A uniform horizontal strut weighs 350 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 125 N. The strut is also supported by a cable attached between the end of the strut and the wall. Assuming that the entire weight of the sign is attached at the very end of the strut, find the tension in the cable and the force at the hinge of the strut.



To solve this problem, we need to apply the principles of static equilibrium. Since the system is in equilibrium, the sum of forces in both horizontal and vertical directions must be zero, and the sum of torques (moments) about any point must also be zero. Here's how we can break it down:

### Step 1: Define Variables
- \( W_s \): Weight of the strut = 350 N
- \( W_{sign} \): Weight of the sign = 125 N
- \( T \): Tension in the cable (what we need to find)
- \( F_H \): Force at the hinge (what we need to find)
  - \( F_{Hx} \): Horizontal component of the hinge force
  - \( F_{Hy} \): Vertical component of the hinge force
- \( L \): Length of the strut
- \( \theta \): Angle between the cable and the strut (let’s assume it's known for now)

### Step 2: Calculate the Torque
Choose the hinge as the pivot point to sum torques, as the force at the hinge will exert no torque about this point. The goal is to set the sum of torques equal to zero.

#### Forces contributing to the torque:
- **Weight of the strut**: Acts at the center of the strut, a distance \( \frac{L}{2} \) from the hinge.
  - Torque due to strut weight: \( W_s \cdot \frac{L}{2} \)
- **Weight of the sign**: Acts at the end of the strut, a distance \( L \) from the hinge.
  - Torque due to sign weight: \( W_{sign} \cdot L \)
- **Tension in the cable**: Since the cable is attached to the end of the strut, it creates a torque depending on the angle \( \theta \).
  - Torque due to tension: \( T \cdot L \cdot \sin(\theta) \)

By summing the torques about the hinge and setting them equal to zero:
\[
T \cdot L \cdot \sin(\theta) = W_s \cdot \frac{L}{2} + W_{sign} \cdot L
\]
Simplifying:
\[
T \cdot \sin(\theta) = \frac{W_s}{2} + W_{sign}
\]
Substitute the known values of \( W_s = 350 \, N \) and \( W_{sign} = 125 \, N \):
\[
T \cdot \sin(\theta) = \frac{350}{2} + 125 = 300 \, N
\]
Thus, the tension is:
\[
T = \frac{300}{\sin(\theta)}
\]
So the tension \( T \) in the cable depends on the angle \( \theta \). If the angle is given, we can calculate the exact value of the tension.

### Step 3: Sum of Forces
For static equilibrium, the sum of forces in both horizontal and vertical directions must be zero.

#### Vertical forces:
\[
F_{Hy} + T \cdot \sin(\theta) = W_s + W_{sign}
\]
Substitute the known weights:
\[
F_{Hy} + 300 = 475 \, N
\]
Thus, the vertical component of the hinge force is:
\[
F_{Hy} = 475 - 300 = 175 \, N
\]

#### Horizontal forces:
\[
F_{Hx} = T \cdot \cos(\theta)
\]
This relates the horizontal component of the hinge force to the tension in the cable.

### Final Results:
1. **Tension in the cable**: \( T = \frac{300}{\sin(\theta)} \)
2. **Vertical component of the hinge force**: \( F_{Hy} = 175 \, N \)
3. **Horizontal component of the hinge force**: \( F_{Hx} = T \cdot \cos(\theta) \)

To find the total hinge force \( F_H \), you would combine the horizontal and vertical components using Pythagoras' theorem:
\[
F_H = \sqrt{F_{Hx}^2 + F_{Hy}^2}
\]

Would you like to clarify the angle \( \theta \) or any other details to proceed further?

### Related Questions:
1. How does the length of the strut impact the tension in the cable?
2. What happens to the tension if the weight of the sign increases?
3. How would the force at the hinge change if the cable were attached at a different point along the strut?
4. Can you explain why the torque due to the hinge force is zero?
5. What would happen if there were no cable and the strut was supported only by the hinge?

### Tip:
Always start by summing the torques about a point where unknown forces act, to eliminate those forces from the equation and simplify your calculations.

A uniform horizontal strut weighs 350 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 125 N. The strut is also supported by a cable attached between the end of the strut and the wall. Assuming that the entire weight of the sign is attached at the very end of the strut, find the tension in the cable and the force at the hinge of the strut.

Find the tension in the cable and the force at the hinge of a uniform horizontal strut supporting a sign. This problem explores concepts of torque, static equilibrium, and trigonometry, providing a detailed solution for students in Grades 11-12.

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