 A 39.2 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90o. If the beam is inclined at an angle of θ
 = 33.5° with respect to horizontal.

What is the horizontal component of the force exerted by the hinge on the beam? (Use the `to the right' as + for the horizontal direction.)

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,

A 39.2 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90o. If the beam is inclined at an angle of θ = 33.5° with respect to horizontal. What is the horizontal component of the force exerted by the hinge on the beam? (Use the 'to the right' as + for the horizontal direction.)

Learn how to calculate the horizontal component of the hinge force for a beam supported by a cable at an angle. This solution involves key concepts of statics, torque, and equilibrium, ideal for high school and introductory college physics students.

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