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one is straight and the other angular, and that the weight Q is 100 pounds, the arm DE of the lever 6 feet; its momentum will be 600. Then if we wish to ascertain at what distance Of a weight of 60 pounds must be placed so that it may be in equilibrio with the first, we shall have

Of=2x0d600=10 feet, the distance sought.

P

1265. Reciprocally, to find the effect of a force P placed at the point C of the other arm of the lever at a known distance from the fulcrum, and marked Of, in order to counterpoise Q placed at the distance Of, we have the formula P = 2x0d; and if we apply this

Of

formula to the numbers taken in the preceding example, the question will be, to find a force which placed at the distance of 10 feet from the fulcrum may be in equilibrio with a weight of 100 pounds at the end of the arm of a lever of 6 feet. We must in using the formula divide 600 by 10, and the quotient 60 will indicate the effect with which the force ought to act. If, instead of placing it in C, it is at B, 12 feet from the fulcrum, the force would be 600, which gives 50; and lastly, if we have to place it at a point 15 feet from the fulcrum, the force would be 609 = 40. Thus, in changing the situation of the force to a point more or less distant from the fulcrum, we must divide the momentum of the weight which is to be supported by the distance from the fulcrum taken perpendicularly to its direction.

OF THE CENTRE OF GRAVITY.

1266. The centre of gravity of a body is a certain point within it on which the body, if | freely suspended, will rest in any position; whilst in other positions it will descend to the lowest place to which it can get. Not only do whole bodies tend by their weight to assume a vertical direction, but also all the parts whereof they consist; so that if we suspend any body, whatever be its form, by means of a string, it will assume such a position that the thread produced to the internal part of the body will form an axis round which all the parts will remain in equilibrium. Every time that the point of suspension of a body is changed, the direction of the thread produced exhibits a new axis of equilibrium. But it is to be remarked, that all these axes intersect each other in the same point situate in the centre of the mass of the body, supposing it composed of homogeneous parts but sometimes out of the mass of the body, as in the case of bodies much curved, this point is the centre of gravity. 1267. It is therefore easy to perceive that for a body to be in a state of rest its centre of gravity must be supported by a vertical force equal to the resultant of all the forces that affect it, but acting in a contrary direction. So in figs. 520. and 523., the weight supported by the forces AB and BC which draw or push, will be equally supported by a vertical force represented by the diagonal DB of the parallelogram which expresses the resultant of the forces.

1268. An acquaintance with the method of finding centres of gravity is indispensable in estimating the resistances, strains, and degree of stability of any part of an edifice. There arise cases in which we may cast aside all consideration of the form of a body, especially too when it acts by weight, and suppose the whole figure collected in the centre of gravity. We may also, for the sake of simplifying operations, substitute a force for a weight.

OF THE CENTRE OF GRAVITY OF LINES.

1269. A straight line may be conceived to be composed of an infinite number of points, equally heavy, ranged in the same direction. After this definition, it is evident that if it be suspended by the middle, the two parts, being composed of the same number of equal points placed at equal distances from the point of suspension, will be necessarily in equilibrium; whence it follows that the centre of gravity of a right line is in the middle of its length.

1270. The points in a curve line not being in the same direction, the centre of its volume cannot be the same as its centre of gravity; that is to say, that a curve suspended by the middle cannot be supported in equilibrio but in two opposite situations; one when the branches of the curve are downwards, and the other when they are upwards, so that the curve may be in a vertical plane.

1271. If the curve is the arc of a circle ADB (fig. 533.), it is easy to see that from the uniformity of its curvature, its centre of gravity will be found in the right line DC drawn from the centre C to the middle D; moreover, if we draw the chord AB, the centre of gravity will be found between the points D and E.

D

G

A

Fig. 533.

B

1272. Let us suppose that through all the points of the line DE parallels to the chord AB be drawn, terminated on each side by the curve; and let us imagine that each of these

lines at its extremities bears corresponding points of the curve; then the line DE will loaded with all these weights; and as the portions of the curve which answer to ea parallel AB go on increasing as they approach D, the centre of gravity G will be near the point D than to the point E.

ABXCD
ABD

1273. To determine the position of this point upon the radius CD which divides the a into two equal parts, we must use the following proportion: the length of the are ABD is the chord AB, as the radius CD is to the fourth term x, whose value is That in order to obtain upon the radius DC the distance CG of the centre of gravity fro the centre of the arc of the circle, the chord AB must be multiplied by the radius CD ar divided by the length of the arce ABD.

1274. When the circumference of the circle is entire, the axes of equilibrium bein diameters, it is manifest that their intersection gives the centre of the curve as the centre gravity. It is the same with all entire and symmetrical curves which have a centre, an with all combinations of right lines which form regular and symmetrical polygons.

OF THE CENTRE OF GRAVITY OF SURFACES.

1275. In order that a centre of gravity may be assigned to a surface, we must, as in th case of lines, imagine them to be material, that is, consisting of solid, homogeneous, an heavy particles.

1276. In all plane smooth surfaces, the centre of gra

B

A

Fig. 534.

D vity is the same as that of the volume

of space; thus the centre of gravity
G (figs. 534, 535, 536.), of a square
of a rectangle, or of a parallelogram,
is determined by the intersections of
its diagonals AD, BC.

The centre of gravity of a regular polygon, composed of an equal or unequal number of sides, is the same as that of a circle within which it may be inscribed.

A

Fig. 535.

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D

[graphic]

1277. In order to find the centre of gravity of any triangle, bisect each of the sides, and from the points of bisection draw lines to the opposite angles; the point of intersection with each other of these lines will be the centre of gravity sought; for in the supposition that the surface of the triangle is composed of lines parallel to its sides, the lines AE BF, and CD (fig. 537.) will be the axes of equilibrium, whose intersection at G gives

the centre of gravity. We shall moreover find that this point is at one third of the distance from the base of each of the axes; so that, in fact, it is only necessary to draw a line from the point of bisection of one of the sides to the opposite angle, and to divide it into three equal parts, whereof that nearest the base determines the centre of gravity of the triangle.

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Fig. 536.

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B

K

Fig. 538.

1278. To find the centre of gravity of any irregular rectilinear surface, such as th pentagon, fig. 538., let it be divided into the three triangles, AED, ABC, ADC (fig. 538.) and by the preceding rule determine their centres of gravity F, G, H. Then draw th two lines NO, OP, which form a right angle surrounding the polygon. Multiply th area of each triangle by the distance of its centre of gravity on the line ON, indicated b Ff Gg, Hh, and divide the sum of these products by the entire area of the pentagon, an this will give a mean distance through which an indefinite line IK parallel to ON is to b drawn. Conducting a similar operation in respect of the line OP, we obtain a new mea distance for drawing another line LQ parallel to OP, which will intersect the first in th point M, the centre of gravity of the pentagon.

The centre of gravity of a sector of a circle AEBC (fig. 539.) must be upon the radiu CE which divides the arc into two equal parts. To determine from the centre C, a

what distance the point G is to be placed, we must multiply twice the radius CE by the chord AB, and divide the product by thrice the length of the are AEB. The quotient is the distance CG from the centre C of the circle of the centre of gravity of the sector. 1279. To find the centre of gravity of the crown portion of an arch DAEBF (fig. 540.) comprised between two concentrie axes, we must

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E

Fig. 539.

A

B

of their respective centres of gravity from the common centre C.

Fig. 510.

3. Subtract the smaller product from the greater, and divide the remainder by the area of DAEBF; the quotient will give the distance of the centre of gravity G from the centre C.

E

1280. To determine the centre of gravity of the segment AEB; subtract the product of the area of the triangle ABC (fig. 541.) multiplied by the distance of its centre of gravity from the centre C, from the product of the area of the sector, by the distance of its centre of gravity from the same point C, and divide the remainder by the area AEB; the quotient expresses the distance of the centre of gravity G of the segment from the centre C, which is to be set out on the radius, and which divides the segment into two equal parts.

9

Fig. 541.

