Chapter 9.4, Problem 88AYU

To determine

To find: thetensions in each part of the rope.

Answer to Problem 88AYU

  $T_r=1208.49\text{lbf and }{T}_l=1209.90\text{lbf}$ .

Explanation of Solution

Given:

Weight of the tight rope walker is 150 pounds.

A tight rope walker located at a certain point deflects the rope as shown in given figure:

  

Calculation:

As $\overrightarrow{W}=\overrightarrow{T_l}+\overrightarrow{T_r}$

Then

  $\begin{array}{l}{T}_l\cos 3.8=T_r\cos 2.6\\ W=T_l\sin 3.8+T_r\sin 2.6\end{array}$

First equation,

  $T_l=\frac{\cos 2.6}{\cos 3.8}{T}_r=1.00117T_r$

Second equation,

  $\begin{array}{l}135=\left(1.00117T_r\right)0.06627+T_{r}0.04536=0.11171T_r\\ T_r=\frac{135}{0.1171}=1208.49\text{lbf}\\ T_l=\left(1.00117\right)1208.49=1209.90\text{lbf}\end{array}$

Hence, the tensions in the two cables is $T_r=1208.49\text{lbf and }{T}_l=1209.90\text{lbf}$ .