Static Equilibrium of Suspended Stage Lighting Solution

In live entertainment production, heavy lighting trusses are suspended over stages. Entertainment engineers must calculate force components to ensure that the supporting cable rigs are in static equilibrium and do not exceed the safe load capacity of the rigging cables.

A lighting fixture weighing $1200\text{ N}$ is suspended from a single central ring. The ring is held in static equilibrium by two cables, Cable A and Cable B, attached to overhead beams.

• Cable A pulls upward and to the left at an angle of $45^\circ$ with the horizontal.
• Cable B pulls upward and to the right at an angle of $60^\circ$ with the horizontal.

Determine the tension magnitudes $T_A$ and $T_B$ in each cable.

Step-by-Step Solution:

1. Set Up the Force Vectors: Let the horizontal axis be the $x$-axis (positive to the right) and the vertical axis be the $y$-axis (positive upward). The forces acting on the central ring are:

   • Tension from Cable A: $\vec{T}_A = \langle -T_A \cos(45^\circ), T_A \sin(45^\circ) \rangle = \langle -T_A \frac{\sqrt{2}}{2}, T_A \frac{\sqrt{2}}{2} \rangle$
   • Tension from Cable B: $\vec{T}_B = \langle T_B \cos(60^\circ), T_B \sin(60^\circ) \rangle = \langle T_B \frac{1}{2}, T_B \frac{\sqrt{3}}{2} \rangle$
   • Gravity: $\vec{F}_g = \langle 0, -1200 \rangle$

2. Apply Static Equilibrium in the Horizontal Direction ($\sum F_x = 0$):

   $$-T_A \frac{\sqrt{2}}{2} + T_B \frac{1}{2} = 0 \implies T_B = \sqrt{2} T_A$$

3. Apply Static Equilibrium in the Vertical Direction ($\sum F_y = 0$):

   $$T_A \frac{\sqrt{2}}{2} + T_B \frac{\sqrt{3}}{2} - 1200 = 0 \implies T_A \frac{\sqrt{2}}{2} + T_B \frac{\sqrt{3}}{2} = 1200$$

4. Substitute $T_B = \sqrt{2} T_A$ into the Vertical Equation:

   $$T_A \frac{\sqrt{2}}{2} + (\sqrt{2} T_A) \frac{\sqrt{3}}{2} = 1200$$
   $$T_A \left( \frac{\sqrt{2} + \sqrt{6}}{2} \right) = 1200$$
   $$T_A = \frac{2400}{\sqrt{2} + \sqrt{6}}\text{ N}$$
   Evaluate the exact expression:
   $$T_A \approx \frac{2400}{1.4142 + 2.4495} = \frac{2400}{3.8637} \approx 621.17\text{ N}$$

5. Solve for $T_B$:

   $$T_B = \sqrt{2} T_A \approx 1.4142 \times 621.17 \approx 878.46\text{ N}$$

Meaning of the Answer & Real-Life Application:

Rigging safety codes require rigging equipment to have safety factors of at least 5:1 or 10:1. Determining the tension vectors acting on each cable ensures that riggers select cables rated for loads far exceeding these values (e.g., cables rated for at least $9\text{ kN}$), protecting crew and audiences underneath.