« PreviousContinue »
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
// PS PS
Ps; 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. S, 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.
228. 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 cen
tres 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. 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 Ag — CG & CE: or ag 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 But BF is also = AG; consequently AG-2GD, or AD. In like
manner, co = 2GE or ce.
230. To find the Centre of Gravity of a Trapezium.
DIVIDE the trapezium ABCD into two triangles, by the diagonal BD, aud 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: GE:: triangle BCD : triangle ABD, then & will be the centre of gravity of the trapezium.
231. 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 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.
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.
A I BC
LET ACDE be a vertical section of a bank of earth; and suppose, that if it were not supported, a triangular part of it, as ABE, would slide down, leaving it at what is called the natural slope be; but that, by means of a wall AEFG, it is supported, and kept in its place. It is FN E required to find the force of ABE, to slide down, and the dimensions of the wall AEFG, to support it.
Let н be the centre of gravity of the triangle ABE, 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 ABE, 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 HL represent the force of the triangle ABE in its natural direction HL, HK will denote its force in its direction нк, and PK the same force in the direction PK perpendicular to the lever EK, on which it acts. Now the three triangles EAB, HKL, HKP are all similar; therefore EB: EA ::
(HL: HL HK) the weight of the triangle EAB: w w, which will be the force of the triangle in the direction нк. Then, to find the effect of this force in the direction PK, it will be,
as нк: PK :: EB : AB:
w, the force at K, in direction PK, perpendicularly on the lever EK, which is equal to AE. But 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
m is the force acting at K in di
rection PK. And the effect of this pressure to overturn the the wall, is also as the length of the lever KE or ¦AE*: conEA3. AB2. sequently its effect is 6EB2
m, for the perpendicular force
The principle now employed in the solution of this 45th prop. is a little different from that formerly 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 = 3 AE instead of AE: but every thing else remaining the same as before. Indeed this problem 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 uncertain way in which loose earth may be supposed to act in such a case; which on account of its various circumstances of tenacity, friction, &c. will not perhaps admit of a strict mechanical certainty. On these accounts 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.
against K, to overset the wall AEFG. Which must be balanced by the counter resistance of the wall, in order that it may at least be supported.
Now, if м be the centre of gravity of the wall, into which its whole matter may be supposed to be collected, and acting in the direction мNW, its effect will be the same as if a weight w were suspended from the point N of the lever FN. Hence, if A be put for the area of the wall AEFG, and n its specific gravity; then an will be equal to the weight w, and N. FN its effect on the lever to prevent it from turning about the point F. And as this effort must be equal to that of the triangle of earth, that it may just support it, which was before
EA3. AB2 found equal to m; therefore A. n. FN 6EB2 in case of an equilibrium.
AB 2 -m,
234. But now, both the breadth of the wall FE, and the lever FN, or place of the centre of gravity u, will depend on the figure of the wall. If the wall be rectangular, or as broad at top as bottom; then FN = FE, and the area a AE. FE; consequently the effort of the wall a. 2. FN is 1FE2; AE n; which must be
m, the effort of
the earth. And the resolution of this equation gives the
perp. to FB. So that the breadth of the wall is always proportional to the prep depth AQ of the triangle ABE. But the breadth must be made a little more than the above value of it, that it may be more than a bare balance to the earth. If the angle of the slope E be 45°, as it is nearly in most cases; then FE= AE very nearly.
235. If the wall be of brick, its specific gravity is about 2000, and that of the earth about 1984; namely, m to n as 1984 to 2000; or they may be taken as equal; then.
1 very nearly; and hence FEAE, or AE nearly. That is, whenever a brick rectangular wall is made to support earth, its thickness must be at least or of its height.
But if the
the wall be of stone, whose specific gravity is about 2520;
= AE: that is, when the rectangular wall is of stone, the breadth must be at least of its height.
236. But if the figure of the wall be a triangle, the outer side tapering to a point at top. Then the lever FN FE, and the area a = FE. AÉ consequently its effort a. n. FN is FE2. AE. n; which being put
AQ√ for the breadth
of the wall at the bottom, for an equilibrium in this case also.
-If the angle of the slope E be 45°; then will FE be
AE√. And when this wall is of brick, then Fe
AE nearly. But when it is of stone; then✓ =-447
nearly; that is, the triangular stone wall must have its thickness at bottom equal to 4 of its height. And in like manner, for other figures of the wall and also for other figures of the earth.
237. To determine the Thickness of a Pier, necessary to support a given Arch.
of the arch BCDA, in the direction of gravity, this will resolve into KQ, the force acting against the pier perp. to the joint SR, and LQ the part of the force parallel to the same.