220 



THE POPULAE EDUCATOR. 



Fig. 27. 



centre of gravity, we have the centre of gravity of an " area" 

 or " surface." Also let it be understood that the centre of 

 gravity of a line, straight or curved, means that point for such 

 a line of atoms. 



TO FIND CENTRES OF GRAVITY BY CONSTRUCTION.' 



This is done by the rule for finding the centre of parallel 

 forces, given in Lesson IV. (page 123). We shall commence 



with the most gene- 

 *=** ra l case, namely: 



1. To find the com- 

 mon Centre of Gra- 

 vity of any number 

 of Bodies, the sepa- 

 rate Centres and 

 Weights of which are 

 given. The masses 

 may be anyhow 

 placed, but the ope- 

 ration is the same 

 whether they are all 

 on the same plane, 

 as in the case of the 

 balls on the ground, 



in Fig. 27 above, or are some in that plane, some above, and 

 some below. Let them be four in number and on the same 

 plane, their centres being A, B, C, D ; then four parallel forces, 

 the weights, act at these centres : what has to be done ? Join 

 first A with B, and cut the joining lino at x inversely as the 

 weights at these points. Next connect x and c, and cut c x 

 at Y inversely as the two first weights to that at c. Lastly, 

 Y being joined to D, divide D Y at z inversely, as the weights of 

 the three balls already used are to that of the fourth, D. This 

 last point, z, is the required common centre of gravity. 



You observe that the joining and cutting of the lines is in no 

 way influenced by, or dependent on, the bodies being on the 

 same or in different planes, or of their number. How many 

 soever they be, the operation is the same. Note, also, that a 

 common centre of gravity can be outside the bodies of which it 

 is the centre. 



2. To find the Centre of Gravity of a Eight Line. A mechanical 

 right line being, as we have agreed, a line of atoms of equal 

 size and weight, the case is that which we have considered in 

 Lesson IV., of a number of equal parallel forces acting at equal 

 distances from each other, along a right line. The resultant 

 passes through the middle point of that line ; hence the centre 

 of gravity of a righf- line is its middle point. 



This enables us to find the centre of gravity of a uniform rod. 

 By " uniform," I mean such that the cross sections are of the 



same size 

 and form 

 throughout 

 its length. 

 Such a body 

 may be con- 

 sidered a col- 

 _ lection of 

 sg_J equal mecJia- 

 nical right 

 lines placed 

 side by side, 

 their ends 

 being made 

 flat or level. 

 As the cen- 

 tre of each 

 line is in its 



middle, the centre of the whole bundle is in the cross section 

 at the rod's middle. Aud observe that this holds good of 

 all other bodies, besides mere rods, which can be considered 

 made up of equal parallel lines, such as of a cylinder or uni- 

 form pillar, or_of a beam of timber, a cubical block of stone; 

 fche centres of gravity will be in the cross sections at their 

 middle points. And it makes no difference whether the flat 

 ends of the cylinder, pillar, beam, or block are perpendicular to 

 Hie lines of which it is supposed to be composed, as in c and e 

 {Fig. 28), or oblique to them, as at d and / (Fig. 29) ; the 

 entre of gravity is still in the middle cross section parallel to 



Fig. 28. 



the flat ends. Moreover, as aJl bodies so shaped may foe con- 

 sidered a collection of areas, one atom thick, piled on top of 

 each other, either perpendicularly or with a slope, like cards, or 

 a, pile of sovereigns, the centre of gravity of each must lie also 

 on the line joining the centres of gravity of the two areas which 

 form their ends. The centre itself, therefore, is the point in. 

 which this line pierces the middle cross section, as at c and e, 

 Fig. 28, in the cylinder and cube. But this requires us to be 

 able to find the centre of gravity of such areas, of which take 

 first the triangle. 



3. To find tlie Centre of Gravity of a Triangle. This wo do by 

 considering the triangle made up, as in the triangle a, in Fig. 30, 

 of lines an atom thick, all parallel to the side A B. The centre 

 of gravity of each line is at its middle point. If, therefore, 

 I can satisfy you that the middles of all the lines are on the- 

 line c M, which joins the vertex c with the middle M of A B, 

 the centre for the whole triangle is somewhere on that line, 

 I have, then, to prove that 

 c M bisects, or divides into g. 



two equal parts, every line 

 parallel to A B. Suppose, 

 now, that I cut c M into 

 three equal parts, ex, x y, 

 y M, as in the triangle o, in 

 Fig. 30, and draw paral- 

 lels to A B at the two 

 points of section inside, 

 meeting A c and B c each 

 in two points from which 

 parallels to c M are drawn, 

 meeting A B in four points, 

 two on each side of M. 

 Now, since c M is equally 

 divided, and the white 



Fig. 29. 



figures inside are parallelograms, it is evident that the line 

 parallel to c M marked a, o, on each side, are equal to each other, 

 and to c x, the third part of c M. Hence the three small shaded 

 triangles next to A C are equal to each other, and have equal angles. 

 Their three sides parallel to A B are therefore equal, which shows 

 that A M is cut by the parallels to c M into three equal parts. For 

 the same reason B M is cut into three equal parts ; and since 

 A M is equal to B M, the six parts into which A B is divided are 

 equal to each other. You thus see that the first parallel above 

 A B is made of parts, two on either side of c M, equal to the 

 parts below, and is therefore bisected by c M. The next ahovft 

 is also evidently bisected, being composed of two parts, one 

 on either side. Now, if I divide c M into five parts instead of 

 three, I have four other parallels also bisected by c M ; if into T 

 or any other number, it is the same I can fill the whole triangle 

 with parallels to A B bisected each by the line c M. The centre- 

 of gravity of the triangle is therefore on c M. 



But by a similar reasoning it can be shown that this centre 

 of gravity must be in A L (in triangle a, Fig. 30) bisecting A c^ 

 Hence we have for rule that, in order to find the centre of 

 gravity of a triangle, we 

 must join any two of its 

 vertices with the middle 

 points of the sides opposite 

 to them, and that the in- 

 tersection G of the joining 

 lines is the required point. 

 This centre G is distant 

 from M one-third of C M, 

 and from L one- third of A L. 



The centre of gravity of 

 a parallelogram can now 

 be shown to be the inter- 

 secting of its diagonals 

 A C, B D (see c, Fig. 30) ; 

 for, since the diagonals 

 bisect each other, the line 



B D is the bisector of the common side A c of both the trianglea, 

 ABC and A C D. The centre of each, therefore, is on that 

 line, and therefore the common centre of both thai is, the 

 centre of the parallelogram. But, by the same reason, con- 

 sidering the parallelogram made of the two triangles on BD, 

 the centre is on A C. Being thus on both diagonals, it is at 

 their intersection. 



Fig. 30. 



