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hat may at least be sup

entre of gravity of the wall, into mater may be supposed to be collected, and reetton ww, is effect will be the same as

were aspended from the point N of the ence.e put for the area of the wall sectic gravity; then A. n will be equal TA À .#. FN its effect on the lever to ng about the point F. And as this e to that of the triangle of earth, that it port, which was before found equal to herefore à. 7. FN = {AE3. m, in case of an

I

cath the breadth of the wall FE, and the ne centre of gravity M, will depend a wal. If the wall be rectangular, sottom; then FN=F, and the sequently the effot of the wall which must be = 43 . m, the resolution of this equation

newal 75 = AF√/=. So that the

==

3

wwways proportional to its height, and are te same height, whatever the slope Jut ce reauth must be made a little more than de, art may be more than a bare ba

1 real e fra is agy is about

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en ton as

early; and

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of its height. But if the wall be of stone, whose specific

m

gravity is about 2520; then

<=5125=

3n

nearly that is, when the rectangular wall is of stone, the breadth must be at least of its height.

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of the wall at the bottom, for an equilibrium in this case also. Where again FE is as AE.

And when this wall is

of brick, then FE = 704AEAE nearly. But when it

is of stone; then

m

=627 =

2n

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.

LET ABCD be half the arch, and DEFG the pier. From the centre of gravity K of the half arch draw KL perpendicular to the horizon. Then the weight. of the arch in direction KL will be to the horizontal push at A in direction LA, as KL to LA: for the weight of the arch in direction KL, the hori

E D

:B

zontal push or lateral pressure in direction LA, and the push

in direction KA, will be as the three sides KL, LA, KA, So that if A denote the weight or area of the arch; then AL A will be its force at A in the direction LA; and

KL

AL

KL

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• AG

A its effect on the lever GA to overset the pier,

or to turn it about the point F.

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Again, the weight or area of the pier, is as EF. FG; and therefore EF FG FG, or EF. 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 arch be in equilibrio, these two effects must be equal. Therefore we have EF. FG2 =

AL

KL

. AG. A; and consequently FG = √

is the thickness of the pier as required.

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Example 1. Suppose the arc ABM to be a semicircle; and that DC or AO or OB = 45, BC = 6, and AG 18 feet. Then KL will be found 40, AL 15 nearly, and EF = 69; also the area ABCD or A = 704. Therefore FG =

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Example 2. Suppose, in the segment ABM,

OB = 41, BC 61, and AG = 10. 35, AL = 15 nearly, and ABCD or

AM = 100,

Then EF

A = 842.

58, KL. = Therefore

FG

=24G. AL

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EF. KL

58

35

the thickness of the pier in this case.

ON

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 KL, LA may be measured by the scale.

ON THE CENTRES OF PERCUSSION, OSCILLATION, AND GYRATION.

238. THE CENTRE of PERCUSSION of a body, or a system of bodies, revolving about a point, or axis, is that point, which striking an immoveable object, the whole mass shall not incline to either side, but rest as it were in equilibrio, without acting on the centre of suspension.

239. The Centre of Oscillation is that point, in a body vibrating by its gravity, in which if any body be placed, or if the whole mass be collected, it will perform its vibrations in the same time, and with the same angular velocity, as the whole body, about the same point or axis of suspension.

240. The Centre of Gyration, is that point, in which if the whole mass be collected, the same angular velocity will be generated in the same time, by a given force acting at any place, as in the body or system itself.

241. The angular motion of a body, or system of bodies, is the motion of a line connecting any point and the centre or axis of motion; and is the same in all parts of the same revolving body. And in different unconnected bodies, each revolving about a centre, the angular velocity is as the absolute velocity directly, and as the distance from the centre inversely; so that, if their absolute velocities be as their radii or distances, the angular velocities will be equal.

PROPOSITION XLVII.

242. To find the Centre of Percussion of a Body, or System of Bodies.

And

LET the body revolve about an axis passing through any point s in the line SGO, passing through the centres of gravity and percussion, G and o. Let MN be the section of the body, or the plane in which the axis SGO moves. conceive all the particles of the body to be reduced to this plane, by perpendiculars let fall from them to the plane: a supposition which will not affect the centres G, O, nor the angular motion of the body.

M

Let

202

STATICS.

Aa per

Let a be the place of one of the particles, so reduced; join SA, and draw AP perpendicular to AS, and pendicular to Go: then AP will be the direction of A's motion as it revolves about s; and the whole mass being stopped at o, the body a will urge the point P, forward, with a force proportional to its quantity of matter and velocity, cr to its matter and distance from the point of suspension s; that is, 25 &.SA; and the efficacy of this force in a direc

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simler triangles; also, the effect of this force on the lever,
tion perpendicular to se, at the point P, is as
to turn it about o, being as the length of the lever, is as

A. S. 10 = 4.52. SO SP = A. Sa

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sa, by

A sa

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B.S.50 - B . SB2, and

= 4.5a. So — A. SA. In like manner, the forces of E

c.sc.so - c. sc2, &c.

and c, to turn the system about o, are as

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destroy

But, since the forces on the contrary sides of o
one another, by the definition of this force, the sum of the
positive parts of these quantities must be equal to the sum of
the negative parts,

that is, a. sa. so + B. sb.so + c

SC

.so &=

sc2 &c; and

A. SAB.
A. SAB. sɛ2 + c
sa+B. sb + c
A.

hence so =

SB2+c. Sc2 &c

sc &c.

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which is the distance of the centre of percussion below the axis of motion.

And here it must be observed that, if any of the points a, b, &c, fall on the contrary side of s, the corresponding product A. sa, or B . sb, &c, must be made negative.

943. Coral 1. Since, by cor. 3, pr. 40, A + B + c&c, or the bodrix the distance of the centre of gravity, SG, is = 4. B. $3 + C. sc &c, which is the denominator of the value of so; therefore the distance of the centre of

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SP

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