in which they are cut be specified by reference to two rectangular axes in the plane. Let the plane be YOX: OX, OY, the axes of reference, O the origin of co-ordinates, and A4, A2, A3, &c., the points in which the plane cuts the lines of the forces, P., P2, P3, &c. Thus each of these points will be specified by perpendiculars drawn from it to the axis. Let the co-ordinates Y of the point A, be denoted by Xi Yı; of A,, by X, Y,; and so on; that is, EY ON=x,, N, A=yı; ON=x,, N, A, =y2, &c.; let also the final resultant be K M A, denoted by R, and its co-ordinates by x and y. Find the resultant of P, and P, by joining A, A2, and dividing the line NN, inversely as the forces. Suppose E the point in which this resultant cuts the plane of reference. Then P,XA,E'=P, E'AZ. To find the co-ordinates, which may be denoted by x'y, of the point E' with reference to OX and OY; draw E'N' perpendicular to OX and cutting it in N, and from A, draw A, K parallel to OX, or perpendicular to A, N,, and cutting it in K and E'N' in M. Then (Euclid VI. 2) A, E' : E'A, :: A,M: MK. Hence PAXA,M=P, MK, P.(x'— X;)=P2(x,- x'), whence we get (P,+P2)x'=P,X, +P,x,; and since R'=P,+P2, R'x'=P,x,+P,x2, and similarly, Ry=Py+P,Y2. We may find the resultant of R! and P, in like manner, and so with all the forces. Hence we have for the final resultant, R=P,+P+P + +P. (3) Rx=P,x,+P2x2 + P3x3 + +P, * (4) Ry=P Y,+ Piy,+Pyys + ... + P In. (5) These equations may include negative forces, or negative coordinates. 558. Conditions of equilibrium of any number of parallel forces. In order that any given parallel forces may be in equilibrium, it is not sufficient alone, that their algebraic sum be equal to zero. For, let R=P, +P, + &c.=0. or we have From this equation it follows that if the forces be divided into two groups, one consisting of the forces reckoned positive, the other of those reckoned negative, the sum, or resultant ($ 556) of the former is equal to the resultant of the latter; that is, if ,R and 'R denote the resultants of the positive and negative groups respectively, R='R. But unless these resultants are directly opposed they do not balance one another; wherefore, if »x,y) and (xy) be the co-ordinates of R and 'R respectively, we must have for equilibrium P,X2 1 and y='y; whence we get ,R,*-'R'x=0 and Ry-'R'y=0. But ,R „x is equal to the sum of those of the terms P x1, &c., which are positive, and 'R'x is equal to the sum of the others each with its sign changed: and so for ,R,y and 'R'y. Hence the preceding equations are equivalent to P,x,+P,x,+ + Prixn=0. PY,+Py,+ +P=0. We conclude that, for equilibrium, it is necessary and sufficient that each of the following three equations be satisfied :P,+P,+P8+ ........ +P=0. (6) P.x,+Pox, +P3*, + +PX=0. (7) Py+ P232+P333 + + P=0. (8) 559. If equation (6) do not hold, but equations (7) and (8) do, the forces have a single resultant through the origin of co-ordinates. If equation (6) and either of the other two do not hold, there will be a single resultant in a line through the corresponding axis of reference, the co-ordinates of the other vanishing. If equation (6) and either of the other two do hold, the system is reducible to a single couple in a plane through that line of reference for which the sum of the products is not equal to nothing. If the plane of reference is perpendicular to the lines of the forces, the moment of this couple is equal to the sum of the products not equal to nothing. 560. In finding the resultant of two contrary forces in any case in which the forces are unequal—the smaller the difference of magnitude between them, the farther removed is the point of application of the resultant. When the difference is nothing, the point is removed to an infinite distance, and the construction ( 555) is thus rendered nugatory. The general solution gives in this case R=0; yet the forces are not in equilibrium, since they are not directly opposed. Hence two equal contrary forces neither balance, nor have a single resultant. It is clear that they have a tendency to turn the body to which they are applied. This system was by Poinsot denominated a couple. In actual cases the direction of a couple is generally reckoned positive if the couple tends to turn contrary to the hands of a watch as seen by a person looking at its face, negative when it tends to turn with the hands. Hence the axis, which may be taken to represent a couple, will show, if drawn according to the rule given in $ 201, whether the couple is positive or negative, according to the side of its plane from which it is regarded. 561. Proposition I. Any two couples in the same or in parallel planes are in equilibrium if their moments are equal and they tend to turn in contrary directions. 1°. Let the forces of the first couple be parallel to those of the second, and let all four forces be in one plane. CLA Let the forces of the first couple be Pin AB and CD, and of the second 12P P in A' B' and C'D. Draw any line E FK EN IFM EF across the lines of the forces, cutPA AP' E' and F'; then the moment of the B DÍ B first couple is P.EF and of the second P.E'F'; and since the moments are equal we have P.EF=P.E'F. Of the four forces, P in AB and P in C'D act in similar directions, and P in CD and Pin A' B' also act in similar directions; and their resultants respectively can be determined by the general method ($ 556). The resultant of P in AB, and Pin C'D', is thus found to be equal to P+P', and if HL is the line in which it acts, P.EK=P.KFY. Again, we have P.EF=P.E'F. Subtract the first member of the latter equation from the first member of the former, and the second member of the latter from the second member of the former: there remains P.FK=P.KEY, from which we conclude, that the resultant of P in CD and P in A'B' is in the line LH. Its magnitude is P+P. Thus the given system is reduced to two equal resultants acting in opposite directions in the same straight line. These balance one another, and therefore the given system is in equilibrium. Corollary. A couple may be transferred from its own arm to any other arm in the same line, if its moment be not altered. 