one side, as the base is to half that side. And therefore the weight of the fluid is to the pressure against all the four upright sides, as the base is to half the upright surface. And the same holds true also in any upright vessel, whatever the sides be, or in a cylindrical vessel. Or, in the cylinder, the weight of the fluid, is to the pressure against the upright surface, as the radius of the base is to double the altitude.

Moreover, when the rectangular prism becomes a cube, it appears that the weight of the fluid on the base, is double the pressure against one of the upright sides, or half the pressure against the whole upright surface.

Corol. 3. The pressure of a fluid against any upright surface, as the gate of a sluice or canal, is equal to half the weight of a column of the fluid whose base is the surface pressed, and its altitude the same as the altitude of that surface.

For the pressure on a horizontal base equal to the upright surface, is equal to that column; and the pressure on the upright surface is but half that on the base, of the same area.

So that, ifb be the breadth, and d the depth of such a gate or upright surface; then the pressure against it, is equal to the weight of the fluid whose magnitude is ¿bď = §AB. AD'.

If the fluid be water, a cubic foot of which weighs 1000 ounces, or 62} pounds; and if the depth AD be 12 feet, the breadth AB 20 feet; then the content, or AB. AD is 1440 feet; and the pressure is 1,440,000 ounces, or 90,000 pounds, or 40 tons weight nearly.

PROP. IV.

6. 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 vesse 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.

b B

But if the vessel be wider at top than bottom; 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 abDC than keeping it in its position, by the lateral pressure against aC and bD, which does not alter its perpendicular pressure downwards. And thus the pressure on the bottom is less than the weight of the contained fluid 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 Cc and d D press equally as those in cd, and therefore equally the same

a

A B b

c d

as if the sides of the vessel had gone upright to A and B, the defect of fluid in the parts A Ca and BDb being exactly compensated by the downward pressure or resistance of the sides aC and 6D 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 altitudes be but equal, the bases always sustain the same pressure. And as that pressu: e, 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 the 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 heights. But when both the heights and bases are equal, the pressures are equal in all, though their contents be ever so different.

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.

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.

B

C

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.

PROP. V.

7. 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 a greater density will sink; and one of a less density will ascend 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 was 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 upwards nor downwards, to put the body out of its place. And therefore it will remain wherever

the water upwards at the same depth; therefore the greater force will overcome the less, and push the body upwards to A.

And if the body be heavier, the pressure downwards will be greater than the fluid at the same depth; and therefore the greater force will prevail, and carry the body 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 as much of the fluid, as the immersed part of the body takes up, 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::: S, which proportion will give either of those specific gravitios, the one from the other

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

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,

=

S

its specific gravity ;“

L = weight of the lighter body in air,s its specific gravity;

= weight of the same in water,

C = weight of the compound in air,

c = weight of the same in water,
the specific gravity of water.

1st, (H-h) S = Hw,
2d, (L-l) 8 Lw,
3d, (C—c) ƒ = Cw,
4th, H+ L = C,
5th, h
H

= c,
C

6th, S

+ 8

specific gravity of water.

}si

Then

} its specific gravity

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 gravity of any body, as

8=

Lw

L by dividing the absolute weight of the

body by its loss in water, and multiplying by the

Lw

But if the body L be lighter than water; then I will be negative, and we must divide by L+ instead of L-, and to find we must have recourse to the compound mass C; and because, from the 4th and 5th equations, L—l = C — c-H-h, therefore s = that is, divide the absolute weight of the light body, by the difference between the losses in water, of the compound and heavier body, and multiply by the specific gravity of water. Or thus, s =

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.

the quantities of the two ingredients H and L, in the compound C.

And so

for any other demand.

PROP. VI.

To find the specific gravity of a body.

8. CASE L.-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

Bw
B-b

water. Then, by corol. 6, prop. 4. s = where B is the weight of the body out of water, b its weight in water, & 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 10 lb. but in water only 62 lb., required its specific gravity, that of water being 1000? Ans. 3077.

9. 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 com

874

pound 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,

As the last remainder,

Is to the weight of the light body in air,

So is the specific gravity in water,

To the specific gravity of the body.

That is, the specific gravity s =

Lw

(C — c) — (H — h)

by cor. 6, prop. 4.

Example. Suppose a piece of elm weighs 15 lb, in air; and that a piece of copper, which weighs 18 lb. in air and 16 lb. in water, is affixed to it, and that the compound weighs 6 lb. in water; required the specific gravity of the elm ?

Ans. 600.

10. 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 difference of the two; then say,

As the whole or absolute weight,

Is to the loss of weight,

So is the specific gravity of the solid,

To the specific gravity of the fluid.

That is, the specific gravity w =

B-b
B

s, by cor. 6, prop. 4.

Example. A piece of cast iron weighed 34 ounces in a fluid, and 40 ounces out of it; of what specific gravity is that fluid?

Ans. 1000.

PROP. VIL

11. To find the quantities of two ingredients in a given compound.

TAKE the three differences of every pair of the three specific gravities, namely, the specific gravities of the compound and each ingredient; and multiply each specific gravity by the difference of the other two. Then say,

As the greatest product,

Is to the whole weight of the compound,

So is each of the other two products,
To the weights of the two ingredients,
(S− s) S
(S — 8) ƒ

That is, the one H = by cor. 6. prop. 4.

C; and the other L =

(S― ƒ) $
(S-8) f

C,

Example. A composition of 112 lb. being made of tin and copper, whose specific gravity is found to be 8784; required the quantity of each ingredient, the specific gravity of tin being 7320, and that of copper 9000?

Answer-There is 100 lb. of copper, in the composition.
and consequently 12 lb. of tin,

12. SCHOLIUM.-The specific gravities of several sorts of matter, as found from experiments, are expressed by the numbers annexed to their names in the following Table: