Corollary 3. From the property which the centre of gravity has, of always descending to the lowest point, is derived an easy mechanical method of finding that centre.

For if the body be hung up by any point A, and a plumb line AB be hung by the same point, will pass through the centre of gravity; because that centre is not in the lowest point till it fall in the plumb line. Mark the line AB upon it. Then hang the body up by any other point D, with a plumb line DE, which will also pass through the centre of gravity, for the same reason as before; and therefore that centre must be at C where the two plumb lines cross each other.

Or, if the body be suspended by two or more cords, GF, GH, &c., then a plumb line from the point G will cut the body in its centre of gravity C.

H

58. Likewise, because a body rests when its centre of gravity is supported, but not else; we hence derive another easy method of finding that centre mechanically. For, if the body be laid on the edge of a prism, and moved backwards and forwards till it rest, or balance itself; then is the centre of gravity just over the line of the edge. And if the body be then shifted into another position, and balanced on the edge again, this line will also pass by the centre of gravity, and consequently the intersection of the two will give the centre itself.

PROP. XIII.

59. 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.

lever, when two bodies are in equilibrio about a

fixed point C, they are reciprocally as their distances from that point; there fore A: B: CB: CA.

Corollary 1. Hence AB: AC :: A+ B: B; or, the whole distance be tween the two bodies, is to the distance of either of them from the common centre, as the sum of the bodies is to the other body.

Corollary 2. Hence also, CA. A = CB. B; or, the two products are equal, which are made by multiplying each body by its distance from the centre of gravity.

Corollary 3. As the centre C is pressed with a force equal to both the weights A and B, while the points A and B are each pressed with the respec

tive weights A and B. Therefore, if the two bodies be both united in their common centre C, and only the ends A and B of the line AB be supported, each will still bear, or be pressed by the same weights A and B as before. So that, if a weight of 100 lb. be laid on a bar at C, supported by two men at A and B, distant from C, the one four feet, and the other 6 feet; then the nearer will bear the weight of 60 lb., and the farther only 40 lb. weight,

Corollary 4. Since the effect of any body to turn a lever about the fixed point

C, is as that body and its distance from that point; therefore, if C be the common centre of gravity of all the bodies

B

D

E

A, B, D, E, F, placed in the straight line AF; then is CA. A + CB . B = CD. D+ CE. E+ CF. F; or, the sum of the products on one side, equal to the sum of the products on the other, made by multiplying each body by its distance from that centre. And if several bodies be in equilibrium upon any straight lever, then the prop is in the centre of gravity.

B G

E

C

D

H

Corollary 5. And although the bodies be not situated in a straight line, but scattered about in any promiscuous manner, the same property as in the last corollary still holds true, if perpendiculars to any line whatever af be drawn through the several bodies and their common centre of gravity, namely, that Ca. A + Cb . B = Cd . D + Ce. E+ Cf. F. For the bodies have the same effect on the line af, to turn it about the point C, whether they are placed at the points a, b, d, e, f, or in any part of the perpendiculars Aa, Bb, Dd, Ee, Ff.

PROP. XIV.

60. If there be three or more bodies, and, if a line be drawn from any one body D to the centre of gravity of the rest C; then the common centre of gravity E of all the bodies, divides the line CD into two parts in E, which are reciprocally proportional as the body D to the sum of all the other bodies.

Corollary Hence we have a method of finding the common centre of gravity of any number of bodies; namely, by first finding the centre of any two of them, then the centre of that centre and a third, and so on for a fourth, or fifth, &c.

PROP. XV.

61. If there be taken any point P, in the line passing through the centres of two bodies; then the sum of the two products of each body, multiplied by its

distance from that point, is equal to the product of the sum of the bodies multiplied by the distance of their common centre of gravity C from the same point P.

That is, PA. A + PB . B = PC. A + B.

FOR, by the 38th, CA. A CB. B, that is,

PA PC. A = PC

by adding

PB. B, therefore,

PA. A + PB .B = PC. A+ B.

C

P

B

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

Or, if the line, with the bodies, move about the point P; the sum of the momenta of A and B, is equal to the momentum of the sum S or A + B placed at the centre C.

Corollary. 2. The same is also true of any number of bodies whatever, as will appear by cor. 4, prop. 38, namely, PA. A+ PB. B+ PD. D, &c. = PC . A + B + D, &c., where P is any point whatever in the line AC.

And, by cor. 5, prop. 38, the same thing is true when the bodies are not placed in that line, but any where in the perpendiculars passing through the points A, B, D, &c.; namely, Pa. A+ Pb. B+ Pd. D, &c., = PC x A+ B+ D, &c

that is, equal to the sum of all the forces

Corollary. 3. And if a plane pass through the point P perpendicular to the line CP; then the distance of the common centre of gravity from that plane, is Pa. A+ Pb. B+ Pd. D, &c. PC= A+ B+ D, &c. divided by the sum of all the bodies. Or, if A, B, D, &c., be the several particles of one mass or compound body; then the distance of the centre of gravity of the body, below any given point P, is equal to the forces of all the particles divided by the whole mass or body, that is, equal to all the Pa. A, Pb . B, Pd. D, &c. divided by the body or sum of the particles A, B, D, &c.

