310 OF THE POSITIONS OF EQUILIBRIUM 



* ~ T = ^83 = - 20 ' 4083 



Then, by taking the positive values of x and y respectively, the 

 position of equilibrium indicated by the 

 above results, is represented in the an- 

 nexed diagram, where CE and CH are 

 respectively equal to 20.4083 inches, 

 and consequently, AB the base of the 

 section is parallel to L K the surface of 

 the fluid. 



Bisect the base AB in the point F, and 

 draw the straight lines er, FE and FII ; then because CE is equal to 

 CH, and the angle ECF equal to the angle HCF, it follows, that the 

 line F E is equal to the line F H ; this is one condition that must be 

 satisfied, when the body floats in a state of quiescence ; and another 

 is, that the area of the immersed triangle ECH, is to the area of the 

 whole section ACB, as the fraction 0.53125 is to unity. 



Now, by the property of the equilateral triangle, the area of the 

 section ACB, is expressed by the product of one fourth of the square 

 of its side, drawn into the square root of the number 3, and the same 

 property holding with respect to the area of the triangle ECH; it 

 follows, that in the case of an equilibrium, 



JV3> : i#V3~ : : 0.53125 : 1* 

 or by suppressing the common quantity J\/3, we have 



x 2 : 6 2 : : 0.53125 : 1 ; 



but x z ~ 416.5, and 2 ~ 784; therefore, by substitution, we obtain 

 416.5 :784 : : 0.53125 : 1. 



It is therefore evident, that by the above results, both the condi- 

 tions of equilibrium are satisfied, and consequently, the body floats in 

 a state of equilibrium when placed as represented in the diagram ; 

 that is, with 20.4083 inches of its side immersed, and its base parallel 

 to the plane of floatation. 



393. The adfected quadratic equation a 2 42z:zz 416.5, has 

 obviously two positive roots, each of them less than b the side of the 

 section ; from which we infer, that besides the position of equilibrium 

 above exhibited, the body may have other two, and these will be 

 determined by the resolution of the equation, as follows. 



Complete the square, and we obtain 

 ^ _ 42a; + 21 2 = 416 5 + 441 = 24.5, 



