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Question 1160797: The cable of a suspension bridge hangs in the form of a parabola when the load is uniformly distributed horizontally. The distance
between the two towers is 1500 ft, the points of support of the cable on the towers are 220 ft above the roadway, and the lowest
point on the cable is 70 ft above the roadway. Find the vertical distance of the cable from a point in the roadway 150 ft from the
foot of a tower.
Found 2 solutions by Alan3354, ankor@dixie-net.com:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! The cable of a suspension bridge hangs in the form of a parabola when the load is uniformly distributed horizontally. The distance between the two towers is 1500 ft, the points of support of the cable on the towers are 220 ft above the roadway, and the lowest point on the cable is 70 ft above the roadway. Find the vertical distance of the cable from a point in the roadway 150 ft from the
foot of a tower.
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Make the Origin the point midway between the towers and 70 feet above the roadway.
---> 3 points:
A(-750,220), B(0,0) and C(750,220)
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Find the equation of the parabola.
Then, find the y value at x = 600.
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email via the TY note for help or to check your work.
====================
PS A cable supporting its weight does form a parabola.
It's a catenary curve.
Answer by ankor@dixie-net.com(22740)  (Show Source):
You can put this solution on YOUR website! The cable of a suspension bridge hangs in the form of a parabola when the load is uniformly distributed horizontally.
The distance between the two towers is 1500 ft, the points of support of the cable on the towers are 220 ft above the roadway, and the lowest point on the cable is 70 ft above the roadway.
Find the vertical distance of the cable from a point in the roadway 150 ft from the foot of a tower.
:
let the first support be at x=0, y=220
let the 2nd support be at x=1500, y=220
the lowest point: x=750, y=70 (half way across the bridge)
;
Using the form ax^2 + bx + c = y,
x=0, y=220, we know c = 220
:
x= 750, y= 70
750^2a + 750b + 220 = 170
5625000a + 750b = 70 - 220
5625000a + 750b = -150
:
x=1500, y=220
1500^2a + 1500b + 220 = 220
2250000a + 1500b = 0
:
multiply the 1st equation by 2, subtract from above equation
2250000a + 1500b = 0
1125000a + 1500b = -300
--------------------------subtraction eliminates b, find a
1125000a = 300
a = 300/1125000
a = .0002667
:
Find b
2250000(.0002667) + 1500b = 0
600 + 1500b = 0
1500b = -600
b = -600/1500
b = -.4
The equation: y = .0002667x^2 - .4x + 220
looks like this, green is 70 ft
:
"Find the vertical distance of the cable from a point in the roadway 150 ft from the foot of a tower."
x = 150
y = .0002667(150^2) - .4(150) + 220
y = 6 - 60 + 220
y = 166 ft height at 150 ft from the tower (blue line)
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