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407. We naturally divide Statics into two parts-the equilibrium of a Particle, and that of a rigid or elastic Body or System of Particles whether solid or fluid. The second law of motion suffices for one part--for the other, the third, and its consequences pointed out by Newton, are necessary. In the succeeding sections we shall dispose of the first of these parts, and the rest of this chapter will be devoted to a digression on the important subject of Attraction.

408. By $ 221, forces acting at the same point, or on the same material particle, are to be compounded by the same laws as velo cities. Therefore the sum of their resolved parts in any direction must vanish if there is equilibrium; whence the necessary and sufficient conditions.

They follow also directly from Newton's statement with regard to work, if we suppose the particle to have any velocity, constant in direction and magnitude (and $ 211, this is the most general supposition we can make, since absolute rest has for us no meaning). For the work done in any time is the product of the displacement during that time into the algebraic sum of the effective components of the applied forces, and there is no change of kinetic energy. Hence this sum must vanish for every direction. Practically, as any displacement may be resolved into three, in any three directions not coplanar, the vanishing of the work for any one such set of three, suffices for the criterion. But, in general, it is convenient to assume then in directions at right angles to each other.

Hence, for the equilibrium of a material particle, it is necessary, and sufficient, that the (algebraic) sums of the applied forces, resolved in any one set of three rectangular directions, should vanish.

409. We proceed to give a detailed exposition of the results which follow from the first clause of $ 408. For three forces only we have the following statement.

The resultant of two forces, acting on a material point, is repre

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

в. 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.

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 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. one point, not necessarily in one planeLet MA, MA, MA, MA, repre

D" 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,

DI and MA It will be represented by MD. Then similarly, find MD", the resultant of MD' (the first subsidiary resultant), and , the third force. Lastly, find MD'", the resultant of

A1 MD and MA MD'" represents the resultant of the given forces

Thus, by successive ap- 'ications of the fundamental proposition, the resultant of any numbur of forces in lines through one point can be found.

413. In executing this construction, it is not necessary to describe

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the successive parallelograms, or even to draw their diagonals. It is

D" enough to draw through the given point

a line equal and parallel to the repres D"

sentative of any one of the forces / through the point thus arrived at, to

draw a line equal and parallel to the D'

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.

The several given forces may be taken

in any order, in the construction just 4

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.

In the fig. the order is MA, MA, D


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 D"

equilibrium. If, on the other hand, the

figure is not closed (8 413), the resultant os D

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.

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 con. sidered, 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 repre: sented 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 à convenient specification of the resultant. The instrument employed for this purpose is Trigonometry.

418. A distinction may first be pointed out between two classės 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. Finding components is called Resolution 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 a

Required the magnitude of the resultant,
and its inclination to the line of either

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 B denote the angle DMB, at which it is inclined to MB, the direction of the force .

Given P, Q, and e; required R, and a or ß. We have

MDP = MA* + MB*- 2 MA.MB x cos MAD. Hence, according to our present notation,

REP+ Q* - 2PQ cos (180°-1), Or

R* = PS + Q . +2PQ cosi. Hence R =(P8+ Q* +2PQ cos e)).

(1) To determine a and B after the resultant has been found; we have

sin. DMA - sin MAD


e OT

sin a = and similarly, sin B = sin e

(3) 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 iis acute, both a and B are acute, and there is no ambiguity. If į is obtuse, one of the two

sin ,


COS a=

angles, a, b, 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 i, we find the value, whether acute or obtuse, of the other of the two angles, a, B.

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

MDP + MA - AD cos DMA

2MD.MA R? + P - Q

(4) 2 RP

R+-p and similarly, cos B=


2 RC by successive applications of the elementary trigonometrical formula used above for finding MD. Again, using this last-mentioned for. mula for MDior Rin the numerators of (4) and (5), and reducing,

P + cost we have cos ar

(6) R cos B={+ p cose;


R formulae which are convenient in many cases.

There is no an. 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. in special cases, the negative sign, which with the proper interpre. tation 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 : Let

1 (P+Q)= F,

* (P-Q)=G, which implies that


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