as are terminated by Planes , &c. 
H9 
Prop. 1. 
Let rvm, fig. 1, be a triangle, right angled at r, and pm a 
right line, perpendicular to the plane of the triangle, at the 
angular point m ; it is required to find the attraction of the 
triangle, on the point p, both in quantity and direction. 
Conceive a plane to pass through the point p, parallel to the 
plane of the triangle, and, in it, the lines pg, po, respectively 
parallel to rm, rv. The problem will be solved, if we find the 
actions of the triangle, in the directions of the three rectan- 
gular co-ordinates pm, pg, po. 
Draw ks parallel to rv, and put a — pm, b — rm, T= mk, 
t = kq ; then pq = s/ a 2 T x 4- t\ Let r = tang, vmr, then 
ks = r x km. 
The element of the plane at q is T x t, and its action, on p, 
in the direction pq, is by resolving which, and put- 
ting A, B, C for the actions of the triangle, in the directions 
P m > P g, po, we get 
a =ff 
aTt 
B 
TTt 
=ff 
Ttt 
(a*+T*+t*)* ~~ (a*-\-T z +t z p ^ (a z +T z + t z f< 
in all which expressions, we must first take the fluent, with 
respect to t , from t = 0, to t — rT; and afterwards, with re- 
spect to T, from T= 0, to T = b. To begin with A, — a first 
operation gives 
=/ 
arTT 
(a*+r»)(a*+(;i + r»)r») 
1 _ 
z 
which, if we put / 3 * = 1 + r a , will be changed to 
