MECHANICS. 



201 





the flat end*. Moreover, an all bodies o shaped may be con- 

 M.I. !. I 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 area* which 

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

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

 (Fijr. 'IB), in the cylinder and cube. Bat this requires as to be 

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

 first the triangle. 



15. To find the Centre of Gravity of a Triangle. This we 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 H bisects, or divides into 

 two equal parts, every line 

 parallel to A B. Suppose, 

 now, that I cut c M into 

 three equal parts, ex, xy, 

 y M, as in the triangle 6, 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. 

 Pi 29 Now, since c M is equally 



divided, and the white 



figures inside are parallelograms, it is evident that the line 

 parallel to c M marked a, b, 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 AC are equal to each other,and have equalangles. 

 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 H, 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 above 

 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 7 

 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 a of the joining 

 lines is the required point. 

 This centre o 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 

 AC, BD (see e, 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 triangles, 

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

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

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

 sidering the parallelogram made of the two triangles on B D, 

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

 their intersection, 



Fig. 30. 



Fig. 31. 



4. To find the Centre of Gravity of a Polygon. Let A B c D 

 (Fig. 31) be the polygon, and from the angle A draw the dotted 

 lines A c, A D to the remote angles and D. The polygon it 

 thus oat np into three triangle*. Let o, H, and K be the centre* 

 of gravity of these latter figure* ; there are thus three bodie* 

 whose centre*, o, u, and K, are known, and whose mn*o* are the 

 three areas of the three triangle*. 



Suppose now that you had calcu- 

 lated these areas, and had them 

 written down in number*. Then 

 join o with H and cat a H at z 

 inversely as the numberi ei press- 

 ing the areas of the triangle* ABC, 

 ADC. Connect x now with K, 

 and cat K x at Y inversely, as the 

 quadrilateral ABC D to the triangle 

 A E D ; the point Y is the required 

 centre. If the polygon had more 

 sides than are in Fig. 31, the 

 process is the same, and most be 



continued until all the triangles into which it i* neoe**ary to 

 divide the polygon have been gone over. 



5. To find the Centre of Gravity of the Circumference of a 

 Circle. Let the circumference be taken to be a carved line of 

 atoms, as in a, in Fig. 32, to the right ; and through the centre 

 o of the circle let any line, A o B, be drawn passing through two 

 of them, one on either side. Since these two are of equal weight, 

 and equally distant from a, their common centre of gravity is 

 the middle of A B, that is, the point o. So, likewise, going 

 round the figure, the centre of gravity of every opposite pair of 

 atoms is o, and therefore o is the common centre of all, or of 

 the circumference. 



The centre of gravity of a ring is thus seen to be the centre 

 of the circle in which it is formed, for the ring may be con- 

 sidered a bundle of circles an atom thick, bound together, one 

 above and around the other, so as to have for common centre of 

 gravity the centre of 

 the central circle. A 



The centre of gra- 

 vity of the area of 

 a circle is also the 

 centre of figure of 

 the circle, for the 

 area may be con- 

 sidered as made up 

 of a number of cir- 

 cles of atoms, lying 

 one inside the other, and having the same centre, Q, which, by 

 the above, is therefore their common centre of gravity. 



The centre of gravity of a hollow sphere may, in like manner, 

 be proved, by drawing lines through o to the atoms on its sur- 

 face, to be the centre of figure of the sphere ; and a solid sphere 

 may be considered as consisting of a number of these hollow 

 ones inside one another. 



Fig. 32. 



ANSWERS TO THE QUESTIONS IN LESSON IV. 

 Parallel Forces. 



1. The required parallel centre in this case is on the line joining any 

 vertex of the triangle with the middle point of the side opposite, at a 

 distance of two-thirds of that Hue from the vertex, or one-third from 

 the side. 



2. The required centre is distant one-third of the length of the beam 

 from the end at which the three-pound force acts. 



3. The required centre is 8 inches distant from the end of the rod 

 at which the three-and-a-half pound weight acts. 



4. The strain on the upper hinge is 20 Ibs. 3 oz. nearly, and on the 

 lower 39 Ibs. 13 oz. In doing this question the student most first find 

 the centre of parallel forces for the 23 Ibs. and 37 Ibs. This point will 

 be found by cutting the 7 feet of height of door into 60 parts (the sum 

 of 23 nnd 37). and counting off 23 of these from the bottom. The 

 resultant then acts at the end of the 23rd subdirision. But as the 

 hinges bear the drag of this force, it is dirided between them in the 

 inverse proportion of their distances from this point. Diride then the 

 60 Ibs. into two part*, which hare this proportion, and the above 

 strains will be found. It is better to do thi question by arithnttically 

 calculating the position of the point and magnitudes of statins. The 

 strain caused by the weight of the door is not here taken into coo- 

 Bideration. 



