Chapter 9.4, Problem 86AYU

To determine

Tofind:thetensions in the two cables.

Answer to Problem 86AYU

  $T_1=513.90,T_2=658.44$ .

Explanation of Solution

Given:

A weight of 800 pound is suspended from two cables as shown in figure.

Calculation:

  

From the figure,

  $\begin{array}{l}{T}_1\cos {35}^{\circ }=T_2\cos {50}^{\circ }\\ T_1\left(0.82\right)=T_2\left(0.64\right)\\ T_1=\frac{T_2\left(0.64\right)}{0.82}\\ T_1\sin {35}^{\circ }+T_2\sin {50}^{\circ }=800\\ T_1\left(0.57\right)+T_2\left(0.77\right)=800\\ \frac{T_2\left(0.64\right)}{0.82}\times 0.57+T_2\left(0.77\right)=800\\ \Rightarrow 1.215T_2=800\\ \Rightarrow T_2\approx 658.44\\ T_1=513.90\end{array}$

Hence, the tensions in the two cablesis $T_1=513.90,T_2=658.44$ .