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OAB , and the usual See also:convention as co sign is that the See also:area is to be reckoned See also:positive or negative according as the letters 0, A, B, occur in " See also:counter-clockwise " or " clockwise " See also:order. The sum of the moments of two forces about any point 0 is equal to the moment of their resultant (P. Varignon, 1687). Let AB, AC (fig. 16) represent the two forces, AD their resultant; we have to prove that the sum of the triangles OAB, OAC is regard being had to signs. Since the See also:side OA is See also:common, we have to prove that the sum of the perpendiculars from B and C on OA is equal to the perpendicular from D on OA, these perpen- diculars being reckoned positive or negative according as they See also:lie to the right or See also:left of AO. Regarded as a statement concerning the orthogonal projections of the vectors AB and AC (or BD), and of their sum AD, on a See also:line perpendicular to AO, this is obvious. It is now evident that in the See also:process of reduction of a coplanar See also:system no See also:change is made at any See also:stage either in the sum of the projections of the forces on any line or in the sum of their moments about any point. It follows that the single resultant co which the system in See also:general reduces is uniquely determinate, i.e. it acts in a definite line and has a definite magnitude and sense. Again it is necessary and sufficient for See also:equilibrium that the sum of the projections of the forces on each of two perpendicular directions should vanish, and (moreover) that the sum of the moments about some one point should be zero. The fact that three See also:independent conditions must hold for equilibrium is important. The conditions may of course be expressed in different (but See also:equivalent) forms; e.g.•the sum of the moments of the forces about each of the three points which are not See also:col-linear must be zero. The particular See also:case of three forces is of See also:interest. If they are not all parallel they must be concurrent, and their vector-sum must be zero. Thus three forces acting perpendicular to the sides of a triangle at the See also:middle points will be in equilibrium provided they are proportional to the respective sides, and See also:act all inwards or all outwards. This result is easily extended to the case of a See also:polygon of any number of sides; it has an important application in See also:hydrostatics. Again, suppose we have a See also:bar AB resting with its ends on two smooth inclined planes which See also:face each other. Let G be the centre of gravity (§ iI), and let AG=a, GB=b. Let a, $ be the inclinations of the planes, and 8 the See also:angle which the bar makes with the See also:vertical. The position of equilibrium is determined by the See also:consideration that the reactions at A and B, which are by See also:hypothesis normal tothe planes, must meet at a point J on the vertical through G. Hence JG/a=See also:sin (B—a)/sin a, JG/b=sin (8-1-3)/sin i3, (6) 8 = a cot a — b cot whence cot a+b If the bar is See also:uniform we have a=b, and (7) cot0=a (cot a —cot M. The problem of a See also:rod suspended by strings attached to two points of it is virtually identical, the tensions of the strings taking the See also:place of the reactions of the planes. Just as a system of forces is in general equivalent to a single force, so a given force can conversely be replaced by combinations of other forces, in various ways. For instance, a given force (and consequently a system of forces) can be replaced in one and only one way by three forces acting in three assigned straight lines, provided these lines be not concurrent or parallel. Thus if the three lines See also:form a triangle See also:ABC, and if the given force F meet BC in H, then F can be resolved into two components acting in HA, BC, respectively. And the force in HA can be resolved into two components acting in BC, CA, respectively. A See also:simple graphical construction is indicated in fig. 19, where the dotted lines are parallel. As an example, any system of forces acting on the lamina in fig. 9 is balanced by three determinate tensions (or thrusts) in the three links, provided the directions of the latter are not concurrent. If P, Q, R, be any three forces acting along BC, CA, AB, respectively, the line of See also:action of the resultant is determined by the consideration that the sum of the moments about any point on it must vanish. Hence in " trilinear " co-ordinates, with ABC as fundamental triangle, its See also:equation is Pa+QR+Ry=o. If P : Q : R= a : b : c, where a, b, c are the lengths of the sides, this becomes the " line at infinity," and the forces reduce to a couple. The sum of the moments of the two forces of a couple is the same about any point in the See also:plane. Thus in the figure the sum of the moments about 0 is P . OA—P . OB or P . AB, which is independent of the position of O. This sum is called the moment of the couple; it must of course have the proper sign attributed to it. It easily follows that any two couples of the same moment are equivalent, and that any I number of couples can be replaced by a single couple whose moment is the sum of their moments. Since a couple is for our purposes sufficiently represented by its moment, it has been proposed to substitute the name See also:torque (or twisting effort), as See also:free from the See also:suggestion of any See also:special pair of forces. A system of forces represented completely by the sides of a plane polygon taken in order is equivalent to a couple whose C a 'Ss moment is represented by twice the area of the polygon; this is proved by taking moments about any point. If the polygon intersects itself, care must be taken to attribute to the different parts of the area their proper signs. Again, any coplanar system of forces can be replaced by a single force R acting at any assigned point 0, together with a couple G. The force R is the geometric sum of the given forces, and the moment (G) of the couple is equal to the sum of the moments of the given forces about 0. The value of G will in general vary with the position of 0, and will vanish when 0 lies on the line of action of the single resultant. The formal See also:analytical reduction of a system of coplanar forces is as follows. Let (x,, y,), (x2, y2), . . . be the rectangular co-ordinates of any points A,, See also:A2, . . . on the lines of action of the respective forces. The force at A, may be replaced by its components X,, Y,, parallel to the co- y See also:ordinate axes; that at Az by its components X2, Y2, and A. 1,Y, so on. Introducing at 0 two equal and opposite forces Y, ~x X, in Ox, we see that X, at Al may be replaced by an 0 X, .v equal and parallel force at O together with a couple Y, -y,X,. Similarly the force Yi at A, may be replaced by a force Y, at 0 together with a couple x, Y,. The forces X,, Y,, at 0 can thus be transferred to 0 provided we introduce a couple x,Y,-y,X,. Treating the remaining forces in the same way we get a force X, +X2 + . . . or E(X) along Ox, a force Y, + Y2 + . . . or 1(Y) along Oy, and a couple (x,Y,-y,X,) + (x2 Y2-y2X2) + .. . or E(xY-yX). The three conditions of equilibrium are therefore 1(X) = o, 2(Y) = o, E (xY - yX) = o. (8) If 0' be a point whose co-ordinates are (, rl), the moment of the couple when the forces are transferred to 0' as a new origin will be .11 (x- E) Y- (y-n) X). This vanishes, i.e. the system reduces to a single resultant through 0', provided - E.E(Y)+ n•E(X)+ (xY- yX) = o. (9) If E, rt be regarded as current co-ordinates, this is the equation of the line of action of the single resultant to which the system is in general reducible. If the forces are all parallel, making say an angle 0 with Ox, we may write X, = Pi See also:cos 0, Yi = Pi sin 0, X2 = P2 cos 0, Y2 = P2 sin 0, . . . The equation (9) then becomes. {E(xP) - f.E(P)) sin 0 - {E(yP) - n.E(P)) cos 8= o. (to) If the forces P,, P2, ... be turned in the same sense through the same angle about the respective points A,, Az, . . . so as to remain parallel, the value of 0 is alone altered, and the resultant I(P) passes always through the point (xP) -_ (yP) x = 2-CPT' Y- (p) which is determined solely by the configuration of the points A,, A2, . . . and by the ratios Pi: P2:... of the forces acting at them respectively. This point is called the centre of the given system of parallel forces; it is finite and determinate unless 1(P) = o. A geometrical See also:proof of this theorem, which is not restricted to a two-dimensional system, is given later (§ Is). It contains the theory of the centre of gravity as ordinarily under-stood. For if we have an assemblage of particles whose mutual distances are small compared with the dimensions of the See also:earth, the forces of gravity on them constitute a system of sensibly parallel forces, sensibly proportional to the respective masses. If now the assemblage be brought into any other position relative to the earth, without alteration of the mutual distances, this is equivalent to a rotation of the directions of the forces relatively to the assemblage, the ratios of the forces remaining unaltered. Hence there is a certain point, fixed relatively to the assemblage, through which the resultant of gravitational action always passes; this resultant is moreover equal to the sum of the forces on the several particles. The theorem that any coplanar system of forces can be reduced to a force acting through any assigned point, together with a couple, has an important See also:illustration in the theory of the See also:distribution of shearing stress and bending moment in a See also:horizontal See also:beam, or other structure, subject to vertical extraneous forces. If we consider any vertical See also:section P, the forces exerted across the section by the portion of the structure on one side on the portion on the other F may be reduced to a vertical force F at P and a couple M. The force See also:measures the shearing stress, and the couple the bending moment at P; we will reckon these quantities positive when the senses are as indicated in the figure. If the remaining forces acting on the portion of the structure on either side of P are known, then resolving vertically we find F, and taking moments about P we find M. Again if PQ be any segment of the beam which is free from load, Q lying to the right of P, we find Fr=FQ, Mr-MQ=-F.PQ; (12) hence F is See also:constant between the loads, whilst M decreases as we travel to the right, with a constant gradient -F. If PQ be a See also:short segment containing an isolated load W, we have FQ-Fr=-W, MQ=MP; (13) hence F is discontinuous at a concentrated load, diminishing by i i an amount equal to the load as we pass the loaded point to the ~^ right, whilst M is continuous. Accordingly the graph of F for any system of isolated loads will consist fgl . of a See also:series of horizontal lines, whilst that of M will be a continuous . polygon. To pass to the case of continuous loads, let x be measured horizontally along the beam to the right. The load on an See also:element bx of the beam may be represented by wbx, where w is in general a See also:function of x. The equations (12) are now replaced by SF= - wbx, SM= -FSx, whence x, tM) (14) FQ - Fs = - f p wdx, M Q - M r = - PFdx. (II) graphs due to different distributions The latter relation shows that the bending moment varies as the area cut off by the ordinate in the graph of F. In the case of uniform load we have F = wz+A, m = zwx2-Ax+B, (15) where the arbitrary constants A,B are to be determined by the conditions of the special problem, e.g. the conditions at the ends of the beam. The graph of F is a straight line ; that of M is a See also:parabola with vertical See also:axis. In all cases the of load may be superposed. The figure shows the case of a uniform heavy beam supported at its ends. § 5. Graphical See also:Statics.