of this line from the angle will be the position of the centre of gravity. 8. For a trapezium,-draw the two diagonals, and find the centres of gravity of each of the four triangles thus formed, then join each opposite pair of these centres of gravity, and the two joining lines will cut each other in the centre of gravity of the figure. 9. For the cone and pyramid,-the centre of gravity is in the axis, at the distance of of the axis from the vertex 10. For the arc of a circle, radius of circle X chord of arc length of arc = distance of the centre of gravity from the centre of the circle. 11. For the sector of a circle, 2 x chord of arc × radius of circle 3 × length of arc = distance of the centre of gravity from the centre of the circle. 12. For a parabolic space, the distance of the centre of • gravity from the vertex is of the axis. 13. For a paraboloid,—the centre of gravity is of the axis from the vertex. 14. For two bodies,-if at each end of a bar a weight be hung, the common centre of gravity will be in that point which divides the bar, in the same ratio that the weights of the bodies bear to each other, and this point will be nearest the heavier body. Examples. If the line drawn from the middle of the base of a triangle to the opposite angle be 15, then we have 15 3 X 2 = 10 = the distance of the centre of gravity from the vertical angle. 24 4 If the height of a cone be 24 inches, then we have × 3 = 18 = the distance of the centre of gravity from the vertex. If the length of the arc of a circle be 157.07, and the chord 153.07, and radius 200; then, 200 × 153.07 157.07 = 194.9 = distance of the centre of gravity from the centre of the circle. If there be the sector of a circle of which the chord radius, and length of arc, are the same as in the last ex ample, we have 2 x 153.07 × 200 3 X 157.07 = 129.8 = distance of the centre of gravity from the centre of the circle. 25 5 In a parabolic space, if the axis be 25 inches long, then × 3 = 15 = the distance of the centre of gravity from the centre. 30 In a paraboloid, if the axis be 30, then we have X 2 =20 = the distance of the centre of gravity from the vertex. A bar of wood, 24 feet long, has a weight suspended at each end, that at one end being 16 lbs., and the other 4: then, we have 20: 24: 16: 19.2 and 20 24 :: 4: 4.8 the distances of the weights from the common centre of gravity, the greater weight being least distant. Hence we = see, that 19.2 + 4.8 24, the whole length of the bar; and also 4 x 19.2 16 x 4.8 76.8; so that the principle of virtual velocities, stated before, holds good here also; and here it may be observed, that it is of the greatest importance to trace any leading principle of this kind througn its various applications, as it serves to link together and harmonize the whole, and enables us to apply and remember it with greater facility. It is often necessary to determine the centre of gravity experimentally, as in many cases it cannot be conveniently done by calculation. To maintain the firmness of any body resting on a base, it is necessary that the perpendicular drawn from the centre of gravity of the body, to the base on which it rests, be within that base; and the body will be the more difficult to overset, the nearer that perpendicular is to the centre of the base, and the more ex tensive the base is, compared to the height of the centre of gravity. THE CENTRE OF OSCILLATION.-THE PENDULUM, AND CENTRE OF PERCUSSION. 1. THE centre of oscillation in a vibrating body, is that point in the axis of vibration, in which, if the whole matter The contained in the body were collected, and acted upon by the same force, it would, if attached to the same axis of motion, perform its vibrations in the same time. centre of oscillation is always situated in the straight line which passes through the centre of gravity, and is perpendicular to the axis of motion. It will be seen by these remarks, that the subject of pendulums must be considered here. 2. In theory, a simple pendulum is a single weight, considered as a point, hanging at the lower extremity of an inflexible right line, having no weight, and suspended from a fixed point or centre, about which it vibrates, or oscillates; a compound pendulum, on the other hand, consists of several weights, so connected with the centre of suspension, or motion, as to retain always the same distance from it, and from each other. 3. If the pendulum be inverted, so that the centre of oscillation shall become the centre of suspension, then the former centre of suspension will become the centre of oscillation, and the pendulum will vibrate in the same time: this is called the reciprocity of the pendulum; and it is a fact of the greatest utility, in experimenting on the lengths of pendulums. 4. Of the simple pendulum we may observe, that its length, when vibrating seconds, must in the first place be determined by experiment, as it vibrates by the action of gravity, which force differs at different distances from the pole of the earth. By the latest experiments, the length of the seconds' pendulum in the latitude of London, has been found to be 39.1393 inches, or 3.2616 feet; the length at the equator is nearly 39.027, and at the pole 39.197 inches. The length for the latitude of London may be taken for all places in Britain, without any material error. 5. The times of vibration of two pendulums, are directly proportional to the square roots of the lengths of these pendulums. 6. Thus what will be the time of one vibration of a pendulum of 12 inches long at London? √39·1393: ✅✔✅ 12 :: 1 : 0.5537 If the pendulum be 36 inches long, = time of one vibration. ✔39.1393: ✔ 36 :: 1 : 0·9599 = time of one vibration. 7. The lengths of the pendulums are to each other in versely as the squares of the numbers of vibrations made in a given time. What is the length of a pendulum vibrating half-seconds or making 30 vibrations in a minute? (60): (30) :: 39.1393: 9-7848 = length in inches. The length of a pendulum to make any given number of vibrations in a minute, may be easily found by the following short rule: Thus a pendulum to make 50 vibrations in a minute, will 8. All the rules for simple pendulums may be expressed as follows: The time of one vibration in seconds of any pendulum is or = 1 number of vibrations in one second the length of the pendulum) 39.1393 Exam. If the number of vibrations of a pendulum be '6256, then 1 ⚫6256 = 1.598 = the time of one vibration. Or, if the length of the pendulum be 100 inches, then The length of a pendulum in inches is = 39·1393 × time of one vibration3; Exam.-If the time of one vibration be 1.598; find the length. 39-1393 x 1.5982 = 100, length of pend. Or. if the number of vibrations in a second be as above, 6256, then we have The number of vibrations in a second may be found thus: 39.1393 length of pendulum = number of vibrations; or, the number of vibrations in a second is 1 = time of one vibration If the time of one vibration be, as above, 1.598; then 1 1.598 =6256, number of vibrations; or, if the length of 100, we have When a clock goes too fast or too slow, so that it shal lose or gain in twenty-four hours, it is desirable to regulate the length of the pendulum so that it shall go right. The pendulum bob is made capable of being moved up or down on the rod by means of the screw. If the clock goes too fast, the bob must be lowered, and if too slow, it must be raised; and we have this rule: number of threads in an inch of the screw X the time in minutes that the clock loses or gains in 24 hours; this product divided by 37 will give the number of threads that the bob must be screwed up or down, so that the clock shall go right. Ex.-If the rod have a screw 70 threads in the inch, and the pendulum is too long, so that the clock is 12 minutes slow in 24 hours; then we have 2 x 70 x 12 37 = 45195 threads we must raise the bob, so that the clock shall go right. 9. It is often desirable that a pendulum should vibrate seconds, and yet be much shorter than 39.1393 inches; which may be done by placing one bob on the rod above the centre of suspension, and another below it: then, having the distances of the weights from the centre of suspension, we may find the ratio which the weights should bear to each other by this rule. Call D the distance of the lower, and d the distance of the upper weight, from the centre of suspension; then, |