as are terminated by Planes , &c. 
2^5 
We will now consider the attractions of prisms more gene- 
rally. 
Let prVvrm, fig. 8, be a right prism, whose base is the 
triangle vrm right angled at r. It is required to determine its 
attraction on a point at p, in the directions pm, pr , po ; po 
being parallel to rV. 
If we wish to obtain a solution by means of what has been 
already done, we may conceive the solid under consideration 
to be divided into two pyramids ; viz. the pyramid pmrv with 
the triangular base mrv ; and the other pyramid prVvr, whose 
base is the rectangle rv'vr ; the point p being at the common 
vertex of both. 
Put pm — x, mr = pr 7 = x' ; conceive the diagonal vr 7 to 
be drawn, and put tang, rmv == tang. r 7 pv 7 = r; tang. v 7 rr = 
r' ; tang. vrV = r " ; tang, rpm = m ; tang, rpr 7 = ml. 
We get immediately, from propositions 7 and 10 (putting 
A, B, C for the respective actions in the directions pm, pr 7 , po) 
A = X(p ( 1, m, r) — x arc (tang. = y) -f- x'% (1, m' , r 7 ) + 
rd (1, r, r") ; 
B = ( L r) -f x\ p (1, m! , r 7 ) + x'<p (1, r, r") — x' y ; 
C = xfy ( 1, m , r) x'-]) (1, m r') -{- x'x )i, r, r"). 
In finding these values, the first expressions of Prop. 10 
were made use of ; if we take the others, there will result the 
simpler forms 
A = xcp ( 1 , m, r) — x arc (tang. = yj x'^L.{V 1 -fp-f- r) 
Prop. 20 . 