—A graphical method of reducing a plane system of forces was introduced by C. Culmann (1864). It involves the construction of two figures, a force-See also:diagram and a funicular polygon. The force-diagram is constructed by placing end to end a series of vectors representing the given forces in magnitude and direction, and joining the vertices of the polygon thus formed to an arbitrary See also:pole 0. The funicular or See also:link polygon has its vertices on the lines of action of the given forces, and its sides respectively parallel to the lines See also:drawn from 0 in the force-diagram; in particular, the two sides See also:meeting in any vertex are respectively parallel to the lines drawn from 0 to the ends of that side of the force-polygon which represents the corresponding force. The relations will be understood from the annexed diagram, where corresponding lines in the force-diagram a '‘R (to the right) and the funicular (to the left) are numbered similarly. The sides of the force-polygon may in the first instance be arranged in any order; the force-diagram can then be completed in a doubly See also:infinite number of ways, owing to the arbitrary position of 0; and for each force-diagram a simply infinite number of funiculars can be drawn. The two diagrams being sup-posed constructed, it is seen that each of the given systems of forces can be replaced by two components acting in the sides of the funicular which meet at the corresponding vertex, and that the magnitudes of these components will be given by the corresponding triangle of forces in the force-diagram; thus the force I in the figure is equivalent to two forces represented by Or and 12. When this process of replacement is See also:complete, each terminated side of the funicular is the seat of two forces which neutralize one another, and there remain only two uncompensated forces, viz., those See also:resident in the first and last sides of the funicular. If these sides intersect, the resultant acts through the intersection, and its magnitude and direction are given by the line joining the first and last sides of the force-polygon (see fig. 26, where the resultant of the four given forces is denbted by R). As a special case it may happen that the force-polygon is closed, i.e. its first and last points coincide; the first and last sides of the funicular will then be parallel (unless they coincide), and the two uncompensated forces form a couple. If, however, the first and last sides of the funicular coincide, the two outstanding forces neutralize one another, and we have equilibrium. Hence the necessary and sufficient conditions of equilibrium are that the force-polygon and the funicular should both be closed. This is illustrated by fig. 26 if we imagine the force R, reversed, to be included in the system of given forces. It is evident that a system of jointed bars having the shape of the funicular polygon would be in equilibrium under the action of the given forces, supposed applied to the See also:joints; moreover any bar in which the stress is of the nature of a tension (as distinguished from a thrust) might be replaced by a See also:string. This is the origin of the names " link-polygon " and " funicular " (cf. § 2). If funiculars be drawn for two positions 0,0' of the pole in the force-diagram, their corresponding sides will intersect on a straight line parallel to 00'. This is essentially a theorem of projective See also:geometry, but the following statical proof is interesting. Let AB (fig. 27) be any side of the force-polygon, and construct the corresponding portions of the two diagrams, first with 0 and then with 0' as pole. The force corresponding to AB may be replaced by the two components marked x, y; and a force corresponding to BA may be represented by the two components marked x', y'. Hence the forces x, y, x', y' are in equilibrium. Now x, x' have a resultant through H, represented in magnitude and direction by 00', whilst y,y' have a resultant through K represented in magnitude and direction by 0'O. Hence HK must be parallel to 00'. This xvll. 31theorem enables us, when one funicular has been drawn, to construct any other without further reference to the force-diagram. The complete figures obtained by See also:drawing first the force-diagrams of a system of forces in equilibrium with two distinct poles 0, 0', and secondly the corresponding funiculars, have various interesting relations. In the first place, each of these figures may be conceived as an orthogonal See also:projection of a closed plane-faced See also:polyhedron. As regards the former figure this is evident at once; viz. the polyhedron consists of two pyramids with vertices represented by 0, 0', and a common See also:base whose perimeter is represented by the force-polygon (only one of these is shown in fig. 28). As regards the funicular diagram, let LM be the line on which the pairs of corresponding sides of the two polygons meet, and through it draw any two planes w, co'. Through the vertices A, B, C, . . . and A', B', C', . . of the two funiculars draw normals to the plane of the diagram, to meet w and ei' respectively. The points thus obtained are evidently the vertices of a polyhedron with plane faces. B To every line in either of the See also:original figures corresponds of course a parallel line in the other; moreover, it is seen that concurrent lines in either figure correspond to lines forming a closed polygon in the other. Two plane figures so related are called reciprocal, since the properties of the first figure in relation to the second are the same as those of the second with respect to the first. A still simpler instance of reciprocal figures is supplied by the case of concurrent forces in equilibrium (fig. 29). The theory of these reciprocal figures was first studied by J. Clerk See also:Maxwell, who showed amongst other things that a reciprocal can always be drawn to any figure which is the orthogonal projection of a plane-faced polyhedron. If in fact we take the pole of each face of such a polyhedron with respect to a paraboloid of revolution, these poles will be the vertices of a second polyhedron whose edges are the " conjugate lines " of those of the former. If we project both polyhedra orthogonally on a plane perpendicular to the axis of the paraboloid, we obtain two figures which are reciprocal, except that corresponding lines are orthogonal instead of parallel. Another proof will be indicated later (§ 8) in connexion with the properties of the linear complex. It is II convenient to have a notation which shall put in See also:evidence the reciprocal See also:character. For this purpose we may designate the points in one figure by 'letters A, B, C, . . . and the corresponding polygons in the other figure by the same letters; a line joining two points A, B in one figure will then correspond to the side common to the two polygons A, B in the other. This notation was employed by R. H. See also:Bow in connexion with the theory of frames(§ 6, and see also APPLIED See also:MECHANICS below) where reciprocal diagrams are frequently of use (cf. DIAGRAM). When the given forces are all parallel, the force-polygon consists of a series of segments of a straight line. This case has important See also:practical applications; for instance we may use the method to find the pressures on the supports of a beam loaded in any given manner. Thus if AB, BC, CD represent the given loads, in the force-diagram, we construct the sides corresponding to OA, OB, OC, OD in the funicular; we then draw the closing line of the funicular polygon, and a parallel OE to it in the force diagram. The segments DE, See also:EA then represent the upward pressures of the two supports on the beam, which pressures together with the given loads constitute a system of forces in equilibrium. The pressures of the beam on the supports are of course represented by ED, AE. The two diagrams are portions of reciprocal figures, so that Bow's notation is applicable. A graphical method can also be applied to find the moment of a force, or of a system of forces, about any assigned point P. Let F be a force represented by AB in the force-diagram. Draw a parallel through P to meet the sides of the funicular which correspond to OA, OB in the points H, K. If R be the intersection of these sides, which is the moment of F about P. If the given forces are all parallel (say vertical) OM is the same for all, and the moments of the several forces about P are represented on a certain See also:scale by the lengths intercepted by the successive pairs of sides on the vertical through P. Moreover, the moments are compounded by adding (geometrically) the corresponding lengths HK. Hence if a system of vertical forces be in equilibrium, so that the funicular polygon is closed, the length which this polygon intercepts on the vertical through any point P gives the sum of the moments about P of all the forces on one side of this vertical. For instance, in the case of a beam in equilibrium under any given loads and the reactions at the supports, we get a graphical See also:representation of the distribution of bending moment over the beam. The construction in fig. 30 can easily be adjusted so that the closing line shall be horizontal; and the figure then becomes identical with the bending-moment diagram of § 4. If we wish to study the effects of a movable load, or system of loads, in different positions on the beam, it is only necessary to shift the lines of action of the pressures of the supports relatively to the funicular, keeping them at the same distance apart; the only change is then in the position of the closing line of the funicular. It may be remarked that since this line joins homologous points of two " similar " rows it will envelope a parabola. The " centre " (§ 4) of a system of parallel forces of given magnitudes, acting at given points, is easily determined graphically. We have only to construct the line of action of the resultant for each of two arbitrary directions of the forces; the inter-section of the two lines gives the point required. The construction is neatest if the two arbitrary directions are taken at right angles to one another. § 6. Theory of Frames.