=

OQOD, and PR = PB; then will and R be the centres of gravity of these triangles (Art. 55). Join Qand R by a straight line; the centre of gravity of the trapezoid must be on this line (Art. 52). Hence, it is at G where the line QR cuts OP.

(Art.

Centre of Gravity of a Polygon.

58. Let ABCDE be any polygon, and a, b, c, d, e, the middle points of its sides. The weights of the sides will be proportional to their lengths, and may be represented by them. Let it first be required to find the centre of gravity of the perimeter; join a and b, and find a point o, such that

G

ao ob: BC: BA;

E

Fig. 87.

then will o be the centre of gravity of the sides AB and BC. Join o and c, and find a point o', such that

oo' o'c:: CD : AB + BC;

then will o' be the centre of gravity of the three sides, AB, BC, and CD. Join o' with d, and proceed as before, continuing the operation till the last point, G, is found; this will be the centre of gravity of the perimeter.

B

To find the centre of gravity of the area, divide it into the least number of triangles possible, and find the centre of gravity of each triangle. The weights of these triangles will be proportional to their areas, and may be represented by them. (Art. 52.) Let ABCDEA be any polygon, and O, O', O", the centre of gravity of the triangles into which it can be divided. Join O and O', and find a point O", such that

0'0" 00"" :: ABC: ACD;

Fig. 38.

C

then will O' be the centre of gravity of the two triangles ABC and ACD.

Join O" and "", and find a point G, such that

0"

E

Fig. 38.

O""GO"G:: ADE: ABC + ACD;

then will G be the centre of gravity of the given polygon. Every curvilinear area may be regarded as polygonal, the number of sides being very great. Hence, the centres of gravity of their perimeters and areas may be found by the methods given.

Centre of Gravity of a Pyramid.

59. Any triangular pyramid may be regarded as made up of infinitely thin layers parallel to either of its faces. If a straight line be drawn from either vertex to the centre of gravity of the opposite face, it will pass through the centres of gravity of all the layers parallel to that face. We may regard the weight of each layer as being applied at its centre of gravity, that is, at a point of this line; hence, the centre of gravity of the pyramid is on this line (Art. 52). Let ABCD be a pyramid, and K the middle point of DC. Draw KB and KA, and lay

KB, and KO' Then will be the centre of gravity of the face DBC, and O' that of the face CAD. Draw 40 and BO' intersecting in G. Because the centre of gravity of the pyramid is upon both AO and BO', it is at their intersection G. Draw 00'; then KO and KO' being respectively third parts of KB and KA, 00' is parallel to AB, and

D.

Fig. 39.

A

the triangles OGO' and AGB are similar, consequently

their homologous sides are proportional. But 00' is onethird of AB, consequently OG is one-third of GA, or onefourth of AO.

Hence, the centre of gravity of a triangular pyramid is on a line drawn from its vertex to the centre of gravity of its base, and at one-fourth of the distance from the base to the vertex.

Either face of a triangular pyramid may be taken as the base, the opposite vertex being considered as the vertex of the pyramid.

To find the centre of gravity of a polygonal pyramid; let A-BCDEF, represent any pyramid, A being the ver tex. Conceive it divided into tri

Be

angular pyramids, having a common vertex at A. If a plane be passed parallel to the base, and at one-fourth of the distance from the base to the vertex, it follows, from what has just been shown, that the centres of gravity of all the partial pyramids will lie in this plane. We may regard each pyramid as having its weight concentrated at its centre of gravity; hence, the centre of gravity of the entire pyramid must lie in this plane (Art. 52). But it may be shown, as in the case of the triangular pyramid, that the centre of gravity lies somewhere in the line drawn from the vertex to the centre of gravity of the base; it must, therefore, lie where this line pierces the auxiliary plane:

Fig 40.

Hence, the centre of gravity of any pyramid whatever lies on a line drawn from its vertex to the centre of gravity of its base, and at one-fourth of the distance from the base to the vertex.

A cone is a pyramid having an infinite number of faces: Hence, the centre of gravity of a cone is on a line drawn from the vertex to the centre of gravity of the base, and at one-fourth of the distance from the base to the vertex.

Centre of Gravity of Prisms and Cylinders.

60. Any prism whatever may be regarded as made up of layers parallel to the bases. If a straight line be drawn between the centres of gravity of the two bases, it will pass through the centres of gravity of all these layers. The centre of gravity of the prism will, therefore, lie somewhere in this line, which we may call the axis of the prism. We may also regard the prism as made up of material lines parallel to the lateral edges of the prism. If a plane be passed midway between the two bases and parallel to them, it will bisect all of these lines, and consequently their centres of gravity, as well as that of the entire prism, will lie in it. It must, therefore, be at the point in which the plane cuts the axis of the prism, that is, at its middle point.

Hence, the centre of gravity of a prism is at the middle point of its axis.

When the bases of the prisms become polygons having an infinite number of sides, the prism will become a cylinder, and the principle just demonstrated will still hold good:

Hence, the centre of gravity of a cylinder with parallel bases is at the middle point of its axis.

Centre of Gravity of Polyhedrons.

61. If any point within a polyhedron be assumed, and this point be joined with each vertex of the polyhedron, we shall thus form as many pyramids as the solid has faces: the centres of gravity of these pyramids may be found by the rules for such cases. If the centres of gravity of the first and second pyramid be joined by a straight line, the common centre of gravity of the two may be found by a process entirely similar to that used in finding the centre of gravity of a polygon, observing that the weights of the partial pyramids are proportional to their volumes, and that they may be represented by their volumes. Having compounded the weights of the first and second, and found its point of application, we may, in like manner, compound this

with the weight of the third, and so on, till the centre of gravity of the entire pyramid is determined.

Any solid body bounded by a curved surface may be regarded as a polyhedron whose faces are extremely small, and its centre of gravity may be determined by the rule just explained.

Experimental determination of the Centre of Gravity.

63. We know that the weight of a body always passes through its centre of gravity, no matter what may be the position of the body. If we attach a flexible cord to a body at any point and suspend it freely, it must ultimately come to a state of rest. In this position, the body is acted upon by two forces: the weight, tending to draw the body towards the centre of the earth, and the tension of the cord, which resists this force. In order that the body may be in equilibrium, these forces must be equal and directly opposed. But the direction of the weight passes through the centre of gravity of the body; hence, the tension of the string, which acts in the direction of the string, must also pass through the same point. This principle gives rise to the following method of finding the centre of gravity of a body.

Let ABC represent a body of any form whatever. Attach a string to any point, C, of the body, and suspend it freely; when the body comes to a state of rest, mark the direction of the string; then suspend the body by a second point, B, as before, and when it comes to rest, mark the direction of the string; their point of intersection, G, will be the centre of gravity of the body.

Fig. 41.

Instead of suspending the body by a string, it may be balanced on a point. In this case, the weight acts vertically downwards, and is resisted by the reaction of the point; hence, the centre of gravity must lie vertically over the point.