Let B'g' cut F'B in the point C, and join AC. Then because AB = AB', and AC is common to the triangles ABC, ABC, and the ∠B = ∠B', therefore AC bisects the angles BCB', BAB': hence the resultant of the two equal forces F', g' lies in AC produced; and that of the equal forces F, g lies in CA produced. But, because AF is parallel to CF' and Ag parallel to Cg', therefore the LFAg = ∠F'Cg'; and consequently the resultant of the forces F', g' is equal to that of the forces F, g: we have just proved also that they act in opposite directions; therefore, the four forces F, g, F', g' balance each other, and may be removed.

There is then left only the couple f, f', which is the same as if the arm of the original couple had been turned through the ∠BAB'.

49. The effect of a couple is not altered by removing it to any other part of the rigid body, provided its new plane be parallel to its original plane. (fig. 11.)

Let F, F' be a couple acting upon a rigid body in the plane HH; let AB be its arm. Let KK be any other plane, in the rigid body, parallel to HH; and in this plane draw the line ab, parallel and equal to AB. At a and b apply pairs of opposite forces f, g; f, g': each force being equal and parallel to the forces F, F'. These pairs of forces balance each other, and therefore produce no effect. Draw Ab, Ba; they, being in the plane which contains AB and ab, necessarily intersect in some point C. In fact, if A, a, and B, b were joined, Aab B would be a parallelogram, and therefore Ab, Ba, being its diagonals, bisect each other in C. Draw PCQ parallel to AF.

Then F: g' :: bC : AC.

Hence F and g' (by Art. 40) have a resultant F + g', which acts at C in the line CP. Similarly it may be shewn, that F' and g have a resultant, F' + g acting at C in the line CQ. Now F + g' = F' +g and CP is opposite to CQ, therefore the four forces F, g', F', g balance each other, and may be removed. There remains then only the couple f, f, which is the same as if the original couple had been removed into the new plane KK, retaining its arm ab parallel to AB; but we may now turn the arm ab through any angle without altering the effect of the couple. Hence the effect of a couple, &c.

50. The effect of a couple is not altered by removing it to any other part of the rigid body in its own plane.

The demonstration in the last Article will serve for this, using fig. 12 instead of fig. 11.

51. A couple may be changed for any other couple acting upon the same rigid body, provided the moments of the two couples be equal, their planes parallel, and they be both of the same kind, i. e. both positive or both negative.

Let HH (fig. 11) be the plane of the couple F, F'; and in any other plane KK of the rigid body draw, parallel to AB, a line ab of any proposed length: at a, b apply pairs of equal and opposite forces f, g; f', g'; of such magnitude that

F. AB = f. ab,

these balance each other, and therefore produce no effect.

Now AB and ab being parallel, the lines Ab, Ba lie in one plane, and intersect in some point C: and because AB is parallel to ab, the ∠CAB = ∠Cba, and the CВА = ∠Cab; consequently the two triangles ACB, bca are similar.

Now F: g':: ab : AB

:: Cb: CA;

therefore, by Art. 40, the resultant F + g' of the two forces F, g' acts at C in the direction CP. In a similar way it may be shewn that F + g, the resultant of F', g, acts at C in the direction CQ. Now F = F' and g' = g, and therefore F + g' = F' + g; consequently the four forces F, g', F', g are in equilibrium, and may be removed; which being done, the original couple is replaced by the equivalent couple f, f whose arm is ab. This couple f, f may now be turned through any angle in the plane KK, and thus the proposition is established.

52. Any number of couples act upon a rigid body in parallel planes; to find their resultant.

Change all the couples into others of the same moment, but all having their arms of the length b. Then if F1, F2, F3... F be the forces; and a1, a2, a3...an the arms of the original couples; and P1, P2, P3,...P the forces of the corresponding equivalent couples, we shall have

P1b = F1a, P2b = F2a2,...Pnb = Fnan.

Now since the new couples act in parallel planes and have equal arms, they may be removed into the same plane, and then turned round and transposed so as to make all their equal arms exactly coincide; in which position the system of couples is reduced to one couple, the arm of which is b, and the forces of which are equal to

P1 + P2 + P3 + ... + P.

Hence the moment of the resultant couple

= (P1 + P2 + P3 + ... + P2).b

= F1a1 + F2A2 + F3A3 +

...

+ Fnan

= the sum of the moments of the original couples. Whence, the moment of the 'resultant couple is equal to the sum of the moments of the original couples.

The reader will be careful to remark, that if any of the couples are of a negative character, their moments are to be accounted negative in taking this sum.

53. COR. If all the n couples be equal, the moment of their resultant couple is n times the moment of one of them; and as the effect of n equal couples must be n times the effect of one of them, it follows that the moment of a couple is a proper measure of its effect in producing or destroying equilibrium. Whenever, therefore, we have occasion to speak of the magnitude of a couple, we shall do so by stating its moment ; thus, the couple G will signify the couple whose moment is G. It will lead to no inconvenience that the magnitude of the forces which compose the couple are not stated, seeing that the effects of all couples of equal moments acting in the same plane, whatever be the magnitudes and directions of their forces, are the same. It will also be observed, that it is not necessary to state the precise plane in which a couple is situated; it will be sufficient to know its moment, and the position of some line to which its axis is parallel.

54. It will be observed, that all equivalent couples have their axes equal and parallel.

55. If from a point two straight lines be drawn parallel and equal to the axes of two couples, and upon them a parallelogram be described, the diagonal drawn from the same point will be parallel and equal to the axis of the resultant couple.

As the planes (HOA, HOB, suppose) of the couples are not parallel, let them intersect in the line HO (fig. 13). Change the couples into two equivalent couples having their forces FF', ff all equal; place these new couples so that one extremity of their arms OA, OB shall be at O, and the forces F, f which act there, shall act in the line OH, as in the figure. Complete the parallelogram OADB, and draw the diagonals OD, AB, bisecting each other in C. Then because F' and f are equal and act in the same direction, they are equivalent to a resultant F'+ f' acting at C (Art. 40). But such a force at C would likewise be the resultant of the same forces F', f' acting at D, O. We may therefore transpose F' to D, and f' to 0, which being done, f' and fat O balancing may be removed; and there will only remain Fat O and F' at D, forming a couple F, F' whose arm is OD, which is therefore the resultant of the two original couples.

Now the forces of the two component couples and of their resultant being equal, their axes which are proportional to their moments, are in this case proportional to their arms OA, OB, OD; we may therefore consider OA, OB, OD as being equal to the axes. If therefore from O in the plane OBA, we draw three lines respectively perpendicular and equal to OA, OB, OD, they will be the axes of the three couples, and will then take the same position as if the parallelogram OADB were turned through a right angle about the fixed line OH. This figure OADB so turned is the parallelogram stated in the enunciation of the proposition to possess the property which we have just proved belongs to it.

56. Two couples act upon a rigid body in planes which are at right angles to each other; to find their re

sultant.

From any point O, draw OА, ОВ 0

equal and parallel to the axes of the two couples. Complete the rectangle OACB, and draw its diagonal OC. OC is equal and parallel to the axis of the resultant couple.

B

A

C

Let L, M, G be the moments of the two component couples and of their resultant. = COA the angle at which

the axis of G is inclined to that of L.

Then because OA = OC cos e, and OB = OC sin 0;

.. L = G cos e, and M = G sin 0,

which equations determine both the magnitude of the re

sultant couple, and the position of its axis.

57. If it should be required to resolve a given couple whose moment is G into two components acting in planes at right angles to each other, we must use the equations L = G cos 0, M = G sin 0.