Question

the cable of a uniformly loaded suspension bridge hang in the form of a parabola .The roadway which is horizontal and 100 meter long is supported by vertical wires attached to the cable , the longest wire being 30 meter and the shortest being 6 meter .Find the length of a supporting wire attached to the roadway 18 meter from the middle

Answer 1

Answer on Question #53801 – Math– Algebra

Question

The cable of a uniformly loaded suspension bridge hang in the form of a parabola. The roadway which is horizontal and 100 meter long is supported by vertical wires attached to the cable, the longest wire being 30 meter and the shortest being 6 meter. Find the length of a supporting wire attached to the roadway 18 meter from the middle

Solution

Suppose that the bridge is a parabolic arc with the vertex $O$ (fig.1). The following information is given:

$AB = L = 100m$ is the length of the roadway;

$PB = OP / 2 = L / 2 = 50m;$

$OP = h = 6m$ is the shortest vertical wire;

$SB = H = 30m$ is the longest vertical wire;

$PQ = I = 18m$

Fig. 1

Let's write coordinates of the points R and S:

$$R (1 8, y), S (5 0, y ^ {\prime}).$$

Taking into account that $OP = O'B$ , we find the longest vertical wire above $x$ -axis:

$$S O ^ {\prime} = S B - O ^ {\prime} B = H - h = 3 0 - 6 = 2 4 m.$$

Hence, from (1), (2) we obtain $SO' = y' = 24m$ and

$$R (1 8, y), S (5 0, 2 4).$$

The equation of the parabola has the form

$$y = a x ^ {2}.$$

Find the parameter $a$ . Since the parabola passes through the point $S(50,24)$ , (4) gives that

$$S (5 0, 2 4): \quad 2 4 = a \cdot (5 0) ^ {2} \Rightarrow$$ $$a = \frac {2 4}{(5 0) ^ {2}} = \frac {2 4}{2 5 0 0} = \frac {6}{6 2 5}.$$

Further using (4) and (5) the ordinate $y$ of the point $R(18, y)$ on the parabola will be

$$R(18, y): \quad y = a \cdot (18)^2 = \frac{6}{625} \cdot 324 = \frac{1944}{625}.$$

The length of a supporting wire attached to the roadway 18m from the middle is given by

$$RQ = y + h = y + 6.$$

Substituting $y$ from (6) into (7) we get the required length:

$$RQ = \frac{1944}{625} + 6 = 3.1104 + 6 \approx 9.11m.$$

Answer: $RQ \approx 9.11m$ is the length of a supporting wire attached to the roadway 18 meter from the middle.

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