B

It would, from want of space, be inconvenient to give the strict demonstrations of the above rules; nor, indeed, is it absolutely necessary for the architectural student. Those who wish to pursue the subject au fond, will, of course, consult more abstruse works on the matter. We will merely observe, that whatever the figure whose centre of gravity is sought, it is only necessary to divide it into triangles, sectors, or segments, and proceed as above described for the pentagon, fig. 588.

OF THE CENTRE OF GRAVITY OF SOLIDS.

1281. It is supposed in the following considerations, that solids are composed of homogeneous particles whose weight in every part is uniform. They are here arranged under two heads, regular and irregular.

1282. Regular solids are considered as composed of elements of the same fgure as their base, placed one upon the other, so that all their centres of gravity are in a vertical line, which we shall call the right axis. Thus parallelopipeds, prisms, cylinders, pyramids, cones, conoids, spheres, and spheroids have a right axis, whereon their centre of gravity is found.

1283. In parallelopipeds, prisms, cylinders, spheres, spheroids, the centre of gravity is in the middle of the right axis, because of the similarity and symmetry of their parts equally distant from that point.

1284. In pyramids and cones (figs. 542, 543.), which diminish gradually from the base to the apex, the centre of gravity is at the distance of one fourth of the axis from the base.

1285. In paraboloids, which diminish less on account of their curvature, the centre of gravity is at the height of one third the axis above the base.

To find the centre of a pyramid or of a truncated cone (figs. 542, 543.), we must first multiply the cube of the entire cone or pyramid by the distance of its centre of gravity from the vertex. 2. Subtract from this product that of the part MSR which is cut off, by the distance of its centre of gravity from the apex. 3. Divide this remainder by the cube of the truncated pyramid or cone;

Fig. 542.

S

[graphic]
[graphic]

Fig. 543

the quotient will be the distance of the centre of gravity G of the part of the truncated

rone or pyramid from its apex.

1286. The centre of gravity of a hemisphere is at the distance of three eighths of the radius from the centre.

1287. The centre of gravity of the segment of a sphere (fig. 544.) is found by the following proportion: as thrice the radius less the thickness of the segment is to the diameter less three quarters the B thickness of the segment, so is that thickness to a fourth term which expresses the distance from the vertex to the centre of gravity, set off on the radius which serves as the axis.

1288. Thus, making r the radius, e the thickness of the segment, and = the distance sought, we have, according to La Caille,

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L

Fig. Di4.

Suppose the radius to be 7 feet, the thickness of the segment 3 feet, we shall have 8x7x3-3x9

x=

12x7-3x4'

which gives x=1+3=1 + 23, equal the distance of the centre of gravity from its vertex on the radius.

E

H

K

1289. To find the centre of gravity of the zone of a sphere (fig. 545.), the same sort of operation is gone through as for truncated cones and pyramids; that is, after having found the centre of gravity of the segment cut off, and that in which the zone is comprised, multiply the cube of each by the distance of its centre of gravity from the apex A, and subtract- B ing the smaller from the larger product, divide the remainder by the cube of the zone. Thus, supposing, as before, the radius AC=7, the thickness of the zone 2, and that of the segment cut off = 1}, we shall find the distance from the vertex of the centre of gravity of this last by the formula x= which in this case gives x= +x2x7x13x; and pursuing the investigation, we have x=18},

8re-See
4 Sr-e)

103

L

Fig. 545.

That of the centre

4(21--13) which will be the distance of the centre of gravity from the vertex A. of gravity of the segment in which the zone is comprised will, according to the same formula, be r= 8x7x31-3x12, which gives x=2+ for the distance of the centre of gravity

4.3x7-3

from the same point A.

1290. The methods of finding the solidities of the bodies involved in the above inves tigation are to be found in the preceding section, on Mensuration.

OF THE CENTRE OF GRAVITY OF IRREGULAR SOLIDS.

