directions in which the forces act, and have their lengths proportional to the magnitudes of the forces.

26. If from a point two lines be drawn representing two forces which act upon a point; and if upon these lines a parallelogram be constituted, the diagonal drawn from the same point will represent the resultant of the two forces. This property is usually cited as "the parallelogram of forces."

p

We shall first prove that the diagonal represents the direction of the resultant force. This part of the proposition is evidently true when the two given forces are equal; let us assume that p, q and r are three forces, such that this is true for Ρ and and also for and r. At the point A apply p in the direction AB: and q, r, both in the direction AD. Take AB, AC, CD to represent the respective magnitudes of these forces. Complete the parallelograms, and draw the diagonals as in fig. 2. The resultant of p and q acts in the direction AE, by hypothesis; and we may by Art. 21, suppose it to act at E; and we may there resolve it into its original components p and q; the latter acting in the line EF, and the former in the line CE produced; this we may suppose by Art. 21 to act at C, and the former at F; also the force r may be supposed to act at C. We have now two forces p,r at C represented by the lines CE, CD; their resultant by hypothesis acts in the line CF, and therefore we may suppose at F. The three forces p, q, r, which originally acted at A, are by this process reduced to forces acting at F. F is therefore in the line in which their resultant acts when they are applied at A, (Art. 21.) Now AD, AB, represent two forces q + r and p; and we have just shewn that their resultant acts in the direction of the diagonal AF. If then our proposition be true for the two forces p, q; and also for the two forces p, r; it is also true for the forces p and q + r. Now it is true when p, q, r are all equal; and hence it is true for p and 2p: and because it is true for p, p; and also for p, 2p; therefore it is true for p, 3p;... and by following the same mode of reasoning it is true for p and mp, m being any whole number.

Again, because it is true for mp and p, and also for mp and p; therefore it is true for mp and 2p; and as before for mp and np, n being an integer. Hence our proposition respecting the direction of the resultant is true for any two commensurable forces mp, np.

If the proposed forces P, Q be incommensurable, by taking p extremely small and the integers m, n correspondingly large, we can make mp differ from P, and np from Q by less than any quantities which can be assigned; and we may then use mp and np, instead of P and Q, without any sensible error; and therefore the proposition is true of P and Q.

7 We shall now prove that the diagonal represents the magnitude of the resultant force.

Let AC, AB (fig. 3) represent the magnitudes and directions of two forces acting on a point. Complete the parallelogram: its diagonal AE represents the direction of the resultant; and it also represents its magnitude. For in EA produced take AF to represent its magnitude; then AB, AC, AF represent three forces which balance each other: wherefore completing the parallelogram AFGB, its diagonal AG represents the direction of the resultant of AF, AB, and is consequently in the same line with AC (Art. 20). AGBE is a parallelogram, and therefore AE = GB that is, AE represents the magnitude of the resultant of AB, AC.

=

Hence
AF;

27. If three forces acting on a point are represented by the sides of a triangle taken in order, they will balance each other. And conversely; If three lines, forming a triangle, be parallel to the directions of three forces which, acting on a point, balance each other, the sides of the triangle taken in order will represent the forces.

For let AB, BE, EA (fig. 3) represent the forces P, Q, R which act on a point. Complete the parallelogram BC; then because AC is drawn parallel and equal to BE, it also represents the force Q. Hence the resultant of P and Q is represented by AE; which being equal, and in a contrary direction, to EA which represents R, there is equilibrium.

Conversely, let the three forces P, Q, R, acting on a point balance each other: and suppose ABE (fig. 4) to be the triangle whose sides are respectively parallel to the directions in which P, Q, R act. Two, at least, (AB, BE suppose) represent the directions of the corresponding forces; and we are at liberty to suppose that one of these (AB suppose) represents also the magnitude of its force P: if BE do not represent the magnitude of the other force Q, take BE' to represent it, and join AE'. Then (by the former case) P, Q and a force represented by E'A will balance each other. But P, Q, R balance each other; and therefore R is represented by E'A both in magnitude and direction; which is impossible (because EA represents R in direction by hypothesis) unless E' coincide with E. Therefore E' does coincide with E, and therefore the forces are represented by AB, BF EA, which are the sides of the triangle taken in order.

28. If three forces, acting upon a point, balance each other, their directions lie in a plane; and their magnitudes are respectively proportional to the sine of the angle between the directions in which the other two act.

Let the forces be P, Q, R (fig. 5) acting in the directions OA, OB, OC. In OA, OB, take points A, B, such that OA, OB represent the magnitudes as well as the directions of P, Q. Complete the parallelogram AOBD, and join OD. Then three forces represented by OA, AD, DO, acting on a point will balance each other (Art. 27); that is, P, Q and a force represented by DO balance each other; but P, Q, R balance each other; therefore R is represented by DO: and consequently DOC is a straight line, and OA, OB, OC lie in the same plane. Also

PQR :: OA : AD : DO

29. Two forces act upon the same point in directions at right angles to each other, to find their resultant. (fig. 5*).

Let the forces be X, Y acting upon the point O in the directions Ox, Oy. Take OM, ON to represent the forces; complete the rectangle OMPN, and draw the diagonal OP. This line by Art. 26 represents the resultant of X and Y. Let R be the resultant, and the angle POM which its direction makes with the direction of the force X.

which determines the value of R: and then the equation

30. COR. If a force R be given and it be required to resolve it into two components acting in directions at right angles to each other, we must employ the equations X-R cose, and Y = R sin 0,

=

which are derived from the equations

OM = OP cos 0, and MP = OP. sin 0.

31. Any number of forces act upon a point in given directions in a plane; to find their resultant.

Let F1, F2, F...F, be the forces, and 0 (see fig. of Art. 29) the point upon which they act. In the plane in which are the lines in which the forces act, draw any two lines Ox, Oy at right angles to each other; and denote by a, a, as...a, the angles which the directions of F1, F2, F...F, respectively make with Ox.

Then the components of these forces are respectively

Ficos a 1, F cos a2, F3 cos a.....
...... F2 cos an

n

in the direction Ox; and

F, sin a,, F, sin a, F, sin a.......F, sin a,

in the direction Oy.

3

Let us replace (see Art. 15) the original forces by these two sets of components. These components are respectively equivalent to two forces acting in the lines Ox, Oy (Art. 23), and being equal to

and F, sin a1 + F2 sin a2 + F ̧ sin + + F2 sin a,.

1

2

n

Let R be the resultant of the original forces, and suppose that is the angle which the line in which it acts makes with Ou. Then since R is equivalent to the original forces, it is also equivalent to the two components of them which have just been found hence

F1 cos a1 + F2 cos a2 + + F2 cos an

1

2

n

and R sin = F, sin a, + F, sin a + ... + F sin an

From which two equations both R and 0 may be found.

REMARK. The sum of a number of quantities of the same form is often, for brevity, represented by prefixing the symbol to a term representing the general form. Upon this principle the above equations may be written thus:

R cos 0

=

Σ(F cos a), or Σ. F cos a,

and R sin 0Σ (F sin a), or Σ. F sin a;

=

.. R2 = (Σ. F cos a)2 + (Σ. F sin a) ̊,

and tan

1

=

Σ. F sin a

Σ. F cos a

32. To find the conditions of equilibrium of a point acted on by forces in any directions in a plane.

Let F1, F2...F be the forces; a1, a...a, the angles which their directions make with a line Ox; then, proceeding as in the last article, we have the equations there found