A uniform rod is attached to a wall by a hinge at its base. The rod has a mass of 1.5 kg, a length of 1.3 m, is at an angle of 35° above the horizontal, and is held in place by a horizontal cord attached to the other end of the rod and bolted to the wall above the base of the rod.

To analyze this scenario, we’ll look into the forces and torques involved. We have:

1. **Weight of the rod (W)**: The gravitational force acts at the center of mass of the rod, which is located at the midpoint of its length. This force acts vertically downward.

2. **Tension in the cord (T)**: The horizontal cord exerts a force on the rod at its end, which counteracts the torque due to the weight of the rod.

3. **Reaction at the hinge (H)**: The hinge at the base provides both horizontal and vertical components of reaction forces.

We need to find:
- The tension \( T \) in the cord.
- The components of the hinge reaction force \( H_x \) and \( H_y \).

### Step 1: Calculate Weight of the Rod

The weight \( W \) of the rod can be calculated using:
\[
W = mg
\]
where \( m = 1.5 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
\[
W = 1.5 \times 9.8 = 14.7 \, \text{N}
\]

### Step 2: Set Up Torque Equation

To maintain equilibrium, the sum of torques about any point must be zero. Choosing the hinge as the pivot point allows us to ignore the hinge forces in the torque calculation.

1. **Torque due to the weight (W)**:
   - This force acts at the center of the rod, which is at a distance of \( \frac{L}{2} \) from the hinge.
   - The angle between the weight and the rod is \( 35^\circ \), so the perpendicular distance from the pivot is:
     \[
     d_W = \frac{L}{2} \cos(35^\circ)
     \]
   - Torque due to weight, \( \tau_W \):
     \[
     \tau_W = W \cdot d_W = 14.7 \times \frac{1.3}{2} \cos(35^\circ)
     \]

2. **Torque due to the tension (T)**:
   - The tension acts horizontally, so the full length of the rod (1.3 m) is the perpendicular distance.
   - Torque due to tension, \( \tau_T \):
     \[
     \tau_T = T \cdot L \sin(35^\circ)
     \]

Since the system is in equilibrium:
\[
\tau_T = \tau_W
\]
Solving this equation will give the value of \( T \).

### Step 3: Calculate Hinge Reactions

After finding \( T \), we can set up force balance equations for both the horizontal and vertical directions:

1. **Vertical Forces**:
   \[
   H_y - W = 0
   \]
2. **Horizontal Forces**:
   \[
   H_x - T = 0
   \]

Let’s go through each calculation systematically. Would you like the details calculated for each step? 

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### Related Questions
1. What if the angle was changed? How would this affect the tension?
2. How would we calculate the moment if the rod wasn't uniform?
3. What if the cord was at an angle instead of horizontal?
4. How would a longer rod affect the system’s equilibrium?
5. How would the reaction forces change if the hinge was frictionless?

**Tip**: Always use torque about a point where unknown forces are present to simplify your calculation.

Learn how to calculate the tension in a cord and hinge reactions in a 1.5 kg uniform rod at a 35° angle, held by a hinge at its base and a horizontal cord at the end. This problem involves torque equilibrium, trigonometry, and force balance, suitable for advanced high school physics students.

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