88. Accordingly, every thing which we have said hitherto upon the common centre of gravity of several bodies, considered as points, is equally applicable to bodies of whatever figure, if we take, in estimating the moments, instead of the distance of each body, the distance of its particular centre of gravity.

89. Hence, finally, if several bodies, of whatever figure, have their particular centres of gravity in the same straight line, or in the same plane; their common centre of gravity will, in the former case, be in the given straight line, and in the latter in the given plane.

Application of the Principles of the Centre of Gravity to particular

Problems.

90. Let AB be a straight line uniformly heavy. It will be Fig. 30. seen at once without the aid of any demonstration, that the middle point P, of its length will be its centre of gravity. But in order to illustrate and confirm the theory of moments, developed in the preceding articles, let us seek the centre of gravity according to the principles of this method.

We imagine this straight line divided into an infinite number of points, of which Pp represents one; and that each is multiplied by its distance from a fixed point, as the extremity A for example. We then take the sum of these products, and divide it by the sum of the parts of which Pp is one, that is, by the line AB. Accordingly, if we call AB, a; AP, x; we shall have Ppdx; and the moment of Pp will be equal to x dx, which Cal. 7. must be integrated to obtain the sum of the moments. This sum

x2
2

therefore will be equal to and in order to have it for the Cal. 82.

whole extent of the line, we must suppose x = a, which gives

a2

2

for the entire sum of the moments. Dividing this by the sum

Fig. 31.

91. Hence, (1.) In order to have the centre of gravity of the perimeter of a polygon, it is necessary, from the middle of each of the sides, to let fall perpendiculars upon two fixed lines AB, AC, taken in the plane of this polygon; and, considering the 86. weight of each side as united in the middle of this side, to seek the common centre of gravity of these weights in the manner already explained.

92. (2.) The centre of gravity of the surface of a parallelogram is the middle point of the line which joins the middle of two opposite Fig. 32. sides. For, by considering the parallelogram as composed of

material lines, parallel to these two sides, each will have its centre of gravity in the line which passes through the middle of these same sides. The common centre of gravity, therefore, of all the lines will be in the bisecting line. It will, moreover, be in its middle point, since this line, considered as sustaining the 90. weights of all the other lines, is uniformly heavy.

Fig. 33.

77.

93. (3.) To find the centre of gravity of a triangle ABC; we draw from the vertex A to the middle D of the opposite side BC, the straight line DA, and from the point D we take DG = DA.

Indeed, the straight line DA, which divides BC into two equal parts at the point D, divides also into two equal parts every other line LN, parallel to BC; accordingly, if we consider the surface of the triangle as an assemblage of material lines parallel to BC, the line DA, which passes through the particular centres of gravity of all these lines, will also pass through their common centre of gravity, that is, through the centre of gravity of the triangle. For the same reason, the line CE, which passes through the middle of AB, will in like manner pass through the centre of gravity of the triangle. This centre is consequently at the point of intersection G of the two lines CE and DA. Now, if we join ED, it will be parallel to AC, since it divides into two Geom. equal parts the sides AB, BC. The two triangles EGD, AGC, are accordingly similar, as well as the triangles ABC, EBD; we Geom. have, therefore,

199.

202.

DG: AG: DE: AC :: BD : BC :: 1:2 that is, DG is half of AG, and therefore one third of AD.

94. Hence, In order to find the centre of gravity G of a trapezoid, we draw KL through the middle points of the two parallel Fig. 34. sides, and from these same points K, L, we draw the lines KA, LD, to the vertices of the opposite angles A, D; then having taken

KE = 1 KA, LF = LD,

we join EF, which will cut KL in G, the point sought.

For, by reasoning as we have done in the case of the triangle, we shall see that the centre of gravity G must be in KL. Moreover, since E, F, are the centres of gravity respectively, of the triangles CAD, ADB, which compose the trapezoid ABDC, the 98. common centre of gravity of the two triangles or of the trapezoid must be in EF; it follows, therefore, that it is at the intersection G.

