OF THE POSITIONS OF EQUILIBRIUM. 311 



extract the square root of both sides, and we get 



x 21 = dby 7 2475 = =t 4.95 nearly; 

 consequently, by transposition, we obtain 



x :r21 -f 4.95 == 25.95 inches, and x = 2l 4.95 16.05 inches; 

 and the corresponding values of y are 



Now, the positions of equilibrium supplied by the above values of 

 a; and y, are as exhi- 

 bited in the subjoined 

 diagrams, where LK is 

 the surface of the water, 

 ABC the position of the 

 body corresponding to 

 x equal 25.95 inches, 

 and y equal 16.05 

 inches; abc being the 

 position which the solid 



assumes, when the values of x and y reverse each other ; that is, when 

 y equal 16.05 inches and y equal 25.95 inches. 



Bisect A B and a b in the points F and /, and draw the straight lines 

 FE, FH andyi,yD to meet the surface of the water in the points E, H 

 and i, D, the points in which the plane of floatation intersects the im- 

 mersed sides of the solid ; then are the lines FE, FH andyi,yD equal 

 among themselves, and the areas ECH, ICD, are respectively to the 

 whole areas ABC, a b c as the number 0.53125 to unity. 



PROBLEM LVII. 



394. Suppose that a solid homogeneous body, in the form of 

 a triangular prism, floats upon the surface of a fluid of greater 

 specific gravity than itself, in such a manner, that two of its 

 edges shall fall below the plane of floatation : 



It is required to determine its position, when it has attained 

 a state of perfect quiescence. 



Let ABC represent a section perpendicular to the axis of a solid 

 homogeneous triangular prism, floating in a state of quiescence on a 

 fluid whose horizontal surface is IK; A DEB and DCE being respec- 

 tively the immersed and extant portions. 



