Chapter 7.5, Problem 7.147P

The 10-ft cable AB is attached to two collars as shown. The collar at A can slide freely along the rod; a stop attached to the rod prevents the collar at B from moving on the rod. Neglecting the effect of friction and the weight of the collars, determine the distance a.

Fig. P7.147

To determine

Find the distance a.

Answer to Problem 7.147P

The distance a is $\underset{\_}{3.50\text{ft}}$.

Explanation of Solution

Given information:

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

The value of angle $\theta$ is $30°$.

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 the 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 as follows:

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

The length of the portion CB is as follows:

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

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 $\theta =30°$, find the angle $\theta$ in the equation (5). The calculated value of angle $\theta$ and the given value of $\theta =30°$ 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 $30°$ for $\theta$ in Equation (1).

$\begin{array}{c}{x}_A=1.60\times {\sinh }^{-1}\left(\tan \left(90°-30°\right)\right)\\ =2.107\text{ft}\end{array}$

Substitute 1.60 ft for c and 2.107 ft for $x_A$ in Equation (2).

$\begin{array}{c}{x}_B=1.60\times {\sinh }^{-1}\left[\frac{10}{1.60}-\sinh \left(\frac{2.107}{1.60}\right)\right]\\ =3.541\text{ft}\end{array}$

Substitute 1.60 ft for c and 2.107 ft for $x_A$ in Equation (3).

$\begin{array}{c}{y}_A=1.60\times \cosh \left(\frac{2.107}{1.60}\right)\\ =3.20\text{ft}\end{array}$

Substitute 1.60 ft for c and 3.541 ft for $x_B$ in Equation (4).

$\begin{array}{c}{y}_B=1.60\times \cosh \left(\frac{3.541}{1.60}\right)\\ =7.404\text{ft}\end{array}$

Substitute 2.107 ft for $x_A$, 3.541 ft for $x_B$, 3.20 ft for $y_A$, and 7.404 ft for $y_B$ in Equation (5).

$\begin{array}{c}\tan \theta =\frac{7.404-3.20}{3.541+2.107}\\ \theta =36.658°\end{array}$

The calculated value of $\theta =36.658°$ is not equal to the given value of $\theta =30°$

Trial 2:

Consider a trial value of 1.70 ft for c.

$c=1.70\text{ft}$

Substitute 1.70 ft for c and $30°$ for $\theta$ in Equation (1).

$\begin{array}{c}{x}_A=1.70\times {\sinh }^{-1}\left(\tan \left(90°-30°\right)\right)\\ =2.239\text{ft}\end{array}$

Substitute 1.70 ft for c and 2.239 ft for $x_A$ in Equation (2).

$\begin{array}{c}{x}_B=1.70\times {\sinh }^{-1}\left[\frac{10}{1.70}-\sinh \left(\frac{2.239}{1.70}\right)\right]\\ =3.622\text{ft}\end{array}$

Substitute 1.70 ft for c and 2.239 ft for $x_A$ in Equation (3).

$\begin{array}{c}{y}_A=1.70\times \cosh \left(\frac{2.239}{1.70}\right)\\ =3.40\text{ft}\end{array}$

Substitute 1.70 ft for c and 3.622 ft for $x_B$ in Equation (4).

$\begin{array}{c}{y}_B=1.70\times \cosh \left(\frac{3.622}{1.70}\right)\\ =7.257\text{ft}\end{array}$

Substitute 2.239 ft for $x_A$, 3.622 ft for $x_B$, 3.40 ft for $y_A$, and 7.257 ft for $y_B$ in Equation (5).

$\begin{array}{c}\tan \theta =\frac{7.257-3.40}{3.622+2.239}\\ \theta =33.352°\end{array}$

The calculated value of $\theta =33.352°$ is not equal to the given value of $\theta =30°$

Trial 3:

Consider a trial value of 1.803 ft for c.

$c=1.803\text{ft}$

Substitute 1.803 ft for c and $30°$ for $\theta$ in Equation (1).

$\begin{array}{c}{x}_A=1.803\times {\sinh }^{-1}\left(\tan \left(90°-30°\right)\right)\\ =2.374\text{ft}\end{array}$

Substitute 1.803 ft for c and 2.374 ft for $x_A$ in Equation (2).

$\begin{array}{c}{x}_B=1.803\times {\sinh }^{-1}\left[\frac{10}{1.803}-\sinh \left(\frac{2.374}{1.803}\right)\right]\\ =3.694\text{ft}\end{array}$

Substitute 1.803 ft for c and 2.374 ft for $x_A$ in Equation (3).

$\begin{array}{c}{y}_A=1.803\times \cosh \left(\frac{2.374}{1.803}\right)\\ =3.606\text{ft}\end{array}$

Substitute 1.803 ft for c and 3.694 ft for $x_B$ in Equation (4).

$\begin{array}{c}{y}_B=1.803\times \cosh \left(\frac{3.694}{1.803}\right)\\ =7.110\text{ft}\end{array}$

Substitute 2.374 ft for $x_A$, 3.694 ft for $x_B$, 3.606 ft for $y_A$, and 7.11 ft for $y_B$ in Equation (5).

$\begin{array}{c}\tan \theta =\frac{7.11-3.606}{3.694+2.374}\\ \theta =30°\end{array}$

The calculated value of $\theta =30°$ is equal to the given value of $\theta =30°$

Therefore, the value of c is 1.803 ft.

Substitute 3.606 ft for $y_A$, and 7.11 ft for $y_B$ in Equation (6).

$\begin{array}{c}a=7.11-3.606\\ =3.50\text{ft}\end{array}$

Therefore, the distance a is $\underset{\_}{3.50\text{ft}}$.