Binomial Theorem

Question
CBSEENMA11015160

The cable of a uniformly loaded suspension bridge hang in the form of a parabola.
The roadway which is horizontal and 100 m long, is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to a roadway 18 m from the middle.

Solution

Let AOB be the cable of a uniformly loaded suspension bridge as shown in the figure.
Let AL and BM be the longest wires of length 30 m each. Let OC be the shortest wire of length 6 m and LM be the road way as shown in the figure.
                     
The longest supporting wire = AL = BM = 30 m
The shortest supporting cable = OC = 6 m
          Length of the roadway  = LM = 100 m
∴                           LC = CM = 50 m
Let O be the vertex and OY be the axis of the parabola
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#6 {main}</pre>                                 ...(i)
                            ON = CM = 50
                            BN = BM - MN = BM - OC = 30 - 6 = 24
∴   Co-ordinates of point B are (50, 24)
Since it lies on the parabola (i)
∴                               left parenthesis 50 right parenthesis squared equals 4 left parenthesis straight a right parenthesis space left parenthesis 24 right parenthesis space or space straight a equals 2500 over 96 equals 625 over 24
∴   From (i), the equation of the parabola is: straight x squared equals 4 open parentheses 625 over 24 close parentheses straight y
or                            straight x squared equals 625 over 6 straight y                                          ...(ii)
Let the length of the supporting wire PQ at a distance of 18 m be h.
∴                             OR = CQ = 18 m
                                PR = PQ - RQ = PQ - OC = h - 6
rightwards double arrow    Co-ordinates of point P are (18, h - 6)
Since it lies on parabola (ii)
∴                        space space open parentheses 18 close parentheses squared space equals space 625 over 6 left parenthesis straight h minus 6 right parenthesis
rightwards double arrow                       space space space space 324 cross times 6 space equals space 625 space straight h space minus space 3750
rightwards double arrow                       1944 + 3750 = 625 h
rightwards double arrow                                         straight h equals 5694 over 625 equals 9.11 space straight m     

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