and the specific gravities of the prism and fluid s and s'; then it is easily found that

whose roots will give all the positions of equilibrium which the prism can take.

690.

Let a be the length of the straight line, and x, y,

and ay the three portions required; then the pressures on

691. Let x be the side of the cube; then the depth of

the part immersed is x

100, and the volumes of the whole

iceberg and of the part immersed are

x3 and x x (x - 100).

:. (Vince prop. XVI)

x3: x3× (x - 100):: 1.0263.9214

The same process of reasoning as is used in (687,) will shew whether the position be stable or unstable.

692.

Let s ounces be the specific gravity of the wood; that of water being 1000 ounces; then the specific gravity of lead is 11,000, and the magnitudes of the wood and lead are

respectively. Hence, since the water displaced weighs slb., we have

693.

This is nothing more than "Required the greatest segment of a sphere which can be described in a given cone.” If r denote the radius of the sphere, and x the depth of the segment in the cone, the volume of that segment is

where a the altitude of the cone, and b the radius of its base.

which, by substituting in (1), will give the radius of the required

sphere.

HYDRODYNAMICS.

000000

DISCHARGE OF FLUIDS.

694. THE velocity of the water issuing through the orifice ∞ √ (depth of the orifice in the fluid). See Vince prop. XXXVIII; and the quantity of the fluid injected ∞ time × velocity. But the cubes being equal, these quantities are the same, and the time c Hence the times required are as

1

vel.

695.

Let z be the distance from the surface of the water at which the orifice must be made; then the velocity with which the fluid begins to describe the parabola is that acquired down 2 (Vince prop. XXXIX), and the parameter of the parabola is 4z. Hence by the property of the parabola, and the question,

(32) 4z X (32—z)

and z =

696.

Generally, if A denote the area of the orifice, a the depth of the fluid to be discharged, and X the variable area of the descending surface; then the time of discharging the fluid is (Vince's Fluxion, p. 268)

t =

1

X

ANG SXdr

integrated between x = x, and x = 0.

√x

Hence, if r be the radius of the base of the cylinder, when its axis is horizontal, we have

X = 2r × 2(2rx — x2o)3 — 4r. (2rx — x2)1

DISCHARGE OF FLUIDS.

697.

By the preceding problem it readily appears that

the time of emptying any depth is

t = 21
× √x ∞ √x.
A√g

or x ∞ ť.

Hence

1 foot.

4 16: 1 space described in the last hour 22 space described in the two last hours 4 feet. Similarly (39) feet is the space described in the three last

42 16

hours; so that the spaces required are

1697, 94 = 54

13 and 1.

By 696, the time of discharging the depth x is

698.

a being the axis of the cylinder, and r the radius of its base.

3

3

Let x = 2r and r; then we get t: t':: (2): r* . (2o − 1),

699. Let p be the principal parameter of the generating parabola, P that corresponding to the diameter at the extremity of which the orifice is made.

Also let ẞ be the inclination of the axis to the horizon or to the surface of the fluid; then it may easily be shewn that the descending surface for any depth of the fluid x is an ellipse, whose semiaxes are

Again, if D be the given I distance from the surface to the base, we have

the whole depth of the fluid; which being substituted in the above expression, will give the time required.

700. Let l, a, r, be the slant side, axis and radius of the base of the given cone, and a the altitude of the surface of the fluid filling any portion of the cone; then, since this surface is a parabola, whose axis and base are

× f(2ar — lx) × √(2ar — lx2) dx