1. 1
The Catenary and Parabolic Cables
The Catenary
Assume that one has a cable which
hangs under its own weight with only
tangential forces acting along the length of
the cable. The cable is assumed to have
uniform density, w0 , along its length.
Arrange the coordinate system so that the y-
axis passes through the lowest point of the
cable. Consider a free-body diagram of that
portion of the cable above the interval [0,x].
Suppose that the length of this portion of the
cable is s. Then its weight is w0s . In the
free-body diagram this is a force acting
downward on that portion of the cable. There is also a tangential force T0 acting
horizontally to the left in the free-body diagram. This force is due to the cable to the left
of the figure. Similarly, there is a force T acting tangentially at the right in the free-body
diagram. This force is due to the cable to the right of the figure. Since the cable is
assumed static, the sum of the horizontal and vertical forces must be zero. This leads to
two equations.
T cos(!) = T0 and T sin(!) = w0s
Dividing we eliminate T and get the following.
tan(!) =
w0s
T0
We rewrite this in terms of the derivative.
dy
dx
= a ! s where a =
w0
T0
We take the derivative of this expression with respect to x to get the following
expression.
d2
y
dx
2 = a !
ds
dx
= a ! 1 +
dy
dx( )
2
because
ds
dx
= 1+
dy
dx( )
2
Now let p =
dy
dx
in the above expression to get
dp
dx
= a ! 1+ p
2
.
w0s
T0
T
x ! axis
y ! axis
Figure 1
"

2. 2
This can be solved by separating variables.
dp
1+ p2! = a " dx! and thus arcsinh(p) = a ! x + C1
By the choice of axes, when x = 0 , p = 0 . Thus C1 = 0 and we get the following.
p = sinh(a ! x) and thus
dy
dx
= sinh(a ! x)
We now find y by integrating.
y = 1
a
cosh(a ! x) + C2
Notice that the constant is determined by the y-intercept.
C2 = y(0) !
1
a
This material is adapted from Calculus Gems by George Simmons (McGraw-Hill,
1992), pp. 256-259.
The Parabolic Cable
Suppose now that the cable does not
have uniform density along its length, but has
uniform density along the x-axis. This would
happen if it were supporting a heavy
horizontal span of a bridge, for instance. The
free-body diagram for this is given in Figure
2. We go through the calculations to see
what shape this curve has. The explanation
for each calculation is the same as for the
catenary replacing s with x. However, the
integration is easier and can be done in one
step.
T sin(!) = w0 x and tan(!) =
w0 x
T0
dy
dx
= a ! x where a =
w0
T0
T0
T
x ! axis
y ! axis
Figure 2
"
w0 x

3. 3
y =
a ! x2
2
+ C
Thus, this curve is a parabola. In 1645 Galileo thought that the catenary was a
parabola. Christiaan Huygens showed this false in 1646. James Bernoulli asked in 1690
what was the formula for the curve of the catenary. John Bernoulli showed in 1691 that
the catenary had the form y = 1
a
cosh(a ! x) + C2 . Christian Huyhgens and Gottfried
Leibniz also derived correct formulas for the catenary.