1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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##
    $$\cos \alpha=\frac{30.7}{32}\rightarrow \alpha=16.1^0$$
    $$\beta=(30^0-\alpha)+20^0=33.9^0$$
    Now i refer to drawing B:
    $$x_{COM}=26.5 \surd,\ y_{COM}=17.8 \otimes$$
     

    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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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
     
  5. Apr 22, 2015 #4
    Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Center of mass of an inclined triangle
Loading...