Question
Replace the two forces by an equivalent resultant force F(R) and couple moment M(RO) at point O .Replace the two forces by an equivalent resultant force and couple moment at point . F = 94N , P = 94N , and L = 160mm .Determine M(RO).
Explanation / Answer
00 00 ro g] act at and make angles ( a i ft. n)> ( A*, y*), ... (n A> yn) with the axes > let each of them be replaced by its three components along Ox, Oy, Oz Fig. 228. 2 NON-COPLANAR FORCES. [199. and if 2J, ST, 2Z denote the sums of the components along the axes, we shall have 2J = P 1 cosa 1 + P 2 cosa 2 + ... + P n cosa 7l , ^ + P n cosy n , , and the whole system of forces will be replaced by the three forces, 2X, 27, and 2# along the axes of a?, y, and z. But the resultant of three forces in these directions is the diagonal of the parallelepiped determined by them. Hence, R being the magnitude of this resultant, (Sr)H(2^ (2) and if 0, , //, are the direction-angles of R, ^ X 2 Y ^ / COS = yp-j COS= ry-3 008^ = - (3) ./t _Zl ffm ^n) are the co-ordinates of the extremities 200.] TBANSFOBMATION OP -COUPLES. 3 A l9 A 2 ,...A n of the forces acting on the particle, it is clear that where or, ^, 5 are the co-ordinates of G, the centre of mass of equal masses placed at the extremities of the forces. Hence by equations (1) of Art. 198, JB = * . OG, and COB 9=^-, con* = JL, which show that the resultant is represented in magnitude and direction by n. 00. 200.] Transformation of Couples. To what has been given in Chapter V on the transformation of couples it is necessary to add a few propositions relating to couples in different planes. (a) A couple acting on a rigid body may be transferred to any plane parallel to its own. Let AB (Fig. 229) be the arm of a couple (P, P) and let A' If be any line parallel and equal to AB. At A' introduce two equal and opposite forces, P and P', parallel to AP, and at IS introduce f * the same forces. The introduction of these forces will not disturb the * t p f " -V" I *.-""* O j. state of the body. Draw Aff &"*""" ^B and A'B, which will bisect each I other at 0. Then the force P ^ 229t at A and the force P' at -S' will give a resultant 'equal to 2P at 0; and P at Ji and P 7 at A / will give a resultant equal and opposite to this at the same point. Hence there remain the forces P at Jf and P at T$ f ; that is, the couple (P, P) with arm AB has been moved to any plane parallel to its own. From Chapter V it is now clear that the only essential properties of a couple are (1) the constancy of its moment and (2) the direction of its plane ; or, in other words, the constancy of the magnitude and direction of its axis the actual position of the axis in space is of no consequence, but only its direction ; two couples whose axes are of equal length and in the same direction are absolutely identical. 4 NON-COPLANAR FORGES. [200. Hence the axis of a couple is what is called a vector, or directed line of constant magnitude but not localized and we shall always, as in the representation of forces, suppose the axis to be marked by an arrow-head. (/3) Convention with regard to the sense of tlie axis of a couple. The following convention for representing the magnitude and sense of the moment of a couple by means of an axis is adopted by common consent for the purpose of enabling us to compound and resolve couples in any planes: Hold a watch with its plane parallel to the plane of the couple. Then, according as the motion of the hands is contrary to, or along with, the sense in which the couple tends to produce rotation, draw the axis of the couple through the face or through the back of the watch. (y) Two couples result in a single couple whose axis is found from the axes of the component couples by the parallelogram law. Let the planes of the couples intersect in the line AB (Mg. 230) and the arm of each be made AB, by moving each couple in its own plane, and then suitably altering the forces of each couple (Art. 79, Chap. V). Let P, P be the forces of one couple, and Q, Q those of the other. At B draw * Bp per- pendicular to the plane PABP and proportional to the mo- 330. ment of the couple (P, P). We may evidently take Bp = P, since the couples have a common arm. Draw Bq perpendicular to the plane QABQ and equal to Q. Now evidently the forces P and Q at B compound a resultant, JV, equal and parallel to the resultant of P and Q at A. Hence the two couples compound a single couple. Again, draw Jir perpendicular to the plane RABlt and equal to 11. Bp, Bq t and Br are then 'the axes of the couples (P, P), (Q> Q)> and (11, H). But it is manifest that the figure Bprq is * According to the convention (0) the couples in this figure are both negative, and the axes Bp and Hq should be drawn downwards. This inaccuracy in the figure was detected too late for correction. 200.] TRANSFORMATION OF COUPLES. 