rest in equilibrio. But, by the nature of the lever, when two bodies are in equilibrio about a fixed point c, they are reciprocally as their distances from that point; therefore

A B CB CA.

97. Corol. 1. Hence AB AC :: A+ B: B; or, the whole distance between 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.

98. Corol. 2. Hence also, CA. ACB. B; or the two products are equal, which are made by multiplying each body into its distance from the centre of gravity.

99. Corol. 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 respective 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 weight a and в as before. So that, if a weight of 100lb. be laid on a bar at c, supported by two men at a and в, distant from c, the one 4 feet, and the other 6 feet; then the nearer will bear the weight of 60lb. and the farther only 40lb. weight. This should be noted as a principle of extensive application.

B
A

C

D E

F

100. Corol. 4. Since the effect of any body to turn a lever about the fixed point c, is as that body and as its distance from that point; therefore, if c be the common centre of gravity of all the bodies A, B, D, E, F, placed in the straight line AF; then is CA. A+ CB. B = CD. D + CE. ECF. 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 into its distance from that centre. And if several bodies be in equilibrio on any straight lever, then the prop is in the centre of gravity.

the several bodies, and their common centre of gravity, namely, that ca. A + cb. в

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 aа, вb, dd, Ee, Ff.

102. PROP. 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.

That is, CE ED :: D: A + B, &c. For, suppose the bodies A and B to be collected into the common centre of gravity c, and let their sum be called s. Then, by the last prop. CE: EDDS or A + B, &c.

A

B

E

D

Corol. 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.

103. PROP. 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 into its distance. from that point, is equal to the product of the sum of the bodies multiplied into the distance of their common centre of gravity c from the same point P.

That is, PA .APB. B PC. A + B.

For, by art. 98th, CA. A=CB. B, that is, (PA-PC). A=(PC± PB). B; therefore, by adding,

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

C P

Р

B

104. Corol. 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.

105. Corol. 2. The same is also true of any number of bodies whatever, as will appear by cor. 4, art. 100. namely, PA. A + PB. B+ PD. D, &C. PC. (A+B+D, &c.) where P is in any point whatever of the line ac.

And, by cor. 5, art. 101, 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. (A + B + D, &c.)

106. Corol. 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

PC =

Pa. Apb. B+ Pd. D, &c.

A+B+D, &c.

of all the moments 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 ra. A, Pb. B, rd. D, &c. divided by the body or sum of particles A, B, D, &c.

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

Through any point r draw

a plane, and let pa, rb, rd, &c. be the distance of the bodies A, B, D, &c. from the plane; then, by the last cor. the distance of the common centre of gravity from the plane, will be

PC =

Pa

Pɑ . A + Pb . B + PD . D, &c.

A+B+D, &c.

Q

108. Or, if b be any body, and QPR any plane; draw PAR &c. perpendicular to QR, and through A, B, &c. draw innu. merable sections of the body b parallel to the plane QR. Let s denote any one of these sections, and d=PA, or PB, &c, its distance from the plane QR. Then will the distance of the centre of gravity of the body from the plane be sum of all the ds And if the

PC=

b

distance be thus found for two intersecting planes, they will give the point

in which the centre is placed.

Р

R

A

B

D

CE

b

F

G

109. But the distance from one plane is sufficient for any regular body, because it is evident that, in such a figure, the centre of gravity is in the axis, or line passing through the centres of all the parallel sections.

Thus, if the figure be a parallelogram, or a cylinder, or any prism whatever; then the axis or line, or plane Ps, which bisects all the sections parallel to QR, will pass through the centre of gravity of all those sections, and consequently through that of the whole figure. C. Then, all the sections s being equal, and the body b PS. s, the distance of the centre will be PC =

PA.S+ PB.s+&c.

b

PA+PB+PD &c.

PS. S

But PA PB+ &c. is the sum of an arithmetical progression, beginning at 0, and increasing to the greatest term PS, the number of the terms being also equal to ps; therefore the sum PA + PB + &c. = PS. PS; and consequently =rs; that is, the centre of gravity is in the middle of the axis of any figure whose parallel sections are equal.

PC=

PS. PS

PS

110. In other figures, whose parallel sections are not equal, but varying according to some general law, it will not be easy to find the sum of all the PA. 8, PB. s', PD . s“, &c. except by the general method of Fluxions; which case therefore will be best reserved till we come to treat of that doctrine. It will be proper, however, to add here some examples of another method of finding the centre of gravity of a triangle, or any other right-lined plane figure.

111. PROP. 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 & 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 sec

B

D

tions or component parts of the figure; and consequently the centre of gravity of the whole figure lies in the line AD. For the same reason, it also lies in the line CE. Consequently it is in their common point of intersection G.

112. Corol. The distance of the point e, is AG = AD, and CG = 4CF 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

But the

equal, because AE BE; consequently AG BF. triangles CDG, CBF are also equiangular, and CB being =2CD, therefore BF = 2GD. But BF is also = ᎪᏀ ; consequently AG = 2GD or AD. In like manner, co = 26E or CE.

E

113. PROP. To find the centre of gravity of a trapezium. Divide the trapezium ABCD into two A 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 Er be divided, in e, in the alternate ratio of the two triangles, namely, EGGF triangle BCD: triangle ABD, then G will be the centre of gravity of the trapezium.

B

H

I

114. 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 1 of these two triangles be likewise 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.

115. 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.

PROBLEMS FOR EXERCISE.

1. Find, geometrically, the centre of gravity of a trapezoid.

2. Find, geometrically, the centre of gravity of a triangular pyramid.

3. Infer, thence, the centre of gravity of any pyramid. 4. Find, algebraically, the centre of gravity of the frus. tum of a pyramid.

5. Let a sphere whose diameter is 4 inches, and a cone whose altitude is 8 inches, and diameter of its base 3 inches, be fastened upon a thin wire which shall pass through the centre of the globe and the axis of the cone; let the vertex of the cone be toward the sphere, and let its distance from the sphere's surface be 12 inches. Required the place of their common centre of gravity.