:KE OF GRAVITY OF A Ti 17 



' 



lo 



i, by 

 I or triangles, 



ce : CE :: Ae : . 

 U 

 .e ce : CE :: be 



forc<*-fo. 

 ry straight line parallel to /?(?. 1 



foiv r;i.-h ,,f t!i- -trip-i Miuiiar t 0sL int., win- h w- m: 



^le to b 11 balance on 



cntre of gravity must be in tho straight In. 



Bisect AC at F and join thin Oti 



, as bci ro of gravity most U 



-o in AE\ and therefore 6* is the centre of grav 



cause CE= CF-AF, there- 



is parallel to AD and ^B 



triaii.:! If, 



: AO : AD, therefor.- I'JJ^^AO. 



v of a triangle, bisect any 

 the opposite angle, and th<> 



of gravity lies a third of the way up this straight 



/ of any plane polygon may be found 

 by dividing it e centre of gr 



by Art 66 deducing the centre of 

 gravity of the whole figure. 



y observe that the crntre of gravity of a triangle 



itre of gravity of three equal particles 



I at the angular points ot ! tho 



of m\ irce equal particles placed at 



iesp-< < 'It ami bisect it at ^ - the 



of gravity of the particles at C and 



le AE*l G so 

 tiiat /.''' may bo to AO as the mass of the one particle at 



that is, as 1 is to 2; ti 

 ree equal particles. 1 

 iously also the centre of gravity of the triangle 



