sented in direction and magnitude by the diagonal, through that point, of the parallelogram described upon lines representing the forces. 410. Parallelogram of forces stated symmetrically as to the three forces concerned, usually called the Triangle of Forces. If the lines representing three forces acting on a material point be equal and parallel to the sides of a triangle, and in directions similar to those of the three sides when taken in order round the triangle, the three forces are in equilibrium. Let GEF be a triangle, and let MA, MB, MC, be respectively equal and parallel to the three sides EF, FG, GE of this triangle, and in directions similar to the consecutive directions of these sides in order. The point M is in equilibrium. B M C E H D F 411. [True Triangle of Forces. Let three forces act in consecutive directions round a triangle, DEF, and be represented respectively by its sides: they are not in equilibrium, but are equivalent to a couple. To prove this, through D draw DH, equal and parallel to EF, and in it introduce a pair of balancing forces, each equal to EF. Of the five forces, three, DE, DH and FD, D E are in equilibrium, and may be removed; and there are then left two forces, EF and HD, equal, parallel, and in dissimilar directions, which constitute a couple.] 412. To find the resultant of any number of forces in lines through D" one point, not necessarily in one plane Let MA, MA,, MA, MA, repre sent four forces acting on M, in one plane; required their resultant. Find by the parallelogram of forces, the resultant of two of the forces, MA, and MA, It will be represented by MD'. Then similarly, find MD", the resultant of MD' (the first subsidiary resultant), and MA,, the third force. Lastly, find MD"", the resultant of MD" and MA, MD"" represents the resultant of the given forces. Thus, by successive applications of the fundamental proposition, the resultant of any number of forces in lines through one point can be found. 413. In executing this construction, it is not necessary to describe D" D" the successive parallelograms, or even to draw their diagonals. It is enough to draw through the given point a line equal and parallel to the representative of any one of the forces; through the point thus arrived at, to draw a line equal and parallel to the representative of another of the forces, and so on till all the forces have been taken into account. In this way we get such a diagram as the annexed. A A2 D' The several given forces may be taken in any order, in the construction just described. The resultant arrived at is necessarily the same, whatever be the order in which we choose to take them, as we may easily verify by elementary geometry. A5 M D D In the fig. the order is MA, MA„ MA, MA, MA ̧. 414. If, by drawing lines equal and parallel to the representatives of the forces, a closed figure is got, that is, if the line last drawn leads us back to the point from which we started, the forces are in equilibrium. If, on the other hand, the figure is not closed (§ 413), the resultant is obtained by drawing a line from the starting-point to the point finally reached; (from M to D): and a force represented by DM will equilibrate the system. D D" 415. Hence, in general, a set of forces represented by lines equal and parallel to the sides of a complete polygon, are in equilibrium, provided they act in lines through one point, in directions similar to the directions followed in going round the polygon in one way. 416. Polygon of Forces. The construction we have just considered, is sometimes called the polygon of forces; but the true polygon of forces, as we shall call it, is something quite different. In it the forces are actually along the sides of a polygon, and represented by them in magnitude. Such a system must clearly have a turning tendency, and it may be demonstrated to be reducible to one couple. 417. In the preceding sections we have explained the principle involved in finding the resultant of any number of forces. We have now to exhibit a method, more easy than the parallelogram of forces affords, for working it out in actual cases, and especially for obtaining a convenient specification of the resultant. The instrument employed for this purpose is Trigonometry. 418. A distinction may first be pointed out between two classes of problems, direct and inverse. Direct problems are those in which the resultant of forces is to be found; inverse, those in which com ponents of a force are to be found. The former class is fixed and determinate; the latter is quite indefinite, without limitations to be stated for each problem. A system of forces can produce only one effect; but an infinite number of systems can be obtained, which shall produce the same effect as one force. The problem, therefore, of finding components must be, in some way or other, limited. This may be done by giving the lines along which the components are to act. To find the components of a given force, in any three given directions, is, in general, as we shall see, a perfectly determinate problem. Finding resultants is called Composition of Forces. 