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the circumference described by the power, let this circumference be what it may.

Example 1.

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Given the distance OP at which the power acts 3 feet, and the distance between two of the threads 2 inches, to find what weight a man would be able to sustain when he acts at P with a force of 150 lbs.

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Now 6 feet 72 inches, and 72 x 3·1416 226.1952 inches the circumference described by the power, therefore, by Art. 71,

=

P: W:2: 226-1952.

But P 150 lbs. hence W =

16964-64 lbs.

226-1952 x 150
2

When the screw is applied as in Fig. 37, it is called an endless screw; we have here also a combination of wheels, and, to obtain an equilibrium, we must have recourse to Art. 58, from which we have,

PX2AC x 3·1416 x radii of all the wheels distance of two threads x radii of all the pinions.

Example 2.

W ×

If the endless screw A B (Fig. 37) be turned by a handle A C of 20 inches, the threads of the screw being distant half an inch each; and the screw turns a toothed wheel E, the pinion L of which turns another wheel F, and the pinion of this another wheel G, to the pinion or barrel of which is hung a weight W; it is required to determine what weight a man will be able to sustain who acts at the handle CD with a force of 150 lbs. supposing the diameters of the wheels to be 18 inches, and those of the pinions and barrel 2 inches.

From what has been premised, we have 150 × 40 × 3.1416 x 183 W x & x 23.

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ON THE CENTRE OF GRAVITY.

72. The centre of gravity of any body or system of bodies is that point about which the body or system, acted upon only by the force of gravity, will balance itself in any position whatever: or it is that point, which being supported, the body or system will be supported, however it may be situated; and that centre will always tend to descend to the lowest place to which it can get, when it is not the point of suspension.

73. The centre of gravity of a body is not always within the body itself. Thus, the centre of gravity of a ring is not in the substance of the ring, but in the centre of its circumscribing circle; and the centre of gravity of a hollow staff or bone is somewhere in its imaginary axis.

74. If a line drawn perpendicular to the horizon, from the centre of gravity, fall within the base of any body, it will rest in that position; but if the perpendicular fall without the base, the body will fall down.

For when the perpendicular falls within the base, the body can be moved in no manner whatever but the centre of gravity will rise. And if the perpendicular fall without the base towards any side, the body cannot be moved towards that side but the centre of gravity descends, and therefore the body will fall that way.

Cor. 1. If a perpendicular, drawn from the centre of gravity, fall just on the extremity of the base, the body may stand; but the least force whatever will cause it to fall that way and the nearer the perpendicular is to any side, or the narrower the base is, the easier it will be made to fall, or be pushed over that way.

Cor. 2. Hence, if the centre of gravity of a body be supported, the whole body is supported; and the place of

the centre of gravity may, in many inquiries, be accounted the place of the body; for we may consider the whole mass or quantity of matter to be concentrated in that point, and therefore all the force also with which it endeavours to descend.

Cor. 3. If a body be laid upon a plane G B (Fig. 38), and one end A be gradually raised up, the body will slide down the plane if the perpendicular C D fall within the base; but if the perpendicular fall without the base, the body will roll down the plane.

It may be remarked here, that an equilibrium may take place when the centre of gravity is at the highest point to which it can ascend; but then this is only a tottering equilibrium, which the least motion will destroy; and the body or system, after the equilibrium is destroyed, will vibrate till the centre of gravity has obtained the lowest place of its descent.

75. The common centre of gravity C of any two bodies A, B, divides the line joining their centres into two parts, which are reciprocally as the bodies. AC: BC :: B: A. (Fig. 39.)

For if the centre of gravity C be supported, the two bodies A and B will be supported, and will rest in equili brio. But by the property of the lever, when two bodies are in equilibrio about C as a fulcrum, we have A x A C =Bx B C, or AC: BC:: B: A.

Cor. Hence, by composition, A + B: B::AB: A C.

To find the centre of gravity of any number of bodies connected together by inflexible right lines without weight (Fig. 40).

76. Let A, B, C, D, E, represent the bodies, the centre of gravity of which is required. Join any two of them, as A and B, by the right line A B; which divide in E, so that A + B : B :: A B: AE; and by Cor. to the last Art. E will be the centre of gravity of A and B. We must now suppose the sum of the bodies A and B to be collected in

E, and join E, C, by the right line E C, and divide it in F, so that A + B + C : C :: EC: EF; then will F be. the centre of gravity of A, B, C.

Lastly, join F, D, and suppose the sum of the bodies A, B, C, to be collected in F; then divide F D in G, so that A+B+C+D:D:: FD: FG, and G will be the centre of gravity of the four bodies A, B, C, D. And in the very same manner, the centre of gravity of any number of bodies may be found.

77. Let A, B, be any two bodies, and C their centre of gravity. If any point P be taken in the line A B, then PAXA+PBX BPC. (A + B), if the bodies are on the same side of the point P; and PA x A— P B × B = P C. (A + B), when they are on opposites sides of that point. (Fig. 41.)

=

=

=

For, by Art. 75, A x A CBX B C; and when the bodies are both on the same side of the point P, we have A C PA P C, and B C PC — PB; therefore A. (P APC) B. (PC-P B). Or PA x A -PCX APC × B-PB x B, transpose and PA × A+ PBxB=PCxA+PC x B = PC. (A + B). But if they are on contrary sides, then A CPAP C, and B CPB + PC; therefore A. (P A PC) — = B. (B P+P C), or PA x A-PCXA-PBX B + PCX B, and by transposition PAXA-PBX B = PCX A+ PC x B = P C. (A + B).

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Cor. 1. Hence the bodies A and B have the same force to turn the lever A P about the point P, as if they were both placed at C, their centre of gravity.

Cor. 2. In a manner similar to the above, it may be shown that whatever may be the number of bodies A, B, D, &c. we have PA x A+ PBX B+ PD × D, &c. = P C. (A+B+ D, &c.), when the bodies are all on the same side of the point P; but if on contrary sides, then, the difference of the products must be taken.

To find the centre of gravity of the area of a parallelogram (Fig. 42).

78. Bisect A B in E, and A D in G; draw E F parallel to A D, and G H parallel to A B; their intersection O is the centre of gravity required. For E F bisects every right line that can be drawn parallel to A B; therefore the centre of gravity must be somewhere in E F G H also bisects every right line that can be drawn parallel to A D ; therefore the centre of gravity must be somewhere in the line G H. But it can only be in both these lines when it is at O, their point of intersection; therefore O is the centre of gravity of the area of the parallelogram A B C D.

To find the centre of gravity of a triangle (Fig. 43).

79. Let A B C be any triangle, and from any two of its angles B and C draw the right lines C D and B E to bisect the opposite sides in D and E; then will their intersection G be the centre of gravity of the triangle.

For since C D bisects A B, it will bisect all the right lines that can be drawn parallel to A B; that is, all the parallel sections of the figure; therefore the centre of gravity of the triangle lies in C D. For the same reason, it also lies in B E; consequently it is in G, their common point of intersection.

Cor. Join DE, which will be parallel to B C, and equal to one half of it, and the triangles B G C and E G D are similar; therefore since D E = B C, D G = &D C, or CGCD. In the same manner, B G =

BE.

To find the centre of gravity of a trapezium (Fig. 44). 80. Divide the trapezium A B C D into two triangles by the diagonal B D, and find the centres of gravity E and F of these two triangles; then the centre of gravity of the trapezium will lie in the line E F, joining them. If we suppose the weight of each triangle to be collected in E

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