PROPOSITION XLIV.

230. To find the Centre of Gravity of a Trapezium.

therefore if Er be divided, in G, in the alternate ratio of the two triangles,

C

namely, EG: Gr:: 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 I of these two triangles be also found; then the centre of gravity of the trapezium will also lie in the line Hr.

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 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.

PROPOSITION XLV.

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 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 required to find the force of ABE, to slide down, and the dimensions of the wall AEFG; to support it.

GAI BC

D

τω

Let

Let u 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 perp. 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 HK, 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: HK ::) w the weight of the triangle EAB : EA w, which will be the force of the triangle in the direc

EB

tion HK. Then, to find the effect of this force in the direc

EA

tion PK, it will be, as HK: PK::EB:AB:: w:

EB

the force at x, in direction PK, perpendicularly on the lever EK, which is equal to AE. But AB. 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

2EB2

m is the force acting at k in

direction PK. And the effect of this pressure to overturn the wall, is also as the length of the lever KE Or AE*: con

* 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, раrallel 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 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.

sequently sequently its effect is

EA3. AB2

6EB2

-m, for the perpendicular force 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 u be the centre of gravity of the wall, into which its whole matter may be supposed to be collected, and acting in the direction MNW, 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 A n will be equal to the weight w, and A 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 found equal to

AE3. AB2 6EB2

EA3. AB2
6EB2

m; therefore A.n. FN =

m, in case of an equilibrium.

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234. But now, both the breadth of the wall Fe, and the lever FN, or place of the centre of gravity M, 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. n. FN is = m, the effort of the earth. And the resolution of this equation gives the breadth of the wall FE =

FE. AE.n; which must be =

AB. AE

EB

AE3. AB2
6EB2

m 3n

AQ

m

3n

-, drawing

AQ perp. to EB. So that the breadth of the wall is always proportional to the perp. 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;

235. If the wall be of brick, its specific gravity is about 2000, and that of earth about 1984; namely, m to nas 1984

to 2000; or they may be taken as equal; then = 1 very

n

nearly; and hence FE = AE, Or AE nearly. That is, wherever a brick rectangular wall is made to support earth, its thickness must be at least of its height. But if

or

the

the wall be of stone, whose specific gravity is about 2520;

m

m

then ==, and = = 895; hence FE = 358 AE

n

n

= 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.AE; consequently its effort A. n. FN is = FE.AE.n; which being put AE3. AB2

6BE2 AB. AF

EB

-m, the equation gives FE =

m

A

BC

F NE

D

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

m

m

= AE. And when this wall is of brick, then √2 2n

n

m

12

=

FE = AE nearly. But when it is of stone; then V
*447 = nearly: that is, the triangular stone wall must
have its thickness at bottom equal to + of its height. And
in like manner, for other figures of the wall and also for
other figures of the earth.

PROPOSITION XLVI.

237. To determine the Thickness of a Pier, necessary to support a Given 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. Now Ko de

FG.FG, or DF. FG2, is its effect on the lever FG, to prevent the pier from being overset; supposing the length of the pier, from point to point, to be no more than the thickness of the arch.

But that the pier and the arch may be in equilibrio, these two efforts must be equal. Therefore we have DE. FG2 =

KQ.GP.A

KL

-, an equation, by which will be determined the thickness of the pier FG; A denoting the area of the half arch BCDA *.

Example 1. Suppose the arc ABM to be a simicircle; and that CD or on or ob = 45, BC = 7 feet, AF = 20. Hence AD= 52, DF = GE = 72. Also by measurement are found OK = 50.3, KL = 40.6, Lo = 29.7, TD = 30.87, KQ = 24, the area BCDA = 750 = A; and putting FG = x the breadth of the pier.

Then TE = TD + DE = 30.874x, and KL: LO::TÈ:
Ev = 22.58 + 0.73x,

then GE - EV=GV =49.42

lastly OK: KL :: GV: GP = 39.89

73x,

59x.

These values being now substituted in the theorem DF.

FG2 =

KQ. GP. A

KL

-, give 36x2 = 17665 – 261·5x, or x2+

* Note. As it is commonly a troublesome thing to calculate the place of the centre of gravity k of the half arch ADCB, it may be easily, and sufficiently near, found mechanically in the manner described in art. 211, thus: Construct that space ADCB accurately by a scale to the given dimensions, on a plate of any uniform flat substance, or even card paper; then cut it nicely out by the extreme lines, and balance it over any edge or the side of a table in two positions, and the intersection of the two places will give the situation of the point k; then the distances or lines may be measured by the scale, except those depending on the breadth of the pier FG, viz. the lines as mentioned in the examples.

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