MECHAN'ICAL PHILOSOPHYSTATICS. [OB >xrKictt SOLIDS. 



And AD-AO + DG-2DO + DO-3DG. 

 Or DO -i AD. 



EXAMPLE II. To find the ernlre of graritv of a plate of 

 uniform thicknett and drntily bounded by parallel 

 planet, who* vpptr and lover turfacet an paralleh- 



Let the parallelogram A B C D (Fig. 87) represent the 

 per surface of tie plate, A' B' (7 IX the lower ; the 



u; ;.,-r 



other boundary planes, such as B C C' B', being perpen- 

 dicular to A B C D. 



Join B D and A C meeting in E, B' D and A' C' meet- 

 ing in E'. AC will bisect all lines drawn through the 

 parallelogram parallel to B D, and B D will bisect all 



similar lines drawn parallel to A C. 



Then we may conceive, as in the last example, the 

 plate as made up of a number of parallel plates, each 

 parallel to the plane B D D' B', each of which will 

 balance about a line drawn through the centres of A A 

 and OCX. 



Again, we may conceive the plate as composed of a 

 number of parallel plates parallel to A A' C 1 C, which 

 will each balance about a line drawn through the centre 

 of D I) and B B'. 



G, the intersection of these lines, or the centre of 

 E E*, will therefore be the centre of gravity of the plate. 



To find the centre of gravity of the surface of any 

 rectilineal figure, we have only to divide it into triangles, 

 find the centre of gravity of each triangle, suppose 

 weights acting at each of these centres of gravity propor- 

 tional to the surface of the triangle at whoso centre of 

 gravity it is supposed to act, and then find the centre of 

 gravity of these weights, which will be tho centre of 

 gravity of the figure. 



It will be found, on calculation, that the centre of 

 gravity of any triangular surface is the same as that of 

 three spheres, or any other body, whose weights are 

 equal, placed with their centres of gravity in the angular 

 points of the triangle. This also applies to a parallel- 

 ogram, but not to all four-sided figures. 

 EXAMPLE III. To find the centre of gravity of a homoge- 

 neous tolid of uniform density, whose form it that of a 



pyramid on a triangular base. 



Let B C D (Fig. 88) be the base, and A the vertex of 

 the pyramid. 



Bisect one of the aides of the base D C in E. Join 

 A K. i: K 



Take EG,, one-third of BE, and EG.,, one-tliird of AE. 



Join AG, and BG, meeting in G, Gj will bo the 

 centre of gravity of the 

 pyramid. 



In A C take any point <-, 

 through c draw c d parallel 

 to C D cutting A K in , 

 and be parallel to II C. 

 Join 6 d and 6 e cutting 

 AU, in/. 



Then the plane b c d is 

 evidently parallel to the 

 plane BCD, and the line 

 I parallel to the In,. 



Hence by similar tri- 

 angles AEG,, 



~ . p lid ty similar tri- 



Tliorefore Q- - fi but E G, - 3 B E 



Hence e/ J of be. 



Again, because c d is parallel to C D, and C D U 

 bisected in E, cdia also bisected by A I 



/ therefore U the centre of gravity of the triangle bed. 



In a similar manner it may be shown that thi- 

 ef gravity of any triangular slice of the pyramid parallel 

 to B C D must lie in the line A G, ; and since we may 

 conceive the pyramid made up of an intiiiitu innnl- 

 such parallel slices, the centre of gravity of the whole 

 pyramid must lie in AC,. 



By similar reasoning it may be shown that the centra 

 of gravity of the pyramid lies in the line 1: 



Consequently, G 3 , the intersection of the lines A G t 

 and B G a , will be the centre of gravity of the pyramid. 



Join G, G,, then since E G, = i A E, and E G, - J 

 BE, G, G, is parallel to AB. Consequently the tri- 

 angle G, G, G, is similar to the triangle G a A B, and 

 the triangle E G, G, to the triangle E B A. 



GKQ, 



A i; 



,G, G. EG, 

 = BE 



-J. 



Therefore 





, = i and G, G, = i A G. 



, - } AG,. 



Hence the centre of gravity of the pyramid is found 

 to be in the line joining the vertex and the centre of 

 gravity of the base, at a distance of a fourth of its 

 length. 



EXAMPLE IV. To find the Centre of Gravity of any 

 Pyramid of uniform density, whose base is any Polygon. 



Let A be the vertex of the pyramid, B C D E F H its 

 polygonal base (Fig. 89). Let G t be the centre of 

 gravity of this base ; join AG U and in A G, take G a so 

 that G, G 2 { A G,. G a will be the centre of gravity 

 of the pyramid. 



Join G , with each of the angular points of the base, 

 through G draw the plane 

 bcdefh parallel to the 

 base, and cutting A B in 

 6, A C in c, &c., and join 

 G a c, G a d, etc. 



Let jr, bo the centre of 

 gravity of the triangle 

 B G! C, join A </] cutting 

 the triangle 6 G, c in g,. 



Then since the triangle 

 b Gj c is parallel to the 

 triangle BG. C and G, G, 

 = ^ A G, ; hence g, g t = 

 } A <;, and jr., is the centre 

 of gravity of tho pyramid 

 A B C Gj on a triangular 

 base G, B C, and it lies in the plane bcdefh. In a 

 similar manner it may be shown tliat the centres of 

 gravity of every one of the triangular pyramids A G! 

 C D, A G, D E, <fcc., into which the whole pyramid may 

 be divided, all lio in tho plane b e d efh. 



It follows, therefore, that the centre of gravity of the 

 whole pyramid must lie in this plane. 



Again the plane bcdefh is in every respect similar 

 to the base, and G, its centre of gravity will therefore 

 lie in the line A G,. In like manner it may be shown 

 that the centre of gravity of every section of tho pyra- 

 mid parallel to the base will lie in AG, ; hence the 

 centre of gravity of the whole pyramid must lie in A G,. 

 We have shown, therefore, that the centre of gravity 

 of the whole pyramid must be in tho piano b c d e fh, and 

 also in the line A G,. Hence tho centre of gravity 

 must be their intersection G.. 



Tho centre of gravity of any pyramid of uniform 

 density on a polygonal base, will lie in the line 



