Inclined Plane. 119 



200. If therefore, the body be urged only by two forces, it 

 is necessary; (1.) That the two forces should be in the same 

 plane ; (2.) That this plane should be perpendicular to that on 

 which the body rests ; (3.) That the resultant (which must be 

 always perpendicular to this last plane), should not leave all the 

 points of contact on the same side ; and if one of these forces be 

 gravity, it is necessary, moreover, that this plane should be verti- 

 cal and pass through the centre of gravity of the body. 



201. Let us now see what ratio must exist between two forces 

 which hold a body in equilibrium upon a plane. Let F q, Fp, 



be the directions of these two forces, and AE the intersection ofFig.m 

 the plane of these forces with that upon which the body rests ; 

 having drawn the perpendicular FH upon AE, let us suppose 

 that on this line, as a diagonal, and upon F q, Fp, as sides, that the 

 parallelogram FEDC is constructed. In order that the resultant 

 of the two forces q and p may be directed according to FD orFH, 

 it is necessary that the two forces q and p should be to each 

 other as FC to FE ; and then the two forces p and q, and the 

 pressure which they exert upon the plane, and which I shall 

 represent by p, will be such as to give the proportion 



q : p : p : : FC : FE : FD. 



38. 



202. According to article 40, we have likewise 



q : p : p : : sin EFD : sin CFD : sin EFC. 



203. From the two points A, E, taken arbitrarily in AE, we 

 let fall upon the directions of the two forces q,p, the perpendicu- 

 lars AG, EG. The triangle AEG having its sides perpendicular 

 respectively to those of the triangle FDE, the two triangles will 

 be similar ; hence 



AG : EG : AE : : DE or FC : FE : FD : : q : p : g 



accordingly 



AG : EG : AE : : q : p : g . 



But Trig. 32. 



AG : EG : AE : : sin AEG : sin EAG : sin AGE, 

 therefore 



