always half that at another, the thickness of the chain at the former point need only be half that at the latter; a saving in material and in weight is thus effected.

We have learnt that when a body is in equilibrium, the forces exerted on any portion of it by the adjacent portions counteract the remaining forces acting on the portion in question. As, however, there is an infinite number of systems of forces, each of which counteracts a given system, we cannot as a rule determine which system is the one actually exerted, without going beyond the limits of Elementary Statics. If, for instance, a rope composed of several fibres be taut, though we may know the tension of the rope itself, i.e. the sum of the tensions of the different fibres, we cannot say how it is distributed among them. This can only be ascertained when the elasticity of each fibre is known.

When a beam is merely stretched, i. e. when the external forces all act along it, the only internal forces called into play will be between particles arranged in lines along the beam. If then the beam be supposed to consist of two parts A and B, the action of B on A will be the sum of the forces exerted by particles of B on the adjacent particles of A, all such forces being in the same direction along the beam. This action is equal to the resultant of the forces acting on the portion A, and which are also external to the beam. It is clear that the greater this action becomes, the more likely is the beam to be pulled asunder at the point of junction of A and B; the action therefore measures the tendency of the beam to break at that point.

79. When the external forces on the beam are not all along it, the action of one portion on another is not so simple as in the above case. Take the following case. Let ABCD be a rectangular beam which is firmly fixed at the end AB in a vice; along DC let a force S be applied: it will of course be perpendicular to the beam. Consider

the equilibrium of the portion CDPQ, where PQ is an imaginary section perpendicular to the beam's length.

The forces in action are S and the innumerable forces due to ABQP, acting at every point of the section PQ.

Let the latter be resolved along PQ and at right angles to it: the sum of the former components must be equal and opposite to S, and will with it form a couple. The components perpendicular to PQ must therefore be equivalent to a couple, equal and opposite in sign to the former. This shews that the forces near P must be in the direction PA, and those near in the opposite direction: and therefore that the tendency of S is to stretch the fibres near P and crush those near Q. It must follow too, that the magnitudes of the components perpendicular to QP depend on the moment of S about Q, and not on the magnitude of S simply. Hence the greater the moment of S about Q the more likely are the fibres along PQ to give way and the rod to bend at PQ.

Since PQ is supposed small compared with QC, the numerical sum of the forces along PQ must be very much greater than S, i.e. a force is far more likely to bend a rod, when applied at right angles to it, than to pull it asunder when applied along it.

Similar reasoning will apply to a beam under the action of any system of forces. We can shew that the tendency to bend at any point is measured by the algebraical sum of the moments about that point, of the forces external to the rod and acting on one of the parts into which the beam is divided by the point. This tendency to bend is also termed the bending moment.

Ex. 1. A light beam is supported in a horizontal position at its ends, and a weight w is hung from its middle point. Find the bending moment at a point distant x from one end. Ans. wx.

Ex. 2. If a heavy uniform rod be supported at its middle point, shew that the bending moment at any point varies as the square of its distance from the rearer end.

Ex. 3. A uniform rod AB of weight w and length a is supported in a horizontal position at A and B; from a point distant x from A a weight w' hangs: find the bending moment at a point distant y from A.

Ex. 4. A uniform rod of weight w and length a, can turn freely about a hinge at one end, and rests with its other end against a smooth vertical wall, distant b from the hinge. Prove that the bending moment at a point whose distances from the two ends are x, y, respectively, is wxyba ̄2.

Ex. 5. A uniform rod of length a rests horizontally on two pegs, one at one end of the rod, find where the other peg must be placed so that the bending moment at a point distant x from the first peg may be zero. Ans. a2/(2a-x) from the first peg.

80. When each of the bodies forming a system in equilibrium is acted on by forces that reduce to three, the problem of finding the position of each of the bodies

A

B

Fig.53.

E

or of ascertaining the different forces, can often be easily solved by constructing a series of triangles each of which is the triangle of forces corresponding to one of the bodies. For instance, let us consider the case of a number of particles of equal weight fastened at intervals along a weightless string, the ends of which are attached to fixed points. Let A, B, C, D &c. be the positions of the particles, when in equilibrium. Any particle, B for instance, is kept in equilibrium by three forces, its weight vertically_downwards, and the tensions of the strings BA, BC. Draw a triangle Obc, having its sides ba, aO, Ob respectively parallel to the lines of action of these forces: then by the

Pro

triangle of forces these lines are proportional to the forces, to whose directions they are parallel: i.e. weight of A: tension of AB : tension of BC= ab : a0 : Ob. duce ab downwards, and mark off bc, cd, de &c., each equal to ab; join Oc, Od, Oe &c. Then Ob, bc represent in every way the tension of BC, on C, and the weight of C respectively, so that cO must represent the tension of CD. Similarly do represents the tension of DE, eO that of EF, and so on.

Draw OM perpendicular to abc : then the tangents of the angles that aÔ, b0, co &c. make with the horizon are

hence the tangents of the angles which the strings make with the horizon form an arithmetic series. Also the horizontal resolved part of the tension of each string is represented by OM, and is therefore the same for all.

Such a figure which is drawn to enable us to solve the problem is called a Force-Diagram.

The above results can be obtained very easily by equating to zero the algebraical sums of the resolved parts in a horizontal and vertical direction of the forces that act on each particle separately.

81. This Graphic' method can be applied to prove the following important proposition.

Prop. If a weightless string be stretched across a smooth surface, the tension is everywhere the same.

Let ABCD &c. be the string: then any small portion

B

Fig.55.

of it AB is kept in equilibrium by the tensions at its ends, and the resultant of the pressures of the surface on it: as the pressures along AB all act along the normals to