system should not close. Shew how to determine graphically the moment of the couple.

17. In order that a rigid body turning upon a fixed point and subject to a series of forces situate in one plane may remain in equilibrium, it is necessary and sufficient that any funicular, whose first side passes through the fixed point, should close.

18. In order that a rigid body touching a fixed straight or curved line xy in a single point P may be in equilibrium, it is necessary and sufficient that (1°) the resultant of the forces applied to the body, as determined by means of the polygon of forces, should be parallel to the normal to xy at P; (2) that any funicular, whose first side passes through P, should close.

19. In order that a rigid body touching two fixed straight or curved lines in the points P and Q respectively may be in equilibrium, it is necessary and sufficient that (1°) any funicular, whose first side passes through the point O of intersection of the normals at P and Q, should close; (2°) that the components of the resultant force applied to the body in the directions OP and OQ should tend to press it against the fixed lines.

20. A rigid body turns on a fixed point P and also rests on a single point Q of the fixed straight or curved line xy; shew how to find the reactions at P and Q, whether the forces acting on the body reduce to a single resultant or a couple.

21. Shew how to determine the three reactions of a body resting with a point on each of three fixed straight or curved lines, by the sole aid of the force and funicular polygons.

22. Find a point in a given triangle ABC, from which rays drawn to the three angular points shall divide the whole triangle into parts AOB, AOC, and BOC of areas proportional to the given numbers m, n, and r (§ 52).

23. Find a point O within a given irregular quadrangle ABCD from which rays drawn to the angular points shall divide the quadrangle into four triangles, AOB, BOC, COD, and AOD, such that the areas of three of the triangles may be proportional to the given numbers m, n, and r (§ 52).

CHAPTER IV

CENTRES OF GRAVITY

DEFINITION. The centre of gravity of a materialised surface is the point traversed by the resultant of the weights of its separate parts, no matter in what direction such weights or forces may be supposed to act.

58. The Triangle.-Imagine the area of the triangle.

(Fig. 83) to be divided into

a series of parts by lines
parallel to the base AC;
then the centre of gravity of
each of those parts must lie
on the line BD, bisecting all
lines parallel to the base.
Similarly, the centre of gravity A
must lie on the line CE,

bisecting the side AB. Now

D

E

Fig. 83.

B

(§ 5), if the transversal CE be drawn intersecting the three sides, or sides produced, of the triangle ABD, we have

59. The Trapezium.—The centre of gravity of the trapezium ABCD (Fig. 84) must lie in the line EF joining the centres of the parallel sides AB and CD. If, moreover, we imagine the trapezium resolved into two triangles by the diagonal BD, the centre of gravity of the triangle ABD will lie in the horizontal line through f, or upon the line through ƒ parallel to the sides AB and DC, where

Eƒ=EF.

Similarly the centre of gravity of the triangle BCD will lie upon the horizontal line through e, where

But the weights of the component triangles ABD and BCD are respectively proportional to the bases AB and CD. Wherefore, by the principle of the lever, the distances of the centre of gravity O from ƒ and e are inversely proportional to AB and CD, or to EB and FC. If, therefore, fc and eb be set off parallel to the bases and equal respectively to FC and EB, the line cb will intersect EF in O, the required centre. Further, by a simple geometric deduction, it can be shewn that, if AG be set off equal to CD upon the continuation of BA, and if, upon the continuation of DC, CH be set off equal to AB, the line GH will intersect EF in the required centre of gravity.

60. A Circular Arc.-In Fig. 85 the element of the chord CC' or h is

dh=ds cos 0;

whilst the perpendicular let fall from the centre of gravity

of the infinitesimal arc ds upon the diameter parallel to the chord is

But, if y be the perpendicular distance of the centre of gravity of the circular arc from the axis or diameter through O parallel to the chord, we have

Wherefore, if the arc be rebatted upon the tangent at the crown in BB', and the points B and B' be joined to O, the centre of gravity will lie upon the line cc', where c and c' are the points of intersection of the vertical lines drawn through C and C' with the lines OB and OB'. The required centre, therefore, lies at G, or at the point of intersection of cc' with the vertical through O.

61. Irregular Area.-Divide the irregular rail-section (Fig. 86, Pl. II.) into a series of laminæ by lines parallel to the base. If we consider the areas of these lamina to be so many forces applied at their respective centres of gravity