To find the distance LG, which we shall have occasion to use hereafter, we draw the lines EH, FI, parallel respectively to AB;

and since

we shall have

or

KEKA, and LF = } LD,

EHAL, and FI= KD,

EH = } AB, and FI= CD.

77.

that is,

EH: GH:: FI: GI;

EH + FI : GH + GI or HI :: FI : GI;

and, because LG = LI+ GI, if we substitute for LI and GI the

values above found, we shall have

Fig. 35.

93.

86.

} KL × (AB + 2 CD)
AB+ CD

We remark in passing, that if the height KL of the trapezoid were infinitely small, and the difference of the two sides AB, CD, were infinitely small, these sides must be considered as equal, so KL × 3 AB that the distance LG would reduce itself to 2 AB KL; that is, the centre of gravity in this case is equally distant from the two opposite bases.

or

95. To find the centre of gravity of the surface of any polygon, we divide it into triangles, and having found the centre of gravity of each triangle in the manner above shown, we determine the common centre of gravity of all the triangles, by considering them as so many masses proportional to their surfaces, and concentrated each at its particular centre of gravity, agreeably to the method already adopted.

It will hence be seen how we should proceed in determining the centre of gravity of the surface of any solid figure terminated by plane surfaces.

96. In fine, it is not always necessary to have recourse to moments in finding the centre of gravity. If it were proposed for example, to determine the centre of gravity of the perimeter Fig. 36, of a regular pentagon ABCDE, I should draw from the vertex of

one of its angles A, to the middle of its opposite side, a straight line AH; likewise from the vertex of another angle E, to the middle of its opposite side, a straight line EI, and the intersection G of these lines will be the centre of gravity.

Indeed the common centre of gravity of the two sides AB, AE, is evidently the middle c, of the line ba, which passes through their middle points. The common centre of gravity of the two sides BC, DE, is for the same reason the middle of the line IK which passes through their middle points; and the side CD has its centre of gravity in H. Now it is manifest that the line AH passes through the middle points c, e, and H; it accordingly passes through the common centre of gravity of the five sides. It may be shown, in like manner, that IE also passes through the

centre of gravity; therefore this centre is at the intersection G of AH and IE.

97. By pursuing the same kind of reasoning which we adopted in the case of the triangle, it might be demonstrated that the point G is the centre of gravity of the surface of a regular pentagon.

In general, it may be shown, by the same method, that the centre of gravity of the perimeter, as well as of the surface of any regular polygon, of an odd number of sides, is the point of intersection of two straight lines, each of which is drawn from the vertex of one of the angles to the middle of the opposite side. And when the number of sides is even, the centre of gravity, Fig. 37. both of the perimeter and of the surface, is the point of intersection of two straight lines drawn through the middle points of two pairs of opposite sides. We might also extend this mode of reasoning to the circle, by regarding it as a polygon of an infinite number of sides, and we should find that the centre of gravity of the circumference, and of the surface, is the centre.

When the number of lines, surfaces, bodies, &c., is not considerable, the centre of gravity may be found by the method of articles 53, 54. Let the three points A, B, C, for example, be the Fig. 38 centres of gravity of three lines, or three surfaces, or three bodies, whose weights are represented by the masses m, n, o. Having joined two of these points, as B and C, by the line BC, we divide BC at G', in such a manner as to give

or

n: o :: CG': BG'

n+on:: CB : CG';

masses n, o, united

and the point G' thus found, will be the centre of gravity of the two weights n, o. We now draw G'A, and supposing the two in G', we divide, in the same way, G'A in the inverse ratio of the two masses m and n + 0, that is, so as to give

or

n+o: m :: AG: G'G,

n+o+m: m :: AG': G'G;

and the point G will be the common centre of gravity of the three weights m, n, o. We might proceed in a similar manner with a greater number of bodies.