The 4.0 m long bar shown in the figure is fixed so that it can rotate freely around point O at the end of the bar. Three forces are applied along the bar at the positions and angles shown. Use the convention that a counter-clockwise spin is positive. Which of the following forces create a counterciockwise (positive) torque? Mark all that apply. 27 N 16 N 30 45° +3.0 m→ .4.0 m. 33 N

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### Analyzing Torque on a Rotating Bar

The diagram illustrates a 4.0-meter long bar that is fixed so it can rotate freely around point O located at one end of the bar. Three forces are applied at various points and angles along the bar as specified in the figure.

#### Forces and Angles:
1. **16 N Force**: Applied at 3.0 meters from point O at an angle of 30° to the left (counterclockwise direction).
2. **27 N Force**: Applied at 4.0 meters from point O and directed straight upwards.
3. **33 N Force**: Applied at point O, at an angle of 45° to the right (clockwise direction).

#### Objective:
To determine which of the forces create a counterclockwise (positive) torque about point O.

#### Detailed Analysis:
- **16 N Force**: 
  - Torque is calculated as \( \tau = r \cdot F \cdot \sin(\theta) \)
  - Here \( r = 3.0 \) meters, \( F = 16 \) N, and \( \theta = 30^\degree \)
  - The direction is counterclockwise, resulting in a positive torque.

- **27 N Force**: 
  - Applied perpendicularly at a distance of 4.0 meters
  - Equation: \( \tau = r \cdot F \cdot \sin(90^\degree) \)
  - Here \( r = 4.0 \) meters, \( F = 27 \) N
  - The force causes an upwards push, leading to a counterclockwise (positive) torque.

- **33 N Force**:
  - Applied at the pivot point (O) which means \( r = 0 \) meters.
  - Regardless of the angle, the torque is zero since distance \( r \) is zero.
  - Hence, it does not create any torque.

#### Conclusion:
The forces that create a counterclockwise (positive) torque are the:
- 16 N force
- 27 N force

#### Options:
- [ ] The 33 N force
- [x] The 16 N force
- [x] The 27 N force
- [ ] None
Transcribed Image Text:### Analyzing Torque on a Rotating Bar The diagram illustrates a 4.0-meter long bar that is fixed so it can rotate freely around point O located at one end of the bar. Three forces are applied at various points and angles along the bar as specified in the figure. #### Forces and Angles: 1. **16 N Force**: Applied at 3.0 meters from point O at an angle of 30° to the left (counterclockwise direction). 2. **27 N Force**: Applied at 4.0 meters from point O and directed straight upwards. 3. **33 N Force**: Applied at point O, at an angle of 45° to the right (clockwise direction). #### Objective: To determine which of the forces create a counterclockwise (positive) torque about point O. #### Detailed Analysis: - **16 N Force**: - Torque is calculated as \( \tau = r \cdot F \cdot \sin(\theta) \) - Here \( r = 3.0 \) meters, \( F = 16 \) N, and \( \theta = 30^\degree \) - The direction is counterclockwise, resulting in a positive torque. - **27 N Force**: - Applied perpendicularly at a distance of 4.0 meters - Equation: \( \tau = r \cdot F \cdot \sin(90^\degree) \) - Here \( r = 4.0 \) meters, \( F = 27 \) N - The force causes an upwards push, leading to a counterclockwise (positive) torque. - **33 N Force**: - Applied at the pivot point (O) which means \( r = 0 \) meters. - Regardless of the angle, the torque is zero since distance \( r \) is zero. - Hence, it does not create any torque. #### Conclusion: The forces that create a counterclockwise (positive) torque are the: - 16 N force - 27 N force #### Options: - [ ] The 33 N force - [x] The 16 N force - [x] The 27 N force - [ ] None
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