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PROPOSITION XLI.

220. If there be taken any Point P, in the Line passing througla

the Centres of two Bodies ; then the sum of the two Producis, of each Body multiplied by its Distance from that Point, is equal to the Product of the Sum of the Bodies mul. tiplied 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-PB . B;

C P

А. therefore, by adding, PA.A + PB. B = PC A + B.

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221. 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 sums or A + B placed at the centre c.

222. Corol. 2. The same is also true of any number of bodies whatever, as will appear by cor. 4, prop. 39, namely, PAA + PB , B + PD . D &c. = PG. A + B + D &c. where P is in any point whatever in the line ac.

And, hy cor. 5, prop 39, 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.

PC =

223. 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 PA. A + PB, B+ Pd.D &c

that is, equal to the sum A+ B + D &c of all the forces 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 hy 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 particles A, B, D, &c.

PROPOSITION PROPOSITION XLII.

224. To find the Centre of Gravily of any Body, or of any sys.

tem of Bodies.

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225. Or, if b be any body, and QPR any plane ; draw PAB &c perpendicular to ar, and through A, B, &c. draw innumerable sections of the body b parallel

P

R to the plane qn: Let denote any of Qthese sections, and d = PA, or PB, &c. its distance from the plane QR. Then

A will the distance of the centre of gra

B vity of the body from the plane be

D sum of all the d's

And if the b

F distance be thus found for two inter-,

G secting planes, they will give the point in which the centre is placed.

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PC

Q PR

226. 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 i hrough 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 sec- A lions parallel to ar, will pass through the

B centre of gravity of all those sections, and consequently through that of the whole figure -C c. Then, all the sections s being equal, and the body b PS . 8, the distance of the centre will be pc =

IS

PA.S+ PB.s + &c

PA + PB + PD &c.

X8=

PA + PB + &c b

But

PS

PS :

PS

PC a

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PS

But pa + PR + &c. is the sum of an arithmetical prom gression, beginning at 0, and increasing to the greatest term Ps, the number of the terms being also equal 10 PS ; therefore the sum PA + PB + &c. = PS . Ps; and consequently

jps; that is, the centre of gravity is in the middle of the axis of any figure whose parallel sections are equal.

227 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 . :', PD .8", &c, except by the general inethod 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.

PROPOSITION XLII.

228. To find the Centre of Gravity of a Triangle.

F
E

G

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; there

B D fore ad passes through the centres of gravity of all the parallel sections or component parts of the figúre ; 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.

229. Corol. The distance of the point g, is AGAD, and cg = CE: or ag = 2GD, and cg = 262,

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 = 3F. But the triangles CDG. CBF are also equiangular, and ce being > 200, therefore BF = 2GD. But BF is also = AG; cons#quently ae = 260 or fad. In like manner, co = 2GE or 'CE.

PROPOSITION

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PROPOSITION XLIV. 230. To find the Centre of Gravity of a Trapezium. Divide the trapezium ABCD into А two triangles, by the diagonal BD, and

E find E, F, the centres of gravity of

I these iwo triangles ; then shall the centre of gravity of the trapezium lie in the line EF connecting them. And

H

B therefore if EF be divided, in g, in the alternate ratio of the two triangles, namely, EG : GE :: triangle BCD : triangle

ABD, then G will be the centre of gravity of the trapezium.

231. Or, having found the two points E, F, if the trapé zium be divided into two other triangles BAC, DAC, by the other diagonal ac, and the centres of gravity H and 1 of these iwo 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.

232. 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.

Of the use of the place of the centre of gravity, and the nature of forces, the following practical problems are added ; viz. to find the force of a bank of the earth pressing against a wall and the force of the wall to support it; also the push of an arch, with the thickness of the piers necessary to support it ; also the strength and stress of beams and bars of timber and metal, &c.

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233. To determine the Force with which a Bank of Earth, or

such like, presses against a Wall, and the Dimensions of the Wall necessary to Support it.

LET ACDe be a vertical section of a bank of earth ; and suppose, that if it G A I BC were not supported, a triangular part of it, as Ale, would slide down, leaving

H it at what is called the natural slope BE; MIK but that, by means of a wall AEFG, it is supported, and kept in its place. It is required to find the force of ABE, FNE

D to slide down, and the dimensions of the wall AEFG, to support it.

Let

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HKP

EA

Let u be the centre of gravity of the triangle age, through which draw Khi parallel to the slope face of the earth BE. Now the centre of gravity h may be accounted the place of the triangle ale, or the point into which it is all collected. Draw hl parallel, and KP perpendicular to AE, also kl prep. to IK or BE Then if nl represent the force of the triangle ABE in its natural direction HL, HK will denote ils force in its direction hk, and pk the same force in the direction PK, perpendicular to the lever Ek, on which it acts. Now the three triangles EAB, AKL are all similar ; therefore EB : EA ::(AL: HK : :) w the weight of the triangle EAB ;

w, which will be the force of the triangle in the direction AK.

Then, to find the effect of this force in the direction pk, it will be, as He: PK :: EB : AB :: <w: the force at k, in direction PK, perpendicularly on the lever EK, which is equal to AE. But į AE , AB is the area of the triangle ABE; and if m be the specific gravity of the earth, then & AE AB

m is as its weight. Therefore

EA2 AB2
AE . AB = -m is the force acting at & in

2E B2
direction PK. And the effect of this pressure to overturn
the wall, is also as the length of the lever ki orj Ar* : con-

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EA

EA. AB

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EB

EB2

ar

EA. AS

om

.

EB2

The principle now employed in the solution of this 45th prop. is a little different from that formly used ; viz. by considering the triangle of earth ABE as acting by lines IK, &c. parallel to the face of the slope BE, instead of acting in directions parallel to the horizon AB ; an alteration which gives the length of the lever ek, only the half of what it was in the former way, viz. Ek=ŠAE instead of fae: but every thing else remaining the same as before. Indeed this prob. lem has formerly been treated on a variety of different hypotheses, by Mr. Muller, &c. in this country, and by many French and other authors in other countries. And this has been chiefly owing to the urr certain way in which loose earth may be supposed :o act in such a case ; which on account of its various circumstances of tenacity, friction, &c. will not perhaps admit of it strict mechanical certainty. On these acounts it seems probable that it is to good experiments only, made on different kinds of earth and walls, that we may probably hope for a just and satisfactory solution of the problem.

The above solution is given only in the most simple case of the problem. But the same principle may easily be extended to any other case that may be required, either in theory or practice, either with walls or banks of earth of different figures, and in different situations.

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