1818.] Mr. Adams's Solution of a Problem. 351 



And FD + Dffl = F«=- + 



a c x — a ex + a 



c ' 2 c~ ~ ~~ 2 c * 



The area AEDG=EDxDG= c -^^- x a = W, 



c 



The area FDG = |DF xDG = 4 r x-xa = ^ = w. 



8 3 c 2c 



Hence F»x w + Fm x W = — x — H — x 



3 c 2c 2 c c 



Therefore ( 4r — ,r-) a n = " stabilitating force" of the sec- 



V 2 6 cV ° 



tion A E F G, the specific gravity of which being n. 



eft* C* rt3 



The stability of the supported earth A B E = — - — , m, 



being the specific gravity of the earth, and s, the nat. sin. (rad. 1) 

 of the angle A E B (page 258, ibid.) 



Hence the following equation : 



T -T?) an = — •'• I= ;V ^(mc*s* + n), a simple 

 equation. If the quantity 7 —jr-, be reduced on account of fric- 

 tion, as is usual, then (-3 £-J a« = — 5- , and x = -^~ 



V 7 



2msV + 3 n 



If the angle A E B = 45°, and m = n, 



the last equation would become x = — V c 4 + 3 ; and when 

 c as 5, then x = -^ V 28 = -3527 a = base EF, 



'15 ' 



and -3527 a - -2 a = -1527 a = top A G, 

 almost £ a in brick walls. 



If m = 4, « = 5, c = 5, and s 2 = -|-, we then have x = 



§7\/ 



2 m s» c* + 3 n 



= -£r ^23 = -3197 a = EF 



la 



and -3197 a - -2 a = -1197 a = top A G, 

 about £ a in stone walls. 



At page 260 ibid, (using the same numbers as above) 

 The thickness at the top for brick walls is .... -189 a, 



And for stone walls *159 a, 



which do not agree with the preceding, owing to a mistake in 

 the solution. 



For in completing the square of the equation 



o . 2 a ,2 a* m a* • , • j 



x 4 + — x + — = - x -y, it is printed 



xa+ T x+ 2T == ^ x T + ^' in3tead of 



9 



