0.3 Boundary Value Problems 27
We suppose that the cable is not moving. Then by Newton’s second law,
the sum of the horizontal components of the forces on the segment is 0, and
likewise for the vertical components. IfT(x)andT(x+ x)are the magnitudes
of the tensions at the ends on the segment, we have these two equations:
T(x+ x)cos(
φ(x+ x))
−T(x)cos(
φ(x))
= 0 (Horizontal), (1)
T(x+ x)sin(
φ(x+ x))
−T(x)sin(
φ(x))
−f(x) x= 0 (Ve r t i c a l). (2)In the second equation,f(x)is the intensity of the distributed load, measured
in force per unit of horizontal length, sof(x) xis the load borne by the small
segment.
From Eq. (1) we see that the horizontal component of the tension is the same
at both ends of the segment. In fact, the horizontal component of tension has
the same value — call itT— at every point, including the endpoints where
the cable is attached to solid supports. By simple algebra we can now find the
tension in the cable at the ends of our segment,
T(x+ x)=T
cos(
φ(x+ x)), T(x)= T
cos(
φ(x)),
and substitute these into Eq. (2), which becomes
T
cos(
φ(x+ x))sin(
φ(x+ x))
− T
cos(
φ(x))sin(
φ(x))
−f(x) x= 0or
T(
tan(
φ(x+ x))
−tan(
φ(x)))
−f(x) x= 0.Before going further we should note (Fig. 4) thatφ(x)measures the angle
between the tangent to the centerline of the cable and the horizontal. As the
position of the centerline is given byu(x),tan(φ(x))is just the slope of the
cable atx. From elementary calculus we know
tan(
φ(x))
=
du
dx(x).Figure 4 Section of cable showing forces acting on it. The angles areα=φ(x),
β=φ(x+ x).