A uniform horizontal strut weighs 440.0 N. One end of the strut is attached to a hinged support at the wall, and the other end of the strut is attached to a sign that weighs 150.0 N. The strut is also supported by a cable attached between the end of the strut and the wall at an angle of 30 degrees with respect to the strut. Assume that the entire weight of the sign is attached at the very end of the strut.

(a) Find the tension in the cable (in N).

(b) Find the force (in N) at the hinge of the strut. (Find the magnitude and give the direction in degrees counterclockwise from the +x-axis. Assume that the +x-axis is to the right.)




To solve this problem, we'll use the principles of static equilibrium, which state that the sum of all forces and the sum of all torques (moments) in a system at rest must be zero. 

### Given:
- Weight of the strut (\( W_s \)) = 440.0 N
- Weight of the sign (\( W_{sign} \)) = 150.0 N
- Angle of the cable with respect to the strut (\( \theta \)) = 30 degrees

### Part (a): Finding the Tension in the Cable

1. **Torque Calculation**:
   - Choose the point where the strut is hinged to the wall as the pivot point.
   - The torque due to the weight of the strut (\( W_s \)) acts at the center of the strut, which is at half the length of the strut.
   - The torque due to the weight of the sign (\( W_{sign} \)) acts at the end of the strut.
   - The torque due to the tension in the cable (\( T \)) acts at the end of the strut and is directed upwards.

   The sum of torques about the hinge must be zero for rotational equilibrium:

   \[
   T \cdot L \cdot \sin(\theta) - W_s \cdot \frac{L}{2} - W_{sign} \cdot L = 0
   \]

   Where \( L \) is the length of the strut (which will cancel out). Simplifying:

   \[
   T \cdot \sin(\theta) = \frac{W_s}{2} + W_{sign}
   \]

   \[
   T = \frac{W_s/2 + W_{sign}}{\sin(\theta)}
   \]

2. **Substitute the values**:

   \[
   T = \frac{440.0 \, \text{N}/2 + 150.0 \, \text{N}}{\sin(30^\circ)}
   \]

   \[
   T = \frac{220.0 \, \text{N} + 150.0 \, \text{N}}{0.5}
   \]

   \[
   T = \frac{370.0 \, \text{N}}{0.5} = 740.0 \, \text{N}
   \]

So, the tension in the cable is **740.0 N**.

### Part (b): Finding the Force at the Hinge

1. **Force Equilibrium in the x-direction**:
   - The force at the hinge in the x-direction is \( H_x \).
   - The horizontal component of the tension in the cable is \( T \cos(\theta) \).

   \[
   H_x = T \cos(\theta)
   \]

   \[
   H_x = 740.0 \, \text{N} \cdot \cos(30^\circ)
   \]

   \[
   H_x = 740.0 \, \text{N} \cdot \frac{\sqrt{3}}{2} = 640.69 \, \text{N}
   \]

2. **Force Equilibrium in the y-direction**:
   - The force at the hinge in the y-direction is \( H_y \).
   - The sum of vertical forces: \( H_y \) + the vertical component of the tension in the cable \( T \sin(\theta) \) should balance the weight of the strut and the sign:

   \[
   H_y + T \sin(\theta) = W_s + W_{sign}
   \]

   \[
   H_y + 740.0 \, \text{N} \cdot \sin(30^\circ) = 440.0 \, \text{N} + 150.0 \, \text{N}
   \]

   \[
   H_y + 370.0 \, \text{N} = 590.0 \, \text{N}
   \]

   \[
   H_y = 590.0 \, \text{N} - 370.0 \, \text{N} = 220.0 \, \text{N}
   \]

3. **Resultant Force at the Hinge**:
   - The magnitude of the force at the hinge:

   \[
   H = \sqrt{H_x^2 + H_y^2}
   \]

   \[
   H = \sqrt{(640.69 \, \text{N})^2 + (220.0 \, \text{N})^2}
   \]

   \[
   H = \sqrt{410482.08 + 48400.0} \, \text{N}
   \]

   \[
   H = \sqrt{458882.08} \, \text{N} \approx 677.4 \, \text{N}
   \]

4. **Direction of the Force at the Hinge**:
   - The angle \( \alpha \) with respect to the +x-axis:

   \[
   \alpha = \tan^{-1}\left(\frac{H_y}{H_x}\right)
   \]

   \[
   \alpha = \tan^{-1}\left(\frac{220.0 \, \text{N}}{640.69 \, \text{N}}\right)
   \]

   \[
   \alpha = \tan^{-1}(0.3434) \approx 18.95^\circ
   \]

So, the force at the hinge has a magnitude of **677.4 N** and a direction of **18.95° counterclockwise from the +x-axis**.

---

Would you like further details or have any questions? 

Here are 5 related questions you might consider:
1. How would the tension change if the angle of the cable were different?
2. What if the weight of the sign was increased? How would it affect the forces?
3. How does the length of the strut affect the torque calculations?
4. What assumptions are made in this type of static equilibrium problem?
5. How would the results change if the hinge were not frictionless?

**Tip:** Always start by identifying all the forces and their points of application when solving equilibrium problems. This makes setting up equations for force and torque balances more straightforward.

Learn how to solve static equilibrium problems involving a uniform horizontal strut attached to a hinged support and a sign, including calculating tension in a cable and force at the hinge. Detailed step-by-step solution provided.

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