EQUAL FLUID PRESSURES. 47 



two equations are derived, as well as the specified conditions of the 

 problems, are nearly similar, the difference consisting simply in the 

 position of the dividing line, it being parallel to the diagonal of the 

 parallelogram in the one case, and oblique to it in the other. 



62. Let the quantity 2JZ 2 be added to both sides of the preceding 

 equation, and we shall obtain 



from which, by extracting the square root, we get 



x i$i = + \i V3"; 



therefore, by transposition, we have 



(24). 



The practical rule by which the point of division is to be determined, 

 may be expressed as follows : 



RULE. Multiply the difference between 3 and the square 

 root of 3, by half the length of that side of the parallelogram 

 in which the line of division terminates, and the product will 

 'be the distance of the required point from the lower extremity 

 of the given length. 



63. EXAMPLE 11. Let the numerical data remain precisely as in 

 the preceding case; from what point in the length of the paral- 

 lelogram, must a straight line be drawn to the opposite lower angle, 

 so that the parallelogram may be divided into two parts sustaining 

 equal pressures ; and moreover, if a straight line be drawn from the 

 same point, to the opposite upper angle, what will be the pressure on 

 the triangle thus cut off? 



Here, by proceeding as directed in the above rule, we have 



x 21 (3 V 3) = 26.628 feet. 



In order to find the pressure on the triangle a bf cut off by the 

 line bf, we have afl x, and ae rp \(l x)\ conse- 

 quently, vp^n^(l x) sin.0, where it must be observed, that ep 

 and vp are respectively parallel to a b and sm. 



Now, the pressure perpendicular to the surface of the triangle a bf, 

 is found by multiplying its area into vp, the perpendicular depth of 

 its centre of gravity ; hence, we have 



p' ^bs (I #) 2 sin.^. ; 

 but the value of x, according to equation (24), is 



x = .634 I ; 



consequently, by substitution, we have 



p' = %b Z 2 s (1 .634)* sin.0, from which, by substituting the 

 several numerical values, we obtain 



>' = 4 X 42 s X 62.5 X .366* X .78801 = 46551. 35 Ibs. 



