Corollary. The distance of the point G, is AG = AD, and CG = ‡CE; or AG = 2GD, and CG 2GE.

For, draw BF parallel to AD, and produce CE to meet it in F. Then the triangles AEG, BEF are similar, and also equal, because AE = BE; consequently AG = BF. But the triangles CDG, CBF are also equiangular, and CB being = 2CD, therefore BF = 2GD. But BF is also = AG; consequently 2GD or AD. In like manner, CG = 2GE or "CE.

AG

PROP. XVII.

63. To find the centre of gravity of a trapezium.

DIVIDE the trapezium ABCD into two triangles, by the diagonal BD, and find E, F, the centres of gravity of these two triangles; then shall the centre of gravity of the trapezium lie in the line EF connecting them. And therefore if EF be divided, in G, in the alternate ratio of the two triangles, namely, EG : GF :: triangle BCD : triangle ABD, then G will be the centre of gravity of the trapezium.

64. Or, having found the two points E, F, if the trapezium be divided into two other triangles BAC, DAC, by the other diagonal AC, and the centres of gravity H and I of these two triangles be also found; then the centre of gravity of the trapezium will also lie in the line HI.

So that, lying in both the lines, EF, HI, it must necessarily lie in their intersection G.

65. And thus we are to proceed for a figure of any greater number of sides, finding the centres of their component triangles and trapeziums, and then finding the common centre of every two of these, till they be all reduced into one only.

PROP. XVIII.

66. To find the centre of gravity of a triangular pyramid.

G

Let ABCD be a triangular pyramid, and to the point of bisection of DC draw AH, BH. Take HK=4HA and HI+HB; then K and I will be the centres of gravity of the surfaces ACD and BCD respectively. Join KL, AI, BK. Now if the pyramid be resolved into elements, by means of planes parallel to BCD, it is evident that the line AI must pass through the centre of gravity of the pyramid, since I is the centre of gravity of BCD. For the same reason, the centre of gravity of the pyramid is in the line BK, and because AI and BK are in one plane, the centre of gravity of the pyramid ABCD must be at G, the point of intersection of AI and BK.

By similar triangles AGB and KGI, we have

AG: GI.: AB: KI::AH: KH::3:1

.. AG: AI:3:4 or AG = 4AI.

H

B

Q

Cor. 1. Bisect AB in P and join HG and GP; then if PQ be drawn parallel to AI, we have BQ=QI=IH; but AI=2PQ and AI=4GI; hence PQ=2GI; and therefore HI: HQ :: IG : PQ; whence HGP is a straight line.

Cor. 2. Hence the centre of gravity of a triangular pyramid is the middle of the line joining the points of bisection of any two edges that do not meet.

Cor. 3. A solid bounded by plane surfaces may be divided by planes into a number of triangular pyramids; and if a plane be drawn parallel to the base, at a distance equal to 4 of the altitude of the pyramid, then the centre of gravity of the whole pyramid must be in this plane, for that of each of the triangular pyramids is in this plane. Hence, the line joining the vertex of the pyramid and the centre of gravity of its base will cut the plane in the centre of gravity of the whole pyramid.

PROP. XIX.

67. To find the centre of gravity of any body, or system of bodies.

Let v1, V2, V3, &c., denote the volumes of the material particles which compose the body or volume V; and a1, Y1, Z1; X2, Y2, Z2, &c.; their co-ordinates in reference to three rectangular axes; then if X, Y, Z denote the co-ordinates of the centre of gravity, we have (by Prop. XV., Cor. 3.)

But to adapt these expressions to computation, we shall introduce the prio ciples of the Differential and Integral Calculus, and then the preceding expres

sions will take the form

X = frdv
Sdv

where x, y, z denote the distances of the centre of gravity of du from the three rectangular planes.

By means of these three equations, the determination of the centre of gravity may be effected; and when the figure is a plane surface, two of these equations are only required, since the centre of gravity is in the plane.

I. When the figure is a plane curve.

Here dv = differential of the arc = √/dx2 + dy2 = ds

and if the arc be symmetrical on each side of the axis of x, we have y = 0, and then we have simply

X = fxds

And if the area is symmetrical on each side of the axis of x, we need andy one equation, viz.

X = Syrdr
Sydz

III. For a surface of revolution round the axis of z.

Here one equation only is necessary; and since du = differential of the surface = 2 yds, we have

X = fryds
Jyds

IV. For a solid of revolution round the axis of x.

Here de differential of the volume = wydr, and hence

We shall now apply these formula to a few examples.

EXAMPLES.

