Results 1 to 3 of 3

Math Help - Suspension Bridge Parabola Problem

  1. #1
    Newbie
    Joined
    May 2011
    Posts
    1

    Question Suspension Bridge Parabola Problem

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,583
    Thanks
    1418
    Set up a coordinate system so the x-axis is along the bridge roadway, the y-axis is vertical, and the origin, (0, 0), is half way between the two towers. Since the two towers are 150 m apart, they are at x= 75 and x= -75. The cable attaches to the towers 22 m above the roadway (y= 0) so the cable passes through the points (75, 22) and (-75, 22). Since the parabola is symmetric, we must have y= ax^2+ b. Also, the minimum is at (0, 7). Put x= 74, y= 22, and x= 0, y= 7 into equation to solve for a and b. Then, of course, evaluate y at x= 75- 15= 60.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie Auri's Avatar
    Joined
    May 2011
    From
    Boston Ma
    Posts
    24
    The bridge 150m long, think of this as your X axis. The height of the tower is 22m, that's your Y axis. The lowest point drops down the middle of the bridge which is at (0,7) which is your vertex. That makes your domain (-75, +75) Which also means that at X=-75, Y=22. And at X=75, Y=22. I like to set this up as a vertex Y=a(x-h)^2+k Where h and k is the vertex. We have X,Y so we can do 22=a(75-0)^2+7, solve for a by subtracting 7, squaring 75, then dividing by 75 which will give you a. Then you want to know the what x will be when y is 15, so do 15=a(x-0)^2+7. a will be what you got from the last part, and just solve the equation Algebraically.

    I'm pretty sure that's how you go about doing it. Please correct me if I am wrong!

    -Nichole
    Last edited by Auri; May 2nd 2011 at 10:30 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Catenary differential equation (suspension bridge)
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: July 13th 2011, 10:42 PM
  2. Parabola word problem about a suspension bridge
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 19th 2010, 04:35 PM
  3. Variation of Konigsburg Bridge Problem
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: December 11th 2009, 05:01 PM
  4. Replies: 1
    Last Post: December 9th 2009, 07:00 PM
  5. Suspension Bridge
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 14th 2008, 08:56 PM

Search Tags


/mathhelpforum @mathhelpforum