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The problem presents a disk rotating counterclockwise with four forces acting on it, and asks for the net torque about the axle. Let's solve this step by step:

### Given Data:
- Diameter of the disk = 59.8 cm, so the radius \( r = \frac{59.8}{2} = 29.9 \) cm = 0.299 m
- Distance from the center to the forces (d) = 3.42 cm = 0.0342 m
- Forces:
  - \( F_1 = 45.8 \, \text{N} \)
  - \( F_2 = 68.3 \, \text{N} \)
  - \( F_3 = 68.3 \, \text{N} \)
  - \( F_4 = 45.8 \, \text{N} \)
- Angle for \( F_4 \) is \( 45^\circ \)
- Counterclockwise torque is taken as positive.

### Torque Formula:
The torque (\( \tau \)) due to a force is given by:
\[
\tau = r F \sin(\theta)
\]
where \( r \) is the perpendicular distance from the axis of rotation to the point of application of the force, \( F \) is the force, and \( \theta \) is the angle between the force and the radius.

### Torque due to each force:

1. **Force \( F_1 \)**:
   - Acts vertically downward (angle \( 270^\circ \) relative to positive x-axis)
   - Torque \( \tau_1 = -r F_1 \sin(90^\circ) = -0.299 \times 45.8 = -13.69 \, \text{N}\cdot\text{m} \)

2. **Force \( F_2 \)**:
   - Acts vertically upward (angle \( 90^\circ \) relative to positive x-axis)
   - Torque \( \tau_2 = r F_2 \sin(90^\circ) = 0.299 \times 68.3 = 20.42 \, \text{N}\cdot\text{m} \)

3. **Force \( F_3 \)**:
   - Acts horizontally to the left (angle \( 180^\circ \))
   - No torque contribution because the force is directed along the axis (\( \sin(180^\circ) = 0 \)).

4. **Force \( F_4 \)**:
   - Acts at \( 45^\circ \) from the x-axis
   - Torque \( \tau_4 = r F_4 \sin(45^\circ) = 0.299 \times 45.8 \times \sin(45^\circ) = 9.68 \, \text{N}\cdot\text{m} \)

### Net Torque:
Now, summing up all the torques:
\[
\tau_{\text{net}} = \tau_1 + \tau_2 + \tau_3 + \tau_4
\]
Since \( \tau_3 = 0 \), we have:
\[
\tau_{\text{net}} = -13.69 + 20.42 + 0 + 9.68 = 16.41 \, \text{N}\cdot\text{m}
\]
Thus, the net torque about the axle is \( 16.41 \, \text{N}\cdot\text{m} \) in the counterclockwise direction.

Would you like more details on this solution, or do you have any questions? Here are some related questions to expand on this:

1. What is the significance of the direction of the torque in physical systems?
2. How would the solution change if the disk rotated clockwise instead of counterclockwise?
3. What would happen if the forces were not perpendicular to the radius?
4. How is torque related to angular acceleration in rotational motion?
5. What other physical factors, such as friction, could affect the net torque in real-life situations?

**Tip:** Always ensure to calculate the correct direction of each torque when dealing with multiple forces, as this can affect the overall result!

The 59.8cm diameter disk rotates counterclockwise on an axle through its center. F1=45.8N, F2=68.3N, F3=68.3N, F4=45.8N, and d=3.42cm. What is the net torque about the axle? Let the counterclockwise direction be positive.

Calculate the net torque about the axle for a 59.8cm diameter disk rotating counterclockwise with forces F1=45.8N, F2=68.3N, F3=68.3N, and F4=45.8N applied at different points. This problem involves understanding the torque formula and trigonometric functions to solve the system, suitable for college physics students.

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