Composition and Decomposition of Forces. 27 



point A being at any assumed distance from ,) and form upon 

 AI, another parallelogram AEID ; then for the force AI may 

 be substituted two forces represented by AE, AD ; so that for 

 the single force BG, we shall have the three forces BF, AE, 

 AD, the effect of which will be equivalent to that of EG. 



46. We remark here, that, since there is no other condition 

 required for determining the forces AD, AE, except that they 

 be expressed by the sides AD, AE, of the parallelogram ADIE 

 of which AI is the diagonal, which condition may be fulfilled 

 in an infinite number of ways, whether the parallelogram ADIE 

 be in the same plane with the parallelogram FBLG or in any 

 other plane, we can decompose any force whatever BG into as 

 many others as we please, and which shall be in such planes as 

 we please. We shall see hereafter the use thaf may be made 

 of this method of compounding and resolving forces. 



47. From what is above said, it will be perceived that we 

 can require certain forces to pass through certain given points, 

 and even to be of certain determinate magnitudes, to be parallel 

 to certain given lines, in a word, to satisfy certain given condi- 

 tions. For example, if we had a force represented by the line 



AB, and we would substitute two others of which one should Fig. 10. 

 pass through the point D, (in a direction parallel to a line LX, 

 whose position is given,) and which at the same time should be 

 of a certain magnitude LK, that is, such as would cause a given 

 body to describe LK, in the same time in which the force rep- 

 resented by AB would cause this same body to describe the line 

 AB ; the principles above established will enable us to solve the 

 problem. 



Through the point D, we draw 1C parallel to LX, and meet- 

 ing AB produced in some point /; we take 1C = LK, and 

 IE = AB ; then joining CE, we draw through the point / the 

 line IH parallel to CE, and through E the line HE parallel to/C; 

 /Cwill be the force required, and IH v ill be the force which, 

 combined with 1C, would take the place of IE or of AB. 



The solution we have given will always be applicable, except 

 when the line LX is parallel to AB, and we shall see soon what 

 is to be done in this case. 



