Here the equation of the given involute AB, is cx =y2 where c is the parameter of the axis AD.

y=√/c.x, and j = 4*√— also ÿ ==

4.x X

Hence then

by making

* constant. Consequently the general values of v and u, or of the absciss and ordinate, EF and FC, above given, become,

in that case,

But the value of the art. 75, was found to be

FC or u, is barely = 3.x.

xy

quantity a or AE, by exam. 1 to c; consequently the last quantity,

Hence then, comparing the values of v and u, there is found 3vc 4u√√x, or 27cv2 - 16u3 ; which is the equa◄ tion between the absciss and ordinate of the evolute curve EC, showing it to be the semicubical parabola.

EXAM. 2. To determine the evolute of the common cycloid. Ans. another cycloid, equal to the former.

TO FIND THE CENTRE OF GRAVITY.

82. By referring to prop. 42, &c, in Mechanics, it is seen what are the principles and nature of the Centre of Gravity in any figure, and how it is generally expressed. It there appears, that if PAQ be a line, or plane, drawn through any point, as suppose the vertex of body, or figure, ABD, and if

ގ

any

s denote any section EF of the figure,
d = AG, its distance below PQ, and
b = the whole body or figure ABD;
then the distance AC, of the centre of
gravity below pa, is universally denoted by

E

Q

whether ABD be a line, or a plane surface, or a curve superficies, or a solid.

But

But the sum of all the ds, is the same as the fluent of db, and b is the same as the fluent of b; therefore the general expression for the distance of the centre of gravity, is ac = fluent of xb

fluent of b

fluent xb

; putting x d the variable distance

b

AG. Which will divide into the following four cases.

83. CASE 1. When AE is some line, as a curve suppose.

=ż or √x2 + j', the fluxion of the curve;

fluent of x2+j ̈2

is the distance of the centre of gravity in a curve.

Z

84. CASE 2. When the figure ABD is a plane; then by; therefore the general expression becomes AC = fluent of yxx for the distance of the centre of gravity in a fluent of yx

plane.

85. CASE 3. When the figure is the superficies of a body generated by the rotation of a line AEB, about the axis AH. Then, putting c = 314159 &c, 2cy will denote the circumference of the generating circle, and 2cyż the fluxion of fluent of 2cyxz fluent of yxx fluent of yż

the surface; therefore AC =

fluent of 2cyż

will be the distance of the centre of gravity for a surface generated by the rotation of a curve line 2.

86. CASE 4. When the figure is, a solid generated by the rotation of a plane AEH, about the axis AH.

Then, putting c = 3.14159 &c, it is cy2: the area of the circle whose radius is y, and cyb, the fluxion of the solid; therefore

AC =

fluent of xb

fluent of b

=

fluent of cy2xx fluent of y2xx fluent of cy2x fluent of y2x

is the distance of the centre of gravity below the vertex in a solid.

87. EXAMPLES.

EXAM. 1. Let the figure proposed be the isosceles triangle

ABD.

It is evident that the centre of gravity c, will be some

where

where in the perpendicular AH. Now, if a denote AH, C = BD, x =AG, and y= EF any line parallel to the base BD:

CX

then as a ::: x: y = ; therefore, by the

2d Case, AC=

a

=

fluent yxx_fluent x2x3
fluent yx fluent ***2

xx

I

= 3x = AH, when r becomes = AH: consequently

CH = AH.

In like manner, the centre of gravity of any other plane triangle, will be found to be at of the altitude of the triangle; the same as it was found in prop. 43, Mechanics.

EXAM. 2. In a parabola; the distance from the vertex is 3x, or of the axis.

EXAM. 3. In a circular arc; the distance from the centre where a denotes the arc, c its chord,

of the circle, is

andr the radius.

cr

a

EXAM. 4. In a circular sector; the distance from the centre

2cr

of the circle, is -: where a, c, r, are the same as in exam. 3.

3a

EXAM. 5. In a circular segment; the distance from thẹ centre of the circle is

area, of the segment.

C3 12a

where c is the chord, and a the

EXAM. 6. In a cone, or any other pyramid; the distance from the vertex is år, or 4 of the altitude.

EXAM. 7. In the semisphere, or semispheroid; the distance from the centre is år, or of the radius; and the distance from the vertex of the radius,

EXAM. 8. In the parabolic conoid; the distance from the base is, or of the axis. And the distance from the vertex of the axis.

EXAM. 9. In the segment of a sphere, or of a spheroid; the distance from the base is r; where r is the height

2a X
6a-4x

of the segment, and a the whole axis, or diameter of the sphere.

EXAM. 10. In the hyperbolic conoid; the distance from 2a + x the base is 6a+ 4x and a the whole axis or diameter,

r; where x is the height of the conoid,

PRACTICAL

PRACTICAL QUESTIONS.

QUESTION I.

A LARGE Vessel, of 10 feet, or any other given depth, and of any shape, being kept constantly full of water, by means of a supplying cock, at the top; it is proposed to assign the place where a small hole must be made in the side of it, so that the water may spout through it to the greatest distance on the plane of the base.

Let AB denote the height or side of the vessel; D the required hole in the side, from which the water spouts, in the parabolic curve DG, to the greatest distance BG, on the horizontal plane.

By the scholium to prop. 61, Hydraulics, the distance BG is always equal to 2AD. DB, which is equal to

A

2x(ax) or 2/ax - x2, if a be put to denote the whole height AB of the vessel, and x = AD the depth of the hole,

a. So that the hole D must be in the middle between the top and bottom; the same as before found at the end of the scholium above quoted.

QUESTION II.

If the same vessel, as in Quest. 1, stand on high, with its bottom a given height above a horizontal plane below; it is proposed to determine where the small hole must be made, so as to spout farthest on the said plane.

Let the annexed figure represent the vessel as before, and be the greatest distance spouted by the fluid, DG, on the plane bG.

D

B

Ꮐ

must be a max

imum. And hence, like as in the former question,

x = c = ab. So that the hole D must be made in the

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 BG the given inclined plane), bG = 2√/AD. Db = 2√x(a − x ± %), putting z = Bb, and, as before, σ = AB, and x = AD. Then be must still be a maximum, as also вb, being in a given ratio to the maximum BG, on account of the given angle B. Therefore ar

G

x2±xz, as well as z, is a maximum. Hence, by art. 54 of the Fluxions, ax 2xx + zx = 0, or a 2x + z = 0;

conseq. z= 2x

a; and hence bG = 2x(a x + z) becomes barely 2.r. But, as the given angle GBb is = 30°, the sine of which is; therefore BG = 2вb or 2z, and bG2 = BG2- Bb2 = 3z2=3 (2x-a), or bc = (2x − a)√√3. BG2.

±

bG ±

Putting now these two values of be equal to each other, gives the equation 2x =+ (2x - a)/3, from which is found 3√3a, the value of AD required.

X =

=

4

Note. In the Select Exercises, page 269, this answer is 6+6

brought out

10

a, by taking the velocity proportional

to the root of half the altitude only.

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 diameter 5 inches.

Let ABC represent the cone of the glass, and DHE the ball, touching the sides in the points D and E, the centre of the ball being at some point F in the axis Gc of the cone.