—A See also:frame is a structure made up of pieces, or members, each of which has two joints connecting it with other members. In a two-dimensional frame, each See also:joint may be conceived as consisting of a small cylindrical See also:pin fitting accurately and smoothly into holes drilled through the members which it connects. This supposition is a somewhat ideal one, and is often only roughly approximated to in practice. We shall suppose, in the first instance, that extraneous forces act on the frame at the joints only, i.e. on the pins. On this See also:assumption, the reactions on any member at its two joints must be equal and opposite. This See also:combination of equal and opposite forces is called the stress in the member; it may be a tension or a thrust. For diagrammatic purposes each member is sufficiently represented by a straight line terminating at the two joints; these lines will be referred to as the bars of the frame. In structural applications a frame must be stiff, or rigid, i.e. it must be incapable of deformation without alteration of length in at least one of its bars. It is said to be just rigid if it ceases to be rigid when any one of its bars is removed. A frame which has more bars than are essential for rigidity may be called over-rigid; such a frame is in general self-stressed, i.e. it is ir. a See also:state of stress independently of the action of extraneous forces. A plane frame of n joints which is just rigid (as regards deformation in its own plane) has 2n-3 bars, for if one bar be held fixed the 2(n-2) co-ordinates of the remaining n-2 joints must just be determined by the lengths of the remaining bars. The See also:total number of bars is therefore 2(n-2) + 1. When a plane frame which is just rigid is subject to a given system of equilibrating extraneous forces (in its own plane) acting on the joints, the stresses in the bars are in general uniquely determinate. For the conditions of equilibrium of the forces on each pin furnish 2n equations, viz. two for each point, which are linear in respect of the stresses and the extraneous forces. This system of equations must involve the three conditions of equilibrium of the extraneous forces which are already identically satisfied, by hypothesis; there remain therefore 2n - 3 independent relations to determine the 2n-3 unknown stresses. A frame of n joints and 2n-3 bars may of course fail to be rigid owing to some parts being over-stiff whilst others are deformable; in such a case it will be found that the statical equations, apart from the three identical relations imposed by the equilibrium of the extraneous forces, are not all independent but are equivalent to less than 2n-3 relations. Another exceptional case, known as the See also:critical case, will be noticed later (§ q). A plane frame which can be built up from a single bar by successive steps, at each of which a new joint is introduced by two the triangles OAB, RHK are similar, and if the perpendiculars OM, RN be drawn we have HK . OM = AB . RN =F . RN, STATICS] new bars meeting there, is called a simple frame; it is obviously just rigid. The stresses produced by extraneous forces in a simple frame can be found by considering the equilibrium of the various joints in a proper See also:succession; and if the graphical method be employed the various polygons of force can be combined into a single force-diagram. This See also:procedure was introduced by W. J. M. See also:Rankine and J. Clerk Maxwell (1864). It may be noticed that if we take an arbitrary pole in the force-diagram, and draw a corresponding funicular in the See also:skeleton diagram which represents the frame together with the lines of action of the extraneous forces, we obtain two complete reciprocal figures, in Maxwell's sense. It is accordingly convenient to use Bow's notation (§ 5), and to distinguish the several compartments of the frame-diagram by letters. See fig. 33, where the successive triangles in the diagram of forces may be constructed in the order XYZ, ZXA, AZB. The class of " simple " frames includes many of the frameworks used in the construction of See also:roofs, lattice girders and suspension See also:bridges; a number of examples will be found in the See also:article BRIDGES. By examining the senses in which the respective forces act at each joint we can ascertain which members are in tension and which are in thrust; in fig. 33 this is indicated by the directions of the arrowheads. When a frame, though just rigid, is not " simple " in the above sense, the preceding method must be replaced, or supplemented, by one or other of various artifices. In some cases the method of sections is sufficient for the purpose. If an ideal section be drawn across the frame, the extraneous forces on either side must be in equilibrium with the forces in the bars cut across; and if the section can be drawn so as to cut only three bars, the forces in these can be found, since the problem reduces to that of resolving a given force into three components acting in three given lines (§ 4). The " critical case " where the directions of the three bars are concurrent is of course excluded. Another method, always available, will be explained under " See also:Work " (§ q). When extraneous forces act on the bars themselves the stress in each bar no longer consists of a simple See also:longitudinal tension or thrust. To find the reactions at the joints we may proceed as follows. Each extraneous force W acting on a bar may be replaced (in an infinite number of ways) by two components P, Q in lines through the centres of the pins at the extremities. In practice the forces W are usually vertical, and the components P, Q are then conveniently taken to be vertical also. We first alter the problem by transferring the forces P, Q to the pins. The stresses in the bars, in the problem as thus modified, may be supposed found by the preceding methods; it remains to infer from the results thus obtained the reactions in the original form of the problem. To find the pressure exerted by a bar AB on the pin A we See also:compound with the force in AB given by the diagram a force equal to P. Conversely, to find the pressure of the pin A on the bar AB we must compound with the force given by the diagram a force equal and opposite to P. This question arises in practice in the theory of " three-jointed " structures; for the purpose in See also:hand such a structure is sufficiently represented by two bars AB, BC. The right-hand figure represents a portion of the force-diagram; in particular ZX represents the pressure of AB on B963 in the modified problem where the loads Wi and See also:W2 on the two bars are replaced by loads Pi, Qi, and. P2, Qa respectively, acting on the pins. Compounding with this XV, which represents Qi, we get the actual pressure ZV exerted by AB on B. The directions and magnitudes of the reactions at A and C are then easily ascertained. On See also:account of its practical importance several other graphical solutions of this problem have been devised. § 7. Three-dimensional See also:Kinematics of a Rigid See also:Body.—The position of a rigid body is determined when we know the positions of three points A, B, C of it which are not collinear, for the position of any other point P is then determined by the three distances PA, PB, PC. The nine co-ordinates (Cartesian or other) of A, B, C are subject to the three relations which See also:express the invariability of the distances BC, CA, AB, and are therefore equivalent to six independent quantities. Hence a rigid body not constrained in any way is said to have six degrees of freedom. Conversely, any six geometrical relations restrict the body in general to one or other of a series of definite positions, none of which can be departed from without violating the conditions in question. For instance, the position of a See also:theodolite is fixed by the fact that its rounded feet See also:rest in contact with six given plane surfaces. Again, a rigid three-dimensional frame can be rigidly fixed relatively to the earth by means of six links. The six independent quantities, or " co-ordinates," which serve to specify the position of a rigid body in space may of course be chosen in an endless variety of ways. We may, for instance, employ the three Cartesian co-ordinates of a particular point 0 of the body, and three angular co-ordinates which express the See also:orientation of the body with respect to O. Thus in fig. 36, if OA, OB, OC be three mutually perpendicular lines in the solid, we may denote by B the angle which OC makes with a fixed direction OZ, by the See also:azimuth of the plane ZOC measured from some fixed plane through OZ, and by f the inclination of the plane COA to the plane ZOC. In fig. 36 these various lines and planes are represented by their intersections with a unit See also:sphere having 0 as centre. This s ery useful, although unsymmetrical, system of angular co-ordinates was introduced by L. See also:Euler. It is exemplified in " See also:Cardan's suspension," as used in connexion with a See also:compass-bowl or a gyroscope. Thus in the gyroscope the " flywheel " (represented by the globe in fig. 37) can turn about a See also:diameter OC of a See also:ring which is itself free to turn about a diametral axis OX at right angles to the former; this axis is carried by a second ring which is free to turn about a fixed diameter OZ, which is at right angles to OX. We proceed to See also:sketch the theory of the finite displacements of a rigid body. It was shown by Euler (1776) that any displacement 111111111111111 in which one point 0 of the body is fixed is equivalent to a pure parallel to their original positions. Hence, if a, v' denote the rotation about some axis through O. Imagine two See also:spheres of equal See also:radius with 0 as their common centre, one fixed in the body and moving with it, the other fixed in space. In any displacement about 0 as a fixed point, the former sphere slides over the latter, as in a " See also:ball-and-socket " joint, Suppose that as the result of the displacement a point of the moving sphere is brought from A to B, whilst the point which was at B is brought to C (cf. fig. 1o). Let J be the pole of the circle ABC (usually a, " small circle " of the fixed sphere), and join JA, JB, JC, AB, BC by See also:great-circle arcs. The spherical isosceles triangles AJB, BJC are congruent, and we see that AB can be brought into the position BC by a rotation about the axis OJ through an angle AJB. It is convenient to distinguish the two senses in which rotation may take place about an axis OA by opposite signs. We shall reckon a rotation as positive when it is related to the direction from 0 to A as the direction of rotation is related to that of See also:translation in a right-handed See also:screw. Thus a negative rotation about OA may be regarded as a positive rotation about OA', the prolongation of AO. Now suppose that a body receives first a positive rotation a about OA, and secondly a positive rotation (3 about OB; and let A, B be the intersections of these axes with a sphere described about 0 as centre. If we construct the spherical triangles ABC, ABC' (fig. 38), having in each g case the angles at A and B equal to 2a and 20 respective- ly, it is evident that the first rotation will bring a point from C to C' and that the second will bring it back to C; the result is therefore equiva- See also:lent to a rotation about OC. We See also:note also that if the given rotations had been effected in the inverse order, the axis of the resultant rotation would have been OC', so that finite rotations do not obey the " commutative See also:law." To find the angle of the equivalent rotation, in the actual case, suppose that the second rotation (about OB) brings a point from A to A'. The spherical triangles ABC, A'BC (fig. 39) are " symmetrically equal," and the angle of the resultant rotation, viz. ACA', is 27 — 2C. This is equivalent to a negative rotation 2C about OC, whence the theorem that the effect of three successive positive rotations 2A, 2B, 2C about OA, OB, OC, respectively, is to leave the body in its original position, provided the See also:circuit ABC is left-handed as seen from O. This theorem is due to O. Rodrigues (184o). The See also:composition of finite rotations about parallel axes is a particular case of the preceding; the radius of the sphere is now infinite, and the triangles are plane. In any continuous See also:motion of a solid about a fixed point 0, the limiting position of the axis of the rotation by which the body can be brought from any one of its positions to a consecutive one is called the instantaneous axis. This axis traces out a certain See also:cone in the body, and a certain cone in space, and the continuous motion in question may be represented as consisting in a See also:rolling of the former cone on the latter. The proof is similar to that of the corresponding theorem of plane kinematics (§ 3). It follows from Euler's theorem that the most general displacement of a rigid body may be effected by a pure translation which brings any one point of it to its final position 0, followed by a pure rotation about some axis through O. Those planes in the body which are perpendicular to this axis obviously remaininitial and final positions of any figure in one of these planes, the displacement could evidently have been effected by. (1) a translation perpendicular to the planes in question, bringing v into some position v" in the plane of v-', and (z) a rotation about a normal to the planes, bringing a" into coincidence with Q (§ 3). In other words, the most general displacement is equivalent to ' a translation parallel to a certain axis combined with a rotation about that axis; i.e. it may be described as a twist about a certain screw. In particular cases, of course, the translation, or the rotation, may vanish. The preceding theorem, which is due to See also:Michel See also:Chasles (1830), may be proved in various other interesting ways. Thus if a point of the body be displaced from A to B, whilst the point which was at B is displaced to C, and that which was at C to D, the four points A, B, C, D lie on a See also:helix whose axis is the common perpendicular to the bisectors of the angles ABC, See also:BCD. This is the axis of the required screw; the amount of the translation is measured by the projection of AB or BC or CD on the axis; and the angle of rotation is given by the inclination of the aforesaid bisectors. This construction was given by M. W. Crofton. Again, H._.Wiener and W. See also:Burnside have employed the See also:half-turn (i.e. a rotation through two right angles) as the fundamental operation. This has the See also:advantage that it is completely specified by the axis of the rotation; the sense being immaterial. Successive half-turns about parallel axes a, b are equivalent to a translation measured by See also:double the distance between these axes in the direction from a to b. Successive half-turns about intersecting axes a, b are equivalent to a rotation about the common perpendicular to a, b at their intersection, of amount equal to twice the acute angle between them, in the direction from a to b. Successive half-turns about two skew axes a, b are equivalent to a twist about a screw whose axis is the common perpendicular to a, b, the translation being double the shortest distance, and the angle of rotation being twice the acute angle between a, b, in the direction from a to b. It is easily shown that any displacement whatever is equivalent to two half-turns and therefore to a screw. In mechanics we are specially concerned with the theory of infinitesimal displacements. This is included in the preceding, but it is simpler in that the various operations are commutative. An infinitely small rotation about any axis is conveniently represented geometrically by a length AB measures along the axis and proportional to the angle of rotation, with the convention that the direction from A to B shall be related to the rotation as is the direction of translation to that of rotation in a right-handed screw. The consequent displacement of any point P will then be at right angles to the plane PAB, its amount will be represented by double the area of the triangle PAB, and its sense will depend on the cyclical order of the letters P, A, B. If AB, AC represent infinitesimal rotations about intersecting axes, the consequent displacement of any point 0 in the plane BAC will be at right angles to this plane, and will be represented by twice the sum of the areas OAB, OAC, taken with proper signs. It follows by See also:analogy with the theory of moments (§ 4) that the resultant rotation will be represented by AD, the vector-sum of AB, AC (see fig. 16). It is easily inferred as a limiting case, or proved directly, that two infinitesimal rotations a, 0 about parallel axes are equivalent to a rotation a+(3 about a parallel axis in the same plane with the two former, and dividing a common perpendicular AB in a point C so that AC/CB =0/a. If the rotations are equal and opposite, so that a+(3=o, the point C is at infinity, and the effect is a translation perpendicular to the plane of the two given axes, of amount a . AB. It thus appears that an infinitesimal rotation is of the nature of a " localized vector," and is subject in all respects to the same mathematical See also:laws as a force, conceived as acting on a rigid body. Moreover, that an infinitesimal translation is analogous to a couple and follows the same laws. These results are due to See also:Poinsot. The analytical treatment of small displacements is as follows. We first suppose that one point 0 of the body is fixed, and take this as the origin of a " right-handed " system of rectangular co-ordinates; i.e. the positive directions of the axes are assumed to be so arranged that a positive rotation of 900 about Ox would bring Oy into the position of Oz, and so on. The displacement will consist of an infinitesimal rotation e about some axis through 0, whose direction-cosines are, say, 1, m, n. From the equivalence of a small rotation to a localized vector it follows that the rotation e will be equivalent to rotations E, al, about Ox, Oy, Oz, respectively, provided E = is, n = mE, = nE, (1) and we note that + + 1-2 = E2. (2) Thus in the case of fig. 36 it may be required to connect the infinitesimal rotations E, n, about OA, OB, OC with the See also:variations of the angular co-ordinates 0, >G, cp. The displacement of the point C of the body is made up of S8 tangential to the See also:meridian ZC and sin 0 SIP perpendicular to the plane of this meridian. Hence, re-solving along the tangents to the arcs BC, CA, respectively, we have =SO sin—sin 0S11, cos0,n=SO cos0+sin0S,ksin0. (3) Again, consider the point of the solid which was initially at A' in the figure. This is displaced relatively to A' through a space S4 perpendicular to the plane of the meridian, whilst A' itself is displaced through a space cos 0 SIP in the same direction. Hence i =S4' +cos 0 Sip. (4) To find the component displacements of a point P of the body, whose co-ordinates are x, y, z, we draw PL normal to the plane yOz, and LH, LK perpendicular to Oy, Oz, respectively. The displacement of P parallel to Ox is the same as that of L, which is made up of a nz a,nd — q'y. In this way we obtain the formulae ex =-nz — 1•y, Sy, = 1'x — Sz = Y —nx. (5) The most general case is derived from this by adding the component displacements X, ,u, v (say) of the point which was at 0; thus 3x=a+nz—3Y, Sy=u+ix—Ea, (6) sz=v+Ey—nx. J The displacement is thus expressed in terms of the six independent quantities E, rt, , X, µ, v. The points whose displacements are in the direction of the resultant axis of rotation are determined by 5x: Sy: Sz = E: n: or (a+nz1"Y)/E=Ca+1'x—tz)In=(v+'fy—nx)li'• (7) These are the equations of a straight line, and the displacement is in fact equivalent to a twist about a screw having this line as axis. The translation parallel to this axis is lSx + mhy + nhz = (XE +,un + vf) Is. (8) The linear magnitude which measures the ratio of translation to rotation in a screw is called the See also:pitch. In the See also:present case the pitch is t ()it + See also:tin + ,) /(I + n2.+ 1'2). (9) Since E2 +.82+ -2,orE'-,is necessarily an See also:absolute invariant for all transformations of the (rectangular) co-ordinate axes, we infer that XE + µa1 + i' is also an absolute invariant. When the latter invariant, but not the former, vanishes, the displacement is equivalent to a pure rotation. If the small displacements of a rigid body be subject to one constraint, e.g. if a point of the body be restricted to lie on a given See also:surface, the mathematical expression of this fact leads to a homogeneous linear equation between the infinitesimals i;, n, j', a, µ, v, say Ai+Bn+CI"+FX+G e+Hv=o. (to) The quantities E, n, T, µ, v are no longer', independent, and the body has now only five degrees of freedom. Every additional constraint introduces an additional equation of the type (lo) and reduces the number of degrees of freedom by one. In See also:Sir R. S. Ball's Theory of Screws an See also:analysis is made of the possible displacements of a body which has respectively two, three, four, five degrees of freedom. We will briefly See also:notice the case of two degrees, which involves an interesting generalization of the method (already explained) of compounding rotations about intersecting axes. We assume that the body receives arbitrary twists about two given screws, and it is required to determine the character of the resultant displacement. We examine first the case where the axes of the two screws are at right angles and intersect. We take these as axes of x and y; then if f, n be the component rotations about them, we have X = hE, p =lob v = o, (II) where h, k, are the pitches of the two given screws. The equations (7) of the axis of the resultant screw then reduce to x/E = z (e + n2) = (k — h) En. Hence, whatever the ratio ; : n, the axis of the resultant screw lies on the conoidal surface z(x2+y2) =ex); (13) where c=2(k—h). The co-ordinates of any point on (13) may be written x-r cos 8, y=r sin 8, z=c sin 28; (14) hence if we imagine a See also:curve of sines to be traced on a circular See also:cylinder so that the circumference just includes two complete undulations, a straight line cutting the axis of the cylinder at right angles and From Sir See also:Robert S. Ball's Theory of Screws. meeting this -curve will generate the surface. This is called a cylindroid. Again, the pitch of the resultant screw is p=(X +,an)/(52+n2)=hcos28+k sin28. (15) The distribution of pitch among the various screws has therefore a simple relation to the pitch-conic hx2+ky2=const; ' (16) viz. the pitch of any screw varies inversely as the square of that diameter of the conic which is parallel to its axis. It is to be noticed that the parameter c of the cylindroid is unaltered if the two pitches h, k be increased by equal amounts; the only change is that all the pitches are increased by the same amount. It remains to show that a system of screws of the above type can be constructed so as to contain any two given screws whatever. In the first place, a cylindroid can be constructed so as to have its axis coincident with the common perpendicular to the axes of the two given screws and to satisfy three other conditions, for the position of the centre, the parameter, and the orientation about the axis are still at our disposal. Hence we can adjust these so that the surface shall contain the axes of the two given screws as generators, and that the difference of the corresponding pitches shall have the proper value. It follows that when a body has two degrees of freedom it can twist about any one of a singly infinite system of screws whose axes lie on a certain cylindroid. In particular cases the cylindroid may degenerate into a plane, the pitches being then all equal. § 8. Three-dimensional Statics.