SECT. 2. Co-ordinate Method.

Let P, P', P',... be any number of weights in one plane, and (x, y), (x', y'), (x", y'),... their co-ordinates referred to any axes, rectangular or oblique, in this plane. Then, x, y, being the co-ordinates of the centre of gravity of the weights,

If all the weights lie in one line, then, this line being taken for the axis of x, the centre of gravity will also lie in this line, and its position will be given by the single formula

If the weights be referred to any three axes in space,

(1) A uniform board is composed of a square ABCD, fig. (61), and an equilateral triangle AEB: to find the distance of the centre of gravity of the whole board from the point C or D.

Bisect CD in O, and join EO: the centre of gravity of the whole board will evidently lie in EO.

Let a = the length of a side of the square: then the area of the square is a2 and the distance of its centre of gravity from O is a. Also, the area of the triangle AEB is a2 √3, and the

distance of its centre of gravity from AB is

a

2√3

G being the centre of gravity of the whole board,

Hence,

8+2√3

(2) To find the centre of gravity of the solid included between two right cones on the same base, the vertex of one cone being within the other; and to determine its limiting position if the vertices approach to coincidence.

Let h, h', be the altitudes of the two cones, and let x be the distance of the centre of gravity of the solid from the common base: then, since the volumes of the cones are as their altitudes, and since the distance of the centre of gravity of a cone from its base is equal to a quarter of its altitude, we have

and, when the vertices of the cones approach to coincidence, h' approaches h as its limit, and therefore, ultimately, x is equal to th.

(3) A beetle crawls from one end A of a straight fixed rod to the other end B: to find the consequent alteration in the position of the centre of gravity of the rod and beetle.

Let a the length and P= the weight of the rod, and W= the weight of the beetle.

Then the original distance of the centre of gravity of the rod and beetle from A is equal to

aP
P+W'

and its distance from A, when the beetle arrives at B, is equal to

aP+aW
P+W

hence the centre of gravity has moved through a space equal to

Wa
P+W'

(4) A circular piece is cut out of a rectangular board ABCD, fig. (62), the two sides AB, AD, touching the circumference of the circle to find the position of the centre of gravity of the remaining portion.

=

Produce AB, AD, indefinitely to x, y, Ax, Ay, being taken as axes of co-ordinates. Let AB-a, AD=b, r the radius of the circle. Let x, y, be the co-ordinates of the centre of gravity of the remaining portion of the board. Then

(5) A weight P is supported on a smooth inclined plane by a string, parallel to the plane, which passes over a fixed pully, and is attached to a weight Q; to prove that, when Q is moved vertically, the centre of gravity of P and Q will neither rise nor fall.

Let x be the distance of P and y the distance of Q from the

pully then the depth of the centre of gravity of P and Q below the pully is equal to

Px sin a + Qy
P+Q

act

But, since P is at rest, we have, resolving the forces, which upon it, parallel to the plane,

Q = P sin a.

Hence, c being the length of the string, the depth of the centre of gravity is equal to

which is independent of the depth of Q below the pully.

(6) The inscribed circle being cut out of a right-angled triangle, the sides of which are 3, 4, 5; to find the centre of gravity of the remainder.

Let OA = 3, OB 4, and therefore AB=5. Let OAx, OBy, fig. (63), be taken as the co-ordinate axes.

Since the centre of gravity of the circular area and the remainder of the triangle must coincide with the centre of gravity of the triangle, we have, x, y, being the co-ordinates of the required centre of gravity,

04 x area of triangle = rad. of circle x its area

The value of a shews that the required centre of gravity lies in a line through the centre of the circle parallel to OB.

(7) Three weights, placed at the corners of a triangle, are proportional to the opposite sides; to shew that their centre of gravity coincides with the centre of the circle inscribed in the triangle.

Let a, b, c, be the sides of the triangle, and Xa, xb, λc, the weights at the opposite angles: let x, y, z, denote the distances of the centre of gravity of the weights from the sides a, b, c, respectively: then, p denoting the perpendicular upon the side a from the opposite angle,

where A is the area of the triangle. Similarly

These results shew that the distances of the centre of gravity from the three sides are all equal to the radius of the inscribed circle. The centre of gravity of the weights must therefore coincide with the centre of the circle.

(8) To find the centre of gravity of the periphery of a triangle formed by a piece of uniform wire.

Let ABC, fig. (64), be the triangle. Bisect the sides BC, CA, AB, in a, B, y, respectively, and join By, ya, aß.

The centre of gravity of the triangle ABC will be the same as that of three weights in the proportion of BC, CA, AB, placed respectively at a, B, y: but BC, CA, AB, are proportional to By, ya, aß; hence the centre of gravity of ABC coincides with