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# Centre of gravity of equilateral triangle of length a?

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1. I've calculated the weight is (a^2 sqrt3)/4 but struggling to find the position.

Apparently the answer is that it is a sqrt 3 / 6 from base not sure how to get that
2. (Original post by ErniePicks)
I've calculated the weight is (a^2 sqrt3)/4 but struggling to find the position.

Apparently the answer is that it is a sqrt 3 / 6 from base not sure how to get that
Without more information it is difficult to help, however note that the centre of mass of a uniform triangle is the average of the vector sum of the vertices.
i.e: where the coordinates of the vertices are given by .
3. (Original post by ErniePicks)
I've calculated the weight is (a^2 sqrt3)/4 but struggling to find the position.

Apparently the answer is that it is a sqrt 3 / 6 from base not sure how to get that
You can evaluate x by using Pythagoras' theorem.
Does the diagram below (poorly drawn in MS Paint) help?

4. I did it in a similar way to above, except I drew a straight line from one of the bottom verticies to the centre. Saves the difficult task of dividing by 3 at the end .
5. Ha i was actually doing the exact same question yesterday, my teacher gave me a formula that height of triangle * (1/3) gives the C of G from the base

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