Mr. Knight on the Attraction of such Solids 
Prop. 14. 
Figure 6 representing, in every respect, the same as in 
Prop . 8, it is required to find the action of the pyramid, on the 
point p', in the produced axis. 
As this solid may be divided into others, situated, with re- 
spect to p', like that in Prop. 12, the problem is solved by what 
was shewn there. 
* 
If the attracted point was not in the line p'pm, perpendicular 
to the base, but in some other line Trpp, passing through the 
vertex, and meeting the base in ^ ; draw lines from ^ to all 
the angles of the base, and the solid will be divided in such 
pyramids as were treated of in Prop. 13. 
Prop. 15. 
Let pumv, fig. 7, be any triangular pyramid whatever, and 
let it be any how cut by a plane, whose intersection with the 
pyramid is the triangle u( 3 y ; it is required to determine the 
action of the portion q>( 3 yvu m ( which is cut off' by the plane ) 
on a point at p. 
The attraction, of the solid in question, is the difference of 
the actions of the pyramids pumv and p ufiy, which actions are 
found by Prop. 9. 
Prop , 16, or general Problem 
To find the action of any solid, bounded by planes, on a 
point either within or without it. 
It is plain, that by drawing lines, from the attracted point 
through the solid, this may always be divided, either into such 
pyramids (with the point at the vertex) as were considered in 
