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cor. 3, the pressure of the fluid on the section is equal to the weight of ds; consequently the total pressure on the whole surface is equal to all the weights ds. But, if b denote the whole surface pressed, and g the depth of its centre of gravity below the top of the fluid; then, by art. 108, bg is equal to the sum of all the ds. Consequently the whole pressure of the fluid on the body or surface b, is equal to the weight of the bulk bg of the fluid, that is, of the column whose base is the given surface b, and its height is g the depth of the centre of gravity in the fluid.

249. PROP. The pressure of a fluid, on the base of the vessel in which it is contained, is as the base and perpendicular altitude; whatever be the figure of the vessel that contains it.

If the sides of the base be upright, so that it be a prism of a uniform width throughout, then the case is evident; for then the base supports the whole fluid, and the pressure is just equal to the weight of the fluid.

But if the vessel be wider at top than bot. tom; then the bottom sustains, or is pressed by, only the part contained within the upright lines ac, bD; because the parts Aca, BDb are supported by the sides AC, BD; and those parts have no other effect on the part aboc than keeping it in its position, by the lateral pressure against ac and bD, which

да

bB

C

D

bB

[graphic]

does not alter its perpendicular pressure downwards. And thus the pressure on the bottom is less than the weight of the contained fluid.

a AB b

ca D

And if the vessel be widest at bottom; then the bottom is still pressed with a weight which is equal to that of the whole upright column abDc. For, as the parts of the fluid are in equilibrio, all the parts have an equal pressure at the same depth; so that the parts within ce and do press equally as those in cd, and therefore equally the same as if the sides of the vessel had gone upright to a and b, the defect of fluid in the parts aca and BDb being exactly compensated by the downward pressure or resistance of the sides ac and BD against the contiguous fluid. And thus the pressure on the base may be made to exceed the weight of the contained fluid, in any proportion whatever.

So that, in general, be the vessels of any figure whatever, regular or irregular, upright or sloping, or variously wide and narrow in different parts, if the bases and perpendicular

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altitudes be but equal, the bases always sustain the same pressure. And as that pressure, in the regular upright vessel, is the whole column of the fluid, which is as the base and altitude; therefore the pressure in all figures is in that same ratio.

Corol. 1. Hence, when the heights are equal, the pressures are as the bases. And when the bases are equal, the pressure is as the height. But when both the heights and bases are equal, the pressures are equal in all, though their contents be ever so different.

C

Corol. 2. The pressure on the base of any vessel is the same as on that of a cylinder, of an equal base and height. Corol. 3. If there be an inverted syphon, or bent tube, ABC, containing two different fluids CD, ABD, that balance each other, or rest in equilibrio; then their heights in the two legs, AE, CD, above the point of meeting, will be reciprocally as their densities.

A

AE

D

B

For if they do not meet at the bottom, the part BD balances the part BE, and therefore the part CD balances the part AE; that is, the weight of CD is equal to the weight of AE. And as the surface at D is the same, where they act against each other, therefore AE: CD :: density of CD: density of AE.

So, if CD be water, and AE quicksilver, which is near 14 times heavier; then CD will be 14AE; that is, if AE be 1 inch, CD will be 14 inches; if AE be 2 inches, CD will be 28 inches; and so on.

250. PROP. If a body be immersed in a fluid of the same density or specific gravity; it will rest in any place where it is put. But a body of greater density will sink; and one of a less density will rise to the top, and float.

The body being of the same density, or of the same weight with the like bulk of the fluid, will press the fluid under it, just as much as if its space were filled with the fluid itself. The pressure then all around it will be the same as if the fluid were in its place; consequently there is no force, neither upward nor down. ward, to put the body out of its place. And therefore it will remain wherever it is put.

But if the body be lighter; its

[graphic]

pressure downward will be less than before; and less than the water upward at the same depth; therefore the greater force will overcome the less, and push the body upward

to A.

And if the body be heavier than the fluid, the pressure downward will be greater than the fluid at the same depth; therefore the greater force will prevail, and carry the body down to the bottom at c.

Corol. 1. A body immersed in a fluid, loses as much weight, as an equal bulk of the fluid weighs. And the fluid gains the same weight. Thus, if the body be of equal density with the fluid, it loses all its weight, and so requires no force but the fluid to sustain it. If it be heavier, its weight in the water will be only the difference between its own weight and the weight of the same bulk of water; and it requires a force to sustain it just equal to that difference. But if it be lighter, it requires a force equal to the same difference of weights to keep it from rising up in the fluid.

