Let APB (fig. 70) be any portion of the string in a position of rest; Pp being a small element of its length; x, y, z, and x+dx, y + dy, z+dz, the co-ordinates of P and p respectively; 8 the length of the string reckoned from some assigned point up to P, and s+ds the length up to p; t the tension of the string at P.

The resolved parts, parallel to the axes of x, y, z, of the force exerted upon the point P of the element Pp by the portion AP of the string, will evidently be

and therefore, since each of these three forces must be some function of s, it is plain by Taylor's theorem that the resolved parts of the force exerted on the element Pp by the portion pB of the string, will be

Again, let X, Y, Z, be the sums of the resolved parts of the accelerating forces which act upon the element Pp; p the density of the string at P, and k the area of a section at right angles to its length at that point. Then the mass of the

portion Pp of the string will be kpds, which therefore, for a constant value of ds, will vary as kp; the product kp, which we will represent by m, may be called "the mass of the string at the point P." The resolved parts, parallel to the co-ordinate axes, of the moving force of the element Pp, will be

mXds, m Yds, mZds.

Hence for the equilibrium of Pp we must have, equating to zero the sum of the resolved forces which act upon it parallel

to each of the three axes, and dividing the three resulting equations by ds,

which three equations constitute the conditions of equilibrium of the entire string.

By the elimination of t we readily obtain the three following equations,

any two of which will be differential equations to the required curve of equilibrium.

COR. 1. From the equations (a) we have also

dx

dy

dz

td--fmXds, t--fmYds, td --JmZds;

ds

squaring and adding these equations, and observing that

dx2 dy dz2

dy3,
+ + = 1

ds2 ds2 ds

.....

we obtain for the value of the tension at any point,

(b),

We may obtain also another expression for the tension: differentiating (b) with respect to s, we get

dx dx dy d'y dz dz

+

+
dsdsds ds2 ds ds

....

(c);

hence, multiplying the three equations (a) by dx, dy, dz, in order, and adding the resulting equations, we have, by the aid of (b) and (c),

t = C' − fm (Xdx + Ydy+ Zdz),

where C is an arbitrary constant.

COR. 2. If the whole string lie entirely within one plane, let the plane of xy be so chosen as to coincide with this plane; then the three differential equations to the string will be reduced to the single one

dx fm Yds=dy fmXds...

and the two formulæ for the tension will become

These two formulæ for the tension, and also the differential equation (d) to the string, coincide with those given by Fuss; Mémoires de St. Pétersbourg, 1794, pp. 150, 151.

SECT. 2. Parallel Forces.

(1) A flexible string fixed at any two points A and B, (fig. 71), is acted on by gravity; supposing the mass of the string to vary according to any assigned law as we pass from one point to another, to find the equation to the catenary of rest; and conversely, the curve being known, to determine the law of the mass of the string.

Let the axis of y extend vertically upwards, and let the axis of x be horizontal, the plane x Oy coinciding with the plane which contains the catenary. Then, since

we have, by the first two of the equations (a) of section (1),

T

where C is a constant quantity: let 7 denote the tension at the lowest point of the curve: then evidently T = C, and therefore

supposing a to be the value of s at the lowest point,

If m be given in terms of the variables x, y, s, the form of the catenary may be determined from (d).

Again, differentiating (d), we obtain

a formula by which m may be computed for every point of the string when the form of the catenary is given. Also from (c) we get

which gives the tension at any point of the catenary when its

form is known.

John Bernoulli; Lectiones Mathematica,

Lect. 38, 39, 40; Opera, Tom. III.

(2) A flexible string A OB, (fig. 72), fixed at two points A and B, is acted on by gravity; the mass at any point P varies inversely as the square root of the length OP measured from the lowest point 0; to find the equation to the catenary.

Let the origin of co-ordinates be taken at O, x being horizontal, and y vertical, and the plane of xy coinciding with the plane of the catenary; also let O be the origin of s.

Then, if μ be the mass at the end of a length c from the lowest point,

=

ct

and therefore by (1, d), a being in the present case zero, we have