Chapter 6.3, Problem 89E

To determine

The tension in the cable for each crane.

Explanation of Solution

Given:

The cranes is shown below:

  

The weight of the object is 20240 pounds.

Considering the below diagram

  

Here, triangle is formed with the given information

Here angle ${65.7}^{\circ }={90}^{\circ }-{24.3}^{\circ }$ and ${68.8}^{\circ }={44.5}^{\circ }+{24.3}^{\circ }$

And third angle is obtained ${180}^{\circ }-\left({68.8}^{\circ }+{65.7}^{\circ }\right)={45.5}^{\circ }$

Now apply sine rule for the triangle $\Delta OPQ$ :

  $\begin{array}{c}\frac{OQ}{\sin \left({45.5}^{\circ }\right)}=\frac{OP}{\sin {65.7}^{\circ }}=\frac{20,240}{\sin {68.8}^{\circ }}\\ OQ=\frac{20,240\sin \left({45.5}^{\circ }\right)}{\sin {68.8}^{\circ }},OP=\frac{20,240\sin \left({65.7}^{\circ }\right)}{\sin {68.8}^{\circ }}\\ OQ\approx ~15484,OP\approx ~19,786\end{array}$

Hence, the tensions in the cable of cranes are $T_L=15,484\text{lb},T_R=19,786\text{lb}$ .