﻿the Descent of Glaciers by their Weight only. 367 



•*. whole work of internal shear of trapezoid A c 



_TT „ - JL p^-flW-r 8 ) , b(fj*-.j>){**-&) -[ 



- v *-6pL--E=fi) + -j^i) J' 



v^^Wt-y^+vP+Py+H**-?*) (F+PyW)]- (3) 



If we suppose the bottom abed of the trapezoid A c to be di- 

 vided into an infinite number of equal elements, they will have 

 been sheared over different parallel distances as the trapezoid 

 moves into its present position, and the work of the shear of each 

 element will be represented by the shear multiplied by the cor- 

 responding distance, so that the whole work of the shear of this 

 displacement will be represented by the sum of these products. 

 Bat the sum of these products is equal to the sum of the shears 

 (that is, the whole shear of the base of the trapezoid) multiplied 

 by the distance which the centre of gravity of the base has tra- 

 versed in the act of displacement. 



This distance is represented by 



a Py2 _j_ yy + y 2] 

 Jvt—Jl' J1J jf i = ad. 

 6 a(7 + r ; ) 



Also the whole shear of the base 



pa(y+y) . 

 = 2 ' 



.*. work of shear of base 



=/^ 9 =f (7 a +w'+y s ). 



B„t 7 '=|, 



U 2 =gJ(/3 3 +«,3+« a ) (4 ) 



In the same manner it may be shown that work of shear of the 

 side B b c C of the trapezoid 



••• U*=!03 2 + /3y + y *) (5) 



