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
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
College Physics
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24666b77-75da-49bd-bf84-8c4674dedd2c%2F4667570b-35a3-47c1-966c-95618d2462d0%2Ftntevd.png&w=3840&q=75)
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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