2 9 -] 



DETERMINATION OF CENTROIDS. 



a homogeneous segment PQ. If we imagine the mass of every 

 such element concentrated at its middle point, the homogeneous 

 triangle is replaced by its median CO in 

 which the density is proportional to the dis- 

 tance from the vertex C. 



The centroid of a homogeneous triangular 

 area lies therefore on the median at two- 

 thirds of its length from the vertex ; as this 

 holds for each median, the intersection of the Al 

 three medians is the centroid (see Art. 32). 



(c) For n 2, we have x\l. This gives the position of the 

 centroid of a homogeneous pyramid or cone, by reasoning pre- 

 cisely similar to that used in (b). 



Thus, to find the centroid of any homogeneous pyramid or 

 cone, join the vertex to the centroid of the area of the base ; 

 the required centroid lies on this line at 

 a distance equal to J of its length from 

 the vertex. 



29. Homogeneous Circular Arc (Fig. 4). 

 Let O be the centre, r the radius of the 

 circle; ACB = s the arc, C its middle 

 point. The centroid G must lie on the 

 bisecting radius OC, since this being a 

 line of symmetry, the sum of the mo- 

 ments of the elements of the arc is =o 

 with respect to this line (Art. 17). To 

 find the distance x=OG, we take mo- 

 ments with respect to the diameter per- 

 With OC as axis of x, we have 



Fig. 4. 



pendicular to OC. 



= r -ds = 



ds cos COP = 



Hence, s x r- c, if c be the length of the chord AB. 



If the angle AOB = 2aoi the arc ^ were given, we might 



