CHAP. II.] DIFFERENT POINTS OF APPLICATION. 25 



If then through k we draw a line parallel to S 3 and produce 

 it to d, we have the same point as before, and thus from d, can 

 continue the polygon. 



\_Note that the direction of rotation shows the side ofS^ upon 

 which the point k must fall. S t acts away from a [from 1 to 

 C in ()] hence for rotation as shown by the arrow, g must fall 

 above S u and Si is shifted upwards. I 



24. Order of Forces Immaterial. As in the case of a com- 

 mon point of application, so also here, the order in which the 

 forces are laid off is immaterial. To prove this for two forces 

 is sufficient, as by continued interchange of two and two, we 

 can obtain any desired order. 



Let the two forces be P 4 and P 5 (Fig. 13, PI. 3) existing either 

 alone, or in combination with others preceding and following. 



Taking the forces first in the order P 4 P 5 , we have the equi- 

 librium polygon S 3 S 4 S 5 , (b) giving the point a in the result- 

 ant. Taking them now in reverse order, P g P 4 , we have the 

 polygon S 3 S' 5 S' 4 giving the same point a' in .the resultant. The 

 resultant in the force polygon (a), viz., 5, is of course un- 

 changed in intensity and direction in either case. It is required 

 to prove that in the second case the last string S' 4 is not only 

 parallel to S 5 in the first, but coincides with it. 



This is easy. The resultant of P 4 P 5 goes through &, the in- 

 tersection of S 3 and S 5 . The same resultant in the second case 

 must also pass through the intersection of S 3 and S' 4 . But S 3 is 

 the same in position and direction in both cases. If the second 

 point of intersection does not coincide with a, still it must lie 

 somewhere upon S 8 . Hence as the resultant must pass through 

 both points, it must coincide with this last line ; viz., S 3 . But 

 this is not possible, as the resultant must also pass through d, 

 the point of intersection of the forces, or when these do not 

 intersect must be parallel to them. As therefore S' 4 must be 

 parallel to S 5 (shown by the force polygon), the intersections in 

 each case must coincide, as also the lines S' 4 , S 5 themselves, and 

 the polygon from e on has the same course in either case. 



25. Pole taken upon closing line. We have seen (Art. 

 20) that when any number of forces are in equilibrium both 

 the force and equilibrium polygon must close. There is one 

 exception to this statement. Since the pole may be taken any- 

 where, suppose it taken somewhere upon the line closing the 

 force polygon. This line, as we know, is the resultant, and 



