Chapter 7.5, Problem 7.148P

To determine

Find the distance a.

Answer to Problem 7.148P

The distance a is $\underset{\_}{\text{5}\text{.71}\text{ft}}$.

Explanation of Solution

Given information:

The length of the cable AB is $L=10\text{ft}$.

The value of angle $\theta$ is $45°$.

The collar at A is slides freely and the collar at B is prevented from the moving.

Calculation:

Show the free-body diagram of the cable assembly as in Figure 1.

Refer Equation 7.16 in the textbook.

Write the equation of the catenary cable as follows;

$y=c\cosh \frac{x}{c}$

Differentiate the equation with x;

$\frac{dy}{dx}=\sinh \frac{x}{c}$

The slope at point A is;

$\begin{array}{l}\tan \theta ={\left|\frac{dy}{dx}\right|}_A=\sinh \frac{x_A}{c}\\ \frac{x_A}{c}=\sinh \left(\tan \left(90°-\theta \right)\right)\\ x_A=c{\sinh }^{-1}\left(\tan \left(90°-\theta \right)\right)\end{array}$ (1)

The length of the portion AC is;

$AC=c\sinh \left(\frac{x_A}{c}\right)$

The length of the portion CB is;

$CB=c\sinh \left(\frac{x_B}{c}\right)$

Find the distance $x_B$ using the relation.

$L=AC+CB$

Substitute 10 ft for L, $c\sinh \left(\frac{x_A}{c}\right)$ for AC, and $c\sinh \left(\frac{x_B}{c}\right)$ for CB.

$\begin{array}{l}10=c\sinh \left(\frac{x_A}{c}\right)+c\sinh \left(\frac{x_B}{c}\right)\\ c\sinh \left(\frac{x_B}{c}\right)=10-c\sinh \left(\frac{x_A}{c}\right)\\ \sinh \left(\frac{x_B}{c}\right)=\frac{10}{c}-\sinh \left(\frac{x_A}{c}\right)\\ x_B=c{\sinh }^{-1}\left[\frac{10}{c}-\sinh \left(\frac{x_A}{c}\right)\right]\end{array}$ (2)

Find the distance $\left(y_A\right)$ using the relation.

$y_A=c\cosh \left(\frac{x_A}{c}\right)$ (3)

Find the distance $\left(y_B\right)$ using the relation.

$y_B=c\cosh \left(\frac{x_B}{c}\right)$ (4)

Consider the triangle ABD;

Find the value of $\tan \theta$ using the relation.

$\begin{array}{c}\tan \theta =\frac{\text{Opposite}\text{side}}{\text{Adjacent}\text{side}}\\ =\frac{y_B-y_A}{x_B+x_A}\end{array}$ (5)

Find the distance a using the relation.

$a=y_B-y_A$ (6)

Use the trial and error procedure to find the value of a.

Consider the value of c and for the given value of $\theta =45°$.

Find the angle $\theta$ in the equation (5). The calculated value of angle $\theta$ and the given value of $\theta =45°$ should be equal.

Trial 1:

Consider a trial value of 1.60 ft for c.

$c=1.60\text{ft}$

Substitute 1.60 ft for c and $45°$ for $\theta$ in Equation (1).

$\begin{array}{c}{x}_A=1.60\times {\sinh }^{-1}\left(\tan \left(90°-45°\right)\right)\\ =1.410\text{ft}\end{array}$

Substitute 1.60 ft for c and 1.410 ft for $x_A$ in Equation (2).

$\begin{array}{c}{x}_B=1.60\times {\sinh }^{-1}\left[\frac{10}{1.60}-\sinh \left(\frac{1.410}{1.60}\right)\right]\\ =3.777\text{ft}\end{array}$

Substitute 1.60 ft for c and 1.410 ft for $x_A$ in Equation (3).

$\begin{array}{c}{y}_A=1.60\times \cosh \left(\frac{1.410}{1.60}\right)\\ =2.263\text{ft}\end{array}$

Substitute 1.60 ft for c and 3.777 ft for $x_B$ in Equation (4).

