A cable of a suspension bridge hangs in the form of a parabola when the
load is uniformly distributed horizontally. The distance between two towers is
1500 ft, the points of support of the cable on the towers are 200ft above the road
way and the lowest point on the cable is 70ft above the roadway. Find the
vertical distance to the cable (parallel to the roadway) from a pole whose height
is 122 ft.
please provide detailed solution
✅ Answers
? Best Answer
∴A′B = 200 − 70 = 130 ft
Thus the point B is (750, 130)
The equation of the parabola is x2 = 4ay
Since B is a point on x2 = 4ay
(750)2 = 4a(130)
⇒ 4a =
75 × 750
13
∴ The equation is x2 =
75 × 750
13 y
Let PQ be the vertical distance to the cable from the pole RQ.
RQ = 122, RR′ = 70 ⇒ R′Q = 52
Let VR′ be x1 ∴ Q is (x1, 52)
Q is a point on parabola
x1
2 =
75 × 750
13 × 52
x1 = 150 10
PQ = 2×1 = 300 10 ft.
– Chosen by Asker
Hello jb008, this is the problem given in text book meant for higher secondary in Tamil Nadu.
Wordings are found to be very difficult to understand the problem.
Instead ‘vertical’ it has to be ‘horizontal’
Imagining we get a parabola facing upward. Let origin be at the vertex of the parabola
So equation has to be x^2 = 4 a y
Now to find a we need point on the parabola
Hence the point on the parabola at the extreme will be (-750, 130)
Plug this and we get 4a = 750*750/130
So equation of the parabola is x^2 = [750*75 / 13 ] y
Of course it may be difficult to understand but if you with total peace in mind and simply observe the problem you would definitely get it. All the best dear
Now we have to find the x co ordinate for given y ie 122-70 = 52
x^2 = [750*75 / 13] * 52
So x = 75./40 = 150./10
Hence required cable horizontal length is 2 times this 150./10. So it will be 300./10 ft