—A system of parallel forces can be combined two and two until they are replaced by a single resultant equal to their sum, acting in a certain line.: As special cases, the system may reduce to a couple, or it may be in equilibrium. In general, however, a three-dimensional system of forces cannot be replaced by a single resultant force. But it may be reduced to simpler elements in a variety of ways. For example, it may be reduced to two forces in perpendicular skew lines. For consider any plane, and let each force, at its intersection with the plane, be resolved into two components, one (P) normal to the plane, the other (Q) in the plane. The assemblage of parallel forces P can be replaced in general by a single force, and the coplanar system of forces Q by another single force. (I2) If the plane in question be chosen perpendicular to the direction of the vector-sum of the given forces, the vector-sum of the components Q is zero, and these components are therefore equivalent to a couple (§ 4). Hence any three-dimensional system can be reduced to a single force R acting in a certain line, together with a couple G in a plane perpendicular to the line. This theorem was first given by L. Poinsot, and the line of action of R was called by him the central axis of the system. The combination of a force and a couple in a perpendicular plane is termed by Sir R. S. Ball a wrench. Its type, as distinguished from its absolute magnitude, may be specified by a screw whose axis is the line of action of R, and whose pitch is the ratio G/R. The case of two forces may be specially noticed. Let AB be the shortest distance between the lines of action, and let AA', BB' (fig. 42) represent the forces. Let a, 8 be the angles which AA', BB' make with the direction of the vector-sum, on opposite sides. See also:Divide AB in 0, so that AA'. cos a. AO = BB'. cos 8. OB, (I) and draw OC parallel to the vector-sum. Resolving AA', BB' each into two components parallel and perpendicular to OC, we see that the former components have a single resultant in OC, of amount R=AA' cos a+BB' cos $, (2) whilst the latter components form a couple of moment G=AA'. AB. sin a=BB'. AB. sin 8. (3) Conversely it is seen that any wrench can be replaced in an infinite number of ways by two forces, and that the line of action of one of these may be chosen quite arbitrarily. Also, we find from (2) and (3) that G.R=AA'. BB'. AB. sin (a+s). (4) The right-hand expression is six times the See also:volume of the See also:tetrahedron of which the lines AA', BB' representing the forces are opposite edges; and we infer that, in whatever way the wrench be resolved into two forces, the volume of this tetrahedron is invariable. To define the moment of a force about an axis HK, we project the force orthogonally on a plane perpendicular to HK and take the moment of the projection about the intersection of HK with the plane (see § 4). Some convention as to sign is necessary; we shall reckon the moment to be positive when the tendency of the force is right-handed as regards the direction from H to K. Since two concurrent forces and their resultant obviously project into two concurrent forces and their resultant, we see that the sum of the moments of two concurrent forces about any axis HK is equal to the moment of their resultant. Parallel forces may be included in this statement as a limiting case. Hence, in whatever way one system of forces is by successive steps replaced by an-other, no change is made in the sum of the moments about any assigned axis. By means of this theorem we can show that the previous reduction of any system to a wrench is unique. From the analogy of couples to See also:translations which was pointed out in § 7, we may infer that a couple is sufficiently represented by a " free " (or non-localized) vector perpendicular to its plane. The length of the vector must be proportional to the moment of the couple, and its sense must be such that the sum of the moments of the two forces of the couple about it is positive. In particular, we infer that couples of the same moment in parallel planes are equivalent; and that couples in any two planes may be compounded by geometrical addition of the corresponding vectors. Independent statical proofs' are of course easily given. Thus, let the plane of the See also:paper be perpendicular to the planes of two couples, and therefore perpendicular to the line of inter-section of these planes. By § 4, each couple can be replaced by two forces P (fig. 43) perpendicular to the plane of the paper, and so that one force of each couple is in the line of intersection (B); the arms (AB,BC) will then be proportional to the respective moments. The two forces at B will See also:cancel, and we are left with a couple of moment P.AC in the plane AC. If we draw three vectors to represent these three couples, they will be perpendicular and proportional to the respective sides of the triangle ABC; hence the third vector is the geometric sum of the other two. Since, in this proof the magnitude of P is arbitrary, it follows incidentally that couples of the same moment in parallel planes, e.g. planes parallel to AC, are equivalent. Hence a couple of moment G, whose axis has the direction (1, m, n) relative to a right-handed system of rectangular axes, C (+11 is equivalent to three couples 1G, mG, nG in the co-ordinate planes. The analytical reduction of a three-dimensional system can now be conducted as follows. Let (x1, y1, z,) be the co-ordinates of a point P, on the line of action of one of the forces, whose components are (say) Xi, Y,, Z1. Draw P,H normal to the plane zOx, and HK perpendicular to Oz. In KH introduce two equal and opposite forces X,. The force X, at P, with —X, in KH forms a couple about Oz, of moment Next, introduce along Ox two equal and opposite forces tX1. The force X, in KH with —X, in Ox forms a couple about Oy, of moment z,X,. Hence the force X, can be transferred from PI to 0, provided we introduce couples of moments z,X, about Oy and about Oz. Dealing in the same way with the forces Y,, Z, at P,, we find that all three components of the force at P, can be transferred to 0, provided we introduce three couples L,, MI, Ni about Ox, Oy, Oz respectively, viz. L1 =y1Zi -z1Y,, M, = z1X1 -x1Z1, Ni = x1Y1-ylXl. (5) It is seen that L,, M,, N, are the moments of the original force at P, about the co-ordinate axes. Summing up for all the forces of the given system, we obtain a force R at 0, whose components are X=M(X,), Y=M(Y,), Z=M(Z,), (6) and a couple G whose components are L=1(L,), M=Z(M,), N=M(N,), (7) where r= r, 2, 3. . . Since See also:R2=X2+Y2+Z2, G2=L2+M2+N2, it is necessary and sufficient for equilibrium that the six quantities X, Y, Z, L, M, N, should all vanish. In words: the sum of the projections of the forces on each of the co-ordinate axes must vanish; and, the sum of the moments of the forces about each of these axes must vanish. If any other point 0', whose co-ordinates are x, y, z, be chosen in place of 0, as the point to which the forces are transferred, we have to write x,—x, z,—z for x,, y,, 01, and so on, in the preceding process. The components of the resultant force R are unaltered, but the new components of couple are found to be L'=L—yZ+zY, M'=M—zX+xZ, (8) N' =N —xY +yX. By properly choosing 0' we can make the plane of the couple perpendicular to the resultant force. The conditions for this are L': M': N'=X:Y:Z,or L—yZ+zY M—zX+xZ N—xY+yX (9) = X Y = Z. STATICS] These are the equations of the central axis. Since the moment of the resultant couple is now G' =XL'-1-YM'-F R R Z R~ LX AMY + NZ (IO) = --Tt the pitch of the equivalent wrench is (LX + MY + NZ)/(X2 + Y2 + Z2). It appears that X2+Y2+Z2 and LX+MY+NZ are absolute invariants (cf. § 7). When the latter invariant, but not the former, vanishes, the system reduces to a single force. The analogy between the mathematical relations of infinitely small displacements on the one hand and those of force-systems on the other enables us immediately to convert any theorem in the one subject into a theorem in the other. For example, we can assert without further proof that any infinitely small displacement may be resolved into two rotations, and that the axis of one of these can be chosen arbitrarily. Again, that wrenches of arbitrary amounts about two given screws compound into a wrench the See also:locus of whose axis is a cylindroid. The mathematical properties of a twist or of a wrench have been the subject of many remarkable investigations, which are, however, of secondary importance from a See also:physical point of view. In the " Null-System " of A. F. See also:Mobius (1790-1868), a line such that the moment of a given wrench about it is zero is called a null-line. The triply infinite system of null-lines form what is called in line-geometry a " complex." As regards the configuration of this complex, consider a line whose shortest distance from the central axis is r, and whose inclination to the central axis is B. The moment of the resultant force R of the wrench about this line is — Rr sin 0, and that of the couple G is G cos B. Hence the line will be a null-line provided tan 8=k/r, (II) where k is the pitch of the wrench. The null-lines which are at a given distance r from a point 0 of the central axis will therefore form one system of generators of a hyperboloid of revolution; and by varying r we get a series of such hyperboloids with a common centre and axis. By moving 0 along the central axis we obtain the whole complex of null-lines. It appears also from (II) that the null-lines whose distance from the central axis is r are tangent lines to a system of helices of slope tan—I(r/k) ; and it is to be noticed that these helices are left-handed if the given wrench is right-handed, and See also:vice versa. Since the given wrench can be replaced by a force acting through any assigned point P, and a couple, the locus of the null-lines through P is a plane, viz. a plane perpendicular to the vector which represents the couple. The complex is therefore of the type called " linear " (in relation to the degree of this locus). The plane in question is called the null-plane of P. If the null-plane of P pass through Q, the null-plane of Q will pass through P, since PQ is a null-line. Again, any plane w is the locus of a system of null-lines meeting in a point, called the null-point of w. If a plane revolve about a fixed straight line p in it, its null-point describes another straight line p', which is called the conjugate line of p. We have seen