EXAMPLES. 65 



Let g and g' be the centres of gravity of the triangles 

 AOH and BOK ; draw M# parallel to AH, and #R and MQ 

 perpendicular to HO. Then 



M</ = ^b tan 0, and OM = %b. 



Therefore, the horizontal distance of the centre of gravity 

 g from the centre 



= OR = OM cos + M.g sin 



= $b cos + \b tan sin ; 



and therefore, for a = RL = 20R, we have 



a = %b cos -f %b tan sin 0. (3) 



Substituting (1), (2), and (3), in (3) of Art. 28, and put- 

 ting S = for indifferent equilibrium, we get 



| 2 tan (%b CQS0 + ^ tan sin 0) 



-- ' 6 S1U t7 . U ^ 



or, [(2 + tan 2 0) J 2 24ey] sin = 0. 



.-. sin = 0, (4) 



and tan = \/24ey 2* 2 . (5) 



The angle = 0, in (4), is applicable to the body when 

 in an upright position, and that given in (5) is applicable 

 to the body when floating in an inclined position, and is 

 possible only when J is = or < 2v3ey. 



COR. Let h = the height EC, and p = the density of 

 the body, the density of the fluid being unity, then we have 



y = hp, and = -(! p), 

 which in (5) gives 



tan = A/12/i 2 (1 - p) p =. (6) 