419. Composition of Forces. Required in position and magnitude the resultant of two given forces acting in giving lines on a material point. Let MA, MB represent two forces, P and Q, acting on a material point M. Let the angle BMA be denoted by . Required the magnitude of the resultant, and its inclination to the line of either force. B Q R P Let R denote the magnitude of the resultant; let a denote the angle DMA, at which its line MD is inclined to MA, the line of the first force P; and let ẞ denote the angle DMB, at which it is inclined to MB, the direction of the force Q. Given P, Q, and : required R, and a or B. We have or MD2 = MA2 + MB2 — 2MA.MB × cos MAD. Hence, according to our present notation, Hence R2 = P2 + Q2 − 2 PQ cos (180o — 1), R2: = P2 + Q3 + 2PQ cos i. R = (P2 + Q2 + 2PQ cos 1). (1) To determine a and ẞ after the resultant has been found; we have 420. These formulae are useful for many applications; but they have the inconvenience that there may be ambiguity as to the angle, whether it is to be acute or obtuse, which is to be taken when either sin a or sin ẞ has been calculated. If is acute, both a and B are acute, and there is no ambiguity. If is obtuse, one of the two し angles, a, ß, might be either acute or obtuse; but as they cannot be both obtuse, the smaller of the two must, necessarily, be acute. If, therefore, we take the formula for sin a, or for sin ẞ, according as the force P, or the force Q, is the greater, we do away with all ambiguity, and have merely to take the value of the angle shown in the table of sines. And by subtracting the value thus found, from the given value of , we find the value, whether acute or obtuse, of the other of the two angles, a, ß. 421. To determine a and ẞ otherwise. After the magnitude of the resultant has been found, we know the three sides, MA, AD, MD, of the triangle DMA, then we have by successive applications of the elementary trigonometrical formula used above for finding MD. Again, using this last-mentioned formula for MD3 or R2 in the numerators of (4) and (5), and reducing, we have COS a P+Q cos (6) (7) There is no am formulae which are convenient in many cases. biguity in the determination of either a or ẞ by any of the four equations (4), (5), (6), (7). Remark.—Either sign (+ or -) might be given to the radical in (1), and the true line of action and the direction of the force in it would be determined without ambiguity by substituting in (2) and (3) the value of R with either sign prefixed. Since, however, there can be no doubt as to the direction of the force indicated, it will be generally convenient to give the positive sign to the value of R. But in special cases, the negative sign, which with the proper interpretation of the formulae will lead to the same result as the positive, will be employed. 422. Another method of treating the general problem, which is useful in many cases, is this: Fand G will be both positive if P>Q. Hence, instead of the two given forces, P and Q, we may suppose that we have on the point M four forces;-two, each equal to F, acting in the same directions, MK, ML, as the given forces, and two others, each equal to G, of which one acts in the same direction, MK, as P, and the other in ML', the direction opposite to Q. Now the resultant of the two equal forces, F, bisects the angle between them, KML; and by the investigation of § 423 below, its magnitude is found to be 2Fcos Again, the resultant of the two equal forces, G, is similarly seen to bisect the angle, KML', between the line of the given force, P, and K F the continuation through M of the line of the given force, Q; and to be equal to 2G sin, since the angle KLM' is the supplement of . Thus, instead of the two given forces in lines inclined to one another at the angle, which may be either an acute, an obtuse, or a right angle, we have two forces, 2Fcos and 2 G sin, acting in lines, MS, MT, which bisect the angles LMK and KML', and therefore are at right angles to one another. Now, according to § 429 below, we find the resultant' of these two forces by means of the following formulae: し and tan SMD= 2G sin し R=2Fcos sec SMD, These formulae might have been derived from the standard formulae for the solution of a plane triangle when two sides (P and Q), and the contained angle (-) are given. 423. We shall now investigate some cases of the general formulae. Case I. Let the forces be equal, that is, let Q=P in the preceding formulae. Then, by (1), R2 = 2P2 + 2P2 cos ɩ = 2P2 (1 + COS 1) an important result which might, of course, have been obtained directly from the proper geometrical construction in this case. Also by (2), 1 In the diagram the direction of the balancing force is shown by the arrowhead in the line DM. |