that the wrench may be replaced by two forces, one of which may act in any arbitrary line p. It is now evident that the second force must act in the conjugate line p', since every line meeting p, p' is a null-line. Again, since the shortest distance between any two conjugate lines cuts the central axis at right angles, the orthogonal projections of two conjugate lines on a plane perpendicular to the central axis will be parallel (fig. 42). This See also:property was employed by L. See also:Cremona to prove the existence under certain conditions of " reciprocal figures " in, a plane (§ 5). If we take any polyhedron with plane faces, the null-planes of its vertices with respect to a given wrench will form another polyhedron, and the edges of the latter will be conjugate (in the above sense) to those of the former. Projecting orthogonally on a plane perpendicular to the central axis we obtain two reciprocal figures. In the analogous theory of infinitely small displacements of a solid, a " null-line " is a line such that the lengthwise displacement of any point on it is zero. Since a wrench is defined by six independent quantities, it can in general be replaced by any system of forces which involves six adjustable elements. For instance, it can in general be replaced by six forces acting in six given lines, e.g. in the six edges of a given tetrahedron. An exception to the general statement occurs when the six lines are such that they are possible lines of action of a system of six forces in equilibrium; they are then said to be in involution. The theory of forces in involution has been studied by A. See also:Cayley, J. J. See also:Sylvester and others. We have seen that a rigid structure may in general be rigidly connected with the earth by six links,-and it now appears that any system of forces acting on the structure can in general be balanced by six determinate forces exerted by the links. If, however, the links are in involution, these forces become infinite or indeterminate. There is a corresponding kinematic peculiarity, in that the connexion is now not strictly rigid, an Infinitely small relative displacement being possible. See § 9.967 When parallel forces of given magnitudes act at given points, the resultant acts through a definite point, or centre of parallel forces, which is independent of the special direction of the forces. If P, be the force at (x,, yr, Zr), acting in the direction (1, m, n), the formulae (6) and (7) reduce to X = Z(P). 1, Y = E(P). m, Z = Z(P). n, (12) and (P).(n5—ms), M = (P).(lz — nx), N = E(P).(mz ly), (13) L = = (4) ((P))' provided x z = z((p) • Ej' Y = These are the same as if we had a single force E(P) acting at the point (x, y, z), which is the same for all directions (1, m, n). We can hence derive the theory of the centre of gravity, as in § 4. An exceptional case occurs when E(P) =o. If we imagine a rigid body to be acted on at given points by forces of given magnitudes in directions (not all parallel) which are fixed in space, then as the body is turned about the resultant wrench will assume different configurations in the body, and will in certain positions reduce to a single force. The investigation of such questions forms the subject of " Astatics," which has been cultivated by Mobius, Minding, G. Darboux and others. As it has no physical bearing it is passed over here. § 9. Work.—The work done by a force acting on a particle, in any infinitely small displacement, is defined as the product of the force into the orthogonal projection of the displacement on the direction of the force; i.e. it is equal to F. 6 s cos 0, where F is the force, 6s the displacement, and 0 is the angle between the directions of F and 6s. In the See also:language of vector analysis (q.v.) it is the " scalar product " of the vector representing the force and the displacement. In the same way, the work done by a force acting on a rigid body in any infinitely small displacement of the body is the scalar product of the force into the displacement of any point on the line of action. This product is the same whatever point on the line of action be taken, since the lengthwise components of the displacements of any two points A, B on a line AB are equal, to the first order of small quantities. To see this, let A', B' be the displaced positions of A, B, and let ¢ be the infinitely small angle between AB and A'B'. Then if a, /3 be the orthogonal projections of A', B' on AB, we have Aa—Bj3=AB—a/3=AB(1 —cos ,) =JAB.02, ultimately. Since this is of the second order, the products F.Aa and F.B/3 are ultimately equal. The total work done by two concurrent forces acting on a particle, or on a rigid body, in any infinitely small displacement, is equal to the work of their resultant. Let AB, AC (fig. 46) represent the forces, AD their resultant, and let All be the direction of the displacement 6s of the point A. The proposi- P tion follows at once from the fact that the sum of orthogonal —j --3 projections of AB, AC on All is equal to the projection of AD. It is to be noticed that AI3 need not be in the same plane with AB, AC. It follows from the preceding statements that any two systems of forces which are statically equivalent, according to the principles of §§ 4, 8, will (to the first order of small quantities) do the same amount of work in any infinitely small displacement of a rigid body to which they may be applied. It is also evident that the total work done in two or more successive infinitely small displacements is equal to the work done in the resultant displacement. The work of a couple in any infinitely small rotation of a rigid body about an axis perpendicular to the plane of the couple is equal to the product of the moment of the couple into the angle of rotation, proper conventions as to sign being observed. Let the couple consist of two forces P, P (fig. 47) in the plane of the paper, and let J be the point where this plane is met by the axis of rotation. Draw JBA perpendicular to the lines of action, and let e be the angle of rotation. The work of the couple is P. JA. e—P. JB.e=P. AB. e=Ge, if G be the moment of the couple. The analytical calculation of the work done by a system of forces in any infinitesimal displacement is as follows. For a two-dimensional system we have, in the notation of §§ 3, 4, Z(XSx+YSy) =EIX(X—ye)+Y(µ+xe)l = (X). a +1(Y). µ+E (xY —yX)e (I) =Xa+Yµ+Ne. Again, for a three-dimensional system, in the notation of §§ 7, 8, (Xax+Yay+Zaz) (2) =YIX(a+nz—i-y)+Y+l"x—Ex)+Z(v+Ey—nx)} _ (X).X+Z(Y).µ+E(Z).v(yZ—zY).i;+T(zX—xZ). =XX+Yµ+Zv+Lt-l-Mn+N (xY—yX).l This expression gives the work done by a given wrench when the body receives a given infinitely small twist; it must of course be an absolute invariant for all transformations of rectangular See also:axe. The first three terms express the work done by the components of a force (X, Y, Z) acting at 0, and the remaining three terms express the work of a couple (L, M, N). The work done by a wrench about a given screw, when the body twists about a second given screw, may be calculated directly as follows. In fig. 48 let R, G be the force and couple of the wrench, e,r the rotation and translation in the twist. Let the axes of the wrench and the twist be inclined at an angle B, and let h be the shortest distance between them. The displacement of the point H in the figure, resolved in the direction of R, is r cos 0—eh sin O. The work is therefore R(r cos 0—eh sin B)+G cos B Rel(p+p') cos B—h sin 01, (3) if G = pR, r = p'e, i.e. p, p' are the pitches of the two screws. The See also:factor (p+p') cos 8—h sin 0 is called the virtual coefficient of the two screws which define the types of the wrench and twist, respectively. A screw is determined by its axis and its pitch, and therefore involves five independent elements. These may be, for instance, the five ratios : n: f : X : µ: v of the six quantities which specify an infinitesimal twist about the screw. If the twist is a pure rotation, these quantities are subject to the relation al: +an+v3' = o. (4) In the analytical investigations of line geometry, these six quantities, supposed subject to the relation (4), are used to specify a line, and are called the six " co-ordinates " of the line; they are of course equivalent to only four independent quantities. If a line is a null-line with respect to the wrench (X, Y, Z, L, M, N), the work done in an infinitely small rotation about it is zero, and its co-ordinates are accordingly subject to the further relation LS+Mn+N3'+Xa+Ya+Zv=o, (5) where the coefficients are constant. This is the equation of a " linear complex " (cf. § 8). Two screws are reciprocal when a wrench about one does no work on a body which twists about the other. The See also:condition for this is
X5'+a+i +v3'+a' +µ n+v'1'=G, (6)
if the screws be defined by the ratios : n :1' : X : µ : v and E': n': µ': v',respectively. The theory of the screw-systems which are reciprocal to one, two, three, four given screws respectively has been investigated by Sir R. S. Ball.
Considering a rigid body in any given position, we may eon-template the whole See also:group of infinitesimal displacements which might be given to it. If the extraneous forces are n equilibrium the total work which they would perform in any such displacement would be zero, since they reduce to a zero force and a zero couple. This is (in See also:part) the celebrated principle of virtual velocities, now often described as the principle of virtual work, enunciated by See also: If this vanishes for all values of X, v, E, rl, we must have X, Y, Z, L, M, N = o, which are the conditions of equilibrium. The principle can of course be extended to any system of particles or rigid bodies, connected together in any way, provided we take into account the See also:internal stresses, or reactions, between the various parts. Each such reaction consists of two equal and opposite'forces, both of which may contribute to the equation of virtual work. The proper significance of the principle of virtual work, and of its converse, will appear more clearly when we come to See also:kinetics (§ 16); for the present it may be regarded merely as a compact and (for many purposes) highly convenient See also:summary of the laws of equilibrium. Its special value lies in this, that by a suitable See also:adjustment of the hypothetical displacements we are often enabled to eliminate unknown reactions. For example, in the case of a particle lying on a smooth curve, or on a smooth surface, if it be displaced along the curve, or on the surface, the virtual work of the normal component of the pressure may be ignored, since it is of the second order. Again, if two bodies are connected by a string or rod, and if the hypothetical displacements be adjusted so that the distance between the points of See also:attachment is unaltered, the corresponding stress may be ignored. This is evident from fig. 45; if AB, A'B' represent the two positions of a string, and T be the tension, the virtual work of the two forces t T at A,B is T(Aa—B,3), which was shown to be of the second order. Again, the normal pressure between two surfaces disappears from the equation, provided the displacements be such that one of these surfaces merely slides relatively to the other. It is evident, in the first place, that in any displacement common to the two surfaces, the work