THE EQUILIBRIUM OP BODIES IN CONTACT. 471 



4. The Buttress. 



If e be taken = the trapezoidal mass will assume the position of 

 the ordinary buttress, (Fig. 5), whose line of resistance is determined by the 

 equation 



I as'jtan-g, -tan^a,} + iq^Hana, + s{2:Psin<l) + 1 a'} +2 ± Pk cos (l> 



^ is'^ Hart n. — tan r,A -i- /y» j. V P'nr.o rh ...(,l^j. 



s 



t' jtanai - tanaa} + as. + iPcos^ 



And its greatest possible height by the least root of the equation 



-3 , 3« ~2 + fi ^-Psin <]y - tan a, 2Pcos<t) — ^ a\ 



"' tan a, — tan a^ ' ~' \ tan «, - tan a.J^ | tan a. - 2 tan ai | ■"' 



6 {2 + PAcos'P - «2Pcos<I)| , , 



^tana, — tanoj} Jtana^ — gtanoij ^ 



The best dimensions of the buttress would seem to be tliose which 

 bring the line of pressure to the center of the base. These may be de- 

 termined by assuming in equation 12, y = i {« + Saltan a, + tan a^)\, 

 whence, 



a;l|tan-a|-tan^a2} + 3as|Jtanai+tanaa}-6x3{22Psin$-(tan, + tana2)2Pcos<l>5 



- 6 {22 + PAcos<l> - a2Pcos<l)} = (14). 



To determine the line of pressure in the buttress, assuming G = 

 in equation (2), Ave have 



_ 2Psin 1> + ^ /(y. - y,) rfC + 2 ± Pk cos a> 

 •' 2Pcost> + /(y. ^ y,)dC '-^^'*- 



Also assuming P, k, <t> not to be functions of C, and taking = 0, 

 y, = a + Ctana,, y, = Ctan«,, we have, by equation (5), 



C _ 2y - a 



tana, + tan a. ^ '' 



Performing the integrations indicated in equation (15), substituting 

 this last value of C, and reducing, 



