Pcos a, Pcos B, Pcos y,

acting in directions OX, OY, OZ, respectively.

Similarly, if P' be another force, a', B', y' the angles its direction makes with OX, OY, OZ, it is equivalent to

P' cos a', P' cos B', P' cos y',

in direction of the same lines;

and so on whatever be the number of forces.

The system of forces is equivalent then to three components X, Y, Z, which are severally equal to

Pcos a + P' cos a'+...in direction of OX or Σ (Pcos a)

......OY...Σ (Pcos ß).. (i).

...............OZ...Σ (P cos y)

Now these three components (i) are equivalent to a single resultant R making angles λ, μ, v with the line OX, OY, OZ, provided

R cosλ(Pcos a),

R cos μ= (Pcos B), R cos v = Σ (P cos y)......................... (ii),

.........

and remembering that cos2 + cos3μ + cos2 v = 1,

these give us the magnitude of the resultant, i. e.

R2 = {Σ (Pcos a)}2 + {Σ (Pcos B)}2 + {Σ (Pcos y)}2; and this being known, the equations of (ii) give λ, μ, v, which assign the direction of the resultant.

This analytical mode of finding the resultant of a system of forces applied at a point is of course equivalent to the geometrical construction of Leibnitz noticed in Art. (26).

63. If the forces be in equilibrium the resultant is nil. i.e. R=0;

and .. {(P cos a)}2 + {Σ (Pcos B)}2 + {Σ (Pcos y)}2 = 0;

which requires

Σ (Pcos a) = 0, Σ (P cos B) = 0, Σ (P cos y) = 0,

the three conditions of equilibrium of a system of forces acting through a point.

That is, the sum of the forces resolved in three directions mutually at right angles must be severally zero.

Or we may reason thus:

In considering any system of forces whose directions pass through a point, if they be in equilibrium, we may (as has been observed before) regard any one of the forces as equal and opposite to the resultant of all the rest. Hence, in any case in which we are discussing the conditions of equilibrium of a body or system of bodies acted on by such a system of forces, we may resolve all the forces in a particular direction (any we please)—and perpendicular to the direction so taken: the conditions of equilibrium then will be

(i) The algebraic sum of the former resolved parts must be zero.

(ii) The resolved parts acting in a plane perpendicular to the direction taken, must be in equilibrium inter se,— and must satisfy the condition of equilibrium of forces in one plane and we may apply the principles established in the second chapter in the same way as if these resolved parts had been the only forces acting.

:

And it will in general constitute part of the solution of the problem to shew that these resolved parts are in equilibrium.

64. The remaining articles of this chapter contain demonstrations of two results relating to the tension of strings passing over smooth and rough surfaces referred to in Article (43)—and some properties of a funicular polygon.

They may be omitted by a student whose previous reading has not prepared him for the consideration of small quantities.

65. The tension of a string which passes in one plane over a smooth curve or surface, is the same at every point— the weight of the string being neglected.

Let ps be any finite length of the string in contact with the curve, the normals to which at p, s include an 4 p.

Let ps be divided into n parts, such that the normals at the extremities of consecutive parts include the same angle -so that no = p.

P

P

8

pq the first of these parts, the normals at p, q meeting in O, t1, t...t the tension of the string at p, q,... S.

R the measure of the pressure on the curve at p—then we may regard the resultant pressure of the curve on the element pq of the string as equal to (R+x). arc pq, acting in some direction rV intermediate to Op, Oq, and making angles a,, B, say with p0, q0,-so that a,+B1 = 0, and κ is some small quantity which vanishes in the limit, when is taken smaller and smaller.

P. M.

5

Considering now the equilibrium of the element pq of the string as a rigid body-resolve the forces upon it parallel and perpendicular to r V, and we obtain the equations

t, sin a,+t, sin ẞ1 = (R+x) pq .................(i),

equation (ii) may be written in the form

t1— 2t, sin2=t,— 2t, sin2

(ii),

(1),

and if we write down the corresponding equation for each consecutive element of ps, we shall obtain

adding equations (1), (2) ... (n), we obtain

2

t-2 (t, sin+sin+...+ sin 2)

t,

2

Now if be the greatest of the quantities t, to... tm+

If now n be increased indefinitely, & remaining unchanged, and therefore being indefinitely diminished-the expression

i. e. the tension of the string is the same at every point.

Further, from equation (i)—suppressing the suffixes

a relation which gives the pressure on the curve at any point in terms of the tension and radius of curvature.

Obs. For simplicity we have supposed the string to be all in one plane-the demonstration might without much trouble be extended to shew that the tension is the same at every point in whatever manner the string passes freely along a smooth surface or tube of any form.

66. A string passes in one plane over a rough curve or surface, the tensions of the extremities being such that the string is on the point of motion-to find the relation between these tensions, the weight of the string being neglected.