58. Any number of forces act in parallel directions in one plane at different points of a rigid body; to find their resultant. Let F1, F2...F, be the forces; from any point 0 (fig. 14) of the rigid body, in the plane of the forces draw a line cutting their directions perpendicularly in the points A, B.....H; and put At O apply two opposite forces each equal and parallel to F1; they do not affect the system. In the same way apply at O a pair of forces for each of the remaining forces F, F... F By this means we have n forces acting at O in the direction OR, respectively equal to F1, F2,...F; these are equivalent to a single force R acting at O in the direction OR, and We have, besides, n couples whose arms are x1, x2... which are (by Art. 52) equivalent to a single couple R′, R", whose moment Consequently the given system of forces is equivalent to a single force . F acting at O in the direction OR; and a negative couple whose moment is . Fx. 59. The result just obtained is perfectly general, but it admits of simplification except in the particular case when Σ. F = 0. (1) In the particular case when . F = 0, there is no resultant force acting at O, and therefore the only resultant is the couple whose moment is Σ. Fx. (2) When . F is not = 0, change the couple whose moment = . Fx into an equivalent couple which has its forces equal to R or Z. F, and place it so that its arm OK (fig. 14) shall coincide with the line OH; .. (2. F). OK = 2. Fx......(Art. 50.) By the arrangement the force R at O is balanced by one of the forces of our new couple; and these being removed, there remains only the force R'. F at the point K determined by the equation Consequently, except when Σ. F= 0, the resultant is a single force equal to Σ. F acting at the point just found. 60. COR. If the line OH instead of cutting the directions of the forces at right angles, should cut them in an ▲ a, we should have found that (1) When . F = 0, the resultant is a couple whose moment is (Σ. Fx) sin a: and (2) When . F is not = 0, the resultant is a single force . F acting at the point determined by the same equation as before, viz. 61. Any number of parallel forces act in one plane at different points of a rigid body; to find the conditions that they may balance each other. Let the system of forces be that of Art. 58; then we have to consider the two cases pointed out in the last Article. In the second case the resultant is the force Σ. F acting at K; and there cannot be equilibrium unless this force vanish, or Σ. F 0. But if this be the case, the second case coincides with the first; and the resultant is a couple whose moment =2. Fx: there cannot be equilibrium therefore unless this couple also vanish. Consequently the conditions of equilibrium are = Σ. F= 0 and Σ. Fx = 0; these are both necessary and sufficient for equilibrium. They are necessary, for if the former only be satisfied, there will exist the couple of Case 1: and if the latter only be satisfied, there will exist the resultant force acting at 0. And they are sufficient, for they secure that there shall exist neither the resultant of Case 1, nor that of Case 2. 1 n 62. DEF. The products F11, F2x2, .... Fx are called the moments of the forces F1, F2, .... F, about the point 0; they are also called the moments of the same forces about an axis passing through O at right angles to the plane of the forces. Hence remembering that the point O was arbitrarily chosen, the conditions of equilibrium of parallel forces may be thus enunciated in words: The algebraic sum of the forces, and the sum of the moments of the forces about any point in the plane of the forces or about any axis perpendicular to the plane of the forces, are each equal to zero. 63. Suppose that there is in the plane of the forces a fixed point, or in the body a fixed axis not parallel to the plane of the forces: to find the conditions of equilibrium. If there be a fixed point in the plane of the forces, let that point be taken for O; or if there be a fixed axis it will cut the plane of the forces in a point, which take for 0; then the investigations of Art. 58 apply here. The force . F which acts at O, can produce no effect since it acts on an immoveable point; it is not necessary therefore that Σ.F should be = 0. But the couple whose moment = Σ. Fx, if it exist, will turn the body about 0; and therefore that there may be equilibrium, it is necessary and sufficient that or, in words, Σ. Fx =0; The sum of the moments of the forces about the fixed point, or about that point where the fixed axis cuts the plane of the forces, must be equal to zero. REMARK. The pressure on the fixed point = Σ F, which is the same as if every force were transposed to that point, without altering the direction in which it acts. 64. Any number of forces act, in one plane, at different points of a rigid body: to find their resultant. Let F1, F2, .... F be the forces, and in the plane in which they act, from a point 0 arbitrarily chosen, draw any two lines Ox, Oy at right angles to each other. To these lines as co-ordinate axes refer the given forces and their points of application. (fig. 15.) 1 1 Let a1, a2.... a, be the inclinations of the lines in which the forces act to Ox; x1Y1, x2Y2, .... xy, the co-ordinates of the points of application of the forces; P being that of F1, x1 = OM, y1 = PM. From the point draw OQ perpendicular to F1 P; and at O apply two opposite forces F, F" each equal and parallel to F. By this means we have a force F' acting at O, and a couple (F,F",) whose moment is equal to F. OQ. Or we may say that the force F1 may be transposed to O without altering its direction, if at the same time we also apply to the body a couple whose moment - F1.0Q - F1. (ON – QN) = 1 1 1 if we put X, Y, for the components of F, parallel to the co-ordinate axes Ox, Oy. The same method being applied in succession to each one of the remaining forces of the system, we shall have transposed all the forces to O, each preserving its original direction; but there will be acting on the body besides them a number of couples whose moments are X11- Y11, X2Y2 - Y22 .... Xnyn – Y„an• If G be the resultant of the couples, and R the resultant of the forces at O, we shall have ·G=2(Xy - Yx) .... Art. 52, and R2 = (Σ. F cos a)2 + (Z. F sin a)2.... Art. 31, being the inclination of the line in which R acts to Ox. 65. The result just obtained is perfectly general, but it can be simplified, being reducible to a single resultant, except when R = 0, i.e. except when ZX = 0 and ΣY = 0. (1) When ΣX=0 and ΣY= 0, there is no resultant acting at O, and the only resultant is the couple whose (Xy - Yx). moment = (2) When the two equations ΣX= 0, and ΣY = 0 are not both satisfied, change the couple whose moment is Σ(Xy – Ya) into an equivalent couple which has each of its forces R' R'' equal to R, and place it so that one end of its arm OK (fig. 16) shall be at O, and one of its forces (R") exactly opposite to R. R and R" balance each other and may be removed; and there remains only the force R' acting at the point T such that OT. cos 0 OK; .. R.OT.cos = R. OK, or, since R cos = ΣX (Art. 31), and R. OK = Σ(Xy – Yx), Consequently in this case the resultant is a single force R acting at the point just found. 66. When the forces are reducible to a single resultant, to find the equation of the line in which it acts. Let x'y' be the co-ordinates of any point in the line R'K (fig. 16) in which the resultant acts. Then because D |