Question

A suspension bridge with weight uniformly distributed along its length has twin towers that extend $75m$above the road surface and are $400m$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 $100m$ from the center. (Assume that the road is level.)

Short answer

The height of the cable at a point $100m$from the center is $18.75m$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

Distance between the two towers is $400m$.

The vertex form of the equation of a parabola facing upwards is $f\left(x\right)=a{\left(x-h\right)}^2+k......\left(i\right)$.

Where, $\left(h,k\right)$is the vertex of the parabola.

Draw the function,

Step 2. Substitute 0,0 in equation (i), followed by simplification.

We know that the points vertex of the parabola is located at $\left(0,0\right)$.

The points $\left(200,75\right)$and $\left(-200,75\right)$are on the graph of the cable.

Substitute $\left(0,0\right)$in equation (i),

$$\begin{gathered} f\left(x\right)=a{\left(x-0\right)}^2+0 \\ f\left(x\right)=a{x}^2......\left(ii\right) \end{gathered}$$

Substitute $f\left(200\right)=75$in equation (ii),

$$\begin{gathered} 75=a{\left(200\right)}^2 \\ a=\frac{75}{40000} \\ a=0.001875 \end{gathered}$$

Step 3. Find the required height of the cables.

Substitute the value of a in equation (ii),

$f\left(x\right)=0.001875x^2......\left(iii\right)$

Thus, the equation of the cable is $f\left(x\right)=0.001875x^2$.

Substitute $x=100$in equation (iii),

$$\begin{gathered} f\left(100\right)=0.001875{\left(100\right)}^2 \\ f\left(100\right)=0.001875\times 10000 \\ f\left(100\right)=18.75m \end{gathered}$$

Therefore, at the point $100m$from the center, the height of the cable is$18.75m$.