of the two equal and opposite normal pressures will cancel; moreover if, one of the surfaces being fixed, an infinitely small displacement shifts the point of contact from A to B, and if A' be the new position of that point of the sliding body which was at A, the projection of AA' on the normal at A is of the second order. It is to be noticed, in this case, that the tangential reaction (if any) between the two surfaces is not eliminated. Again, if the displacements be such that one curved surface rolls without sliding on another, the reaction, whether normal or tangential, at the point of See also:con-tact may be ignored. For the virtual work of two equal and opposite forces will cancel in any displacement which is common to the two surfaces; whilst, if one surface be fixed, the displacement of that point of the rolling surface which was in contact with the other is of the second order. We are thus able to imagine a great variety of See also:mechanical systems to which the principle of virtual work can be applied without any regard to K the internal stresses, provided the hypothetical displacements be such that none of the connexions of the system are violated. If the system be subject to gravity, the corresponding part of the virtual work can be calculated from the displacement of the centre of gravity. If W,, Ws, . . . be the weights of a system of particles, whose depths below a fixed horizontal plane of reference are z,, z2, . . . , respectively, the virtual work of gravity is W,Sz, + W26z2 + ... = S(W,z, + W2z2 + • •) (7) = (W, + W2 + ...)H, where 2 is the See also:depth of the centre of gravity (see § 8 (14) and § II (6)). This expression is the same as if the whole See also:mass were concentrated at the centre of gravity, and displaced with this point. An important conclusion is that in any displacement of a system of bodies in equilibrium, such that the virtual work of all forces except gravity may be ignored, the depth of the centre of gravity is " stationary." The question as to stability of equilibrium belongs essentially to kinetics; but we may state by anticipation that in cases where gravity is the only force which does work, the equilibrium of a body or system of bodies is See also:stable only if the depth of the centre of gravity be a maximum. Consider, for instance, the case of a bar resting with its ends on two smooth inclines (fig. 18). If the bar be displaced in a vertical plane so that its ends slide on the two inclines, the instantaneous centre is at the point J. The displacement of G is at right angles to JG; this shows that for equilibrium JG must be vertical. Again, the locus of G is an arc of an See also:ellipse whose centre is in the intersection of the planes; since this arc is See also:convex upwards the equilibrium is unstable. A general criterion for the case of a rigid body movable in two dimensions, with one degree of freedom, can be obtained as follows. We have seen (§ 3) that the sequence of possible positions is obtained if we imagine the " body-centrode " to See also:roll on the " spacecentrode." For equilibrium, the See also:altitude of the centre of gravity G must be stationary; hence G must lie in the same vertical line with the point of contact J of the two curves. Further, it is known from the theory of " roulettes " that the locus of G will be See also:concave or convex upwards according as cos4'_ i I h p+P (8) where p, p' are the radii of curvature of the two curves at J, ¢ is the inclination of the common tangent at J to the horizontal, and h is the height of G above J. The signs of p, p' are to be taken positive when the curvatures are as in the See also:standard case shown in fig. 49. Hence for stability the upper sign must obtain in (8). The same criterion may be arrived at in a more intuitive manner as follows. If the body be supposed to roll (say to the right) until the curves See also:touch at J', and if JJ'=6s, the angle through which the upper figure rotates is Ss/p+Ss/p', and the horizontal displacement of G is equal to the product of this expression into h. If this displacement be less than the horizontal projection of Jr, viz. Ss cos ¢, the vertical through the new position of G will fall to the left of J' and gravity will tend to restore the body to its former position. It is here assumed that the remaining forces acting on the body in its displaced position have zero moment about J'; this is evidently the case, for instance, in the problem of " rocking stones." The principle of virtual work is specially convenient in the theory of frames (§ 6), since the reactions at smooth joints and the stresses in inextensible bars may be left out of account. In particular, in the case of a frame which is just rigid, the principle enables us to find the stress in any one bar independently of the rest. If we imagine the bar in question to be removed, equilibrium will still persist if we introduce two equal and opposite forces S, of suitable magnitude, at the joints which it connected. In any infinitely small deformation of the frame as thus modified, the virtual work of the forces S, together with that of the original extraneous forces, must vanish; this determines S. As a simple example, take the case of a See also:light frame, whose bars form the slides of a rhombus See also:ABCD with the See also:diagonal BD, suspended from A and carrying a See also:weight W at C; and let it be required to findthe stress in BD. If we remove the bar BD, and apply two equal and opposite forces S at B and D, the equation is W .6(21 cos 8) + 2 S . 6(1 sin 8) = o, where 1 is the length of a side of the rhombus, and 8 its inclination to the vertical. Hence S=W tan B=W . BD/AC. (8) The method is specially appropriate when the frame, although just rigid, is not " simple " in the sense of § 6, and when accordingly the method of reciprocal figures is not immediately available. To avoid the intricate trigonometrical calculations which would often be necessary, graphical devices have been introduced by H. See also: Amongst these forces we must include the two equal and opposite forces S which take the place of the stress in the removed bar FC. The above method lends itself naturally to the investigation of the critical forms of a frame whose general structure is given. We have seen that the stresses produced by an equilibrating system of extraneous forces in a frame which is just rigid, according to the criterion of § 6, are in general uniquely determinate; in particular, when there are no extraneous forces the bars are in general free from stress. It may however happen that owing to some special relation between the lengths of the bars the frame admits of an infinitesimal deformation. The simplest case is that of a frame of three bars, when the three joints A, B, C fall into a straght line; a small displacement of the joint B at right angles to AC would involve changes in the lengths of AB, BC which are only of the second order of small quantities. Another example is shown in fig. 53. The graphical method leads at once to the detection of such cases. Thus in the hexagonal frame of fig. 52, if an infinitesimal deformation is possible without removing the bar CF, the instantaneous centre of CF (when AB is fixed) will be at the intersection of AF and BC, and since CC', FF' represent the virtual velocities of the points C, F, turned each through a right angle, C'F' must be parallel to CF. Conversely, if this condition be satisfied, an infinitesimal deformation is possible. The result may be generalized into the statement that a frame has a critical form whenever a frame of the same structure can be designed with corresponding bars parallel, but without complete geometric similarity. In the case of fig. 52 it may be shown that an equivalent condition is that the six points A, B, C, D, E, F should lie on a conic (M. W. Crofton). This is fulfilled when the opposite sides of the hexagon are parallel, and (as a still more special case) when the hexagon is See also:regular. When a frame has a critical form it may be in a state of stress independently of the action of extraneous forces; moreover, the stresses due to extraneous forces are indeterminate, and may be infinite. For suppose as before that one of the bars is removed. If there are no extraneous forces the equation of virtual work reduces to S . as = o, where S is the stress in the removed bar, and bs is the change in the distance between the joints which it connected. In a critical form we have Os o, and the equation is satisfied by an arbitrary value of S; a consistent system of stresses in the remaining bars can then be found by preceding rules. Again, when extraneous forces P act on the joints, the equation is Z(P.3p) +S . as =o, where by is the displacement of any joint in the direction of the corresponding force P. If E(P. Sp) =o, the stresses are merely indeterminate as before; but if (P. Sp) does not vanish, the equation cannot be satisfied by any finite value of S, since as =o. This means that, if the material of the frame were absolutely unyielding, no finite stresses in the bars would enable it to withstand the extraneous forces. With actual materials, the frame would yield elastically, until its configuration is no longer " critical." The stresses in the bars would then be comparatively very great, although finite. The use of frames which approximate to a critical form is of course to be avoided in practice. A brief reference must suffice to the theory of three dimensional frames. This is important from a technical point of view, since all structures are practically three-dimensional. We may note that a frame of n joints which is just rigid must have 3n—6 bars; and that the stresses produced in such a frame by a given system of extraneous forces in equilibrium are statically determinate, subject to the exception of " critical forms." § to. Statics of Inextensible Chains.