OQ fulfils the required condition for those axes only which are situated in a line OP || CQ, and not for any other axis.

Therefore the diagonal OC of the parallelogram A B represents the resultant, and the only resultant, of the forces represented by OA and O B.-Q. E. D.

Second Demonstration.—Suppose a perpendicular to be erected to the plane O AB at the point O, of any length whatsoever; call the other extremity of that perpendicular R; and at R conceive two forces to be applied, respectively equal, parallel, and opposite to OA and OB. Then OR is the arm common to two couples whose axes and moments are represented (in the manner described in Art. 34) by lines perpendicular and proportional respectively to O A and OB. On the lines so representing the couples, construct a parallelogram; then, as shown in Art. 35, the diagonal of that parallelogram represents the resultant couple constituted by the resultant of OA and O B acting at O, and an equal and opposite force at R; and as the parallelogram of couples has its sides perpendicular and proportional to OA and O B, its diagonal must be perpendicular and proportional to OC, which consequently represents the resultant of OA and O B.-Q. E. D.

[There are various other modes of demonstrating the theorem of the parallelogram of forces, all of which may be studied with advantage: especially those given by Dr. Whewell in his Elementary Treatise on Mechanics, and by Mr. Moseley in his Mechanics of Engineering and Architecture.]

52. Equilibrium of Three Forces acting through One Point in One Plane. To balance the forces OA and OB, a force is required equal and directly opposed to their resultant OC. This may be otherwise expressed by saying, that if the directions and magnitudes of three forces be represented by the three sides of a triangle, (such as OA, AC, CO), then those three forces, acting through one point, balance each other.

53. Equilibrium of any System of Forces acting through One Point.— COROLLARY. If a number of forces acting through the same point be represented by lines equal and parallel to the sides of a closed polygon, those forces balance each other. To fix the ideas, let there be five forces acting through the point O (fig. 16), and represented in direction and magnitude by the lines F,, F, F, F., F., which are equal and parallel to the sides of the closed polygon OABC DO ·

F5

F4

F1

Fig. 16.

viz. :-
:-

2

87

Then by the theorem of Art. 52, the resultant of F, and F, is O B; the resultant of F1, F, and F, is OC; the resultant of F1, F2, F3, and F, is O D, equal and opposite to F, so that the final resultant is nothing.

The closed polygon may be either plane or gauche.

H

F

E

54. Parallelopiped of Forces.-The simplest gauche polygon is one of four sides. Let OABCE F G H (fig. 17), be a parallelopiped whose diagonal is OH. Then any three successive edges so placed as to begin at O and end at H, form, together with the diagonal HO, a closed quadrilateral; consequently if three forces F1, F2, F3, acting through O, be represented by the three edges OA, O B, OC, of a parallelopiped, the diagonal OH represents their resultant, and a fourth force F, equal and opposite to OH balances them.

F

F

F3

Fig. 17.

From the theo

55. Resolution of a Force into Two Components. rem of Art. 51, it is evident that in order that a given single force may be resolvable into two components acting in given lines inclined to each other, it is necessary, first, that the lines of action of those components should intersect the line of action of the given force in one point; and secondly, that those three lines of action should be in one plane.

Returning, then, to fig. 15, let OC represent the given force, which it is required to resolve into two component forces, acting in the lines OX, O Y, which lie in one plane with OC, and intersect it in one point O.

Through C draw CA | OY, cutting O X in A, and C B || O X, cutting OY in B. Then will OA and OB represent the component forces required.

Two forces respectively equal to and directly opposed to OA and OB will balance O C.

56. Resolution of a Force into Three Components.-In order that a given single force may be resolvable into three components acting in given lines inclined to each other, it is only necessary that the lines of action of the components should intersect the line of action of the given force in one point.

Returning to fig. 17, let OH represent the given force which it is required to resolve into three component forces, acting in the lines OX, OY, O Z, which intersect Ō H in one point O.

Through H draw three planes parallel respectively to the planes YOZ, ZOX, XOY, and cutting respectively O X in A, O Y in B, O Z in C. Then will OA, OB, OC, represent the component forces required.

Three forces respectively equal to, and directly opposed to OA, OB, and OC, will balance O H.

