HORIZONTAL SECTIONS SUSTAINING EQUAL FLUID PRESSURES. 55 



quantities may be omitted in the equation, and then the element of 

 comparison, or the n th part of the total pressure becomes 



_bl* 

 p -^~n (30). 



Now, according to the principles of Plane Trigonometry, the lines 

 vz, uq, tp, so and rn, are respectively as below, viz. 



$ nz G o sin.^>, and 3' zz G n sin.0 ; 



but the lines GZ, G^, Gp, GO and GW, when expressed in terms of the 

 respective lengths, are as follows, viz. 



G2 = |v; Gq=:v-\-%w; Gp v -f- w -f- J#; 

 Gozzv-j-w-|- a; 4" iy> an d ftwss:* +'IIP + * + y + J; 



therefore, by substitution, the above values of the vertical depths of 

 the respective centres of gravity, become 



d \v sin.< ; d' zz: (v 4- Jw;) sin.^ ; d w zz (w + w 4- Jar) sin.0 ; 

 5 zz (v 4- w 4- a? + Jy) sin.^, and ^ zz: (v 4- w -f x 4- 2/ + |z) sin.0. 



Consequently, by throwing out the common factor sin.0 and 

 neglecting the specific gravity of the fluid, the value of jo, or the 

 pressure sustained by each of the parts, may be expressed as follows, 

 viz. 



The pressure on the part abme, is p~%bv*, (1). 



- em fl y isp~bw (v -\- Jw), (2). 



- If kg, isp=ibx(v-{-w+^, (3). 



- gkhi, isp = by(v-\-w + x+ jy), (4). 



- ihcd, isp = bz (v + w -{- x+y+z).(5). 



Now, according to the conditions of the problem, all these expres- 

 sions for the value of p, are equal to one another, and each of them is 

 equal to the element of comparison, as given in the equation (30) ; 

 hence, from the first of the above equations, or values of p, we have 



or by expunging the common factor, |6, we obtain 



*< = -?!; 

 n 



therefore, by extracting the square root, we have 



