The body in the figure is pivoted at O. Three forces act on it in the directions shown: FA-5.80 N at point A, 9.40 m from O: F-15.0N at point B, 4.80 m from O; and Fe=19.0 N at point C, 1.40 m from O. Taking the clockwise direction to be negative, what is the net torque about O? Number Units
The body in the figure is pivoted at O. Three forces act on it in the directions shown: FA-5.80 N at point A, 9.40 m from O: F-15.0N at point B, 4.80 m from O; and Fe=19.0 N at point C, 1.40 m from O. Taking the clockwise direction to be negative, what is the net torque about O? Number Units
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Transcribed Image Text:### Net Torque Calculation
The body in the figure is pivoted at point O. Three forces act on it in the directions shown:
- \(F_A = 5.80 \, \text{N}\) at point A, which is 9.40 m from O.
- \(F_B = 15.0 \, \text{N}\) at point B, which is 4.80 m from O.
- \(F_C = 19.0 \, \text{N}\) at point C, which is 1.40 m from O.
We are to determine the net torque about point O, considering the clockwise direction as negative.
### Diagram Explanation:
The diagram displayed shows an irregularly shaped body pivoted at point O.
- Point A is on the body with a force \(F_A\) acting in a direction specified in the diagram.
- Point B is on the body with a force \(F_B\) acting in a different direction.
- Point C is on the body with a force \(F_C\) acting in yet another direction.
- Each force is shown with its respective angle and distance from O.
### Torque Calculation:
1. **Torque from \(F_A\) (Counterclockwise - positive direction):**
- \( \tau_A = F_A \times d_A \times \sin(\theta_A) \)
- \( d_A = 9.40 \, \text{m} \)
- Assume \(\theta_A\) is the angle given relative to the lever arm.
- Assume from diagram: \( \theta_A = 130^\circ \)
2. **Torque from \(F_B\) (Counterclockwise - positive direction):**
- \( \tau_B = F_B \times d_B \times \sin(\theta_B) \)
- \( d_B = 4.80 \, \text{m} \)
- Assume \(\theta_B\) is the angle given relative to the lever arm.
- Assume from diagram: \( \theta_B = 180^\circ - 60^\circ \) (120 degrees)
3. **Torque from \(F_C\) (Clockwise - negative direction):**
- \( \tau_C = F_C \times d_C \times \sin(\theta_C) \)
- \( d_C = 1.40 \, \text
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