Chapter 15 : Equilibrium of Strings „„„„„ 337

Resolving the forces on CP vertically and horizontally,
T sin ψ = w s ...(i)
and T cos ψ = H ...(ii)
Dividing equation (i) by (ii),

tan

ws

H

ψ= ...(iii)
Now let us assume that the horizontal pull (H) at C to be equal to the weight of length c of the
string, such that H = w. c.
Now substituting this value of H in equation (iii),

tan

ws s

wc c

ψ== ...(iv)

or s= c tan ψ
This is the intrinsic equation of the catenary. It is thus obvious, that the shape of the cable
under its own weight, is catenary. The above equations may also be expressed in cartesian coordi-
nates.
From equation (iv), we find that
dy s
dx c

= ... tan

dy

dx

⎛⎞

⎜⎟ψ=

⎝⎠

Q

∴

(^22)
2
dy s
dx c
⎛⎞
⎜⎟=
⎝⎠
...(Squaring both sides)
We know that
2
1
ds dy
dx dx
⎡⎤
=+⎢⎥
⎣⎦
2
(^12)
s
c
=+
(^22)
... 2
dy s
dx c
⎡⎤⎛⎞
⎢⎥⎜⎟=
⎢⎥⎣⎦⎝⎠
Q
∴ 2
(^12)
ds
dx
s
c

• 
  

  Integrating the above equation
  –1
  sinh 1
  s
  xc K
  c
  =+
  where K 1 is the constant of integration. We know that at C, x = 0 and c = 0. Therefore K 1 = 0. The
  above expression may now be written as :
  xcsinh–1s
  c

  
  

  ∴ sinh
  sx
  cc
  = ...(vi)
  or sinh
  dy x
  dx c
  = ...
  sdy
  cdx
  ⎛⎞
  ⎜⎟=
  ⎝⎠
  Q
  ∴ sinh
  x
  dy dx
  c

  
  

  Integrating the above equation,
  cosh 2
  x
  yc K
  c
  =+
  where K 2 is the constant of integration. We know that at C, x = 0 and c = 0. Therefore K 2 = 0. The above
  expression may now be written as :
  cosh
  x
  yc
  c