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But if the perpendicular fall without the base, as cb; then the body will tumble over on that side: because in turning on the point a, the centre c descends by describing the descending arc ce.
207. Corol. 1. If a perpendicular, drawn from the centre of gravity, fall just on the extremity of the base; the body may stand; but any the least force 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; because the centre of gravity has the less height to rise: which is the reason that a globe is made to roll on a smooth plane by any the least force. But the nearer the perpendicular is to the middle of the base or the broader the base is, the firmer the body stands.
208. Corol. 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 must be accounted the place of the body for into that point the whole matter of the body may be supposed to be collected, and therefore all the force also with which it endeavours to descend.
209. Corol. 3. From the property which the centre of gravity has, of always descending to the lowest point, is derived an easy echanical method of finding that centre.
Thus, if the body be hung up by any point A, and a plumb line AB be hung by the same point, it 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 on 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.
210. Or, if the body be suspended by two or more cords GF, GH, &c. then a plumb line from the point & will cut the body in its centre of gravity G.
211. 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, or over one side of a table, and moved backward and forward 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.
212. 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.
FOR, if the centre of gravity c be supported, the two bodies a and в will be supported,
and will 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
AB: CB: CA.
213. 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.
214. Corol. 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.
215. 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 weights A and B as before. So that, if a weight of 100lb. be laid on a bar at c, supported by two men at A and B, 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.
D. E F
216. 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 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 on any straight lever, then the prop. is in the centre of gravity.
217, Corol. 5. And though
a straight line, but scattered Pa
ner, the same property as in the
last corollary still holds true,
the several bodies, and their common centre of gravity, name-
218. If there be three or more Bodies, and if a line be drawn
That is, CE ED :: D: A+B &C.
FOR, Suppose the bodies A and B
217. 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.
220. If there be taken any Point x, in the Line passing through the Centres of Gravity 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 Cominon Centre of Gravity c from the same Point P.
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, PA. A PB. B + PD. D &c. PC A + B + D &c. where p is in any point whatever in the line ac.
And, by 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, Pа . ▲ + Pb. B + Pd. D &c. PC. A + B + D &C.
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
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 by the whole mass or body, that is, equal to all the ea . a, pb . b, Pd . D, &c. divided by the body or sum of particles A, B, D, &c.
224. To find the Centre of Gravity of any Body, or of any Sys-
225. Or, if b be any body, and QPR any plane; draw PAB
be PC =
the distance be thus found for two in
tersecting planes, they will give the
226. But the distance from one plane is sufficient for any
Thus, if the figure be a parallelogram, or a cylinder, or any prism whatever; then the axis or line, or plane rs, 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 =