for the determination of their resultant. But because they balance each other by hypothesis, they have no resultant, and therefore

0 = (Σ. F cos a)2 + (Σ. F sin a)2.

But as the right-hand member consists of two terms which, being squares, are essentially positive, their sum cannot be equal to 0 unless each be separately equal to 0;

If the investigation in Art. 31 be examined, it will be seen that the line Or was taken in any direction in the plane of the forces; and hence we may state the signification of the two equations just found, as follows,

If any number of forces act upon a point, that there may be equilibrium,

The sums of the components of the forces parallel to any two lines, at right angles to each other, in the plane of the forces must be separately equal to zero.

The converse is evidently true also.

No other condition is necessary for equilibrium, for if Σ. F cos a = 0, and . F sin a = 0, it follows inevitably that R = 0, and therefore there is equilibrium.

CHAPTER II.

THE

ON FORCES WHICH ACT UPON A PARTICLE, OR UPON
SAME POINT OF A RIGID BODY, IN ANY DIRECTIONS
NOT IN ONE PLANE.

33. Ir three forces acting upon the same point be respectively represented by the three edges of a parallelopiped which meet, the diagonal of the parallelopiped drawn from that point to the opposite corner will represent their resultant.

For let OA, OB, OC (fig. 6) be the edges which represent the three forces, and OE the diagonal of the parallelopiped: draw OD, CE.

Then because OA, OB represent two forces, OD represents a force which is equivalent to them both (Art. 26): hence the three forces represented by OA, OB, OC are equivalent to the two represented by OD, OC, which again are equivalent to the single force represented by OE, for CD is a parallelogram.

34. Three forces act upon a point in directions which are at right angles to each other; to find their resultant.

Let X, Y, Z be the forces, acting upon the point (fig. 7) in the lines Ox, Oy, Ox which make right angles YOZ, XOY, ZOX with each other. From O set off OL, OM, ON to represent the forces. Complete the parallelograms OMQL, OQPN, and join OP; this line, by the last Art., represents the resultant required.

Let R denote the resultant, and a, ß, y the angles POx, POy, POs which its direction makes with the directions of the given forces.

·· X2 + Y2 + Z2 = R2 (cos2 a + cos3ß + cos2 y)

This equation gives the value of R, and then the three equations marked (A) give the angles a, ẞ, y, which fix the line in which R acts.

REMARK. The reader will observe that

cos2 a + cos2ß + cos2 y = 1.

35. COR. If a force R be given, and it be required to resolve it into three components, whose directions are at right angles to each other, we must employ the equations marked (A).

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

Let 0 (fig. 7) be the point upon which the given forces F1, F2, F3... F, act; from O draw three lines Ox, Oy, Ox, arbitrarily taken, at right angles to each other; and denote by a1 B171, a2 ß2 72, aз B3 Yз... an ẞny, the angles which the directions of the forces make with these three fixed lines.

The respective components of the

F1 cos a1, F2 cos α2, ...

1

2

given forces are

F2 cos an

n

in the direction Ox;

Replace now the original forces by these three sets of components; each set is reduced to one force by Art. 23; and we then have three forces

acting in the lines Ox, Oy, Oz.

Let R be the resultant required, and a', B', y' the angles which the line in which it acts makes with Ox, Oy, Oz. Then since R is equivalent to the original forces, it is also equivalent to the three components of them which have just been found; hence

.. R2 = (Σ. F cos a)2 + (Σ. F cos ẞ)2 + (Σ. F cos y)2

This equation gives the value of R; and then the equations (A) will give a', B', y', which fix the direction in which

R acts.

37. To find the equations of the line in which the

resultant acts.

Suppose OP (fig. 7) the line in which the resultant acts; and let x, y, ≈ be the co-ordinates of any point P in it.

Then if OP be taken to represent R, the co-ordinates will represent the components, and therefore by Art. 25,

If the point at which the forces act be not the origin of co-ordinates, let its co-ordinates be a, b, c; then since the line whose equations are required passes through this point,

38.

To find the conditions that the system of forces in Art. 36 may balance each other.

It is evident that there cannot be equilibrium among the forces F1, F2, F3 ...... F2, unless their resultant be evanescent, and therefore we must have

0 = (Σ . F cos a)2 + (Σ . F' cos ß)2 + (Σ . F cos y)2, which for a reason similar to that assigned in Art. 32 resolves itself into the three independent conditions

0 =Σ. F cos a, 0=Z. F cos B, 02. F cosy.

Or, in words, (remembering that the positions of Ox, Oy, Ox were arbitrarily chosen)

The sum of the components of the given forces parallel respectively to any three lines at right angles to each other must separately be equal to zero.