from the base, and since AB,B and ACC, have equal bases their masses are equal and their mass-centres are equidistant from that of ABC; hence the mass-centre of AB,B and ACC, is at M the mass-centre of ABC.

This theorem is of use in the determination of the mass-centre of a quadrilateral.

Let ABCD be the quadrilateral, join AC and BD intersecting at E.

then

Fig. 58.

If p and q be the altitudes of ADC and ABC (see fig.),

Suppose that CE>AE, then from CE cut off CE,=AE, then since AE= E,C, the mass-centre of EDE, is the same as that of ADC. Similarly the mass-centre of EBE, is the same as that of ABC.

Also

area EDE1 DE area ADC

=

=

area BEE, BE area ABC'

Hence the mass-centre of BDE, is the same as that of ABCD, and the problem is reduced to that of finding the mass-centre of a triangle.

88.

Mass-centre of a sector of a circle.

Let AOBC be a circular sector. If we divide the arc ACB into a number of equal parts EF, FG,... and draw

the chords, then by joining the centre to E, F, G... we shall have a number of triangles. The greater the number of points E, F,... the more nearly will the sum of the areas of these triangles coincide with the area of the sector. We may regard the sector therefore as the limiting case of these triangles. Since all the triangles have equal bases and altitudes the mass-centre of each is at the same distance from O, viz. of the altitude. At the limit, the altitude becomes the radius and the mass-centre of the circular sector must be the same as the mass-centre of the arc A,C,B1 of radius that of ACB. The problem is reduced therefore to that of finding the mass-centre of a circular arc.

Since, arc CB: arc C,B1 = OB: OB1, we may proceed thus, from C (OC being the axis of symmetry) set off CP 1 to OC and equal in length to the arc CB. From B1 (where OB1 = 30B) draw B1Q || OC, from Q draw QM 1 OC. The point M thus determined is the mass-centre of the sector OACB.

To calculate OM, we have, if BOC = 0 and OB=r,

89. Negative Masses.

Suppose we have any figure, say a rectangle, let a be its mass and a the position vector of its mass-centre. Suppose a triangular piece is cut out of the rectangle, and let b be the mass and B the position vector of its masscentre. If now the position vector y of the mass-centre of the remaining part is wanted we could proceed as follows: (a – b) y+b3 = aa by (§ 73),

or

(a−b) y= aa – b

= aa +(-b) B.

We see from this equation that the triangle is treated as having a negative mass.

This conception of a negative mass is often very useful, thus a hollow triangle may be regarded as a large triangle having a positive and a small triangle having a negative

mass.

If we have two points A and B, one with a positive mass a and the other with a negative mass b, then our old construction of § 71 becomes modified, inasmuch as the lines representing the masses have to be drawn in the same sense as well as the same direction.

From A set off a length proportional to b, and from B set off in the same direction and sense a length proportional to a, join the points so determined and produce to cut AB in M the mass-centre of the two points.

Evidently the more nearly equal a and b are, the farther off is M and when ab, M is at an infinitely great distance.

90. Mass-centre of a segment of a circle.

A segment of a circle may be considered by the preceding article as a sector of positive and a triangle of negative mass.

First find the mass-centre M, of the sector, then that of the triangle M。.

Area of sector= OB. CB.

Area of triangle = OB. ER,

r

ER being the 1 from E on OA.

Hence the areas are in ratio

2

CP

ER'

Through M, draw MQ-CP in same direction and sense. Through M, draw MS= ERS

Join QS and produce to cut OC in M, then M is the mass-centre of the segment.

Taking moments about O we have

area of sector × OM1 = area of segment × OM

and

+ area of triangle × OM2,

OM,= ar

sin
0

area of sector r20,

OM=ÿ, area of segment r20 — r2 cos 0 . sin 0,

OM2r cos 0, area of triangle r2 cos 0. sin 0.

2

=

..3 sin 0 =ÿ (r20 — r2 cos 0 sin ) +

cos2 . sin

91. Mass-centre of any irregular area.

By aid of the Link Polygon it is possible to find the mass-centre of any arrangement of points or lines; we will now explain an equally general method for finding graphically the mass-centre of any evenly loaded area-no matter how irregular in contour. To determine the position of mass-centres by this method, it is necessary to be able to find areas of figures with a fair degree of accuracy. Instruments called Planimeters have been invented for finding areas simply by going round the contour with a

tracer.

Suppose we require to find the mass-centre of the accompanying area.

Draw any two parallel lines XX, and X'X' in the plane of the area at a convenient distance α apart.

Divide the area up into thin strips by lines parallel to XX. Let AB be one of the very thin strips so formed.

Take any point O in XX, called the

pole.

From the end points of AB, draw parallels to cut X'X' in A'B'. (It is generally most convenient to take these parallels to XX but it is by no means necessary.) Join OA' and OB', these will cut AB (or AB produced) in two points A1, B1.

If we perform this operation on all the strips, AB, CD, ... we shall get a number of points A, B, C, D1, ... which, if the strips are taken very close together, will form a closed curve.