Chapter 6.3, Problem 103E

To determine

To find: The tension in the cable of each wire.

Answer to Problem 103E

  $193,901\text{ lbf and 151,744 lbf}$

Explanation of Solution

Given: The cranes shown in the figure are lifting an object that weighs 20,240 pounds.

  

First divide vertical and horizontal component of each cable.

  

Now, the vertical component is in equilibrium.

  $\therefore T_1\sin {44.5}^{\circ }+T_2\sin {24.3}^{\circ }=20240g\mathrm{...}\left(i\right)$

Now, the horizontal component is in equilibrium.

  $\therefore T_1\cos {44.5}^{\circ }=T_2\cos {24.3}^{\circ }\mathrm{...}\left(ii\right)$

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

  $\begin{array}{l}{T}_1\sin {44.5}^{\circ }+\frac{T_1\cos {44.5}^{\circ }}{\cos {24.3}^{\circ }}\sin {24.3}^{\circ }=20240\left(10\right)\\ T_1=\frac{202400\cos {24.3}^{\circ }}{\cos {24.3}^{\circ }\sin {44.5}^{\circ }+\cos {44.5}^{\circ }\sin {24.3}^{\circ }}\\ T_1\approx 193901\end{array}$

  $\begin{array}{l}{T}_2=193901\cdot \frac{\cos {44.5}^{\circ }}{\cos {24.3}^{\circ }}\\ T_2\approx 151744\end{array}$

Hence, the tension in cranes wire is $193,901\text{ lbf and 151,744 lbf}$ .