Let AZ, DX parallel forces of all the points in A and D. That these may be opposed, there should be equal and contrary forces at these points, viz. AG, and DK. These must arise from the ac tions of the points C and B, according- to the right lines AC and AB on A, and on" D according to DC and DB. Having drawn GI, GH parallel to BA, AC, it is plain that the force AG must be com posed of and AH, of which the first repels any point in A from C, and the second attracts it to B. On account therefore of the equality of action and reaction, the point C will be repelled from A, and B will be attracted : in like manner C will be repelled from D, and B attracted. The point C therefore has two forces, one in the direction AC, and equal to IA drawn into A ; the other equal to DM into D, and in the direction CD : in like manner, B is af fected by the two attractions HA x A, and LD x D. The force resulting at B ought to be equal, and op posite to the resulting force at C. It has therefore the direction BC, when the point C is within the an gle ABC, and the reverse when without it : and to produce the equivalent reaction in CD, we must give C the two opposite forces, equal also to HA X A and LD x D. Wherefore • ,
The point of A has two forces, AI, AH.
The point of D has two forces, DM, DL.
The point of B has two forces, A x AH, D x LD. And C four, A x IA, D x MD, A X HA, D x LD.
Now let the line BC express the magnitude of the force compounded of CN and CR parallel to DB, AB. BN and BR will express the magnitude of these forces as well as their directions, and therefore RC, NC, equal and parallel to them, will express the third and fourth forces of the point C. Produce AC and DC till they meet, in T and 0, the lines RT,• NO drawn parallel to -VF, GZ, or KX, and drop the perpendiculars AF, DE, RS, NQ.
Since IAG, CTR are similar, having their sides parallel, and also CON and MDK, therefore as IG or AH, to CR or BN or A x AH, (that is to say, as 1 is to A,) so is AG to TR, and AI to TC. TR is therefore equal to GA, (or AZ,) drawn into A, and CT=IA x A. The former consequently ex presses the sum of the forces AZ of all the points of A ; the latter the first part of the force of the point C, viz. A x IA. For the same reason, NO will ex press the sum of all the forces DX of all the points to D, and OC the second force of the point C, viz.