C

B

A

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1291. As all species of solids, whatever their form, are susceptible of division inte pyramids, as we have seen in the preceding observations, it follows that their centres of gravity may be found by following out the instructions already given. Instead of two lines at right angles to each other, let us suppose two vertical planes NAC, CEF (fig. 546.), between which the solid G is placed. Carrying to each of those planes the momenta of their pyramids, that is, the products of their solidity, and the distances of their centres of gravity, divide the sum of these products for each plane by the whole solidity of the body, the quotient will express the distance of two other planes BKL, DHM, parallel to those first named. Their intersection will give a line IP, or an axis of equilibrium, upon which the centre of gravity of the solid will be found. To determine the point G, imagine a third plane NOF perpendicular to the preceding ones, that is, horizontal; upon which let the solid be supposed to stand. of this plane let the momenta of the pyramids be found by also multiplying their solidity by the distance of their centres of gravity. Lastly, dividing the sum of these products by the solidity of the entire body, the quotient gives on the axis the distance PG of this third plane from the centre of gravity of the irregular solid.

Mechanically, where two of the surfaces of a body are parallel, the mode of finding the centre of gravity is simple. Thus, if the body be hung up by any point A (figs. 547, 548.), and a plumb line AB be suspended from the same point, it will pass through

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the centre of gravity, because that centre is not in the lowest point till it fall in the plumb line. Mark the line AB upon 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. Therefore the centre of gravity will be at C, where the lines cross each other. 1292. We have, perhaps, pursued this subject a little further than its practical utility in architecture renders necessary; but cases may occur in which the student will find our extended observations of service.

OF THE INCLINED PLANE.

1293. That a solid may remain in a perfect state of rest, the plane on which it stands must be perpendicular to the direction of its gravity; that is, level or horizontal, and the vertical let fall from its centre of gravity must not fall out of its base.

1294. When the plane is not horizontal, solids placed on it tend to slide down or to

overturn.

1295. As the surfaces of bodies are more or less rough, when the direction of the centre of gravity does not fall without their base, they slide down a plane in proportion to their roughness and the plane's inclination.

1296. Thus a cube of hard freestone, whose surfaces are nicely wrought, does not slide down a plane whose inclination is less than thirty degrees; and with polished marbles the inclination is not more than fifteen degrees.

1297. When a solid is placed on an inclined plane, if the direction of the centre of gravity falls without its base, it overturns if its surfaces are right surfaces, and if its surface is convex it rolls down the plane.

1298. A body with plane surfaces may remain at rest after having once overturned if the surface upon which it falls is sufficiently extended to prevent its centre of gravity falling within the base, and the inclination be not so great as to allow of its sliding on.

1299. Solids whose surfaces are curved can only stand upon a perfectly horizontal plane, because one of the species, as the sphere, rests only on a point, and the other, as cylinders and cones, upon a line; so that for their continuing at rest, it is necessary that the vertical let fall from their centre of gravity should pass through the point of contact with and be perpendicular to the plane. Hence, the moment the plane ceases to be horizontal the direction of the centre of gravity falls out of the point or line of contact which serves as the base of the solid, and the body will begin to roll; and when the plane on which they thus roll is of any extent they roll with an accelerated velocity, equal to that which they would acquire in falling directly from the vertical height of the inclined plane from the point whence they first began to roll.

1800. To find the force which is necessary to support a convex body upon an inclined plane, we must consider the point of contact F (figs. 549, 550.) as the fulcrum of an an

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gular lever, whose arms are expressed by the perpendiculars drawn from the fulcrum to the direction of the force CP and the weight CD, which in the case of fig. 549., where the force which draws the body is parallel to the plane,

P: N:: FC: FD.

Now as the rectangular triangle CFD is always similar to the triangle OSH, which forms the plane inclined by the vertical SO and the horizontal line OH, the proportion will stand as follows:

P: N::OS: SH.

In the first case, to obtain an equilibrium, the force must be to the weight of the body as the height OS of the inclined plane to its length SH.

1301. In the case where the force is horizontal (fig. 550.) we have, similarly,

P: N::FA: FD,

and P: N::OS: OH.

In this last case, then, the force must be to the weight of the solid in proportion to the height

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