17.8 \otimes$$

    Attached Files:

  2. jcsd
  3. Apr 21, 2015 #2
    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
  4. Apr 22, 2015 #3


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    Homework Helper

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

  5. Apr 22, 2015 #4
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