and F, then by Cor. to Art. 75, if E F be divided in G, so that EF: F:: EF: EG (that is, the area of the trapezium A B C D : area of the triangle B C D :: EF: E G), then G is the centre of gravity required. Or, having found the centres of gravity E, F, if the trapezium be divided into two other triangles B A C, D A C, by the other diagonal A C, and the centres of gravity H and I of these two triangles be found, then the centre of gravity of the trapezium will lie in the line H I. But it has also been proved that the centre of gravity of the trapezium lies in EF; therefore, since it lies both in EF and H I, it must necessarily lie in G, their common point of intersection. In the same manner, we may find the centre of gravity of a figure of any number of sides. For it may be divided into triangles, the centres of gravity of which may be separately found; after which, we can find the centre of gravity of the whole of the triangles, by considering the whole weight of each to be collected in its centre of gravity. 1. The centre of gravity of a cylinder, or or any other body the parallel sections of which are equal, is in the middle of the axis of that body. 2. For the arc of a circle, as arc : sine of arc :: radius: distance of its centre of gravity from the centre. 3. For the sector of a circle, as arc : chord :: radius : distance of its centre of gravity from the centre. 4. For a parabolic space, the distance of the centre of gravity from the vertex is of the axis. 5. The centres of gravity of the surface of a cylinder, of a cone, and of a conic frustrum, are respectively at the same distances from the origin as are the centres of gravity of the parallelogram, triangle, and trapezoid, which are vertical sections of the respective solids. 6. In a coné, as well as any other pyramid, the distance of the centre of gravity from the vertex is of the axis. 7. In a conic frustrum, the distance on the axis from 3 R2+2 Rr + r2 R+ Rr + pe the centre of the less end is 4 h. where h denotes the height, and R and r the radius of the greater and lesser ends. 8. The same theorem will serve for the frustrum of any regular pyramid, taking R and r for the sides of the two ends. 9. In the paraboloid, the distance from the vertex is of the axis. 10. In the frustrum of the paraboloid, the distance on 2 R2 + 2 the axis from the centre of the less end is 3 h. R2 + r2 where h denotes the height, R and r the radii of the greater and lesser ends.* Example 1. Given the weights of two bodies 50 and 20 lbs. and distance asunder 35 feet; how far from the larger body is their common centre of gravity? By Art. 75, Cor. 50 + 20 : 20 :: 35: 10 feet. Therefore the centre of gravity is 10 feet from the larger body, and 25 feet from the smaller body. Example 2. If three equal bodies be placed at the angles of any triangle, prove that the common centre of gravity of these bodies is in the same point with the centre of gravity of the triangle. The bodies A, B, and C (Fig. 45), being all equal, the centre of gravity of any two of them, suppose A and B, will be in D, the middle of the side A B. The bodies A and B must now be supposed to be both collected in D, and join D, C, by the right line D C; and since A + B *The above are collected from the Mechanics of Emerson and Dr. Gregory. = 2 C, then, by Cor. to Art. 75, CG: DG :: 2 C: C, or CG: DG 2:1. Hence C G 2 D G, or C G = CD. Therefore, by Cor to Art. 79, G is the centre of gravity of the triangle A B C. Example 3. If five bodies, the weights of which are 3, 6, 9, 5, and 4 lbs. are placed at the distance of 2, 3, 4, 5, and 7 feet, respectively, from a given point, and on the same side of it; what is the distance of their common centre of gravity from the given point? By Art. 77, Cor. 2, we have 2 × 3 + 3 × 6 + 4 × 9 + 5×5+7x4 = (3 + 6 + 9 + 5+ 4) multiplied by the distance of their common centre of gravity from the given point. Consequently, that distance is equal to 2 × 3+3 × 6+ 4 × 9 + 5 × 5+7×4 3 + 6 + 9 + 5 + 4 feet, the distance required. Example 4. 113 27 If a homogeneous beam be 16 feet long, each foot weighing 5 lbs. and a weight of 60 lbs. is suspended at one end, what point of the beam will be the centre of gravity? We must consider the whole weight of the beam to be collected in its centre of gravity, which will be in its middle point, as the beam is uniform. Then 16 x 580 lbs. the whole weight of the beam. Then, by Art. 75, Cor. 80 + 60: 60: 8: 3, which is the distance between the centre of gravity and the middle of the beam; consequently, 8 +35 11%, the distance between the centre of gravity and the end. And 16-11-44, the distance between the centre of gravity and the end where the weight is suspended. Or it may be found thus: 80 + 60: 80 :: 8:44. Hence we have the following rule: Multiply the whole weight of the beam by half its length for a dividend. Then add the whole weight of the beam and the weight suspended at the end together for a divisor. And if the above dividend be divided by this divisor, the quotient will give the distance of the centre of gravity from the end where the weight is suspended. Example 5. If the height of a cylinder be double the diameter of its base, what is the angle of inclination of its axis with the horizon when it is just ready to fall over? (Fig. 46.) By Art. 74, Cor. 1, the inclination of the axis of the cylinder must be such that a perpendicular, drawn from the centre of gravity G of the cylinder, will fall just on the extremity A of the base. Produce the axis of the cylinder to B; then the triangles G A B, A E B, and A E G, are all right-angled and similar, by Euclid, Book 6, Prop. 8; and since the altitude of the cylinder is double the diameter of its base, G E 2 A E, consequently the angle G A E twice the angle A G E, whence the angle G A E = 60°, and the angle A G E GA GE = 30°; but the angle G A E is equal to the angle A B E, which is the angle of inclination of the axis of the cylinder with the horizon; therefore the axis makes an angle ABE of 60° with the horizon. Also, since the angle A G E is equal to the angle B A E, the base of the cylinder will make an angle of 30° with the horizon. If the cylinder be oblique (Fig. 47), then C A will be equal to the altitude of the cylinder; but C A is equal to twice H A, therefore G A is equal to twice BA; hence the angle G B A is equal to twice the angle B G A. Therefore the angle G B A, which is the angle of inclination of the axis of the cylinder with the horizon, is 60o, the same as above. H EXAMPLES FOR PRACTICE. Example 1. A weight of 1 lbs. laid on the shoulder of a man, is no greater burden to him than its absolute weight, or 24 ounces; what difference will he feel between the said weight applied near his elbow, at 12 inches from the shoulder, and in the palm of his hand, 28 inches therefrom; and how much more must his muscles then draw to support it at right angles, that is, having his arm extended right out? If a weight W be sustained on a horizontal plane by three props which are not in the same straight line, the pressure on each will be the same as if a single weight were laid on it, so that the sum of all the three weights were equal to W, and their common centre of gravity the same with the centre of gravity of that body. If the props be A, B, C, (Fig. 48) and if W be placed with its centre of gravity at D, and if A DG, B D F, CD E, be drawn, then the pressure When D coincides with the centre of gravity of the tri angle A B C, the pressure on each of the props is the |