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middle between the top of the vessel, and the given plane, that the water may spout farthest.

QUESTION III. But if the same vessel, as before, stand on the top of an inclined plane, making a given angle, as suppose of 30 degrees, with the horizon; it is proposed to determine the place of the small hole, so as the water may spout the farthest on the said inclined plane.

Here again (D being the place of the hole, and by the given inclined plane),

D
bG
2V AD. Db

21 x(a
IX z),

16 putting z = bb, and, as before, a = AB, and x = AD. Then bg must still be a maximum, as also bb, being in a given ratio to the maximum BG, on account of the given angle B. Therefore ax 712 , as well as z, is a maximum. Hence, by art. 54 of the Fluxions, aš 2xi + zi = 0, or a 2x +z=0; conseq. %= 2x - a; and hence bg = 2vx(a - x + z) becomes barely 2.x. But as the given angle gbb is = 30°, the sine of which is į; therefore BG = 2Bb or 22, and bG? BG? Bb2 3z2 = 3 (2 x – a), or bg = .(2.3 a) v3.

Putting now these two values of bg equal to each other, gives the equation 2x = + (2x -a)/3, from which is found Zav 3

3+3

-a, the value of AD required. N 31 Note. In the Select Exercises, page 269, this answer is

6 + 6 brought out a, by taking the velocity proportional

10 to the root of half the altitude only.

4

QUESTION IV. . It is required to determine the size of a ball, which, being let fall into a conical glass full of water, shall expel the most water possible from the glass; its depth being 6, and diaineter 5 inches. Let abc represent the cone of the

B glass, and DHE the ball, touching the sides in the points D and E, the centre of

D the ball being at some points r in the axis Gc of the cone.

E H

С

Put

=a,

Put AG = GB = 21

cc = 0 = b,
AC = NAG? + Gc' = 01 = 0,

AD = FE = FH = x the radius of the ball.
The two triangles ACG and DcF are equiangular; theref.
AG : AC :: DF : FC, that is, eic :: X :

= FC; hence

CX

CX

a

c., GF = GC - FC = -b and GH = GF + FH = b b + x the height of the segment immersed in the water. Then (by rule 1 for the spherical segment, page 51), the content of the said immersed segment will be (6 DF -- 2G11) X Gh? * 5236 = (2x

-)x which must be a maximum by the question; the fluxion of this made = 0, and divided by 2. and the common factors,

2a to gives

-X) - (

24 - 2 – 6)x= x2=0;

abc this reduced gives x =

251, the ra(c - a) x (c + 2a) dius of the ball. Consequently its diameter is 44 inches, as required.

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PRACTICAL EXERCISES CONCERNING FORCES;

THE RELATION BETWEEN THEM AND THE TIME, VELOCITY, AND SPACE DESCRIBED.

WITH

BEFORE entering on the following problems, it will be convenient here, to lay down a synopsis of the theorems which express the several relations between any forces, and their corresponding times, velocities, and spaces, described; which are all comprehended in the following 1 theoreins, as collected from the principles in the foregoing parts of this work.

Let f, F, be any two constant accelerative forces, acting on any body, during the respective tine t, T, at the end of which are generated the velocities v, v, and described the spaces s, s. Ther, because the spaces are as the ti nes and velocities conjointly, and the velocities as the fo.ces and times; we shall have,

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And if one of the forces, as F, be the force of gravity at the surface of the earth, and be called 1, and its time t be = l"; then it is known by experiment that the corresponding space s is = 16feet, and consequently its velocity v=2s 32, which call 2g, namely, g = 16 feet, or 193 inches. Then the above four theorems, in this case, become as here below:

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8.

f

gt2

2gt

4gs And from these are deduced the following four theorems, for variable forces, viz.

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In these last four theorems, the force f, though variable, is supposed to be constant for the indefinitely small time t, and they are to be used in all cases of variable forces, as the forner ones in constant forces; namely from the circumstances of the problem under consideration, an expression is deduced for the value of the force f, which being substituted in one of these theorems, that may be proper to the ease in hand; the equation thence resulting will determine the corresponding values of the other quantities, required in the problem.

When a motive force happens to be concerned in the question, it may be proper to observe, that the motive force m, of a body, is equal to fq, the product of the accelerative force, and the quantity of matter in it q; and the relation between these three quantities being universally expressed by this equation m = 9f, it follows that, by means of it, any one of the three may be expelled out of the calculation, or else brought into it.

Also, the momentum, or quantity of motion in a moving body, is qv, the product of the velocity and matter.

It is also to be observed, that the theorems equally hold good for the destruction of motion and velocity, by means of retarding forces, as for the generation of the same, by means of accelerating forces.

And to the following problems, which are all resolved by the application of these theorems, it has been thought proper to subjoin their solutions, for the better information and convenience of the student.

8

PROBLEM I.

To determine the time and velocity of a barły descending, by the force of gravity, down an inclined plane; the length of the plane being 20 feet, and its height 1 foot,

Here, by Mechanics, the force of gravity being to the force down the plane, as the length of the plane is to its height, therefore as 20:1::1 (the force of gravity): zo = f, the force on the plane.

Therefore, by theor. 6, vor 4gfs is v/4 161 x tox 20 = 74 x 164 = 2 x 476 or 84'5 feet nearly, the last velocity per second. And,

20

400 20 By theor. 7, 8 or vis V 16,7 x

1612 x zo

1612 4 6 = 429 seconds, the time of descending.

PROBLEM

PROBLEM II.

If a cannon ball be fired with a velocity of 1000 feet per second,

up a smooth inclined plane, which rises I foot in 20: it is proposed to assign the length which it will ascend up the plane, before it stops and begins to return down again, and the time of its ascent.

Heref='s as before.
q2
10002

60000000 Then, hy theor. 5,5=

4gf
4 x 1672 X 76

193 = 31088016, feet, or nearly 59 miles, the distance moved,

1000

120000 And, by theor. 7, t =

2gf

2 x 1612 Xo 193 621" 14} = 1021" 14}, the time of ascént.

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PROBLEM III.

If a ball be projected up a smooth inclined plane, wlich rises 1 foot

in 10, and ascend 100 feet before it stop: required the time of ascent, and the velocity of projection.

First, by theor. 6, v = v 4g fs = N 4 x 1615 x 75 100 = 875 V 10 = 25.36408 feet per second, the velocity.

100

10 And, by theor. 7, t =v

V 10 = 1672 x to

455 192 N 10 = 7.88516 seconds, the time in motion.

PROBLEM IV.

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If a ball be observed to ascend up a smooth inclined plane,

100 feet in 10 seconds, before it stop, to return back again: required the velocity of projection, and the angle of the plane's inclination,

2s 200 FIRST, by theor. 6, v=

= 20 feet per second,

10 the velocity

100

12 And, by theor. 8,f=

That gt? 16 12 X 100 193 is, the length of the plane is to its height, as 193 to 12.

Therefore 193 : 12 :: 100 : 6.2176 the height of the plane, or the sine of elevation to radius 100, which answers to 3° 34', the angle of elevation of the plane.

PROBLEM

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