EDITORIAL APPENDIX.

NOTE (a).

BESIDES its direction defined (Art. 1), we have also to take into consideration, in estimating the effects of a force, its point of application, or the point of the body where it acts, either directly or through the medium of some other body, as a rigid bar, or an inextensible cord in its line of direction; the point on its line of direction towards which the point of application has a tendency to move; and finally the intensity, or magnitude of the force as expressed in terms of some settled unit of measure.

B

P1

NOTE (b).

This result of experiment also admits of the following proof: Let A be the point of appliPcation of a force P, and let this point be invariably connected with another point B, in the line of direction towards which A tends to move from the action of P; suppose now two other forces P, and P,, each equal to P, to be applied; the one at A, in a direction opposite to P, and the other at B, in the same direction as P; the introduction of these two equal forces, acting in opposite directions, will evidently in no wise change the direction or intensity of P; but as P, is equal and opposite to P its effect will be to balance the action of P at A, whilst it leaves P, to exert an action at B precisely the same as P was exerting at A before the introduction of P, and P,.

NOTE (c).

Suppose two forces P, and P, applied to the same point A,

3

B

F P2

R

the direction of the one being AB, that of the other AC; no was these forces make an angle with each other, it is evident, as the point of application can move but in one direction, and as it is solicited to move towards B and C at the same time, that it must move in some direction which is coincident with neither of these; this direction, it is equally evident, must be in the same plane as the directions AB and AC, for there is no argument in favor of a direction assumed exterior to the plane and on one side of it which would not equally apply to a symmetrical direction assumed on the other side;" it is also evident that this direction must be some one AF within the angle formed by AB and AC, for the point, if solicited by P, alone, would take the direction AB, and as it cannot take a direction to the left of BD, as there is no force that solicits it on that side, and, for like reasons, cannot take one to the right of CE, it must therefore take the one assigned somewhere within the angle BAC.

Now suppose further that P, and P, are equal, it is evident that the direction assigned to their resultant, or that of the motion of their point of application, must be the one which bisects the angle BAC, for the argument for any direction on the left of this line would be equally cogent for the like position on the other side.

P

C

If P, and P, are unequal then will the direction of their resultant divide the angle BAC unequally, the smaller portion being next to the greater force; for suppose P, divided into two portions, one Pa of which P shall be equal to P.; P and P, can be replaced by their resultant R,, the direction of which AF bisects the angle BAC; we shall then have two forces R, and the remaining portion of P,, the resultant of which R must lie somewhere within the angle BAF, and therefore nearer to P, than to P,; but R is the resultant of the two forces P, and P,. Therefore, &c.

B

P1

R

R/

Hence it is seen that two forces whose directions form an angle between them and meet, 1st, have a resultant; 2nd, that the direction of this resultant lies in the plane of the two forces; 3d, that it passes through the point where the directions meet, and lies within the angle contained between them; 4th, that it bisects this angle when the forces are equal; 5th, that when the forces are unequal it divides this angle unequally, the smaller portion being next to the greater force.

Now as the two forces P, and P, can be rep laced by their resultant R, and as the effect of this will be the same if applied at any point F in its line of direction as at the point of application of the two forces, it is evident, if we transfer P, and P. also to F, preserving their new parallel to their original directions, that they, in turn, can be made to replace R. It thus appears that the point of application of two forces may be transferred to any point of the line of direction of their resultant without changing the effects of these forces, provided their new directions are kept parallel to their original ones.

It is upon the preceding propositions, in themselves selfevident, that the mode of demonstration, known as Duchayla's, of the proposition, termed the parallelogram of forces, or of pressures, is based.

NOTE (d).

When two parallel forces are applied to two points invariably connected their resultant can be found by applying the propositions in (Arts. 1, 2, 3).

B P
P+P

Let P, and P, be two parallel forces applied at the points A and B invariably connected, as by a rigid bar. Let two equal forces Q, and Q2 be so applied, the one at A the other at B, as to act in opposite directions along AB. These two will evidently PB have no effect to change the action of P, and P. Now the two forces P, and Q applied at A will have a resultant R, the intensity and direction of which can be found by the preceding method. In like manner the resultant R, of P, and Q, can be obtained. Now the forces being replaced by their resultants, the equilibrium will still subsist, and the effect will remain the same whether R, and R, act at A and B, or at o their point of meeting. But as R, and R, can each be replaced by their components at any point of their direction, let these components be transferred to the point o. In this position Q, and Q, will destroy each other, whilst P, and P, will act in the same direction along OC and parallel to their original ones, with an intensity equal to their sum P1+P„.

1

Now from the similar triangles AoC, rom; and BoC, son, there obtains,

om: mr :: oC: CA, or P, : Q, :: oC : CA.
ns: on :: CB: ¿C, or Q, : P, :: CB : oC.

Multiplying the two last proportions, there obtains,

and

P1 : P, : : CB : CA,

2

P1 : P, : P1+ P, :: CB : CA: CB+CA or AB.

1

From this we see that two parallel forces acting in the same direction, 1st, have a resultant which is equal to their sum; 2nd, that the direction of this resultant is parallel to that of the forces; 3d, that it divides the line joining the points of application of the two forces into parts reciprocally proportional to the forces; 4th, that either force is to the resultant as the portion of the line between the resultant and the other force is to the total distance between the points of application; 5th, that the foregoing propositions hold true for any position of the line AB with respect to the two parallel forces and their resultant.

Q

P2

B

R2

When the two forces act in opposite directions at the points A and B, by following a like process, we obtain the two resultants R, and R., which being prolonged to their point of meeting o we can again replace them by their components P1, Q1, and P1, Q; of which P, and P, acting parallel to their ori ginal positions but in opposite directions, will have for their resultant P.-P..

Now prolonging the direction of this resultant until it meets AB prolonged at C, there obtains as in the preceding case, from the similar triangles AoC, rom, and BoĈ, son,

hence,

om: mr :: oC: CA, or P, Q, :: oC : CA,
ns: on :: CB: oC, or Q, : P, :: CB: oC,

P1: P: P-P, :: CB: CA: CB-CA or AB.