and therefore 'v' is parallel to CB. Moreover, the triangles Bay, Caß, are equal: hence the centre of gravity of the two triangles will lie in the middle point between μ' and v', that is, in the intersection of 'v' with Aa. Hence the required centre of gravity lies in Aa at a distance from a equal to Aα.

(7) To prove the following geometrical construction for the centre of gravity of any quadrilateral. Let E be the intersection of the diagonals, and F the middle point of the line which joins their middle points: draw the line EF and produce it to G, making FG equal to one third of EF: then G shall be the centre of gravity required. Tod: 149.10.

Let K, H, fig. (59), be the middle points of the diagonals AC, BD. Join KH, AH, BK, CH, DK. Take PK=}BK, QK=DK, HS=}AH, HR={CH. Join PQ, RS, cutting AC, BD, in M, N, respectively; join MN. Let G be the intersection of PQ, RS; join EG cutting MN, HK, in 0, F, respectively.

Since the centres of gravity of the triangles ABC, ADC, are at P, Q, respectively, the centre of gravity of the whole quadrilateral must be in the line PQ: by like reasoning, it must also be in the line RS: hence it must be at G, the intersection of these two lines.

Again, since PK : QK :: BK : DK, PQ is parallel to BD: similarly, RS is parallel to CA: hence EMGN is a parallelogram.

Again

EM: EK:: BP : BK :: 2 : 3,

and EN: EH :: AS : AH :: 2 : 3 ;

and therefore

EM EK EN: EH,

and consequently MN is parallel to KH: hence, MN being bisected in O, HK is bisected in F.

Again

and therefore

EO: EF:: EM : EK :: 2 : 3,

EO=EF, and FO=}EF.

Hence

FG-GO-FO

=EO-FO

=}EF—}EF= }EF.

(8) How long a piece must be cut off from one end of a rod, of length 2a, in order that the centre of gravity of the rod may approach towards the other end through a distance b?

The length of the piece must be 26.

(9) If a quadrilateral be divided by one of its diagonals into two equal triangles, shew that it will balance about that diagonal.

(10) If the sides of a triangle be taken, two and two, to represent forces, acting in each case from the angle made by the sides, prove that each of the three pairs will balance about the centre of gravity of the triangle.

(11) If the lengths of the sides of a right-angled triangle be 3, 4, 5 feet, find the distance of the centre of gravity from each side.

The distances of the centre of gravity from the three sides

(12) A square and a rectangle of uniform density are joined together in one plane at a common side: find the length of the rectangle in order that the two may balance about that side, the thickness of the square being double that of the rectangle.

the

The length of the rectangle must be equal to a diagonal of square.

(13) If ABC be an isosceles triangle, having a right angle at C, and D, E, be the middle points of AC, AB, respectively, prove that a perpendicular from E upon BD will pass through the centre of gravity of the triangle BDC.

(14) If the centre of gravity of a four-sided figure coincide with one of its angular points, shew that the distances of this point and the opposite angular point from the line joining the other two angular points are as 1 to 2.

(15) Draw a line through one angle of a square area, cutting off a triangle, so that the remaining quadrilateral, when suspended from the obtuse angle, may rest with one side vertical.

Upon the side BC of the square construct the equilateral triangle BMC, fig. (60): draw MV at right angles to BC: take MN equal to half a side of the square: draw NO, parallel to VB, cutting AB in 0: join DO. Then DO is the required line drawn through the angle D of the square, and cutting off the required triangle ADO.

(16) If ABC be a triangular board, having B an obtuse angle and AB less than BC; prove that the board may stand, when AB is placed on a horizontal plane, and will certainly stand when BC is so placed.

(17) A plane quadrilateral ABCD is bisected by the diagonal AC, and the other diagonal divides AC into two parts in the ratio p q; shew that the centre of gravity of the quadrilateral lies in AC and divides it into two parts in the ratio 2p+q: 2q+p.

(18) O is any point within the area of a triangle ABC: another triangle is formed by joining the centres of gravity G, H, K, of the triangles BOC, COA, AOB: prove that the area of the triangle GHK is one ninth of the area of the original triangle and similar to it.

(19) A uniform wire is bent into the form of three sides AB, BC, CD, of an equilateral polygon, and its centre of gravity is at the intersection of AC, BD: prove that the polygon must be a regular hexagon.

(20) ABCD is a plane quadrilateral figure and a, b, c, d, are respectively the centres of gravity of the triangles BCD, CDA, DAB, ABC: prove that the quadrilateral abcd is similar to the quadrilateral ABCD.

(21) Prove that the straight lines joining the middle points of opposite edges of a tetrahedron intersect and bisect each other.

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,

(Px)
, y
Σ (Ρ)

(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

G being the centre of gravity of the whole board,

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 has 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.