CHAP. V.] MOMENT OF ROTATION PARALLEL FORCES. 41 



41. Many other problems will readily occur, which may in a 

 similar manner be solved. The weights may have any position, 

 number and intensities desired ; in any and every case we have 

 only to construct with assumed pole distance the corresponding 

 equilibrium polygon, and we obtain at once the moments at 

 every point. By the use of convenient scales, numerical results 

 may be obtained which may be checked by calculation, and 

 the practical value and accuracy of the method thus demon- 

 strated. 



The above principles will be sufficient for the solution of any 

 such problem which may arise, and we shall therefore content 

 ourselves with the above general indication of the method of 

 procedure, and pass on to the consideration of a few cases 

 where the above needs slight modification, and which, from 

 their practical importance, and the ease with which they may 

 be treated graphically, seem worthy of special notice. 



1ST. BEAM OR AXLE LOAD INCLINED TO AXIS.* [Fig. 25, PI. 6.] 



"We have here simply to draw the " closing line "AC paral- 

 lel to the beam or axle. From d draw d B parallel to the force 

 P, then draw A B in any direction at pleasure, and join B C. 

 We have thus the equilibrium polygon ABC, the ordinates tc 

 which, as d B, parallel to the force P, will give the moments, 

 provided we know the corresponding pole distance. 



But this can easily be found. As we have already seen, the 

 force polygon being given, the equilibrium polygon may be 

 easily constructed. Inversely, the equilibrium polygon being 

 given, the force polygon may be constructed. Thus from A 

 draw A c equal and parallel to P, and then draw c C t parallel 

 to B C. A a and b c are the vertical reactions P x and P 2 ; a b 

 is the horizontal component of the force which must be resisted 

 at one or both of the ends ; and the moments at any point are 

 given by the ordinates parallel to P multiplied by the perpen- 

 dicular distance from C t to A c. If we suppose the force P, a8 

 in the Fig., as causing two opposite vertical forces, instead of 

 acting directly upon the axis, we have only to prolong A B to 

 B! and join BH B-j, and then the ordinates of A B! B^ C parallel 

 to P or A c, multiplied by H (perpendicular distance from C, 

 to A c) will give the moments. 



* See Der Constructeur, Eeuleaux. 



