Chapter 9.4, Problem 87AYU

To determine

To find: thetensions in each part of the rope.

Answer to Problem 87AYU

  $T_1=513.90,T_2=658.44$ .

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:

Let the trensions in string be $F_1\text{ and }{F}_2$

And the tension in tightrope walker is $F_3$

  $\begin{array}{l}{F}_1=\left|F_1\right|\left(\cos {3.7}^{\circ }i+\sin {3.7}^{\circ }j\right)\\ F_1=\left|F_1\right|\left(0.99i+0.064j\right)\end{array}$

Also,

  $\begin{array}{l}{F}_2=\left|F_2\right|\left(\cos {175.8}^{\circ }i+\sin {175.8}^{\circ }j\right)\\ F_2=\left|F_2\right|\left(-0.99i+0.07j\right)\end{array}$

Also,

  $F_3=-150j$

For the static equilibrium sum of all forces should be 0.

  $\begin{array}{l}{F}_1+F_2+F_3=0\\ \left|F_1\right|\left(0.99i+0.064j\right)+\left|F_2\right|\left(-0.99i+0.07j\right)-150j=0\\ \left\{\text{equating }i\text{ and }j\text{ components equal to zero}\right\}\\ \left|F_1\right|0.997915-\left|F_2\right|0.997314=0\\ \left|F_1\right|0.06453-\left|F_2\right|0.073281=0\end{array}$

Solving for $\left|F_1\right|\text{ and }\left|F_2\right|$

  $\begin{array}{l}\left|F_1\right|=1088.4lb\\ \left|F_2\right|=1089.1lb\end{array}$

Hence, the tensions in the two cables is $\left|F_1\right|=1088.4lb\text{ and }\left|F_2\right|=1089.1lb$ .