181] FLOATING BODY. 473 



the body would be acted upon by a couple equal to mg times the 

 horizontal distance between the verticals through G and B, and tending 

 to increase the angular dis- 

 placement. If 1) denote the ^ M 

 length GB which now makes 

 an angle d co with the vertical, 

 this horizontal distance is w 



, and the couple mgbdco. 



\ 



j- 



Ju 



7 _ . j.-, i> _L-^" \ """^-r' 



But this is not the only couple, \ B ) 



for the immersed part is now \^__ ^ 



different from that formerly Fig 151 



immersed by the volume of 



the two wedges of small angle tfco, the wedge of immersion EOE 1 

 and the wedge of emersion DOD'. The buoyancy added by the wedge 

 of immersion and that lost by the wedge of emersion both produce 

 moments in the direction tending to decrease the displacement. These 

 moments must accordingly be subtracted from that previously found, 

 to obtain the whole moment tending to overset the body. 



It is evident that if no vertical force is to be generated, the 

 volumes of the wedges of immersion and emersion must be equal. 

 Since the wedges are infinitely thin we may take for the element 



of volume 



at = zdxdy = yo&dxdy. 



The condition for equal volumes is then 



/ I zdxdy = dG> I I 



47) 



the integral being taken over the area of the plane of flotation. This 

 will be the case if the axis taken through the center of mass of the 

 area of flotation. The moment due to the wedges is 



48) L'=g$ I I j ydx = gQ I j 



sydxdy 



/ / 



y*dxdy = 



where K X is the radius of gyration of the area of flotation about the 

 X-axis. 



In like manner 



49) M' = gg I I I xdt = g$ I I xzdxdy 



= ggdco I I xydxdy. 