Corol. 2. The weights lost, by immerging the same body in different fluids, are as the specific gravities of the fluids. And bodies of equal weight, but different bulk, lose, in the same fluid, weights which are reciprocally as the specific gravities of the bodies, or directly as their bulks.

Corol. 3. The whole weight of a body which will float in a fluid, is equal to the weight of as much of the fluid, as the Immersed part of the body displaces when it floats. For the pressure under the floating body, is just the same as so much of the fluid as is equal to the immersed part; and therefore the weights are the same.

Corol. 4. Hence the magnitude of the whole body, is to the magnitude of the part immersed, as the specific gravity of the fluid, is to that of the body. For, in bodies of equal weight, the densities, or specific gravities, are reciprocally as their magnitudes.

Corol. 5. And because, when the weight of a body taken in a fluid, is subtracted from its weight out of the fluid, the difference is the weight of an equal bulk of the fluid; this therefore is to its weight in the air, as the specific gravity of the fluid is to that of the body.

Therefore, if w be the weight of a body in air,

w its weight in water, or any fluid,

s the specific gravity of the body, and

s the specific gravity of the fluid;

then ww: w:: 8: s, which proportion will give either of those specific gravities, the one from the other

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So that the specific gravities of bodies, are as their weights in the air directly, and their loss in the same fluid inversely.

s its spec. gravity;

s its spec. gravity;

Corol. 6. And hence, for two bodies connected together, or mixed together into one compound, of different specific gravities, we have the following equations, denoting their weights and specific gravities, as below, viz. H = weight of the heavier body in air, h weight of the same in water, Lweight of the lighter body in air, 1= weight of the same in water, cweight of the compound in air, c = weight of the same in water, w=the secific gravity of water. 1st, (h) s = 11w, 2d, (L - 1) s = LW, 3d, (cc)f=cw, 4th, a + L = C, 5th, h

H

6th, +

C,

f its spec. gravity;

Then, From which equations may be found any of the above quantities, in terms of the rest.

Thus, from one of the first three equations, is found the specific gra.

dividing the absolute

LW

vity of any body, as $== by LI -l' weight of the body by its loss in water, and multiplying by the specific gravity of water.

But if the body L be lighter than water; then I will be negative, and we must divide by L + ĺ instead of L 1, and to find we must have recourse to the compound mass c; and because, from the 4th and 5th equations, L-1=c C

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the absolute weight of the light body, by the difference be tween the losses in water, of the compound and heavier body, and multiply by the specific gravity of water. Or thus,

}=

CS

S/L
Hf

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, as found from the last equation.

Also if it were required to find the quantities of two ingredients mixed in a compound, the 4th and 6th equations would give their values as follows, viz.

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(f-s) s (s -8)f

c, and L =

(s-f)
(s- -s)f

C,

the quantities of the two ingredients H and L, in the compound c. And so for any other demand.

PROP. To find the specific gravity of a body.

251. CASE I.-When the body is heavier than water: weigh it both in water and out of water, and take the difference, which will be the weight lost in water. Then, by corol. 6, where B is the weight of the body out of

art. 250, s =

BW

B- -b

water, b its weight in water, s its specific gravity, and w the specific gravity of water. That is,

As the weight lost in water,

Is to the whole or absolute weight,
So is the specific gravity of water,
To the specific gravity of the body.*

EXAMPLE. If a piece of stone weigh 10lb, but in water only 6 lb, required its specific gravity, that of water being

1000?

Ans. 3077.

252. CASE II.-When the body is lighter than water, so that it will not sink: annex to it a piece of another body, heavier than water, so that the mass compounded of the two may sink together. Weigh the denser body and the compound mass, separately, both in water, and out of it; then find how much each loses in water, by subtracting its weight in water from its weight in air; and subtract the less of these remainders from the greater. Then say, by proportion,

As the last remainder,

Is to the weight of the light body in air,

So is the specific gravity of water,

To the specific gravity of the body.

That is, the specific gravity is s = by cor. 6, art. 250.

(c- c)

LW

(H — h)'

EXAMPLE. Suppose a piece of elm weighs 15lb. in air; and that a piece of copper, which weighs 18lb. in air and 161b. in water, is affixed to it, and that the compound weighs 6lb. in water; required the specific gravity of the elm?

Ans. 600.

253. CASE III.-For a fluid of any sort.-Take a piece of a body of known specific gravity; weigh it both in and out of the fluid, finding the loss of weight by taking the differ ence of the two; then say,

In the Lectures on Natural Fhilosophy, in the Royal Mil. Academy, Coates's Hydrostatic steelyard is employed for this purpose. It is an improvement upon the one described in Gregory's Mathematics for Practical Men,

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