The particle shown below is at rest, where F₁ = 6.95 N, and F₂ = 10.4 N. Find the magnitude and direction of F. magnitude direction F₁ 30.0° 45.0° counterclockwise from the +x-axis F x

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### Finding the Magnitude and Direction of the Resultant Force

**Problem Statement:**
The particle shown below is at rest, where \( F_1 = 6.95 \, \text{N} \) and \( F_2 = 10.4 \, \text{N} \). Find the magnitude and direction of the resultant force \( F \).

**Given:**

- \( F_1 = 6.95 \, \text{N} \)
- \( F_2 = 10.4 \, \text{N} \)

**Diagram Description:**

In the provided diagram:

- There is a coordinate system with the \( y \)-axis oriented vertically and the \( x \)-axis oriented horizontally.
- A particle is positioned at the origin.
- There are two forces acting on the particle, \( F_1 \) and \( F_2 \):

  1. \( F_1 \): This force is applied at an angle of \( 30.0^\circ \) counterclockwise from the negative \( x \)-axis.
  2. \( F_2 \): This force is applied at an angle of \( 45.0^\circ \) clockwise from the positive \( x \)-axis.

**Graphical Representation:**
The diagram features arrows representing the forces:

- \( F_1 \) is directed towards the left, slanting upward from the horizontal line.
- \( F_2 \) is directed towards the left, slanting downward from the horizontal line.
- The resultant force \( F \) is indicated, but its magnitude and direction are unknown.

**Calculations:**
To find the magnitude and direction of \( F \), you would typically resolve \( F_1 \) and \( F_2 \) into their vector components, sum these components, and then determine the magnitude and direction of the resulting vector.

**Magnitude Calculation:**
\[ F = \sqrt{(F_{1x} + F_{2x})^2 + (F_{1y} + F_{2y})^2} \]

**Direction Calculation:**
\[ \theta = \text{atan2}(F_{1y} + F_{2y}, F_{1x} + F_{2x}) \]

Where \( F_{1x}, F_{1y}, F_{2x}, \) and \( F_{2y
Transcribed Image Text:### Finding the Magnitude and Direction of the Resultant Force **Problem Statement:** The particle shown below is at rest, where \( F_1 = 6.95 \, \text{N} \) and \( F_2 = 10.4 \, \text{N} \). Find the magnitude and direction of the resultant force \( F \). **Given:** - \( F_1 = 6.95 \, \text{N} \) - \( F_2 = 10.4 \, \text{N} \) **Diagram Description:** In the provided diagram: - There is a coordinate system with the \( y \)-axis oriented vertically and the \( x \)-axis oriented horizontally. - A particle is positioned at the origin. - There are two forces acting on the particle, \( F_1 \) and \( F_2 \): 1. \( F_1 \): This force is applied at an angle of \( 30.0^\circ \) counterclockwise from the negative \( x \)-axis. 2. \( F_2 \): This force is applied at an angle of \( 45.0^\circ \) clockwise from the positive \( x \)-axis. **Graphical Representation:** The diagram features arrows representing the forces: - \( F_1 \) is directed towards the left, slanting upward from the horizontal line. - \( F_2 \) is directed towards the left, slanting downward from the horizontal line. - The resultant force \( F \) is indicated, but its magnitude and direction are unknown. **Calculations:** To find the magnitude and direction of \( F \), you would typically resolve \( F_1 \) and \( F_2 \) into their vector components, sum these components, and then determine the magnitude and direction of the resulting vector. **Magnitude Calculation:** \[ F = \sqrt{(F_{1x} + F_{2x})^2 + (F_{1y} + F_{2y})^2} \] **Direction Calculation:** \[ \theta = \text{atan2}(F_{1y} + F_{2y}, F_{1x} + F_{2x}) \] Where \( F_{1x}, F_{1y}, F_{2x}, \) and \( F_{2y
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