5 merely the figure J3PJRQ turned round in its own plane through a right angle. Hence J3r is the diagonal of the parallelogram determined by the axes of the component couples. Conversely, any couple may be resolved into two couples whose axes are determined from the axis of the given couple by the parallelogram law ; and, as in the case of forces acting at a point, any couple may be resolved into three couples whose axes are determined from the axis of the given couple by the parallele- piped law. All this follows as in Art. 198. It is well to remark that the axis of a couple represents the moment of the forces of the couple round any line in space parallel to the axis. (b) To find the resultant of any number of couples acting in any planes on a rigid body. Let the axes of the couples be all drawn, each in its proper sense according to the rule (), at the same point, (Fig. 228), and let each axis be resolved into three components along rectangular axes OM, Oy, Oz, drawn through 0. Let L = the sum of the axes in the direction Ooc ; then L is the axis of the component of the resultant couple in the plane yz. Let M and N be the sums of the axes in the directions Oy and Oz, re- spectively. Then, if G is the resultant axis, C = V1F+M*+~N*, (1) and if A, /x, v are the direction angles of G, cosA=~, cos/x = , cosy=- , (2) Cr (r Cr equations which are exactly analogous to (2) and (3) of Art. 198. TJie axes of couples are, therefore^ compounded and rewired in the same manner as forces. There is this difference between forces and couples, that, while any number of couples in any planes whatever always result in a single couple, any number of forces cannot, in general, be replaced by a single force, and this difference results from the rectorial nature of the axis of a couple. (e) A force and a couple acting on a rigid body cannot produce equilibrium. For, let the couple be so transferred that one of its forces, P, acts at a point on the line of action of the force .R. Then R 6 NON-COPLANAR FORCES. and P at this point compound a single force which, in general, does not intersect the other force of the couple. Therefore, &c. A force and a couple acting in the same plane are, of course, equivalent to a single force. 201.] Virtual Work of a Couple. Let a couple act on a rigid body which receives, or is imagined to receive, any small displacement whatever. It is required to find the work done by the couple in the displacement. It will be shown subsequently that any displacement of the body may be produced by a motion of translation which is the same for all its points, accompanied by a motion of rotation round an axis through an angle which is the same for all its points. Now since the forces of the couple are equal and in opposite senses, it is obvious that the sum of their works in any motion of translation is zero. Again, resolve the motion of rotation into one round an axis perpendicular to the plane of the given couple, and one round an axis in the plane of the couple. It is obvious that the latter displacement will not be productive of work done by the couple, since the forces constituting it may be supposed to act at the points in which they intersect the axis of this component rotation. There remains only the rotation round an axis perpendicular to the plane of the couple. Suppose (Fig. 88, Art. 79) to be the point in which this axis intersects the plane of the couple, and let 80 be the angular rotation round the axis, measured in the sense of Hie rotation of the couple. Then the displacements of m and n are Omx0 and 0^x60, respectively, so that the work done by the forces is P (Om . b + On . 80), i.e. P. A ^6, or G . 60, where G (= P . //) = the moment of the couple. 202.] Theorem. /I force acting on a rigid lody in a given right line can always le replaced by an equal force acting at any chosen point together with a couple. Let a force P (Fig. 231) act at a point A, and let be the chosen point. At introduce two forces, P and P', opposite to each other and each equal and parallel to P. Then P at A and P' at may be taken to constitute a couple whose 203.] THEOREM. moment is P/;, p being tlie perpendicular from on the line of action of P at A. There remains, then, the force P at ; and this force together with the couple may replace P at A. Let the axis of this couple be drawn at ; let #, y, z be the co-ordinates of A with respect to a rectangular system of axes drawn through ; and let a, ft, y be the angles which the direction of P makes with the axes of #;,y, r, respectively. Fig. 231. The direction-cosines of OJ are -> -> -> where CL4 = r, and r r r it is easy to prove that the direction-cosines of the axis of the couple (which is at once at right angles to OA and to P) are y cos y z cos ft ~ ~ Hence, the axis of the couple being equal to Py;, the projections of the axis on the axes of x, y> and z are P(ycosy ^eos/3), P (rcosa ^cosy), P (# cos ft y cos a) ; but it is clear from (y), Art. 200, that these are the axes of the component couples in the planes y:> #, and %y, into which the couple Pp can be resolved. Putting P cos a = X, P cos ft = JT, P cos y = #, we sec that the three couples are Zy-Ys, Xz-Zv, Yx-Xy. (l) The force P at may also be replaced by its three components, X,T,Z. (2) * f There is another way in which the reduction is sometimes effected. Let Pat A be resolved into its three components, T, F, Z, let the line of Z meet the plane (xy) in N 9 and let Z at A be transferred to N. Let fall Nn perpendicular to Ox ; at n introduce two opposite forces Z" and Z' n ', each equal and parallel to Z and at in- troduce two opposite forces, Z and -', each equal and parallel to Z. Now the senses of positive rotation in the planes wy, 2/-> zx being those indicated by the arrows, the forces Z at N and Z'" at w. form a couple whose moment is Zy parallel to the plane yz ; V 232. 