$\begin{array}{c}{y}_B=1.60\times \cosh \left(\frac{3.777}{1.60}\right)\\ =8.551\text{ft}\end{array}$

Substitute 1.410 ft for $x_A$, 3.777 ft for $x_B$, 2.263 ft for $y_A$, and 8.551 ft for $y_B$ in Equation (5).

$\begin{array}{c}\tan \theta =\frac{8.551-2.263}{3.777+1.410}\\ \theta =50.483°\end{array}$

The calculated value of $\theta =50.483°$ is not equal to the given value of $\theta =45°$

Trial 2:

Consider a trial value of 1.70 ft for c.

$c=1.70\text{ft}$

Substitute 1.70 ft for c and $45°$ for $\theta$ in Equation (1).

$\begin{array}{c}{x}_A=1.70\times {\sinh }^{-1}\left(\tan \left(90°-45°\right)\right)\\ =1.498\text{ft}\end{array}$

Substitute 1.70 ft for c and 1.498 ft for $x_A$ in Equation (2).

$\begin{array}{c}{x}_B=1.70\times {\sinh }^{-1}\left[\frac{10}{1.70}-\sinh \left(\frac{1.498}{1.70}\right)\right]\\ =3.891\text{ft}\end{array}$

Substitute 1.70 ft for c and 1.498 ft for $x_A$ in Equation (3).

$\begin{array}{c}{y}_A=1.70\times \cosh \left(\frac{1.498}{1.70}\right)\\ =2.404\text{ft}\end{array}$

Substitute 1.70 ft for c and 3.891 ft for $x_B$ in Equation (4).

$\begin{array}{c}{y}_B=1.70\times \cosh \left(\frac{3.891}{1.70}\right)\\ =8.472\text{ft}\end{array}$

Substitute 1.498 ft for $x_A$, 3.891 ft for $x_B$, 2.404 ft for $y_A$, and 8.472 ft for $y_B$ in Equation (5).

$\begin{array}{c}\tan \theta =\frac{8.472-2.404}{3.891+1.498}\\ \theta =48.388°\end{array}$

The calculated value of $\theta =48.388°$ is not equal to the given value of $\theta =45°$

Trial 3:

Consider a trial value of 1.8652 ft for c.

$c=1.8652\text{ft}$

Substitute 1.8652 ft for c and $45°$ for $\theta$ in Equation (1).

$\begin{array}{c}{x}_A=1.8652\times {\sinh }^{-1}\left(\tan \left(90°-45°\right)\right)\\ =1.644\text{ft}\end{array}$

Substitute 1.8652 ft for c and 1.644 ft for $x_A$ in Equation (2).

$\begin{array}{c}{x}_B=1.8652\times {\sinh }^{-1}\left[\frac{10}{1.8652}-\sinh \left(\frac{1.644}{1.8652}\right)\right]\\ =4.064\text{ft}\end{array}$

Substitute 1.8652 ft for c and 1.644 ft for $x_A$ in Equation (3).

$\begin{array}{c}{y}_A=1.8652\times \cosh \left(\frac{1.644}{1.8652}\right)\\ =2.638\text{ft}\end{array}$

Substitute 1.8652 ft for c and 4.064 ft for $x_B$ in Equation (4).

$\begin{array}{c}{y}_B=1.8652\times \cosh \left(\frac{4.064}{1.8652}\right)\\ =8.346\text{ft}\end{array}$

Substitute 1.644 ft for $x_A$, 4.064 ft for $x_B$, 2.638 ft for $y_A$, and 8.346 ft for $y_B$ in Equation (5).

$\begin{array}{c}\tan \theta =\frac{8.346-2.638}{4.064+1.644}\\ \theta =45°\end{array}$

The calculated value of $\theta =45°$ is equal to the given value of $\theta =45°$

Therefore, the value of c is 1.8652 ft.

Substitute 2.638 ft for $y_A$, and 8.346 ft for $y_B$ in Equation (6).

$\begin{array}{c}a=8.346-2.638\\ =5.708\text{ft}\\ \approx \text{5}\text{.71}\text{ft}\end{array}$

Therefore, the distance a is $\underset{\_}{\text{5}\text{.71}\text{ft}}$.