Chapter 9, Problem 13P

To determine

The tension in the two wires supporting the traffic light shown in figure.

Answer to Problem 13P

Solution:

195.6 N and 260.8 N

Explanation of Solution

Given:

Mass of traffic light is 33 kg

Formula used:

From second law of motion, equilibrium condition is:

  ${\displaystyle \sum F=0}$

Weight, $W=mg$

Here m is mass of traffic light and g is acceleration due to gravity.

Calculation:

Using free body diagram

  

  $\begin{array}{c}mg=33\times 9.8\\ =323.4\text{N}\end{array}$

(vertically downwards) The hanging traffic light in the free body diagram is in equilibrium with three forces: $T_1,T_2\text{ and }W$ .

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ T_1\cos 37°=T_2\cos 53°\\ T_1=\frac{\cos 53°}{\cos 37°}{T}_2\\ {\displaystyle \sum F_y}=0\\ T_1\sin 37°+T_2\sin 53°=W\\ \frac{\cos 53°}{\cos 37°}{T}_2\times \sin 37°+T_2\sin 53°=323.4\\ 0.45T_2+0.79T_2=323.4\\ T_2=\frac{323.4}{1.24}\\ T_2=260.8\text{N}\\ T_1=\frac{\cos 53°}{\cos 37°}{T}_2\\ T_1=0.75\times 260.8\\ T_1=195.6\text{N}\end{array}$

Conclusion:

Thus, $T_1=195.6\text{N}$ and $T_2=260.8\text{N}$ .