GEOMETRICAL SOLIDS.] 



MECHANICAL PHILOSOPHY. STATICS. 



717 



and since Gi is the position of their resultant, we have 

 (Prop. XI.) 



W-BG = W'-A^; 

 but by the construction of the instrument, 

 W-BC= W'-CD. 



Hence -^ 

 Therefore C 



A t G. 



CD ' 



D- B C 



.BC + CG, A,C CG,. 



BC 



CI> 



OrCGr(CD-r-BC) = (A,C CD)BC- 

 AndCGrBD = AiD-BC. 



A,D-BC 

 Or C Gt = g-g 



When the centre of the sphere is in the position A 2 , 

 we have a weight W acting at B, and a weight W' at A 2 , 

 in parallel directions. Hence (Prop. XI.) he have 

 \V-BG 2 = W'-A, G a . 

 But W-BC = W'-CD. 



BG 3 A,('., 

 Hence g-~- = "cU"' 



Therefore (Euc., B. VI., P. II.), the line C G a is parallel 

 to A 2 D, and B C G t and B D A a are similar triangles ; 



and consequently 



/-^ c* T> r* T? t*i 



VjJ2 JJVy *^AT\ 



A.D = BD' orCG2 "ED' AaD ' 



But A, D = Ai D. Hence C G 



A t P BO 

 BD ' 



The 



ame value which we have before obtained for C G t . 



Fiff M- Hence for every position 



A, of the centre of the 

 movable sphere out of the 

 line BCD, we shall have 

 Fig. 85. 



application of Prop. XVII. to find the centre of gravity 

 of a body of uniform or variable density, in most 

 instances requires the aid of the differential a:id integral 

 calculus before we can arrive at a solution. As this 

 would lead us beyond the limits proposed for the mathe- 

 matical department of our work, we give some instances 

 of the determination 

 of the centre of gravity 

 of a few geometrical 

 forms and solids of uni- 

 form density, depending 

 upon the properties of the 

 centre of gravity demon- 

 strated in Prop. XIX. 



EXAMPLE I. To find the 

 centre of gravity of a 

 triangular plate of uni- 

 form thickness and den- 

 sity. 



Let A B C (Fig. 86) be 

 the surface of the triangular plate, bounded by two 

 parallel triangular faces, and by parallelograms perpen- 

 dicular to these faces. 



Bisect B C in D . Join A D. 



Through 6 any point in A B draw bdc parallel to B C, 

 and cutting A D in d. 



Then the triangle A 6 d, by construction, is similar to 

 the triangle A B D, and the triangle A d c to the tri- 

 angle ADC. 



Hence (Eu,, B. VI., P. IV.),^ - and^ = jfc 



Therefore 



but B D = D C. 



C G, parallel to D A,, and equal to C GL 



As AI therefore describes a circle round D, Gi will 

 describe a circle round C in the same direction, and 

 uniformly with it ; the radius of this circle being equal 

 to the product of the radius of the circle described by 

 the centre of the sphere, and the distance of B from C 

 divided by B D. 



Whi-n the centre of the sphere is in the position A,, 

 the hand will rest in such a position that C Gi shall be 

 vertical (Prop. XIX.), as shown in Fig. 84. When A 2 

 represents the position of the centre of the sphere, 

 C G a will be vertical, and the hand will rest in the posi- 

 tion shown by Fig. 85. Hence it is evident that, as the 

 watch-movement concealed in the hand causes the centre 

 of the sphere to move from A, to A a , the extremity of 

 the pointer of the hand will move through a similar arc 

 in the opposite direction round C as its centre. 



M \TMIv VfATICAL DETERMINATION OF THE 

 CENTRE OF GRAVITY OF A HEAVY BODY. The 



Hence 6 d = dc ; and so it may be 

 shown that every line drawn through 

 any point in A B parallel to B C will 

 be bisected by A D. 



Now we may conceive the trian- 

 gular plate, whose thickness is uni- 

 form, to be made up of a number 

 of extremely thin slices, cut perpen- 

 dicular to the surface, and parallel 

 toBC. 



Each of these slices, being sym- 

 metrical, will balance about its 

 centre, and therefore every one of 

 them will balance about a line drawn 

 parallel to A D, and passing through 

 the centre of the thickness of the 

 plate. 



The whole triangular plate will 

 therefore balance about this line. 



Next bisect AB in E, join EC 

 cutting C D in G. 



It may be shown, as before, that 

 the triangular plate ABC will ba- 

 lance about a line parallel to E C, 

 passing through the centre of the 

 thickness of the plate. 



Since the centre of gravity of the plate liea in both 

 the lines passing through the centre of the thickness 

 parallel to C E and A D, the intersection of these two 

 lines must be the centre of gravity. 



Hence a point in the middle of the thickness of the 

 plate below G will be the centre of gravity of the plate, 

 provided its thickness and density be uniform throughout. 

 To find G, join DE. 



Then because A B is. bisected in E, and B C in D, 

 Therefore (Euc., B. VI., Props. II. and IV.), E D is 

 parallel to A C, and E D = J A C. 



Hence the angle E D G = angle G C A, the angle at G 

 is common, and therefore the triangle E G D is similar 

 to the triangle AGO. 



Therefore (Eue., B. VI., P. IV.), ^ = 



OrDG 



AGDE , AG-AC 

 f ' 



AC 



AC 



AG. 