562. Proposition I. 2o. All four forces in one plane, but those of one couple not parallel to those of the other. Produce their lines to meet in four points; and consider the parallelogram thus formed. The products of the sides, each into its perpendicular distance from the side parallel to it, are equal, each product P being the area of the parallelogram. Hence, since the moments of the two couples are equal, their forces are proportional to the sides of the parallelogram along which they act. And, since the couples tend to turn in opposite directions, the four forces represented by the sides of “a parallelogram act in similar directions relatively to the angles, and dissimilar directions in the parallels, and therefore balance one another. Corollary. The statical effect of a couple is not altered, if its arm be turned round any point in the plane of the couple. 563. Proposition I. 3o. The two couples not in the same plane, but the forces equal and parallel. Let there be two couples, acting re PA spectively on arms EF and E'F', which P+P E are parallel but not in the same plane. Join EF and E'F. These lines bisect E one another in 0. PEP VP Of the four forces, P on F and P on E', act in similar directions, and their resultant equal to P+P', may be substituted for them. It acts in a parallel line through 0. Similarly P on E and P' on F' have also a resultant equal to P+P through 0; but these resultants being equal and opposite, balance, and therefore the given system is in equilibrium. Remark 1.-A corresponding demonstration may be applied to every case of two couples, the moments of which are equal, though the forces and arms may be unequal. When the forces and arms are unequal, the lines EF', E'F cut one another in O into parts inversely as the forces. Remark 2.—Hence as an extreme case, Proposition I, 1°, may be brought under this head. Let EF be the arm of one couple, E'F' of the other, both in one straight line. Join FE', and divide it inversely as the forces. Then FK: KE :: EF: E'F' and EF' is divided in the same ratio. Corollary. Transposition of couples. Any two couples in the same or in parallel planes, are equivalent, provided their moments are equal, and they tend to turn in similar directions. 564. Proposition II. Any number of couples in the same or in parallel planes, may be reduced to a single resultant couple, whose moment is equal to the L. P. algebraic sum of their moments, and whose plane is parallel to their planes. Reduce all the couples to forces acting on arm AB, which may be denoted by a. Then if P1, P2, P3, &c., be the forces, the mo P ments of the couples will be Pja, P,a, Pga, &c. Thus we have P1, P2, P3, &c., in AK, reducible to a single force, their sum, and similarly, a single force P,+P, + &c., in BL. one 1 These two forces constitute a couple whose moment is (P,+P, +P, + &c.) a. But this product is equal to P,a+Pa+Pga t &c., the sum of the moments of the given couples, and therefore any number of couples, &c. If any of the couples act in the direction opposite to that reckoned positive, their moments must be reckoned. as negative in the sum. 565. Proposition III. Any two couples not in parallel planes may be reduced to a single resultant couple, whose axis is the diagonal through the point of reference of the parallelogram described upon their axes. 1°. Let the planes of the two couples cut the plane of the diagram M perpendicularly in the lines AA and BB respectively; let the planes of B L the couples also cut each other in a line cutting the plane of the diagram in 0. Through O, as a point of reE ference, draw OK the axis of the first A A couple, and OL the axis of the se cond. On OK and OL construct the parallelogram OKML. Its diagonal B' OM is the axis of the resultant couple. Let the moment of the couple acting in the plane BB', be denoted by G, and of that in AA', by H. For the given couples, substitute two others, with arms equal respectively to Ğ and H, and therefore with forces equal to unity. From OB and OA measure off OE=G, and OF=H, and let these lines be taken as the arms of the two couples respectively. The forces of the couples will thus be perpendicular to the plane of the diagram: those of the first, acting outwards at E, and inwards at 0); and those of the second, outwards at 0, and inwards at F. Thus, of the four equal forces which we have in all, there are two equal and opposite at 0, which therefore balance one another, and may be removed ; and there remain two equal parallel forces, one acting outwards at E, and the other inwards at F, which constitute a couple on an arm EF. This single couple is therefore equivalent to the two given couples. 2°. It remains to be proved that its axis is OM. Join EF. As, by construction, OL and OK are respectively perpendicular to OA, and OB, the angle KOL is equal to the angle AOB. Hence, MLO the supplement of the former is equal to EOF, the supplement of the latter. But OK is equal to OE; each being equal to the moment of the first of the given couples; and therefore LM, which is equal to the former, is equal to OE. Similarly OL is equal to OF. Thus there are two triangles, MLO and EOF, with two sides of one respectively equal to two sides of the other, and the contained angles equal : therefore the remaining sides OM, EF are equal, and the angles LOM, OFE are equal. But since OL is perpendicular to OF, OM is |