35-] 



DETERMINATION OF CENTROIDS. 



vertices together with a fifth particle of equal but negative 

 mass, placed at the intersection of the diagonals. 



35, In the particular case of a homogeneous trapezoid (Fig. 7), 

 it may be noticed that the figure can be divided into rectilinear 

 elements by lines drawn parallel to the parallel sides of the 

 trapezoid. Every such element has its centroid at its middle 

 point ; the locus of all these points is the so-called median ; and 

 the centroid G of the trapezoid must lie on this median, i.e. on 

 the line joining the middle points E, F of the parallel sides. 



To find the ratio in which G divides the length EF> we use 

 again the method of taking moments. We divide the trapezoid 



into two triangles by the diagonal BC and remember that the 

 distance of the centroid of a triangle from its base is equal to 

 one-third of its height ; then taking moments with respect to the 

 two parallel sides AB = a, CD=b, denoting the height of the 

 trapezoid by h, and the distances of G from a and b by y 

 we obtain 



Dividing, we find 



This gives the following construction : Make AE f = b on the 

 prolongation of a, and DF 1 =a on the prolongation of b, in the 

 opposite sense ; then E'F' will intersect EF in G. 



PART II 2 



