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Centre of gravity of equilateral triangle of length a? watch

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    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
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    (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: \bar{x}=\frac{1}{3}(x_1+x_2+x_3)  , \ \bar{y}=\frac{1}{3}(y_1+y_2+y_3)  . where the coordinates of the vertices are given by (x_i,y_i).
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    (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?

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    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 .
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    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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