262 MECHANICS 



center of gravity of a triangle lies on a line drawn from 

 a vertex to the middle point of the opposite side, and at a 

 distance from that side equal to one-third the length of the 

 line; or it is at the intersection of lines drawn from the ver- 

 texes to the middle points of the opposite sides. 



To find the center of gravity of a trapezoid, Fig. 4, lay off 

 BF = DC and DE = AB; the center of gravity is at the inter- 

 section of EF with Mi Mz, the line joining the middle points 

 of the parallel sides. GM\ can also be determined by the 

 formula m(bi+2bz) 



The center of gravity of any quadrilateral may be determined 

 as follows: First divide it, with a diagonal, into two triangles 

 and join with a straight line the centers of gravity of the two 

 triangles; then, with the second diagonal, divide the figure into 

 two other triangles and join the centers of gravity of these 

 triangles with a straight line. The center of gravity of the 

 quadrilateral is at the intersection of the lines joining the 

 centers of gravity of the two sets of triangles. 



For an arc of a circle, the center of gravity lies on the radius 

 drawn to the middle point of the arc (an axis of symmetry) 

 and at a distance from the center equal to the length of the 

 chord multiplied by the radius and divided by the length of the 

 arc. 2r 



For a semicircle, the distance from the center = = .6366 r, 

 when r = the radius. 



For the area included in a half circle, the distance of the 

 center of gravity from the center is 

 4r 



= .4244r 

 3* 



For a circular sector, the distance of the center of gravity 

 from the center equals two-thirds of the length of the 

 chord multiplied by the radius and divided by the length 

 of the arc. 



For a circular segment, let A be its area and C the length 

 of its chord; then the distance of the center of gravity from 



C 3 



the center of the circle is equal to - . 

 12 A 



