To solve this problem, we’ll analyze the forces and use equilibrium conditions, given that the beam is in static equilibrium (no net force or moment acting on it).

Given Information:

• Mass of the beam, $m = 39.2 \, \text{kg}$
• Angle of inclination of the beam with respect to horizontal, $\theta = 33.5^\circ$
• The cable makes a 90° angle with the beam.

Steps:

1. Calculate the Weight of the Beam: The weight $W$ of the beam is: $W = mg$ where $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity). $W = 39.2 \times 9.81 = 384.552 \, \text{N}$

2. Identify Forces:

   • Weight $W$ acts vertically downward at the beam’s center of mass.
   • Tension $T$ in the cable acts vertically upward (since it’s perpendicular to the beam).
   • Reaction forces at the hinge include:
     • $F_h$: horizontal component of the hinge force.
     • $F_v$: vertical component of the hinge force.

3. Torque Calculation: Since the system is in equilibrium, we’ll set the sum of torques around the hinge equal to zero (to eliminate hinge forces).

   Placing the hinge as the pivot point:

   • Distance of the center of mass of the beam from the hinge: $\frac{L}{2}$
   • Torque due to the weight: $\tau_W = W \cdot \frac{L}{2} \cdot \cos(\theta)$
   • Torque due to tension (counteracting the weight’s torque): $\tau_T = T \cdot L \cdot \sin(\theta)$

   Since $\sum \tau = 0$: $T \cdot L \cdot \sin(\theta) = W \cdot \frac{L}{2} \cdot \cos(\theta)$ Simplifying by dividing both sides by $L$: $T = \frac{W \cdot \frac{1}{2} \cdot \cos(\theta)}{\sin(\theta)}$ Substituting $W = 384.552 \, \text{N}$ and $\theta = 33.5^\circ$: $T = \frac{384.552 \cdot 0.5 \cdot \cos(33.5^\circ)}{\sin(33.5^\circ)}$

4. Horizontal Component of the Hinge Force $F_h$: Since there is no horizontal force component in the weight and the vertical component of the tension balances the vertical component of the hinge reaction,