68. Er. 1. To find the centre of gravity of a circular arc.

Here the curve is symmetrical on each side of the axis of x, and the equation is y2 = 2ax —; hence we have

that is, the distance of the centre of gravity of a circular arc from the vertex ay, and therefore the distance of the centre of gravity from the

Ex. 2. To find the centre of gravity of a cone.

Let z, z, represent any two parts of the axis of the cone, measuring from he vertex, and y1, y, the radii of the circular sections of the cone correspondng to the altitudes z,, z,; then, if z denote any variable part of the axis, and the radius of the corresponding circular section, we have

fx dz

= distance from vertex.

But integrating between the limits a, and a, we have

3

=

This expression gives the distance of the centre of gravity of the frustum of a cone from the greater end; hence, if R, r represent the radii of the greater and less ends of the frustum, and h its altitude, we have the distance R2 + 2Rr + 3r2. ; and when r = 0, we have R2 + Rr + 2 h 1

h
4

of the centre of gravity = the distance of the centre of gravity of a cone from its base = = onefourth of the altitude as found above.

Ex. 3. Four bodies, whose weights are w1, w, wз, w, pounds, are placed at the successive angles of a square whose side is 2a inches; required the position of their common centre of gravity, the square being considered without weight.

Take O, the centre of the square, as the origin of co-ordinate reference, and the two rectangular axes parallel and perpendicular to the sides; then we have X = (w2 + w3 — w, w1) w1 + w2- W3 · -WA a, w1 + w2 + w3 + W1 w1 + w2 + ws + w1 Wi = =3, w2 = 4, w3=5, w1=6; and 2a= 12 inches;

Thus if

and y =

a.

.. X=0 and y = - 6=- -14= OG = distance of

4 18

centre of gravity below O on the axis of y.

WI

#2

PROBLEMS FOR EXERCISE.

69. Ex. 1. Find the centres of gravity of

(1.) The common parabola and the paraboloid.
(2.) A semicircle, and the segment of a circle.

(3.) A hemispheroid, and a hemisphere.

(4.) The sector of a circle, and a spheric sector.

(5.) The surface of a spheric segment, and that of a cone.

Ex. 2. Two cones are placed with their equal bases in contact, and the altitude of the one is three times that of the other; find the position of their common centre of gravity.

Ex. 3. The surface generated by a plane line or curve revolving about an axis in the plane of the figure, is equal to the product of the generating line or curve, and the path described by its centre of gravity.

Ex. 4. The volume of the solid generated by the revolution of a plane figure about an axis in the plane of the figure, is equal to the product of the generating surface, and the path described by its centre of gravity.

Ex. 5. From a given rectangle ABCD of uniform thickness, to cut off a triangle CDO, so that the remainder, ABCO when suspended at O, shall hang with AB in a vertical position.

of the prismatic mass of earth which lies above the surface of a bank that would be itself supported. But this prismatic mass is partly supported by friction, and we must therefore ascertain how much of the horizontal thrust is counteracted by friction.

Suppose a weight W to be placed on a plane, inclined to the vertical at an angle i; and let H be the horizontal force, which, with the friction, just sustains the weight W. Resolve each of the forces W, H into two others, the one parallel and the other perpendicular to the plane; and those parallel to the plane act in opposite directions. while those perpendicular to the plane concur in direction; hence we have

W

H

force parallel to the plane

= W cos i

H sin i

force perpendicular to the plane = W sin i + H cos i. And the first of these forces must be precisely equal to the friction; that is, equal to a force that will just support the weight upon the plane; hence W cos iH sin i = ƒ W sin i +fH cos i

1- ftan i. W
tani+f

.. H =

cos if sin i w =
sin if cos i

the weight W against the wall would be

1-ftan i
tan i+f

W.

If then the weight W were sustained by a wall, the horizontal thrust of

Now to apply this to the investigation of the horizontal thrust of the prism BCH, we shall put BC= a, a variable part Cb = x, bb' = dx, and s the specific gravity of the earth. Then the area of bb'hh' = xdx tan i, and its weight = sxdx tan i; hence the horizontal thrust against bb' will be

1 + f cot

; hence, integrating, we have

f.sMxdx = fa2s M = whole horizontal thrust of triangle BCH.

=

*

Again, the length of the lever Bb ax, and the moment of the thrust of the element bb'hh' = sMx (a — x) dx = as Mrdr

sMx dx; hence

• If lines be expressed numerically, the product of a force acting on a lever, and the perpendicu lar from the axis of motion on its direction is called the moment of that force,