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

2msac cos. ß 24

s'ac 2m cos. a X 23 +


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

[ocr errors]
[ocr errors]

690. Let a be the length of the straight line, and x, y, and a 2 + y the three portions required; then the pressures on them are

a-x+y +

a – Х


[ocr errors]


[merged small][merged small][ocr errors][merged small]

691. Let 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)

23 : 2 x (x – 100) :: 1.0263 : .9214 which gives

x=978.36 feet.

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

12 x 16

cubic feet,

[blocks in formation]

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

[ocr errors][subsumed][merged small][subsumed][merged small][merged small]
[ocr errors]

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

x3 # (rx2

(3r – X).

[ocr errors]
[ocr errors]



[merged small][ocr errors][merged small][merged small][ocr errors][merged small]

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

:. by the question


= maximum.
Hence by the rule we get

2ab. (b + Na + b)

36(b + n ap + b*) + a? which, by substituting in (1), will give the radius of the required sphere.




694. The velocity of the water issuing through the orifice OC V (depth of the orifice in the fluid). See Vince prop. XXXVIII; and the quantity of the fluid injected o time x velocity. But the cubes being equal, these quantities are the same, and the time o

Hence the times required are as vel.


[ocr errors][merged small]

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 z (Vince prop. XXXIX), and the parameter of the parabola is 4z. Hence by the property of the parabola, and the question,

(32)2 = 42 x (32 — )

32 and % =

= 16. 2

696. Generally, if A denote the area of the orifice, x 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)


x integrated between x = x, and x = 0.

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

X= 2r x 2(2r« — 2)* = 4r. (2x - 2)


and when the axis is vertical

X' = Tops
x 4r S v 2r- x x dx=

sr_ {(3r) —(25—2)}} ANI

3A / 8

(2r)}, when x = 2r ;

dx 27r2
and t =

AW g

AJ 9

V2r, when x= 2r.
::t:ť ::

: *r :: S: 37r.

[ocr errors]
[ocr errors]


697. By the preceding problem it readily appears that the time of emptying any depth is

t = 271

x V x a ↑x.

[ocr errors]

or x oc ť". Hence

48 : 16 :: 1 : space described in the last hour = 1 foot. 42 : 16 :: 22 : space described in the two last hours = 4 feet. Similarly (3 = 9) feet is the space described in the three last hours; so that the spaces required are

16 9 = 7, 9 – 4 = 5 4 - 1 = 3 and 1.


By 696, the time of discharging the depth x is
Ana S 2a v(21 – x) dx

x {(?r)} - (2r — 2)}}

3ANS a being the axis of the cylinder, and r the radius of its base.

Let x = 2r and r; then we get t :ť :: (2,13 : %.(2} – 1), and t : t-t:: 23 : 1

the proportion required.


[ocr errors]
[ocr errors]

in Pp


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


and b =

sin. B

sin. Hence (see 696) X = mab =

sin. ß
w Pp

x V xdx
A sin. BN g
22 VPp

X 3A sin. Bu g Again, if D be the given I distance from the surface to the base, we have

2 cos. ß

= D
sin. B
D sin. B

4P cos." B the whole depth of the fluid ; which being substituted in the above expression, will give the time required.

and ta

700. Let l, a, i, be the slant side, axis and radius of the base of the given cone, and x 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 1.


2* 12
and 2(2r.

we have (see 696)
2 1





X =


[ocr errors]
[ocr errors]
[ocr errors]


[merged small][merged small][merged small][ocr errors]
« ForrigeFortsett »