57. Rectangular Components.—The rectangular components of a force are those into which it is resolved when the directions of their lines of action are at right angles to each other.

For example, in fig. 17, suppose O X, OY, O Z, to be three axes of co-ordinates at right angles to each other. Then OH is resolved into three rectangular components simply by letting fall from H perpendiculars on OX, OY, O Z, cutting them at A, B, C, respectively.

To express this case algebraically, let FOH denote the force to be resolved. Let

a=XOH, ß=≤YOH, Y=ZOH,

be the angles which its line of action makes with the three rectangular axes. Then, as is well known, those three angles are connected by the equation

Let

cos" a + cos2 3 + cos2 = 1,

F1=0A, F2=0B, F,=OC,

be the three rectangular components of F; then

.(1.)

(2.)

In order to distinguish properly the direction of the resultant F as compared with the directions of the axes, it is to be borne in

mind that

positive.

the cosine of an

{

acute

obtuse

} angle is {negative.}

From a well known property of right-angled triangles (also embodied in equation 1), it follows that

To express algebraically the case in which a force is resolved into

FORCES ACTING IN ONE PLANE

39

two rectangular components in one plane with it, let the plane in question be that of OX and OY. Then the angles are subject to the following equations :—

y = a right angle; a + 6 = a right angle;

cos y = 0; cos ẞ = sin a; cos α = sin ß............................(4.)

and consequently the equations 2 and 3 are reduced to the following:

In using these equations, the rule respecting the positive and negative signs of cosines is to be observed; and it is also to be borne in mind, that the angle a is reckoned from OX in the direction towards Y, and the angle 3 from O Y in the reverse direction, that is, towards X, and that

If a system of forces acting through one point balance each other, their resultant is nothing; and therefore the rectangular components of their resultant, which are the resultants of their parallel systems of rectangular components, are each equal to nothing; a case represented as follows:

2F0; F,=0; 2. F, 0.............(6.)

=

SECTION 2.-Inclined Forces Applied to a System of Points.
Let any

58. Forces acting in One Plane.—Graphic Solution. system of forces whose lines of action are in one plane, act together on a rigid body, and let it be required to find their resultant.

Assume an axis perpendicular to the plane of action of the forces at any point, and let it be called OZ. According to the principle of Art. 42, let each force be resolved into an equal and parallel force acting through O, and a couple tending to produce rotation about OZ; so that if a force F be applied along a line whose perpendicular distance from O is L, that force shall be resolved into F = and || F

acting through O, and a couple whose moment is

and which is right or left-handed according as O lies to the right or left of the direction of F.

The magnitude and direction of the resultant are to be found by forming a polygon with lines equal and parallel to those representing the forces, as in Art. 53, when, if the polygon is closed, the forces have no single resultant; but if not, then the resultant is equal, parallel, and opposite to that represented by the line which is required in order to close the polygon. Let R be its magnitude if any.

The position of the line of action of the resultant is found as follows:

Let M be the resultant of the moments of all the couples M, distinguishing right-handed from left-handed, as in Arts. 27 and 32. If M = 0, and also R = 0, then the couples and forces balance completely, and there is no resultant. If 2. M = 0, while R has magnitude, then the resultant acts through O. If 2 M and R both have magnitude, then the line of action of the resultant R is at the perpendicular distance from O given by the equation

and the direction of that perpendicular is indicated by the sign of 2. M.. If R = 0, while 2. M has magnitude, the only resultant of the given system of forces is the couple .M.

59. Forces acting in One Plane.—Solution by Rectangular Co-ordinates. Through the point O as origin of co-ordinates, let any two axes be assumed, OX and OY, perpendicular to each other and to O Z, and in the plane of action of the forces; and in looking from Z towards O, let Y lie to the right of X, so that rotation from X towards Y shall be right-handed. Let F, as before, denote any one

of the forces; let a be the angle which its line of action makes to the right of O X; and let x and y be the co-ordinates of its point of application, or of any point in its line of action, relatively to the assumed origin and axes. Resolve each force F into its rectangular components as in Art. 57,

and the angle a, which it makes to the right of O X is found by the