CHAP. V.] MOMENT OF ROTATION PARALLEL FORCES. 37 



moment due to Vj, must be diminished by that due to P l5 since 

 these forces act in opposite directions, and rotation from left 

 to right upon the left of any point is considered positive. We 

 see at once from the force polygon that P x is resolved into S 

 and Si or into a b and b c. Hence the moment at o due to PI 

 is m- n multiplied by H. The total moment at o is then m o 

 m n = n o, multiplied by H. This is but a special case of the 

 principle of the preceding Article. 



Hence we see that the ordinates to the equilibrium polygon 

 from the closing line a g, are proportional to the total mo- 

 ments / while the ordinate at any point between any two adja- 

 cent sides of this polygon, prolonged, represents the moment at 

 that point of a force acting in the vertical through the inter- 

 section of these two sides. 



[The reader should make the construction, changing the order in which 

 the weights are taken, and thus satisfy himself that the order is a matter 

 of indifference.' As to the direction of the reactions Vi, V 2 , it must be re- 

 membered that a b is to be replaced by Vi and H, hence Vi must be opposed 

 to O 0, the direction obtained by following round in the force polygon the 

 triangle 1 O. Force and distance scales should also be assumed. Thus the 

 ordinates to the equilibrium polygon scaled off say in inches, and multiplied 

 by the number of tons to one inch, and then by the " pole distance " taken 

 to the assumed scale of distance, will give the moments of any point.] 



The resultant of any two or more forces must pass through 

 the intersection of the outer sides of the equilibrium polygon 

 for those forces (Art. 16). Thus, the resultant of P x and P 2 

 must pass through the intersection of a b and c d. Of V x and 

 P 1? through the intersection of a g and b c } ' of P! P 2 and P 3 , 

 through intersection of a b and d e, and so on. In every case 

 the intensity and direction of action of the resultant is given 

 directly by simple inspection of the force polygon. 



Thus from the force polygon we see that the resultants k 2 

 and k 3 of Y! P x P 2 and V x P x P 2 P 3 , act in different directions. 

 Their points of application are at the intersection of cd and d e 

 respectively with a g, or upon either side of d in the equilibrium 

 polygon. At d the ordinate and hence the moment is greatest, 

 and at this point the tangent to the polygon is parallel to a g. 

 If we had a continuous succession of forces ; if a g, for in- 

 stance, were continuously or uniformly loaded ; the equilibri- 

 um polygon would become a curve, and the tangent at d would 

 then coincide with the very short polygon side at that point 



