Macmillan Learnin Dima pulls directly backward with a force F = 111 N on the end of a 2.00 m-long oar. The oar pivots about its midpoint. At the instant shown, the oar is completely in the yz-plane and makes a 0 = 39.0° angle with respect to the water's surface. Derive an expression for the torque vector 7 about the axis through the oar's pivot. Express the torque using ijk vector notation. 7 = Txi+ Ty j + T₂ k 7 = N.m Side view Aerial view 0 F

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**Problem Description:**

Dima pulls directly backward with a force \( F = 111 \, \text{N} \) on the end of a \( 2.00 \, \text{m} \)-long oar. The oar pivots about its midpoint. At the instant shown, the oar is completely in the yz-plane and makes a \( \theta = 39.0^\circ \) angle with respect to the water's surface.

**Objective:**

Derive an expression for the torque vector \( \vec{\tau} \) about the axis through the oar's pivot. Express the torque using \( ijk \) vector notation.

**Equation Format:**

\[
\vec{\tau} = \tau_x \, \mathbf{i} + \tau_y \, \mathbf{j} + \tau_z \, \mathbf{k}
\]

\(\vec{\tau} =\) \(\boxed{\phantom{}} \, \text{N}\cdot\text{m}\)

---

**Diagram Explanation:**

1. **Side View:**
   - Shows the oar positioned in the yz-plane.
   - The angle \( \theta = 39.0^\circ \) is formed between the oar and the horizontal plane (water's surface).
   - Axes are indicated with \( z \) pointing upwards and \( y \) running horizontally parallel to the surface.

2. **Aerial View:**
   - Displays the oar from a top-down perspective.
   - The force \( F = 111 \, \text{N} \) is applied downward along the oar.
   - The axes \( x \) and \( y \) are marked, with \( y \) being the lengthwise direction of the oar.

Given these views, use the torque formula \( \vec{\tau} = \vec{r} \times \vec{F} \) to calculate the torque vector components \( \tau_x, \tau_y, \) and \( \tau_z \).
Transcribed Image Text:**Problem Description:** Dima pulls directly backward with a force \( F = 111 \, \text{N} \) on the end of a \( 2.00 \, \text{m} \)-long oar. The oar pivots about its midpoint. At the instant shown, the oar is completely in the yz-plane and makes a \( \theta = 39.0^\circ \) angle with respect to the water's surface. **Objective:** Derive an expression for the torque vector \( \vec{\tau} \) about the axis through the oar's pivot. Express the torque using \( ijk \) vector notation. **Equation Format:** \[ \vec{\tau} = \tau_x \, \mathbf{i} + \tau_y \, \mathbf{j} + \tau_z \, \mathbf{k} \] \(\vec{\tau} =\) \(\boxed{\phantom{}} \, \text{N}\cdot\text{m}\) --- **Diagram Explanation:** 1. **Side View:** - Shows the oar positioned in the yz-plane. - The angle \( \theta = 39.0^\circ \) is formed between the oar and the horizontal plane (water's surface). - Axes are indicated with \( z \) pointing upwards and \( y \) running horizontally parallel to the surface. 2. **Aerial View:** - Displays the oar from a top-down perspective. - The force \( F = 111 \, \text{N} \) is applied downward along the oar. - The axes \( x \) and \( y \) are marked, with \( y \) being the lengthwise direction of the oar. Given these views, use the torque formula \( \vec{\tau} = \vec{r} \times \vec{F} \) to calculate the torque vector components \( \tau_x, \tau_y, \) and \( \tau_z \).
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