Chapter 6.3, Problem 104E

To determine

To find: The tension in the cable.

Answer to Problem 104E

  $15,917\text{ lbf and 38,251 lbf}$

Explanation of Solution

Given: As shown in figure.

  

First divide vertical and horizontal component of each cable.

  

In $\Delta ACD$

  $\begin{array}{l}\cos \alpha =\frac{24}{\sqrt{{24}^2+{10}^2}}=\frac{24}{26}=\frac{12}{13}\\ \sin \alpha =\frac{10}{13}\end{array}$

In $\Delta BCD$

  $\begin{array}{l}\cos \beta =\frac{24}{\sqrt{{24}^2+{20}^2}}=\frac{24}{\sqrt{976}}=\frac{12}{13}\\ \sin \beta =\frac{10}{\sqrt{976}}\end{array}$

Now, the vertical component is in equilibrium.

  $\therefore T_1\cos \alpha +T_2\cos \beta =5000g\mathrm{...}\left(i\right)$

Now, the horizontal component is in equilibrium.

  $\therefore T_1\sin \alpha =T_2\sin \beta \mathrm{...}\left(ii\right)$

Using $\left(i\right)\text{ and }\left(ii\right)$

  $\begin{array}{l}{T}_1\cdot \frac{12}{13}+T_1\cdot \frac{10}{13}\cdot \frac{\sqrt{976}}{10}\cdot \frac{12}{13}=5000\left(10\right)\\ T_1=\frac{50000\cdot 13\cdot 13\cdot 10}{12\cdot 10\cdot 13+10\cdot 12\cdot \sqrt{976}}\\ T_1\approx 15917\end{array}$

  $\begin{array}{l}{T}_2=15917\cdot \frac{10}{13}\cdot \frac{\sqrt{976}}{10}\\ T_2\approx 38251\end{array}$

Hence, the tension in cranes wire is $15,917\text{ lbf and 38,251 lbf}$ .