display | more...
The curve made by a hanging string can be modeled by considering the gravitational and tensile forces acting on an infinitesimal segment of the string. This segment, of length dL, has weight (mass per unit length)*(gravitational acceleration)*dL - we'll take mass per unit length as a constant and define

q = (mass per unit length)*(gravitational acceleration).

In many problems you can assume that the tension is constant over the length of the string, but when the string's own mass is in consideration in the problem you can no longer do this, as we'll see. So we'll say that, over the length dL, the tension on the string changes from T to T+dT. We'll also say that the angle the string makes with the horizontal goes from A to A+dA.

Now since this segment of string is immobile (i.e. it has no acceleration), all forces on it must cancel each other. We'll balance the forces of tension and gravity in the horizontal, then the vertical direction.

In the horizontal direction the only forces we have are the tensions on each end. Tension, by definition, is a force pointing in the same direction as the string does. The horizontal component of the tension therefore involves the cosine of the string's angle with the horizontal. The force equation is this:

Fhoriz = 0 = -T cos(A) + (T+dT) cos(A+dA)

The addition rule for the cosine is

cos(x+y) = cos(x) cos(y) - sin(x) sin(y)

so

cos(A+dA) = cos(A) cos(dA) - sin(A) sin(dA) = cos(A) - dA sin(A)

Here we took the first-order approximations for trig functions of small angles: the cosine of a small angle equals one and the sine of a small angle equals the angle itself. Now back in our force equation we have

0 = -T cos(A) + T cos(A) + dT cos(A) - T dA sin(A) - dT dA sin(A)

The first two terms cancel and the last term, with the product of two infinitesimals, is negligible compared to the other remaining terms, leaving us with

dT cos(A) = T dA sin(A)

(Incidentally, this is where things would fall apart if we took T to be constant over the whole string:
we'd have only 0 = T dA sin(A), leaving us puzzled as to which of those factors is supposed to always be zero.)
Separating the two variables,

dT/T = tan(A) dA

Integrating both sides

ln(T) = -ln(abs(cos(A))) + C

T = CT sec(A)

In the last step we took the exponential of both sides, cancelling the logarithms and redefined the integration constant so that it'd be proportional to the tension. This constant basically just depends on how the string is hung.

Now in the vertical direction, we have gravity as well as tension:

Fhoriz = 0 = -q dL - T sin(A) + (T+dT) sin(A+dA)

We'll again expand according to the addition rule for the sine:

sin(x+y) = sin(x)cos(y) + cos(x)sin(y)

Using the first-order approximations again, this gives

0 = -q dL - T sin(A) + T sin(A) + dT sin(A) + T dA cos(A) + dT dA cos(A)

Cancelling and removing the negligible double-infinitesimal term we get

q dL = dT sin(A) + T dA cos(A)

Now we'll substitute in the solution we obtained for the tension. Differentiating, we have

dT = CT tan(A) sec(A) dA

so

q dL = CT tan2(A) dA + CT dA = CT dA (tan2(A) + 1)

so

q dL = CT sec2(A) dA

At this point we could solve this differential equation and obtain the inclination angle as a function of length along the string. But to find the shape of the actual curve that the string forms, it's better to substitute Cartesian coordinates. A and dL are both definable in terms of differentials of Cartesian coordinates:

A = tan-1(dY/dX)

so

dA = cos2(tan-1(dY/dX)) d(dY/dX)

(look it up in a table of derivatives if you don't believe me... ) and

dL = sqrt((dX)2 + (dY)2) = sqrt(1 + (dY/dX)2) dX

We'll define the slope dY/dX as a new variable S for notational convenience and take X as the independent variable. Our new force equation is this:

q sqrt(1+S2) dX = CT sec2(tan-1(S)) cos2(tan-1(S)) dS

The cosines and secants nicely cancel, leaving

(q/CT) dX = dS/sqrt(1+S2)

The nicest way to integrate the right half of this equation is to substitute a hyperbolic sine for S. The hyperbolic sine and cosine have the properties that each one is the derivative of the other and

cosh2(a) - sinh2(a) = 1

so

sqrt(1+sinh2(a)) = cosh(a)

So if we substitute S = sinh(a) we get

dS/sqrt(1+S2) = cosh(a) da/cosh(a) = da

This leaves the equation, after integration and resubstitution, as a fairly simple

(q/CT) X = sinh-1(dY/dX) + CX

sinh((q/CT) X - CX) dX = dY

As noted before, the hyperbolic cosine and sine are each other's derivatives, so the integral is simple and leaves us with an expression for the curve the string forms in Cartesian coordinates:

Y = (CT/q) cosh((q/CT) X - CX) + CY

See any mistakes? message me.
Know of a better way to format math in e2? message me also.