Question

A traffic light of mass m=12.5kg hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal (see figure below). Find the tension in the three cables.

Answer 1

The free-body diagram is as follows

According to sine law,

$$ \begin{aligned} \frac{m g}{\sin A} &=\frac{T_{2}}{\sin B}=\frac{T_{1}}{\sin C} \\ \frac{m g}{\sin \left(180^{\circ}-53^{\circ}-37^{\circ}\right)} &=\frac{T_{2}}{\sin \left(90^{\circ}+37^{\circ}\right)}=\frac{T_{1}}{\sin \left(90^{\circ}+53^{\circ}\right)} \\ \frac{m g}{\sin 90} &=\frac{T_{2}}{\sin 127}=\frac{T_{1}}{\sin 143} \end{aligned} $$

From the above equation,

$$ \begin{aligned} \frac{m g}{\sin 90} &=\frac{T_{1}}{\sin 143} \\ T_{1} &=\frac{m g(\sin 143)}{\sin 90} \\ &=\frac{(12.5 \mathrm{~kg})(\sin 143)}{\sin 90} \\ &=7.52 \mathrm{~N} \\ \frac{T_{2}}{\sin 127} &=\frac{T_{1}}{\sin 143} \\ \frac{T_{2}}{\sin 127} &=\frac{7.52 \mathrm{~N}}{\sin 143} \\ T_{2} &=9.98 \mathrm{~N} \end{aligned} $$

The tension in the third cable is,

$$ T_{3}=(12.5 \mathrm{~kg})(\sin 143)=7.52 \mathrm{~N} $$