30 CENTRE OF GRAVITY. [CHAP. m. 



polygon by laying them off one after the other. Choose a pole 

 and draw lines from it to the beginning and end of each force. 

 These lines will give the sides of the funicular or equilibrium 

 polygon. Anywhere in the plane of the figure, draw a line 

 parallel to the first of these pole lines (S ). Produce it to inter- 

 section with the first force (P^, prolonged if necessary. From 

 this intersection draw a parallel to the second pole line (S x ), and 

 produce to intersection with second force (P 2 ). So on to last 

 pole line, which produce to intersection with first pole line. 

 Through this point the resultant must pass, and of course it 

 must be parallel to the forces. 



Now suppose the parallel forces all revolved say 90, the 

 points of application remaining the same. Evidently the new 

 force polygon will be at right angles to the first, as also the 

 new pole lines, each to each. It is unnecessary then to form 

 the new force polygon. The directions of the new pole lines 

 are given by the old, and this is all that is needed. 



Anywhere then in the plane of the figure, draw a line (S' ) 

 perpendicular to the first pole line (S ) previously drawn, and 

 prolong to intersection with new direction of first force (P/). 

 Through this point draw a perpendicular (S/) to second pole 

 line, to intersection with new direction of second force (P 2 ') 

 and so on. We thus find a point for new resultant, parallel to 

 new force direction. Prolong this resultant to intersection 

 with first and the centre of gravity is determined. 



[NOTE. If the area given has an axis of symmetry, that can 

 of course be taken as one resultant, and it is then only necessary 

 to make one construction in order to find the other.] 



The given area of irregular outline must, as remarked above, 

 be divided by parallel sections into areas so small that the out- 

 lines of these areas may be considered as practically straight 

 lines. The forces are then taken as acting at the centres of 

 gravity of these areas. This division will give us generally a 

 number of triangles and trapezoids. 



It is therefore desirable to reduce graphically to a common 

 base the area of these triangles and trapezoids, and for this pur- 

 pose thf following principles will prove of service : 



32. Reduction of Triangle to equivalent Rectangle of 

 given Rae. Let b be the base and h the height. Then area 



= . Take a as the given reduction base, and let x represent 



