(^338) „„„„„ A Textbook of Engineering Mechanics
This is the cartesian equation of the catenary, in which c is called parameter of the catenary ;
X -X axis as directrix and point c as vertex of the catenary. The above equation may also be written as :
cosh
yx
cc
= ...(vii)
From equation (vi) and (vii) we find that
2
2
2 sinh
sx
c c
= and
2
2
2 cosh
yx
c c

∴
22
22 –1
ys
cc
= ...(Q cosh^2 θ – sinh^2 θ = 1)
or y^2 – s^2 = c^2
y^2 =c^2 + s^2 = c^2 + c^2 tan^2 ψ ...(Q s = c tan ψ)
=c^2 (1 tan^2 ψ) = c^2 sec^2 ψ
∴ y =c sec ψ ...(viii)
From equation (ii) we find that
T cos ψ = H
or
cos
H
T=
ψ
= H sec ψ = w c sec ψ ...(Q H = w. c)
∴ (^) csec T
w
ψ=
Substituting the value of c sec ψ in equation (viii),
T
y
w
= or T = w y
From the above equation, we find that tension at any point (P) is equal to the weight of the
string of length equal to the height of the point from the directrix. We know that the horizontal
coordinate of P,
x = 2.3 c log (sec ψ + tan ψ)
Example 15.9. A heavy string ABCDE 10 m long hangs over two smooth pegs B and D as
shown in Fig. 15.15.
Fig. 15.15.
Locate the position of vertex C from the peg B along the string.
Solution. Given Total Length of string (L) = 10 m
Let l = Length of the string BC.
We know that length of string BCD
= 10 – 2.5 – 2.0 = 5.5 m
and length of the string CD = 5.5 – l