34 SECTIONS OF RECTANGULAR PARALLELOGRAMS 



p \ b(l a;) 8 s sin.0, andprrr bx(l \ x} s sin.0, 

 but by the conditions of the problem, these pressures are equal ; 

 hence we get 



J(Z *) 2 = a? (/ I *), 

 and this, by reduction, becomes 



** 2/3? = i/ 2 ; 

 consequently, the root of this equation is 



and more elegantly, by collecting the terms, it becomes 

 x = l(l \ V~)' 



51. This is manifestly the same result as would arise, by subtract- 

 ing the value of x in equation (16), from the whole length of the 

 parallelogram ; and the rule for performing the operation is simply 

 as follows : 



RULE. From unity subtract one half the square root o/2 ; 

 then multiply the remainder by the length of the parallelo- 

 gram, and the product will be the distance of the point 

 required from the lower extremity of the immersed dimension. 



Therefore, by taking the length of the parallelogram, as proposed 

 in the preceding example, we shall have for the distance from its 

 lower extremity, through which the line of division passes, 

 x = 28 (1 i ^/2) = 8.2012 feet. 



COROL. It is manifest from the equations (16) and (17), that the 

 solution is wholly independent of the breadth of the parallelogram, its 

 inclination to the horizon, and the specific gravity of the fluid ; these 

 elements, therefore, might have been omitted in the investigation ; 

 but since it became necessary to express the pressure either absolutely 

 or relatively, we thought it better to exhibit the several quantities, of 

 which the measure of the pressure is constituted. 



PROBLEM VII. 



52. A given rectangular parallelogram is immersed in a fluid, 

 in such a manner, that one side is coincident with the surface, 

 and its plane tending obliquely downwards at a given inclination 

 to the horizon : 



It is required to draw a straight line parallel to one of the 

 diagonals, so that the pressures on the parts into which the 

 parallelogram is divided, may be equal to one another. 



