Parallelogram of Forces.-If two forces acting on a point be represented by straight lines, as above described, and a parallelogram be constructed having these lines as adjacent sides, the resultant of the two forces will be represented in direction and intensity by the diagonal of the parallelogram drawn from the point. The parallelogram so constructed is called the parallelogram of forces.

In determining the truth of the above proposition, we shall first consider the case of equal forces, and secondly, that of unequal forces.

1. Equal Forces. Let AB, AC (fig. 9) represent two equal forces acting upon A. Now, since AB and AC are equal, half their effect to move A in their respective directions is neutralised. That is to say, the force AB prevents the B force AC from moving A in the direction A C; and in the same degree AC prevents A B from moving A in the direction AB. Therefore the resultant must be in the direction of the line bisecting the angle BAC. And as the sides of the parallelogram are equal, the diagonal AD bisects the angle B A C.

Fig. 9.

The diagonal is, therefore, in the direction of the resultant.

Let us now show that the diagonal AD represents the resultant in intensity.

Let ABCD (fig. 10) again represent the в parallelogram of the forces A B, A C, acting upon A. From A draw A F equal to the supposed resultant A D. Complete the parallelogram AF EB, and join A E.

As AF is equal to the resultant of A B and AC, and acts opposite to them, the system of forces A B, AC, and A F about A is in equilibrium. One of the forces,

E

Fig. 10.

A C, also

balances the others, A B and AF; it must therefore be equal and opposite to their resultant. This resultant is in the direction A E, because BAFE is a parallelogram. Therefore A C and AE must be in the same straight line. Therefore A EBD is a parallelogram, and AD is equal to E B. But AF is equal to BE. Therefore A F is equal to AD; and AF being assumed to be equal to the resultant, AD is that resultant. The diagonal AD, therefore, represents in intensity the resultant of the two forces A B, AC. Let one

B

F

2. Unequal Forces -First.

force, AB (fig. 11), be double the other, AC. Complete the parallelogram A D, bisect A B in E, and draw E F parallel to A C.

Join A F and ED.

The force A B is equal to the two equal forces A E, EB. The resultant of à C, A E, is in the direction A F, and the reFig. 11. sultant of EF, EB, is in the direction ED. By resolving the force A F at the point F it is equivalent to the two forces CF, EF. The force CF passes through the point D in the resultant of the two forces EF, EB. If therefore the point D remains immovable, the forces CF, EF, and E B are destroyed. But CF and EB are equal to the force A B, and EF is equal to the force AC. Therefore the point D is in the resultant of the forces AC and A B. The point A is another point in this resultant. Wherefore a line joining A and D is in the direction of the resultant of the two forces A C and AB.

Second. Let one force be a multiple of the other, and let the force A B (fig. 12) contain four units, and the force A C three units. Complete the parallelogram ABDC, and from the points a, b, and c, draw a a', bb', cc' parallel to A C, and from d and e draw dd' and ee parallel to AB. Because the forces A a and Ad are equal, their resultant lies in the direction A E;

and because Ad and de are equal, the resultant of A a and Ae is in the direction

AF.

And since e F is equal

to e C their resultant is in the direction e a'; therefore the resultant of A a and A C is in the direction A a'.

e

C

F

Fig. 12.

In like manner it can be proved that the resultant of a b and a a' is in the direction a b'; therefore the resultant of Ab and AC is in the direction Ab', and so on until the resultant of AB and AC is proved to be in the direction A D.

C

G R

Fig. 13.

B

Third. Let the forces A B, AC (fig. 13) be unequal and also incommensurable. Their resultant lies in the direction AD. If it be possible, let A G, intermediate between A C and AD, be the direction of the resultant. Let a measure of AC be taken, less than DG, and measured into AB as often as possible to the point P, leaving a remainder, PB, smaller than DG. Draw PR parallel to B D. Since the forces AP and AC are commensurable, their resultant lies in the direction AR; which resultant, compounded with the remainder P B, will give a final resultant of the forces AB and A C lying in the angle BAR, and therefore, a fortiori intersecting the line CD in the point D outside the point G. Consequently AG is not the resultant of the forces A B and A C; and it can be shown that the resultant can have no other direction than AD, which therefore is the direction of the resultant of A B and A C.

Method of Resolving Forces.-Having proved that the resultant of two forces not in the same straight line but acting upon a given point is in the direction of the diagonal of the parallelogram, and that its intensity

is represented by the length of the diagonal, we are in a position to consider the simpler problems of the decomposition and resolution of forces.

To resolve a force, A B (fig. 14), into two forces acting upon the same point as the given force and in given directions.

B

Let AB (fig. 14) be the given force. From A draw A C in the direction of one of the component c forces, and from B draw BD in the direction opposite to that of the other component force. BD and AC intersect in E. Complete the parallelogram AEBF. The forces AF, A E, have as their resultant A B; and they act in the given directions, and are therefore the required components.

Fig. 14.

The resultant can always be determined by Trigonometry; but as it is our wish to avoid the use of that science in this work, we shall confine our examples to those cases that can be solved by easier methods.

1. When the angle of the forces is a right angle. Let AB (fig. 14a) be 6 lbs., A C, 8 lbs., and let the forces act at

right angles. Complete the parallelogram, and draw the diagonal AD, which will represent the resultant. Now AD2=A B2+ AC2. Therefore, A D2 = 62+ 82 -100. Consequently A D=√100=10.

2. When the angle of the forces is 60°. Let AB (fig. 146) =4 lbs., A C-3 lbs., and the angle B A C 60°. Complete the parallelogram, and join A D. From D draw D F, at right angles to AC produced to F. Because B A C = 60°, DCF-half an equilateral triangle. Therefore CF-half

CD=2. Wherefore DF2=42-22-12. Now AD2-DF2 + A F2=12+25=37. Therefore A D=6.08.

3. When the angle of the forces is 30°. Let A B (fig. 14c)

=4 lbs., A C 3 lbs., and the angle BAC-30°. By the same construction as in the last case, DF=2, CF=√12 = 3·46. Then A D2 =D F2 + A F2 = 4 + 6·462 = 46 nearly. Therefore A D equals 6.78 nearly.

L

F

G

หา C

Fig. 15.

E

K

Polygon of Forces.-We will now suppose the point A to be acted upon by several forces, AB, A C, A. D, and AE (fig. 15). On A B take any distance, AF; through F B draw FG parallel to A C, and of such a length that FG: AF force AC force AB. Again, through G draw GH parallel to AD, and so proportioned that GH: GF as force AD force A C. So also draw HK parallel to AE, so that HK: HG as force AE force AD, and join KA. A single force acting in the direction KA or AL, and having the same ratio to each of the other forces as AK has to that side of the polygon which is parallel to that force, will produce an effect on A equal and opposite to the combined effects of the forces B CDE. This may be proved by finding the resultant of the first two forces, and then the resultant of the next force and that resultant, and so on, whatever be the number of forces.

Wherefore:-If a number of forces acting on a point are parallel and proportional to all the sides of a polygon taken in order, except one; a single force, in the same