Suspension Bridge Parabola Problem


by lenial
Tags: bridge, parabola, suspension
lenial
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#1
May1-11, 03:12 PM
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Please help me. I can't figure out how to solve this problem.

The cable of a suspension bridge hangs in the form of a parabola when the load is evenly distributed horizontally. The distance between the two towers is 150m, the points of support of the cable on the towers are 22m above the roadway, and the lowest point on the cable is 7m above the roadway. find the vertical distance to the cable from a point in te roadway 15m from the foot of the tower.
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HallsofIvy
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#2
May1-11, 03:24 PM
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Set up a coordinate system in which the x-axis is horizontal, along the bridge road way, the y-axis is vertical, and (0, 0) is at the center of the bridge. The cable is clearly symetric about that point so y(x)= ax^2+ b. You are told that the towers are 150 m apart so x= 75 and -75. You are told that the cable is attached at 22 feet above the road way at those towers so when x= 75, y= 22. Finally, you are told that the lowest point of the cable, which, because of symmetry, is at x= 0, y= 7.

Use x= 75, y= 22 and x= 0, y= 7 to solve for a and b in y= ax^2+ b. Then find y when x= 75- 15= 50.
QuarkCharmer
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#3
May1-11, 03:40 PM
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Darn, I just made a graph to help visualize!

Halls of Ivy did a good job explaining it, but now I have this graphic that would otherwise go to waste. They want distance d. (Note: the graph is shifted differently than Halls' but the process is the same.)


HallsofIvy
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#4
May2-11, 07:11 AM
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Suspension Bridge Parabola Problem


Excellent graph! Yes, you can set up your coordinate system however you choose and QuarkCharmer chose to take the origin at one of the cable ends. Of course, the answer to the question will be exactly the same.
spartanboy
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#5
May14-11, 05:02 AM
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please.... can you explain how to use x=75, y=22 and x=0, y=7 to solve for a and b in y=ax^2 +b ??? I can't understand...
SteamKing
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#6
May14-11, 08:35 AM
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You are trying to determine the value of a and b. You know the parabola passes thru two separate points (75, 22) and (0, 7). Substitute the values of x and y from these two points into the equation y = ax^2 + b. You will obtain two equations with a and b as the unknowns, which then may be solved simultaneously.
andonrangelov
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#7
May15-11, 12:57 PM
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I am sorry to disappoint you but the Catenary problem is well known and the shape of the suspended bridge is not a parabola but hyperbolic cosine function (see for example http://en.wikipedia.org/wiki/Catenary)
SteamKing
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#8
May15-11, 01:23 PM
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andonrangelov: If you read the section on Suspension Bridges in your linked Wikipedia article, you will find that when the suspension cables are supporting a distributed horizontal load, the shape of the suspension cable is NOT a catenary, but a parabola. A catenary forms only when the loading is distributed evenly along the length of the cable or chain.
andonrangelov
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#9
May15-11, 02:15 PM
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SteamKing you are right I havenít pay enough attention here is the problem set and the solution if someone is interested in it, about the point of the parabola I think the others give good explanation how to find itÖ.
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