8 NON-COPLANAR FORCES. [203. and the forces Z' at and Z" at n form a couple whose moment is Zx parallel to the plane zx and in addition to these there is the force Z at 0. Similarly, the force X at A can be replaced by X at together with two couples, Xz and Xy, parallel to the planes zx and xy, respec- tively ; and the force Y at A can be replaced by Y at together with the couples Yx and Yz parallel to the planes xy and yz. Hence P at A is replaced by the forces A 7 ", Y, Z at and the couples ZyYz, XzZx^ and YxXy, parallel to the planes yz, zx, and xy, respectively. 203.] Parallel Forces. Suppose a rigid body to be acted on by any number of parallel forces applied at given points in the body. Take any origin, 0, of co-ordinates, and through it draw three rectangular axes, that of z being parallel to the common direction of the forces. Then the force P 19 acting at (ss l9 y ly r,) may be replaced, as in last Art., by P l at along Or, and the couples ^V iin( i P^ l parallel to the planes yz and r#. Replace each force in this manner : then the whole system will be equivalent to a force l + P 2 + ...+P n , or SPat 0, together with the couple parallel parallel to the plane yz, and the couple -Pi^-JVa-. ..-/' to the plane zx. or -2 These two couples compound a single couple whose axis is found by drawing OL = SPy, on any scale, and OM (in the negative sense of the axis ot y) = 2P#, on the same scale, and completing the parallelogram OLGM ( Fi ff- 333)- If OG> th diagonal, is denoted by (7, and E = 11 being the resultant force. 204.] Centre of Parallel Forces. Since the resultant of two parallel forces, P l and P 2 , acting at the points A^ and A^ divides 205.] EXAMPLES. 9 A a P the line A^A^ in a point g such that -f- = -p 2 , and since, by elementary geometry (see Art. 84), the distance of g from any plane = - 1 ~^ > where ^ and # 2 are the distances of A l and // 2 from this plane, it follows, by repeating this construction, that the distances, x, y, z, of the centre of parallel forces from the planes yz, ;x, and xy are given by the equations = _ *- 205.] Conditions of Equilibrium of a System of Parallel Forces. A system of parallel forces has been reduced (Art. 203) to a single force, 7?, and a single couple, G. Now since these cannot in combination produce equilibrium (e, Art. 200), we must have R = 0, and G = 0, separately. Since G cannot be = unless SPj; = and 2Py = 0, the con- ditions of equilibrium are R = o, (i) SPa?= 0, 2fy= 0. (2) DEF. The moment of a force with respect to a plane to which it is parallel is the product of the force and its perpen- dicular distance from the plane. Hence for the equilibrium of parallel forces TJte sum of tie forces must vanis/i, and Uie sum of their moments irith respect to every plane parallel to tit em vmst also vaninli. EXAMPLES. 1. A heavy triangular table, ABO, is supported horizontally on three vertical props at the vertices ; prove that the pressures on the props are equal. Let P, Q, R be the pressures at A, B, (7, and let a vertical plane through A and the centre of gravity of the table cut the side BC in a, its middle point. For equilibrium the sum of the moments of the forces P, Q, It, and W (the weight of the table) with respect to this plane must = 0. But the moments of P and W are each = 0, since these forces lie in the plane. Hence the moments of Q and R are equal and opposite. Now-Sa . sin AaB is the distance of Q from the plane, and the distance of B = Ca. sin AaC '; and since Ba = Ca, these distances are equal. Therefore Q = R ; and similarly it can be shown that R = P ; therefore, &c. 10 NON-COPLANAR FORCES. [205. 2. A heavy triangular plate hangs in a horizontal plane by means of three vertical strings attached to its vertices ; at what point in its area must a given weight be placed so that the system of strings may be least likely to break ? Ans. At the centre of gravity of the board. For if W = the weight of the board and P the sustained weight, we have P + Q + 2t= W+P, or the sum of the tensions is constant, wherever P is placed. Hence if any one is less than -J- (TF+P), some other must be greater than this value. It is evident, therefore, that the best arrangement makes each tension = ^ ( W+ P) ; but this happens (as proved in last example) when P is placed at the centre of gravity. 3. A heavy elliptic cylinder is sustained in a vertical position by three props applied at three points on the circumference of its base; how should the props be placed in order that the cylinder may be least likely to be upset 1 Let the base of the cylinder have any form, ABC (Fig. 234), and let (1 be the projection of its centre of gravity on the plane of the base. Then, if the props are applied at A , B, and C, the perpendiculars from G on the sides of the triangle ABC must bo all equal when the equilibrium is most stable. For, suppose that the cylinder is about to be upset round the line AB then the moment of the force re- quired to upset it is W.p, where W is the weight of the cylinder and p the perpendicular from G on AIL Again, suppose that the cylinder is about to be upset about AC; then the moment of the force required to upset it is W.