where C is an arbitrary constant: but, by (1, b), t= R, and therefore

but from (1, c), since pc and t = r at the point 0, we see that C=CT; therefore

For the sake of simplicity put =nmkc; then

7 ncdr

the equation to the catenary resulting from the integration of this differential equation will be of three different forms according as n is greater than, equal to, or less than unity.

First, suppose that n is greater than unity; then the integral of the equation will be, supposing that 00 when r=c,

=

Secondly, suppose that n=1; then the equation to the catenary will be, if 0=0 when r = c,

Thirdly, let n be less than unity; then, if as before 0 = 0 when r = c, the equation will be

three catenaries, corresponding to the three values of n, have all of them asymptotes passing within a distance from the

nc

n+1

centre of force. Put r in the equations to the three curves, and we get for the inclinations of the pair of asymptotes of each to the line SO,

(3) To find the equation to a uniform catenary AOB (fig. 78), acted on by a central force tending to S, the intensity of which varies as the uth power of the distance; the tension at O being equal to the (1+)th part of the weight of a length SO of the string, each element of which length is supposed to be acted on by a constant force equal to that at O and towards S.

The notation remaining the same as in (2), the equation to the catenary will be

(4) To find the equation to a uniform catenary SA OB, (fig. 79), acted on by a central repulsive force emanating from S, at which the two ends of the string are fastened, the intensity of this force varying inversely as the μth power of the distance; the tension at O being equal to the (1)th part of the weight of a length SO of the string, each element of which length is supposed to be acted on by a constant force equal to that at and from S.

The notation remaining the same as before, the equation to the curve will be

-2

= cos (μ-2) 0.

SECT. 4. Constrained Equilibrium.

(1) A flexible string ab, (fig. 80), acted on by gravity, rests on the arc of a curve APB in a vertical plane; to find the tension of the string and the pressure on the curve at any point.

Let P, p, be any two points of the curve very near to each other; PO, pO, normals at these points, the point being the centre of curvature when p approaches indefinitely near to P: let ax, ay, be the axes of x, y, the former being horizontal, the latter vertical; aP=s, Pp=ds; t = the tension at P and t + dt at p; R= the pressure on the curve at P, m = the mass at any point of the string, POp = 4, PO=p.

Then, resolving forces, which act on the element Pp of the string, parallel to the tangent at P, we have

or, neglecting infinitesimals of higher orders than the first,

dt = mgdy;

integrating and observing that t is equal to zero when y = 0, we

get

t = mgy

which gives the tension at any point of the string.

(1),

Again, resolving the forces on the element Pp parallel to the normal OP,

or, neglecting infinitesimals of orders higher than the first,

(2) Two equal weights Q, Q, are suspended at the extremities of a flexible string hanging over a smooth curve in a vertical plane; to find the pressure at any point of the curve, the weight of the string being reckoned inconsiderable.

=

Let APB (fig. 81) be the curve; OP, Op, normals at two consecutive points P, p; @ the inclination of OP to some assigned line in the plane of the curve, POp de; P0=p, AP=s, Pp ds; p = the pressure at the point P; t the tension of the string at P, t+ dt = the tension at p.

=

Then for the equilibrium of the element Pp of the string we have, resolving forces at right angles to the tangent at P,

(t + dt) sin d0 = pds = ppd0,

and therefore, retaining infinitesimals of the first order,

td0=ppd0, t=pp...

Again, resolving forces parallel to the tangent at P,

(t+dt) cos de- t = 0,

and therefore, retaining infinitesimals of the first order,

dt0, t constant;

.(1)

but evidently at A the tension is equal to Q; hence t = Q. Hence from (1) we have

COR. The whole pressure on the curve AB is equal to

If the tangents at the points where the string leaves the curve be vertical, we have TQ for the whole pressure along the curve; if they be not vertical there will of course be pressures at the points A, B, in addition to the pressure along the curve.

Euler; Nov. Comment. Petrop. 1775, p. 307.

Poisson; Traité de Mécanique, Tom. I. ch. 3.

(3) To find the pressure on a curve AB, (fig. 82), when two weights Q, R, balance each other over it by means of a fine string, the friction between the string and the curve being taken into account; and the weight Q being considered as much greater than R as is consistent with equilibrium.

Let be the coefficient of friction; the rest of the notation μ being the same as in the preceding problem. Then the friction on the element Pp will be upds, and will act nearly in the direction of the tangent at P. Hence, resolving forces on the element Pp parallel to PO, we have

(t+dt) sin do = pds = ppd0 ;

and therefore in the limit

td0=ppd0, t=pp......

again, resolving forces parallel to the tangent at P,

(t + dt) cos de- t + μpds = 0,

.(1);