A force of F = 22i is applied to the end of a rod extending from the origin (xo, Yo) (0, 0) meters to point A with coordinates (XA, YA) = (8.6, 3.2) meters. Calculate the moment arm associated with the moment of force produced by F. Give the answer t

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**Problem Statement: Calculating the Moment of Force**

A force of **F = 22i** is applied to the end of a rod extending from the origin \((x_0, y_0) = (0, 0)\) meters to point A with coordinates \((x_A, y_A) = (8.6, 3.2)\) meters. Calculate the moment arm associated with the moment of force produced by **F**. Give the answer to 1 decimal place.

**Solution: Evaluate the Moment Arm**

1. **Identify the Given Parameters:**
   - **Force vector:** \( \mathbf{F} = 22i \, \text{N} \)
   - **Position vector:** From the origin \((0,0)\) to point A \((8.6, 3.2)\) meters.

2. **Calculate the Moment of Force:**
   - The moment (torque) \(\mathbf{M}\) is given by the cross product of the position vector **r** and the force vector **F**.
   - \(\mathbf{r} = (8.6i + 3.2j) \, \text{m}\)
   - Cross product: \(\mathbf{M} = \mathbf{r} \times \mathbf{F}\)
   - \(\mathbf{M} = (8.6i + 3.2j) \times 22i\)

3. **Determine the Moment about the Origin:**
   - Since the force is purely in the \(i\) direction (22i), the j component of the position vector contributes to the moment.
   - \(\mathbf{M} = 3.2 \times 22 = 70.4 \, \text{Nm}\)

**Final Result:**
- The moment arm associated with the force \( \mathbf{F} \) is **70.4 Nm**.

Note: The exercise involves understanding vectors and applying the principles of cross products to find the moment of force.
Transcribed Image Text:**Problem Statement: Calculating the Moment of Force** A force of **F = 22i** is applied to the end of a rod extending from the origin \((x_0, y_0) = (0, 0)\) meters to point A with coordinates \((x_A, y_A) = (8.6, 3.2)\) meters. Calculate the moment arm associated with the moment of force produced by **F**. Give the answer to 1 decimal place. **Solution: Evaluate the Moment Arm** 1. **Identify the Given Parameters:** - **Force vector:** \( \mathbf{F} = 22i \, \text{N} \) - **Position vector:** From the origin \((0,0)\) to point A \((8.6, 3.2)\) meters. 2. **Calculate the Moment of Force:** - The moment (torque) \(\mathbf{M}\) is given by the cross product of the position vector **r** and the force vector **F**. - \(\mathbf{r} = (8.6i + 3.2j) \, \text{m}\) - Cross product: \(\mathbf{M} = \mathbf{r} \times \mathbf{F}\) - \(\mathbf{M} = (8.6i + 3.2j) \times 22i\) 3. **Determine the Moment about the Origin:** - Since the force is purely in the \(i\) direction (22i), the j component of the position vector contributes to the moment. - \(\mathbf{M} = 3.2 \times 22 = 70.4 \, \text{Nm}\) **Final Result:** - The moment arm associated with the force \( \mathbf{F} \) is **70.4 Nm**. Note: The exercise involves understanding vectors and applying the principles of cross products to find the moment of force.
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