Outside of a diner you notice a precarious 10 kg sign hanging above the entrance, supported by a 12 kg, 0.9 m long support beam. The beam is attached to the sign at one end, and a cable two-thirds of the way along the beam. 1. Draw a complete rigid body diagram for the support beam. 2. Determine the tension in the cable and the magnitude of force at the wall.

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**Physics Problem: Support Beam Analysis Outside of a Diner**

**Scenario:**
Outside of a diner, you notice a precarious 10 kg sign hanging above the entrance, supported by a 12 kg, 0.9 m long support beam. The beam is attached to the sign at one end, and a cable two-thirds of the way along the beam.

**Tasks:**
1. Draw a *complete* rigid body diagram for the support beam.
2. Determine the tension in the cable and the magnitude of force at the wall.

**Diagram Explanation:**
The diagram features the following components:
- A vertical brick wall on the left side.
- A diagonal support beam extending from the wall to the right, making a 40° angle with the horizontal ground.
- A cable connected from two-thirds along the beam to the wall, providing support.
- A circular sign labeled "Mom's Diner" hanging vertically from the far end of the beam.

### Task 1: Rigid Body Diagram

To draw the rigid body diagram:
1. Identify all forces acting on the beam:
   - Weight of the beam (W_beam = m_beam * g), acting at its center of gravity (midpoint).
   - Weight of the sign (W_sign = m_sign * g), acting at the far end where the sign is attached.
   - Tension in the cable (T), pulling upwards at the point where the cable is attached to the beam.
   - Reaction forces at the wall (vertical and horizontal components: R_vertical and R_horizontal).

2. Represent the beam as a straight line, marking:
   - Points of force application.
   - Directions of each force.

### Task 2: Calculations

#### Determine the Tension in the Cable:

Given:
- Weight of the sign, W_sign = 10 kg * 9.8 m/s² = 98 N.
- Weight of the beam, W_beam = 12 kg * 9.8 m/s² = 117.6 N.
- Length of the beam, L = 0.9 m.

Using static equilibrium conditions:

1. Sum of forces in the vertical direction:
   - \( R_{\text{vertical}} + T \sin(40^\circ) = W_{\text{beam}} + W_{\text{sign}} \)

2. Sum of forces in the horizontal direction:
   - \( R_{\text{horizontal}} = T
Transcribed Image Text:**Physics Problem: Support Beam Analysis Outside of a Diner** **Scenario:** Outside of a diner, you notice a precarious 10 kg sign hanging above the entrance, supported by a 12 kg, 0.9 m long support beam. The beam is attached to the sign at one end, and a cable two-thirds of the way along the beam. **Tasks:** 1. Draw a *complete* rigid body diagram for the support beam. 2. Determine the tension in the cable and the magnitude of force at the wall. **Diagram Explanation:** The diagram features the following components: - A vertical brick wall on the left side. - A diagonal support beam extending from the wall to the right, making a 40° angle with the horizontal ground. - A cable connected from two-thirds along the beam to the wall, providing support. - A circular sign labeled "Mom's Diner" hanging vertically from the far end of the beam. ### Task 1: Rigid Body Diagram To draw the rigid body diagram: 1. Identify all forces acting on the beam: - Weight of the beam (W_beam = m_beam * g), acting at its center of gravity (midpoint). - Weight of the sign (W_sign = m_sign * g), acting at the far end where the sign is attached. - Tension in the cable (T), pulling upwards at the point where the cable is attached to the beam. - Reaction forces at the wall (vertical and horizontal components: R_vertical and R_horizontal). 2. Represent the beam as a straight line, marking: - Points of force application. - Directions of each force. ### Task 2: Calculations #### Determine the Tension in the Cable: Given: - Weight of the sign, W_sign = 10 kg * 9.8 m/s² = 98 N. - Weight of the beam, W_beam = 12 kg * 9.8 m/s² = 117.6 N. - Length of the beam, L = 0.9 m. Using static equilibrium conditions: 1. Sum of forces in the vertical direction: - \( R_{\text{vertical}} + T \sin(40^\circ) = W_{\text{beam}} + W_{\text{sign}} \) 2. Sum of forces in the horizontal direction: - \( R_{\text{horizontal}} = T
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