SOLUTION: A suspension bridge with weight uniformly distributed along its length has twin towers that extend 50 meters above the road surface and are 800 meters apart. The cables are parabo
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Question 1047122: A suspension bridge with weight uniformly distributed along its length has twin towers that extend 50 meters above the road surface and are 800 meters apart. The cables are parabolic in shape and are suspended from the tops of the towers. The cables touch the road surface at the center of the bridge. Find the height of the cables at a point 200 meters from the center. (Assume that the road is level.)
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A suspension bridge with weight uniformly distributed along its length has twin towers that extend 50 meters above the road surface and are 800 meters apart. The cables are parabolic in shape and are suspended from the tops of the towers. The cables touch the road surface at the center of the bridge. Find the height of the cables at a point 200 meters from the center. (Assume that the road is level.)
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3 points on the parabola are given: (-400,50), (0,0) and (400,50)
Find the equation of the parabola.
y = f(x) = ax^2
50 = a*400^2
f(x) = 0.0003125x^2
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Find f(200)
f(200) = 0.0003125*200^2
= 12.5 meters
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PS Hanging cables don't form parabolas.
They from catenary curves.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Assume a point on the roadway at the center of the bridge is the origin of a coordinate system. Each tower is 400 feet from the center, so the coordinates of the points where the cable attaches at the tops of the towers are (-400,50) and (400,50). Since the parabola's vertex is at the origin, the function describing the parabola is:
\ =\ ax^2)
Since we know that \ =\ 50)
when 
, solve
^2\ =\ 50)
for 
to determine the unknown coefficient. Once you have a value for , calculate )
John
My calculator said it, I believe it, that settles it
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