Chapter 8.5, Problem 29E

To determine

To find: the magnitude of tension that is being exerted on each part of cable.

Answer to Problem 29E

760 pounds and 761 pounds

Explanation of Solution

Given information:

Total weight of performer and unicyclist = 155 pound

The figure representing the data of an unicyclist performing on a tightrope at a circus is shown below.

  

Calculation:

Analyze either side of the unicyclist as shown below.

  $\begin{array}{c}\overrightarrow{F_1}\cos 174.5°+\overrightarrow{F_2}\cos 6.2°=0\mathrm{...}\left(1\right)\\ \overrightarrow{F_1}\sin 174.5°+\overrightarrow{F_2}\sin 6.2°-155=0\mathrm{...}\left(2\right)\end{array}$

Solve equation (1) as shown below.

  $\begin{array}{c}\overrightarrow{F_1}\cos 174.5°+\overrightarrow{F_2}\cos 6.2°=0\\ \overrightarrow{F_2}\cos 6.2°=-\overrightarrow{F_1}\cos 174.5°\\ \overrightarrow{F_2}=-\overrightarrow{F_1}\left(\frac{\cos 174.5°}{\cos 6.2°}\right)\mathrm{...}\left(3\right)\end{array}$

Substitute the value of $\overrightarrow{F_2}$ in equation (2) and solve $\overrightarrow{F_1}$ as shown below.

  $\begin{array}{c}\overrightarrow{F_1}\sin 174.5°+\left(-\overrightarrow{F_1}\left(\frac{\cos 174.5°}{\cos 6.2°}\right)\right)\sin 6.2°-155=0\\ \overrightarrow{F_1}\sin 174.5°-\overrightarrow{F_1}\left(\cos 174.5°\right)\left(\tan 6.2°\right)-155=0\\ \overrightarrow{F_1}\left[\sin 174.5°-\left(\cos 174.5°\right)\left(\tan 6.2°\right)\right]=155\\ \overrightarrow{F_1}=\frac{155}{\sin 174.5°-\left(\cos 174.5°\right)\left(\tan 6.2°\right)}\\ =760\text{ pounds}\end{array}$

Substitute the value of $\overrightarrow{F_1}$ in equation (3) and solve $\overrightarrow{F_2}$ as shown below.

  $\begin{array}{c}\overrightarrow{F_2}=-\overrightarrow{F_1}\left(\frac{\cos 174.5°}{\cos 6.2°}\right)\\ =\left(\text{760 pounds}\right)\left(\frac{\cos 174.5°}{\cos 6.2°}\right)\\ \approx 761\text{ pounds}\end{array}$

Thus the magnitude of tension (in pounds) that is being exerted on each part of cable is 760 and 761.