PROP. XVI.

62. To find the centre of gravity of a triangle.

FROM any two of the angles draw lines AD, CE, to bisect the opposite sides; so will their intersection G be the centre of gravity of the triangle.

For, because AD bisects BC, it bisects also all its parallels, namely, all the parallel sections of the figure; therefore AD passes through the centres of gravity of all the parallel sections or component parts of the figure; and consequently the centre of gravity of the whole figure lies in the line AD. son, it lies also in the line CE. of intersection G.

For the same rea

F

B

And consequently it is in their common point

Corollary. The distance of the point G, is AG = JAD, and CG = 3CE; or AG 2GD, and CG = 2GE.

For, draw BF parallel to AD, and produce CE to meet it in F. Then the triangles AEG, BEF are similar, and also equal, because AE = BE; consequently AG = BF. But the triangles CDG, CBF are also equiangular, and CB being = 2CD, therefore BF = 2GD. But BF is also = AG; consequently AG = 2GD or AD. In like manner, CG 2GE or 2CE.

=

PROP. XVII.

63. To find the centre of gravity of a trapezium.

DIVIDE the trapezium ABCD into two triangles, by the diagonal BD, and find E, F, the centres of gravity of these two triangles; then shall the centre of gravity of the trapezium lie in the line EF connecting them. And therefore if EF be divided, in G, in the alternate ratio of the two triangles, namely, EG GF triangle BCD: triangle ABD, then G will be the centre of gravity of the trapezium.

64. Or, having found the two points E, F, if the trapezium be divided into two other triangles BAC, DAC, by the other diagonal AC, and the centres of gravity H and I of these two triangles be also found; then the centre of gravity of the trapezium will also lie in the line HI.

So that, lying in both the lines, EF, HI, it must necessarily lie in their intersection G.

6.5. And thus we are to proceed for a figure of any greater number of sides, finding the centres of their component triangles and trapeziums, and then finding the common centre of every two of these, till they be all reduced into one only.

PROP. XVIII.

66. To find the centre of gravity of a triangular pyramid.

Let ABCD be a triangular pyramid, and to the point of bisection of DC draw AH, BH. Take HK = HA and HI=HB; then K and I will be the centres of gravity of the surfaces ACD and BCD respectively. Join KI, AI, BK. Now if the pyramid be resolved into elements, by means of planes parallel to BCD, it is evident that the line AI must pass through the centre of gravity of the pyramid, since I is the centre of gravity of BCD. For the same reason, the centre of gravity

of the pyramid is in the line BK, and because

AI and BK are in one plane, the centre of gravity of the pyramid ABCD must be at G, the point of intersection of AI and BK.

By similar triangles AGB and KGI, we have

AG: GI: AB: KI:: AH: KH:: 3:
AI.

.. AG: AI::3:4 or AG =

Cor. 1. Bisect AB in P and join HG and GP; then if PQ be drawn parallel to AI, we have BQ=QI=IH; but AI=2PQ and AI=4GI; hence PQ=2GI; and therefore HI : HQ : : IG : PQ; whence HGP is a straight line.

Cor. 2. Hence the centre of gravity of a triangular pyramid is the middle of the line joining the points of bisection of any two edges that do not meet.

Cor. 3. A solid bounded by plane surfaces may be divided by planes into a number of triangular pyramids; and if a plane be drawn parallel to the base, at a distance equal to 4 of the altitude of the pyramid, then the centre of gravity of the whole pyramid must be in this plane, for that of each of the triangular pyramids is in this plane. Hence, the line joining the vertex of the pyramid and the centre of gravity of its base will cut the plane in the centre of gravity of the whole pyramid.

PROP. XIX.

67. To find the centre of gravity of any body, or system of bodies.

Let v1, V2, V3, &c., denote the volumes of the material particles which compose the body or volume V; and x1, y1, Z1; X2, Y2, Z2, &c.; their co-ordinates in reference to three rectangular axes; then if X, Y, Z denote the co-ordinates of the centre of gravity, we have (by Prop. XV., Cor. 3.)

But to adapt these expressions to computation, we shall introduce the prin ciples of the Differential and Integral Calculus, and then the preceding expressions will take the form

X:

where x, y, z denote the distances of the centre of gravity of du from the three rectangular planes.

By means of these three equations, the determination of the centre of gravity may be effected; and when the figure is a plane surface, two of these equations are only required, since the centre of gravity is in the plane.

and if the arc be symmetrical on each side of the axis of x, we have y = 0, and then we have simply

X = Lxds