CuETis — Pressure of Earth against a Retaining Wall. 15 



the magnitude of the pressure for any finite rectangular portion of 

 the wall lying between the top and any horizontal line, that through 

 C, suppose, varies as the square of AC. The force introduced, 

 therefore, by any small given increase in the depth of this hori- 

 zontal line varies as A C, its distance from the surface ; the force, 

 therefore, follows the law of fluid pressure, and the centre of force 

 is found by the same law, and therefore is the point 8, which 

 divides AO, so that AS = 2S0. 



One case is worthy of consideration, on account of the sim- 

 plicity of the solution — that in which the friction of the earth 

 against the wall is neglected, the wall is vertical, and carries a 

 surcharge of earth, the surface of which, except in the vicinity of 

 the wall, is horizontal. 



In this case (fig. 5], if we take CE = CZ, and erect ED 



Fig. 5. 



perpendicular to CB, the line CD cuts off the area CED = area 

 CAZD, and thus solves the problem. To prove this, bisect EB 

 in iV, and the problem is then reduced to showing that ACDJV = 1 

 trapezium CAZB, which can be proved thus : — 



Draw ZQ perpendicular to CB, and produce CA and BZ to 

 meet in L ; then 



CE' + EB' + 2CE.EB=CB'= CZ' -^ ZB' + 2LZ . ZB; 



.•.a.&CE=CZ, EB' + 2CE.EB = ZB' + 2LZ.ZB; 