—The theory of bodies or structures which are deformable in their smallest parts belongs properly to See also:elasticity (q.v.). The case of inextensible strings or chains is, however, so simple that it is generally included in expositions of pure statics. It is assumed that the form can be sufficiently represented by a plane curve, that the stress (tension) at any point P of the curve, between the two portions which meet there, is in the direction of the tangent at P, and that the forces on any linear element as must satisfy the conditions of equilibrium laid down in § r. It follows that the forces on any finite portion will satisfy the conditions of equilibrium which apply to the case of a rigid body (§ 4). We will suppose in the first instance that the curve is plane. It is often convenient to resolve the forces on an element PQ (= Ss) in the directions of the tangent and normal respectively. If T, T + ST be the tensions at P, Q, and S¢ be the angle between the directions of the curve at these points, the components of the tensions along the tangent at P give (T+ST) cos +~—T, or ST, ultimately; whilst for the component along the normal at TS>G, or Tas/p, where p is the radius of curvature. Suppose, for example, that we have a light string stretched over a smooth curve; and let RSs denote the normal pressure (outwards from the centre of curvature) on Ss. The two resolutions give ST = o, TS4' = RSs, or T=const., R=T/p. (1) The tension is constant, and the pressure per unit length varies as the curvature. Next suppose that the curve is " rough "; and let Fas be the tangential force of See also:friction on as. We have ST t FSs = o, Talk = RSs, where the upper or See also:lower sign is to be takenaccording to the sense in which F acts. We assume that in limiting equilibrium we have F =,uR, everywhere, where µ is the coefficient of friction. If the string be on the point of slipping in the direction in which i/i increases, the lower sign, is to be taken; hence ST = FSs = µTS¢, whence T = Toeµ4`, (2) if To be the tension corresponding to See also:Ili = o. This illustrates the resistance to dragging of a rope coiled See also:round a See also:post; e.g. if we put µ= •3, 1//=See also:air, we find for the change of tension in one turn T/To=6.5. In two turns this ratio is squared, and so on. Again, take the case of a string under gravity, in contact with a smooth curve in a vertical plane. Let 1/i denote the inclination to the horizontal, and was the weight of an element as. The tangential and normal components of was are —As sin 1// and —Os cos 1//. Hence 6T =was sin ip, TS¢=wOs cos ¢-{-See also:Ras. (3) If we take rectangular axes Ox, Oy, of which Oy is drawn vertically upwards, we have ay= sin Ss, whence ST =Ay. If the string be uniform, w is constant, and T=wy+const. w(y—yo), (4) say; hence the tension varies as the height above some fixed level (yo). The pressure is then given by the See also:formula R=Td—w cos (5) In the case of a See also:chain See also:hanging freely under gravity it is usually convenient to formulate the conditions of equilibrium of a finite portion PQ. The forces on this reduce to three, viz. the weight of PQ and the tensions at P,Q. Hence these three forces will be concurrent, and their ratios will be given by a triangle of forces. In particular, if we consider a length AP beginning at the lowest point A, then resolving horizontally and vertically we have T cos ¢=To, T sin ,p =W, (6) where To is the tension at A, and W is the weight of PA. The former equation expresses that the horizontal tension is constant. If the chain be uniform we have W = ws, where s is the arc AP: hence ws=To tan >G. If we write To=wa, so that a is the length of a portion of the chain whose weight would equal the horizontal tension, this becomes s=a tan yf.. (7) This is the " See also:intrinsic " equation of the curve. If the axes of x and y be taken horizontal and vertical (upwards), we derive x =a See also:log (sec #+tan ,fr), y"= a sec It/. (8) Eliminating II/ we obtain the Cartesian equation y=a cosh ax (9) of the common See also:catenary, as it is called (fig. 56). The omission of the additive arbitrary constants of integration in (8) is equivalent to a special choice of the origin 0 of co-ordinates; viz. 0 is at a distance a vertically below the lowest point (4' = o) of the curve. The horizontal line through 0 is called the directrix. The relations s.= a sinha, y2=a2+s2, T=To sec = wy, (1 o) which are involved in the preceding formulae are also note-worthy. It is a classical problem in the calculus of variations to deduce the equation (q) from the condition that the depth of the centre of gravity of a chain of given length hanging between fixed points must be stationary (§ 9). The length a is called the parameter of the catenary; it determines the scale of the curve, all catenaries being geometrically similar. If weights be suspended from various points of a hanging chain, the intervening portions will form arcs of equal catenaries, since the horizontal tension (wa) is the same for all. Again, if a chain pass over a perfectly smooth peg, the cate- naries in which it hangs on the two sides, though usually of different parameters, will have the same directrix, since by (to) y is the same for both at the peg. As an example of the use of the formulae we may determine the maximum span for a See also:wire of given material. The condition is that the tension must not exceed the weight of a certain length A of the wire. At the ends we shall have y=A, or A = a cosh (II) a and the problem is to make x a maximum for variations of a. Differentiating (II) we find that, if dx/da=o, x x a tank =1. It is easily seen graphically, or from a table of hyperbolic tangents, that the equation u tanh u = I has only one positive See also:root (u =1.200) ; the span is therefore 2X =2au =2A/ sinh u =1.326 A, and the length of wire is 2s=2A/u = 1.667 A. The tangents at the ends meet on the directrix, and their inclination to the horizontal is 56° 30'. The relation between the sag, the tension, and the span of a wire (e.g. a See also:telegraph wire) stretched nearly straight between two points A, B at the same level is determined most simply from first principles. If T be the tension, W the total weight, k the sag in the middle, and yG the inclination to the horizontal at A or B, we have 2T¢ =W, AB =2pl', approximately, where p is the radius of curvature. Since 2kp=(4AB)2, ultimately, we have k=iW . AB/T. (13) The same formula applies if A, B be at different levels, provided k be the sag, measured vertically, half way between A and B. In relation to the theory of suspension bridges the case where the weight of any portion of the chain varies as its horizontal projection is of interest. The vertical through the centre of gravity of the arc AP (see fig. 55) will then bisect its horizontal projection AN; hence if PS be the tangent at P we shall have AS = SN. This property is characteristic of a parabola whose axis is vertical. If we take A as origin and AN as axis of x, the weight of AP may be denoted by wx, where w is the weight per unit length at A. Since PNS is a triangle of forces for the portion AP of the chain, we have wx/To=PN/NS, or y=w . x2/2Ta, (14) which is the equation of the parabola in question. The result might of course have been inferred from the theory of the parabolic funicular in § 2. Finally, we may refer to the catenary of uniform strength, where the See also:cross-section of the wire (or See also:cable) is supposed to vary as the tension. Hence w, the weight per See also:foot, varies as T, and we maywrite T =wA, where A is a constant length. Resolving along the normal the forces on an element Ss, we find Ta¢=was cos yG, whence ds (15) p=T=A sec From this we derive x=y=A (16) log secyk, where the directions of x and y are horizontal and vertical, and the origin is taken at the lowest point. The curve (fig. 58) has two vertical asymptotes x= =2irX; this shows that however the thickness of a cable be adjusted there is a limit aA to the horizontal span, where A depends on the tensile strength of the material. For a uniform catenary the limit was found above to be I.326A. Z(m GP) =o.' For, take any point 0, and construct the vector E(m.OP OG = I(m) E(m.GP) = (m(O+OP)} = (m).G0+E(m).OP = o. (3) Also there cannot be a distinct point G' such that Z (m. G'P) = o, for we should have, by subtraction, Etm(GP+Pa')}=o, or E(m).GG'=o; (4) i.e. G' must coincide with G. The point G determined by (I) is called the mass-centre or centre of inertia of the given system. It is easily seen that, in the process of determining the mass-centre, any group of particles may be replaced by a single particle whose mass is equal to that of the group, situate at the mass-centre of the group. If through P,, P2, . . . Pn we draw any system of parallel planes meeting a straight line OX in the points M,, M2, . . . M,,, the collinear vectors OM,, OM2 . . . OM,, may be called the " projections " of OP,, OP2i . . . OP,,, on OX. Let these projections be denoted algebraically by x,, x2, . . x,,, the sign being positive or negative according as the direction is that of OX or the See also:reverse. Since the projection of a vector- (12) i. e. the curve must be a " geodesic," and that the normal pressure 0 For investigations See also:relating to the equilibrium of a string in three dimensions we must refer to the textbooks. In the case of a string stretched over a smooth surface, but in other respects free from extraneous force, the tensions at the ends of a small element Ss must be balanced by the normal reaction of the surface. It follows that the osculating plane of the curve formed by the string must contain the normal to the surface, per unit length must vary as the See also:principal curvature of the curve. § Ir. Theory of Mass-Systems.—This is a purely geometrical subject. We consider a system of points P,, P2 . . . , with which are associated certain co-efficients m,, m2, . . . m,,, respectively. In the application to mechanics these coefficients are the masses of particles situate at the respective points, and are therefore all positive. We shall make this supposition in what follows, but it should be remarked that hardly any difference is made in the theory if some of the coefficients have a different sign from the rest, except in the special case where Z (m) = o. This has a certain interest in See also:magnetism. In a given mass-system there exists one and only one point G such that (I) (2) Then sum is the sum of the projections of the several vectors, the equation (2) gives (5) axis or pole. In the case of an axial moment, the square root by the total mass E(m), the result is called the mean square of the distances of the particles from the respective plane, of the resulting mean square is called the radius of gyration of the system about the axis in question. If we take rectangular axes through any point 0, the quadratic moments with respect to the co-ordinate planes are Is=E(mx2), I,, =(my2), I,=(mz2); (9) those with respect to the co-ordinate axes are Iv:=E{m(y2+z2)}, I.s=E{m(z2+x2)}, I.,, m(x'--1-y2)}; (Io) whilst the polar quadratic moment with respect to 0 is 10 = E{m(x2+y2+z2)}• z =E(mx) I(m) ' --> if x be the projection of 0G. Hence if the Cartesian co-ordinates of P2, .. . P„ relative to any axes, rectangular or oblique be (x, y,, z,), (xz, y2, z2), . , (xn, y,,, z°), the mass-centre (x, y, z) is determined by the formulae x=~('nx) I(my) E(mz) E(m)' y— E(m)' z x(m). Jf we write x=x+E, Y=3'+27, z=z-H", so that E, n, t denote co-ordinates relative to the mass-centre G, we have from (6) ~(m) =o, Z(mn) =o, (stir) =o. (7) One or two special cases may be noticed. If three masses a, ti, y be situate at the vertices of a triangle ABC, the mass-centre of t3 and y is at a point A' in BC, such that ti. BA' =y . A'C. The mass-centre (G) of a, t3, y will then divide AA' so that a . AG = (tz+y) GA'. It is easily proved that a : i4 : y=ABGA :OGCA : nGAB; also, by giving suitable values (positive or negative) to the ratios a : p :7 we can make G assume any assigned position in the plane ABC. We have here the origin of the " barycentric co-ordinates " of Mobius, now usually known as " areal " co-ordinates. If a+t3+y =o, G is at infinity; if a=t3=y, G is at the intersection of the median lines of the triangle; if a : t3 : 7=a : b : c, G is at the centre of the inscribed circle. Again, if G be the mass-centre of four particles a, 13, 7, S situate at the vertices of a tetrahedron ABCD, we find a : ti : y : S =tet° GBCD : tet° GCDA : tee GDAB : tet° GABC, and by suitable determination of the ratios on the left hand we can make G assume any assigned position in space. Additional information and CommentsThere are no comments yet for this article.
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