(^328) „„„„„ A Textbook of Engineering Mechanics
15.6. TENSION IN A STRING SUPPORTED AT DIFFERENT LEVELS
Fig. 15.6. Tension in string ACB.
Consider a string or cable ACB, supported at different levels at A and B, and carrying a uniformly
distributed load as shown in Fig 15.6. Let C be the lowest point of the cable.
Let w = Uniformly distributed load per unit length of the span,
l = Span of the string,
yc = Depth of the lowest point of the string C, from the lower
support B,
d = Difference between the levels of the two supports,
l 1 = Horizontal length between A and C, and
l 2 = Horizontal length between C and B.
Since the string is supporting vertical loads only, therefore the horizontal thrust at A, must be
equal to the horizontal thrust at B.
In order to locate position of the lowest point C, let us imagine the portion CB of the string to
be extended to CB 1 , such that the new support B 1 is at the same level as that of A. Similarly, imagnie
the portion AC of the string to be cut short to A 1 C, such that the new support A 1 is at the same level as
that of B.
From the geometry of the figure, we find that the string ACB 1 has a span of 2l 1 and a central dip
of (yc + d) ; whereas A 1 CB has a span of 2l 2 and a central dip of yc.
Now in the string ACB 1 the horizontal thrust
2 2
(2 ) 1
88( )cc
wl wl
H
yyd

• 
  

  ...(i)
  Similarly, in the string A 1 CB, the horizontal thrust,
  2 2
  (2 ) 2
  (^88) cc
  wl wl
  H
  yy
  == ...(ii)
  Since the two horizontal thrusts are equal, therefore equating both the equations (i) and (ii)
  22
  (2 ) 12 (2 )
  8(cc) 8
  wl wl
  yd y

  
  


• 
  

  or
  2 2
  2
  cc
  l l
  ydy

  
  


• 
  

  ∴
  1
  2
  c
  c
  l yd
  ly

  
  


• 
  

  = ...(iii)