32 CENTKE OF GRAVITY. [CHAP. m. 



any quadrilateral may be divided into two triangles which may 

 be reduced separately, or into a triangle and trapezoid. 



It is also easy to reduce any quadrilateral to an equivalent 

 tri<in<jlc, which may then be reduced by Art. 32 to an equiva- 

 lent rectangle of given base. 



Thus we reduce the quadrilateral A B C D [Fig. 18 (c)] 

 to an equivalent triangle by drawing C C t parallel to D B to 

 intersection Ci with A B, and joining C t and D. The triangle 

 D B C t is then equal to D B C, and hence the area A D C t is 

 equal to A B C D. The triangle A D C t can now be reduced 

 to an equivalent rectangle of given base by Art. 82. 



The centre of gravity of the quadrilateral may be found as 

 follows : 



Draw the diagonals A C and B D and mark the intersection 

 E. Make A E! = C E and B E 2 D E, also find the centres 

 GI and O 2 of the diagonals A C and B D. Join O 2 E t and Ox 

 Ea ; the intersection S of these two lines is the centre of gravi- 

 ty required. 



The above is sufficient to enable us to find the centre of gravity 

 of any given area of regular or irregular outline. The method 

 may be applied to finding the centre of gravity of a loaded 

 water-wheel (as given in Der Constructeur, Reuleaux, Art. 47), 

 and many similar problems. The reader will have no difficul- 

 ty, following the general method indicated in Art. 30, in mak- 

 ing such applications for himself. The method itself is BO sim- 

 ple that it is unnecessary to give here any practical examples 

 in illustration. We shall, moreover, have occasion to return to 

 the subject in the consideration of moment of inertia of areas. 



"We pass on therefore to the moment of rotation of forces in 

 a plane. 



