A square metal plate 0.180 m on each side is pivoted about an axis through point O at its center and perpendicular to the plate (Figure 1). Figure 1: A square plate has vertical and horizontal sides, 0.180 meters long each. The center of the square is labeled O. Three forces are applied to the plate. Force F 1 is applied vertically downward to the top right vertex. Force F 2 is applied vertically downward to the top left vertex. Force F 3 is applied upward and to the right to the bottom right corner making an angle of 45 degrees with the horizontal. - A square with three forces applied to its vertices. Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces are F1 = 22.0 N, F2 = 16.7 N, and F3 = 16.7 N. The plate and all forces are in the plane of the page. Take positive torques to be counterclockwise. Express your answer in newton-meters.

To calculate the net torque about the axis at point \( O \) due to the three forces, let's analyze each force and its contribution to the torque. We'll take counterclockwise torques as positive.

### Step-by-Step Solution:

1. **Calculate the Moment Arm (Distance from Axis):**
   The side length of the square is \( 0.180 \, \text{m} \), so the distance from each vertex to the center \( O \) (the axis of rotation) is:
   \[
   r = \frac{0.180 \, \text{m}}{\sqrt{2}} = 0.1273 \, \text{m}
   \]

2. **Torque from \( F_1 \):**
   - \( F_1 \) is applied vertically downward at the top right vertex.
   - The lever arm \( r \) for \( F_1 \) is perpendicular to the force direction, so \( \theta = 90^\circ \).
   - Torque \( \tau_1 = r \cdot F_1 \cdot \sin(90^\circ) \).
   - Since \( F_1 \) is applied downward, it causes a clockwise rotation, so the torque is negative.
   
   \[
   \tau_1 = - (0.1273 \, \text{m}) \cdot (22.0 \, \text{N}) = -2.8 \, \text{Nm}
   \]

3. **Torque from \( F_2 \):**
   - \( F_2 \) is applied vertically downward at the top left vertex.
   - The lever arm \( r \) for \( F_2 \) is perpendicular to the force direction, so \( \theta = 90^\circ \).
   - Torque \( \tau_2 = r \cdot F_2 \cdot \sin(90^\circ) \).
   - Since \( F_2 \) is also downward, it produces a clockwise rotation, making the torque negative.
   
   \[
   \tau_2 = - (0.1273 \, \text{m}) \cdot (16.7 \, \text{N}) = -2.13 \, \text{Nm}
   \]

4. **Torque from \( F_3 \):**
   - \( F_3 = 16.7 \, \text{N} \) acts upward and to the right at a 45° angle to the horizontal.
   - To calculate the effective torque, we need the vertical component of \( F_3 \), as it is perpendicular to the lever arm.
   - Vertical component of \( F_3 = F_3 \sin(45^\circ) \).
   
   \[
   \tau_3 = r \cdot (F_3 \sin(45^\circ)) = (0.1273 \, \text{m}) \cdot (16.7 \, \text{N}) \cdot \sin(45^\circ)
   \]
   Since \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \):
   \[
   \tau_3 = (0.1273 \, \text{m}) \cdot (16.7 \, \text{N}) \cdot 0.707 = 1.5 \, \text{Nm}
   \]
   This torque is positive because it would cause a counterclockwise rotation.

5. **Calculate the Net Torque:**
   Summing up the individual torques:
   \[
   \tau_{\text{net}} = \tau_1 + \tau_2 + \tau_3 = -2.8 \, \text{Nm} - 2.13 \, \text{Nm} + 1.5 \, \text{Nm} = -3.43 \, \text{Nm}
   \]

### Final Answer:
\[
\tau_{\text{net}} = -3.43 \, \text{Nm}
\]

The net torque about the axis is \( -3.43 \, \text{Nm} \), indicating a clockwise direction.

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Would you like a breakdown of how each torque was calculated or have additional questions on torque?

1. What effect does changing the angle of \( F_3 \) have on the net torque?
2. How would the torque change if \( F_1 \) or \( F_2 \) was applied at a different vertex?
3. How does changing the pivot point from \( O \) affect the torque values?
4. What happens to the net torque if \( F_1 \) and \( F_2 \) are reversed in direction?
5. What impact does doubling the side length have on the torque?

**Tip:** Remember, torque is maximized when the force is applied perpendicular to the lever arm.

A square metal plate 0.180 m on each side is pivoted about an axis through point O at its center and perpendicular to the plate. Three forces are applied to the plate: F1 vertically downward at the top right vertex, F2 vertically downward at the top left vertex, and F3 upward and to the right at the bottom right corner at a 45-degree angle with the horizontal. Calculate the net torque about this axis if F1 = 22.0 N, F2 = 16.7 N, and F3 = 16.7 N. Take positive torques to be counterclockwise. Express your answer in newton-meters.

Find the net torque on a square metal plate with three applied forces, pivoted about its center. Forces include F1 = 22.0 N, F2 = 16.7 N, and F3 = 16.7 N. This solution explains torque calculations using trigonometry and force decomposition for Grades 11-12 physics.

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