Chapter 6.3, Problem 86E

To determine

The tension in each cable supporting the load.

Explanation of Solution

Given:

The diagram is given as:

  

The weight is hanging through the point A. The tension in the cable AB is T₁ and the tension in the cable AD is T₂ as shown in figure:

In the triangle ABC:

  $\begin{array}{c}\tan {\theta }_1=\frac{BC}{AC}\\ =\frac{10}{24}\\ {\theta }_1={\tan }^{-1}\left(\frac{5}{12}\right)\\ ={22.62}^o\end{array}$

Similarly, in the triangle ACD:

  $\begin{array}{c}\tan {\theta }_2=\frac{DC}{AC}\\ =\frac{20}{24}\\ {\theta }_2={\tan }^{-1}\left(\frac{5}{6}\right)\\ ={39.81}^o\end{array}$

As the weight is in the equilibrium, the horizontal components of the tension in the each rope will balance each other,

Hence,

  $\begin{array}{l}{T}_1\sin {\theta }_1=T_2\sin {\theta }_2\\ T_1\sin 22.62=T_2\sin 39.8\\ T_1=1.665T_2\mathrm{...............}\left(1\right)\end{array}$

Similarly, the vertical components of the tension in the both cable will balance the weight.

Hence,

  $\begin{array}{l}{T}_1\cos {\theta }_1+T_2\cos {\theta }_2=1000\\ T_1\cos 22.62+T_2\cos 39.81=1000\\ T_1+0.832T_2=\mathrm{1083.42.............}\left(2\right)\end{array}$

Solving the equation 1 and 2:

  $\begin{array}{l}1.665T_2+0.832T_2=1083.42\\ 2.497T_2=1083.42\\ T_2=433.89\text{lb}\end{array}$

Therefore, the result is:

  $\begin{array}{c}{T}_1=1.665\left(433.89\right)\\ =722.43\text{lb}\end{array}$