The main cables of a suspension bridge uniformly distribute the weight of the bridge when in the form of a parabola. The main cables of a particular bridge are attached to towers that are 600 ft apart. The cables are attached to towers at a height of 110 ft above the road. The cables are 10 ft above ground at their lowest points. If vertical support cables are at 50 ft intervals along the level roadway, what are the lengths of these vertical cables?
Let's create the equation of your cables using the following data
let the vertex be at (0,10)
two other points are (300,110) and (-300,110)
the equation must be
y = a(x-0)^2 + 10
= ax^2 + 10
but (300,110) lies on it
110 = a(90000) + 10
a = 100/90000 = 1/900
y = (1/900)x^2 + 10
so when x = 0 , y = 10
when x = 50 , y = (1/900)(2500)+10 = 12.77..
when x = 100 , y = (1/900)(100^2) + 10 = 21.111..
when x = 150 , y = (1/900)(150^2) + 10 = 35
when x = 200 , y = (1/900)(200^2)+10 = 54.44..
when x = 250 , y = (1/900)(250^2)+10 79.444..
when x = 300 , y = (1/900)(300^2)+10 = 110
because of the symmetry, the cables on the left would have the corresponding same lengths
BTW, the cables would not hang as a parabola, they would hang as a catenary with equation
y = (a/b)(e^x + e^-x)
If you have time for some extra reading:
https://www.google.ca/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=catenary%20vs%20parabola
Well, it sounds like this bridge is really "suspenseful" with its parabolic shape! Now, let's calculate those lengths of the vertical support cables.
Since the main cables of the bridge follow a parabolic shape, we can assume it to be simply a quadratic function. The vertex of this parabola will be at the lowest point, where the main cables are 10 ft above ground.
So, let's establish our equation. We know that the vertex of the parabola is (300, 10), since the towers are 600 ft apart and the lowest point is halfway between them.
Using the vertex form of a quadratic equation, we have:
y = a(x - h)^2 + k
where (h, k) represents the vertex. In our case, h = 300 and k = 10.
Now, we need to find the coefficient "a" to complete the equation. Since the main cables are attached to towers that are 600 ft apart and are at a height of 110 ft above the road, we can substitute these points into the equation to solve for "a".
110 = a(0 - 300)^2 + 10
Now, let's simplify this equation:
110 = 90000a + 10
90000a = 100
a = 100/90000
a = 1/900
So, our equation for the main cables becomes:
y = (1/900)(x - 300)^2 + 10
To find the lengths of the vertical support cables, we need to calculate the y-values at different x-values.
Let's start by finding the y-value when x = 0 (the leftmost tower):
y = (1/900)(0 - 300)^2 + 10
y = (1/900)(90000) + 10
y = 100 + 10
y = 110
Now, let's move 50 ft to the right and find the next y-value:
y = (1/900)(50 - 300)^2 + 10
y = (1/900)((-250)^2) + 10
y = (1/900)(62500) + 10
y = 69.44 + 10
y ≈ 79.44
So, the length of the first vertical support cable is 79.44 ft.
We can repeat this process every 50 ft along the roadway until we reach the rightmost tower.
To determine the lengths of the vertical support cables, we need to find the equation of the parabola formed by the main cables of the suspension bridge.
Given:
Length between towers (base of parabola): 600 ft
Height of attachment point to the towers: 110 ft
Lowest point of the main cables above the ground: 10 ft
Using this information, we can determine the equation of the parabola.
Step 1: Find the vertex of the parabola.
The vertex of a parabola in standard form is given by the coordinates (h, k), where h is the x-coordinate of the vertex and k is the y-coordinate of the vertex.
Since the parabola is symmetric and the towers are 600 ft apart, the vertex is located at the midpoint between the towers. Therefore, h = (600 / 2) = 300.
The y-coordinate of the vertex is the highest point of the parabola, which is 110 ft above the road. Therefore, k = 110.
So, the vertex of the parabola is (300, 110).
Step 2: Write the equation of the parabola in vertex form.
The vertex form of a parabola is given by y = a(x - h)^2 + k, where (h, k) is the vertex.
Using the vertex (300, 110), the equation becomes y = a(x - 300)^2 + 110.
Step 3: Use additional information to find the value of 'a'.
We are given that the lowest point of the main cables above the ground is 10 ft.
Substituting the coordinate (0, 10) into the equation, we get:
10 = a(0 - 300)^2 + 110
10 = 90000a + 110
90000a = -100
a = -100 / 90000
a = -1/900
Step 4: Substitute the value of 'a' into the equation.
The equation becomes y = (-1/900)(x - 300)^2 + 110.
Step 5: Find the lengths of the vertical support cables.
The vertical support cables are placed at 50 ft intervals along the level roadway. This means that we need to find the length of the cable at x = 0, x = 50, x = 100, etc., until x = 600 (the end of the base of the parabola).
To find the length of the cable at a given x-value, we substitute the x-value into the equation and evaluate y.
When x = 0:
y = (-1/900)(0 - 300)^2 + 110
y = (-1/900)(90000) + 110
y = -100 + 110
y = 10
So, the length of the cable at x = 0 is 10 ft.
When x = 50:
y = (-1/900)(50 - 300)^2 + 110
y = (-1/900)(25000) + 110
y = -27.78 + 110
y ≈ 82.22
So, the length of the cable at x = 50 is approximately 82.22 ft.
Similarly, we can substitute values for x = 100, x = 150, and so on, until x = 600, and evaluate y to find the lengths of the vertical support cables at those points.
To find the lengths of the vertical support cables, we need to determine the shape of the main cables. Since the main cables are uniformly distributing the weight of the bridge and are in the form of a parabola, we can use the equation of a parabola in vertex form:
y = a(x - h)^2 + k
In this equation, (h, k) represents the coordinates of the vertex (the lowest point of the main cables). We know that the cables are 10 ft above the ground at their lowest points, so the vertex is at (0, 10).
Plugging in the values of the vertex (h, k), the equation becomes:
y = a(x - 0)^2 + 10
y = ax^2 + 10
Now we need to find the value of 'a' to fully define the parabola. To do that, we can use the known points where the cables are attached to the towers.
The cables are attached to towers that are 600 ft apart, and they are attached to towers at a height of 110 ft above the road. We can use the coordinates of the tower attachment points (-300, 110) and (300, 110) to find 'a':
110 = a(-300)^2 + 10
110 = 90000a + 10
90000a = 100
a = 100 / 90000
a ≈ 0.0011111
Now that we have determined 'a', we can find the lengths of the vertical support cables by considering the ascending values of 'y' at 50 ft intervals along the level roadway.
To find the lengths, we need to find the corresponding 'x' values by rearranging the equation:
y = ax^2 + 10
Let's solve for 'x':
ax^2 = y - 10
x^2 = (y - 10) / a
x = sqrt((y - 10) / a)
Now we can find the lengths of the vertical support cables at every 50 ft interval along the road by plugging in different values of 'y' into the equation:
For y = 50:
x = sqrt((50 - 10) / 0.0011111)
x ≈ 15.49 ft
For y = 100:
x = sqrt((100 - 10) / 0.0011111)
x ≈ 31.02 ft
For y = 150:
x = sqrt((150 - 10) / 0.0011111)
x ≈ 47.10 ft
Continuing this process at every 50 ft interval along the road, you can calculate the lengths of the vertical support cables.