Starting at B then 8 is decomposed into B and €1, where is parallel to DB, and we have two quadrilaterals ABED and A,B, E, D, of which four sides AB, BE, ED and DA are parallel to A,B1, BE, E, D, and DA, and one diagonal AE parallel to one diagonal D1B1, then the other two diagonals BD and A11 are parallel.

(4) A Rotor can always be decomposed into a parallel and equal one passing through a given point together with a couple whose momental area is the area swept out by moving the new rotor back to its old position.

This is proved by taking moments about any pole. Let a be the vector of the given rotor, AB its axis, P the given point. Take any pole 0, and let p2 be the position

vector of P, produce p, to cut AB so that p1+ p2 is the position vector of some point in AB, then

[P1+ p2[a] = momental area of given rotor
= [p1a] + [p2a]

=

momental area of couple + momental area of new rotor,

which is true wherever the pole O may be.

The theorem may also be stated as follows:-Any rotor is equivalent to a parallel rotor through some given point together with a couple measured by the momental area of the rotor about the given point.

The couple is called the Couple of Transference.

175. Composition of Rotors.

Any number of rotors are equivalent to a rotor through any given point together with a couple.

...

If a1, d2, az, be the vectors of the rotors, and P1, P2, P... the position vectors of points in their axes from the given point P, then by the preceding theorem the rotor a, is equivalent to a, at P with a couple [p11].

The rotor a is equivalent to a2 at P with a couple [22], and so on for all the rotors.

Hence we get at a point P a number of rotors α, a2 A3 ... whose resultant can be found by the Vector Polygon

Besides this resultant there are the couples of transference [p11], [ρ2α2],... which in the case of coplanar rotors can be added algebraically (since they are like vectors and differ only in magnitude and sense).

The given system is therefore equivalent to a resultant

(Note that M has no particular position, being a vector it may be moved anywhere parallel to itself.)

176.

The Equilibrium of Rotors.

For a system of rotors to be in equilibrium, the sum of the momental areas of all the rotors must be zero for every pole. Suppose now that the sum of the momental areas (say EMA) vanishes for a pole A, then either the rotors are in equilibrium or the resultant passes through A.

If ΣMB = 0 for a pole B, then the rotors are either in equilibrium or the resultant passes through A and B.

C

If also ΣMc=0 for a pole C not in the line AB, then rotors are either in equilibrium or the resultant passes through A, B, and C. The latter is impossible since A, B, and C are not collinear. Hence we have the theorem:-If we know that the sum of the momental

areas vanishes for three poles not in a line, then the rotors must be in equilibrium.

177. As special but important cases of the equilibrium of rotors we must consider

(a) 2 rotors in equilibrium,

(b) 3 rotors in equilibrium.

(a) The two rotors must have the same axis and their vectors must only differ in sense.

(b) The three rotors must pass through a point or be parallel.

For if two of them intersect, then taking moments about the point of intersection the sum of the momental areas must be zero, and therefore the third rotor must pass through the same point.

178. The process of finding a rotor which with a given set of rotors forms a system of rotors in equilibrium is exactly the same as finding the resultant of the given set. All that is further required is to change the sense of the resultant.

SECTION (II). FORCES.

179. Forces are Rotor Quantities. We have seen that the specification for a force like that of a rotor requires four things, magnitude, direction, sense, and a point on the line of action. If then the laws for the combinations of forces, acting on a rigid body, follow those laid down for rotors, then the theorems proved in this chapter are at once applicable to forces.

That this is so can be experimentally verified.

A general experiment may be arranged as follows: A fixed vertical board has a number of pulleys arranged round it, passing over these pulleys are light silk threads attached by means of hooks to holes in a light card, to the other ends of the threads are attached weights.

First suppose the card pinned to the board by two stout pins and then weights attached to the threads.

On removing one pin, the card will probably turn about the other pin and take up some position of equilibrium. By measuring the distances of the various threads from the pin and by multiplying the corresponding forces by these distances, it will be found that the sum of the moments (taking account of sense) is zero.

Fig. 115.

(This is the principle of the lever in its most general form.) Suppose the pin removed, the card will in general begin to move and come to rest in some new position, and the forces acting on the card will then be in equilibrium. We may now imagine the card pinned to the board, at any point we like, and the deduction from the first experiment remains true, viz. that the sum of the moments of the forces about the pin is zero. Since this is true for any point it is true for every point. Hence when any system of coplanar forces is in equilibrium the sum of the moments of the forces about every point in the plane is zero.

That this is a sufficient as well as a necessary condition of equilibrium may be demonstrated by another experi

ment.

Fix the card by two pins to the board.

Draw any convenient closed vector-polygon and arrange the silk threads parallel to the sides of this polygon and attach weights to the threads proportional to the lengths of the sides.

Calculate the sum of the moments of the forces in the strings about three non-collinear points in the card (these will be found all equal).

By means of two parallel threads apply a couple of Forces whose momental area is equal to the calculated sum of the moments but of opposite sense.

The whole set of forces acting on the card will now be such that the sum of the moments of all the forces is zero about any point.

On removing the two pins the card will be found to remain stationary.

From these experiments we may assume that the theory of Rotors holds for forces acting on a rigid body.

180. Loaded Beam. Suppose we have a horizontal beam freely supported at its ends and loaded in any manner and it is required to find the reactions at the supports. Graphically this problem is solved by means of the Link- and Vector-Polygon.

Notation. In future we shall find it more convenient to letter, instead of the ends, the spaces between the rotors. Each rotor will then have two letters or numbers to indicate it, the vector will also have the same two numbers but placed at its ends. Thus for the rotors a number denotes a space, two numbers a line; in the vector-polygon a number denotes a point and two numbers a line.

An example will make the matter clear.

Let AB be the beam and suppose the loads to act in the lines shewn. Number the spaces as indicated, then

Fig. 116.

230

4

5

the loads will be 12, 23, 34, ... and the reactions at A and B 01, and 50.

Set the vectors of 12, 23, ... of the forces off to scale and choose some convenient pole P, and draw the vector and link polygons.

Note, in actual drawings, there is no occasion to draw out the lines from P to 1, 2, 3,

an edge

of a set square should be

placed along P1 and then

moved parallel to itself to cut the lines 01 and 12.