Let m1, M2, M3,.. be very small masses into which the body may conveniently be supposed to be divided; X1 Y1 X1, X2 Y2 82, x3 y3 3... their co-ordinates.

Then the forces which urge them are gm, gm2, gm3, respectively; and therefore, substituting in Art. 121, we

obtain

136. Since, whatever be the position of the plane yx, we always have

x. Σm = Σ(mx),

it appears that the moment, with respect to any plane, of the whole mass collected at its centre of gravity, is equal to the sum of the moments of all the molecules, with respect to the same plane.

137. If the origin of co-ordinates be in the centre of gravity, then (mx) = 0, Σ(my) = 0, and (mx) = 0; for

, y, and z are, in that case, each equal to zero.

138. Since the mass of a body of uniform density is measured by the product of its volume into its density (Art. 127); if P1, P2, P3,.... be the densities, and V1, V2, V3, .... the volumes of the molecules m1, m2, m3, we shall have

m1 = p1 V1, m2 = P2 V2, m3 = P3 V3, ....

the molecules being so small, that every part of each one may be considered of uniform density. Hence, by substitution in the formulæ of Art. 135, we have

139. If the density of the whole system be the same in every part, then =

= P1 P2 P3 and these formulæ are sim

....

But it is to be carefully observed, that these formulæ are only to be applied to such bodies as are of homogeneous materials.

140. The general application of these formulæ depends on the Integral Calculus, but there are a few cases which can be made to depend upon the more simple principles of Art. 133, and with them we shall accordingly commence our series of examples on the subject of finding the position of the centre of gravity in bodies of various forms.

All bodies will be supposed homogenous, or of uniform density, unless the contrary is mentioned.

141. If through any figure a plane can be drawn, so that the figure shall be symmetrical with regard to it; that is, so that the two parts of the figure which are situated on opposite sides of that plane are perfectly similar and equal; the centre of gravity is in that plane.

For the moment of the volume on one side is exactly equal to the moment of that on the other side, with respect to that plane, and these moments will have contrary signs, and therefore their sum will be equal to zero. But this

sum (Art. 139) is equal to the moment of the whole volume, collected at its centre of gravity, with respect to the same plane; which cannot be the case unless the centre of gravity be in that plane.

142. Hence, if we can find two such planes differently situated, the centre of gravity will be in the line of their intersection; and if we can find a third plane, the centre of gravity will be that point where it cuts the line of intersection of the other two; in other words, it will be the common point of intersection of any three planes, by which the figure can be symmetrically divided.

143. It follows, from these properties,

(1) That the centre of gravity of a sphere, or of a spheroid, or of a cube, is its centre.

(2) That the centre of gravity of a parallelopiped is the middle point of one of its diagonals; and of a cylinder the middle point of its axis.

(3) That the centre of gravity of any figure of revolution is some point in the axis.

144. When we speak of the centre of gravity of a line, or of a plane figure, it is to be understood that the line consists of material particles, and the plane figure of a single lamina of particles, or else, that the thickness is every where the same, and inconsiderable.

145. Hence the centre of gravity of a straight line is its middle point; of a circle, or ellipse, or square, its centre; and it will follow, from reasoning precisely similar to that of Art. 141, that if we can draw two straight lines in a plane, by each of which the figure is divided into two equal and symmetrical parts, the centre of gravity is the point of their intersection. This property will enable us to determine at once, by inspection, the centre of gravity of almost all regular plane figures.

146. To find the centre of gravity of a plane triangle.

Let ABC (fig. 26) be the triangle, bisect one of the sides as BC in D, and join AD. Then we may suppose the triangle made up of material particles, arranged in lines parallel to BC; let be be any one of them. Then, by the similar triangles BAD, bAd,

BD : DA :: bd : dA,

and, similarly, DA: DC :: dA: dc,

=

.. BD : DC :: bd : dc.

=

But BD DC, therefore bd de; and, consequently, d is the centre of gravity of bc.

Similarly, the centre of gravity of every other line, parallel to BC, of which the triangle consists is somewhere in AD; consequently the whole triangle would rest in equilibrium on AD, and therefore its centre of gravity is in AD (Art. 133). In the same manner it would appear that the centre of gravity of the whole triangle is in BE, which bisects AC, and hence G, the point of intersection of AD and BE, is the point required.

Join DE, then because CA, CB are divided at E, D in the same proportion, viz. each bisected, therefore DE is parallel to AB; and, therefore, the angle DEG is equal to the angle ABG, and angle EDG to the angle BAG, and consequently the triangles ABG, DEG are similar;

.. AG : DG :: AB : DE

Hence AG = 2DG,

:: AC: EC :: 2 : 1.

and . AD = AG + DG = 3DG;

.. DG = {AD.

147. If three equal bodies have their centres of gravity situated in the three angular points of a triangle, the centre of gravity of these bodies will coincide with that of the triangle.

Let A, B, C be the centres of gravity of the three bodies, then BD being equal to DC, the two B, C will be in equilibrium on D; and therefore the three on a line passing through A, D; in the same manner they will be in equilibrium on BE, and therefore G is their common centre of gravity.

Hence (Art. 130) the distance of the centre of gravity of a triangle from any plane, is the mean of the distances of its angular points from the same plane.

148. To find the centre of gravity of a quadrilateral figure.

trapezium; AC, BD its centre of gravity; draw

Let ABCD (fig. 27) be the diagonals intersecting in E; G its GI, GK parallel to the diagonals. Then, supposing the trapezium to be made up of the two triangles ADC, ABC, we have (Art. 139),

=

(trapezium ABCD). (perpendicular from G on AC)

(▲ ABC). (perpendicular from its centre of gravity on AC) (A ADC). (perpendicular from its centre of gravity on AC) = (ABC). (perpendicular from B on AC)

- }(▲ ADC) · (perpendicular from D upon AC).

Now the triangles ABC, ADC, having a common base AC, are proportional to the perpendiculars from B and D on AC, which are also proportional to BE, DE respectively; hence, in the above equation, instead of the triangles ABC, ADC, and the trapezium, which is their sum, write respectively the quantities BE, DE, and BE + DE, to which they are proportional; and, instead of the perpendiculars from B, D and G, or I, which is equal to it, write respectively BE, DE, and EI, which are proportional to them; and then we have

(BE + DE). EI = } BE2 – }DE2

=

= } (BE + DE) (BE – DE);

.. EI = } (BE – DE).

And, similarly, EK = } (AE – CE).