A telephone pole has three cables pulling as shown from above, with F, = (500.0î + 600.0j) N, F, = -700.0î N, and F = -300.0j N. y WebAssign Plot (a) Find the net force (in N) on the telephone pole in component form. N net (b) Find the magnitude (in N) and direction (in degrees counterclockwise from the +x-axis) of this net force. magnitude N direction ° counterclockwise from the +x-axis

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**Title: Calculating Net Force on a Telephone Pole with Three Cables**

A telephone pole has three cables pulling as shown in the diagram, with:

- \(\vec{F}_1 = (500.0\hat{i} + 600.0\hat{j}) \, \text{N}\)
- \(\vec{F}_2 = -700.0\hat{i} \, \text{N}\)
- \(\vec{F}_3 = -300.0\hat{j} \, \text{N}\)

### Diagram Explanation

In the diagram, the forces are represented as arrows originating from a common point, which is the top view of the telephone pole:

1. **\(\vec{F}_1\)** is directed in the positive x- and y-directions.
2. **\(\vec{F}_2\)** is directed in the negative x-direction.
3. **\(\vec{F}_3\)** is directed in the negative y-direction.

### Problems to Solve

(a) **Find the net force (in N) on the telephone pole in component form.**

\[\vec{F}_{\text{net}} = \, \_\_\_\_\_\, \text{N}\]

(b) **Find the magnitude (in N) and direction (in degrees counterclockwise from the +x-axis) of this net force.**

- Magnitude: \_\_\_\_\_ N
- Direction: \_\_\_\_\_ ° counterclockwise from the +x-axis

**Instructions for Solving:**

1. **Calculate the net force components.**
   - Combine the x-components: \(500.0 + (-700.0)\)
   - Combine the y-components: \(600.0 + (-300.0)\)

2. **Find the overall magnitude and direction.**

Utilize the Pythagorean theorem for the magnitude and trigonometric functions (such as arctangent) to find the direction angle.

This educational example illustrates how to resolve vector components, sum them to obtain net force, and calculate resultant magnitude and direction.
Transcribed Image Text:**Title: Calculating Net Force on a Telephone Pole with Three Cables** A telephone pole has three cables pulling as shown in the diagram, with: - \(\vec{F}_1 = (500.0\hat{i} + 600.0\hat{j}) \, \text{N}\) - \(\vec{F}_2 = -700.0\hat{i} \, \text{N}\) - \(\vec{F}_3 = -300.0\hat{j} \, \text{N}\) ### Diagram Explanation In the diagram, the forces are represented as arrows originating from a common point, which is the top view of the telephone pole: 1. **\(\vec{F}_1\)** is directed in the positive x- and y-directions. 2. **\(\vec{F}_2\)** is directed in the negative x-direction. 3. **\(\vec{F}_3\)** is directed in the negative y-direction. ### Problems to Solve (a) **Find the net force (in N) on the telephone pole in component form.** \[\vec{F}_{\text{net}} = \, \_\_\_\_\_\, \text{N}\] (b) **Find the magnitude (in N) and direction (in degrees counterclockwise from the +x-axis) of this net force.** - Magnitude: \_\_\_\_\_ N - Direction: \_\_\_\_\_ ° counterclockwise from the +x-axis **Instructions for Solving:** 1. **Calculate the net force components.** - Combine the x-components: \(500.0 + (-700.0)\) - Combine the y-components: \(600.0 + (-300.0)\) 2. **Find the overall magnitude and direction.** Utilize the Pythagorean theorem for the magnitude and trigonometric functions (such as arctangent) to find the direction angle. This educational example illustrates how to resolve vector components, sum them to obtain net force, and calculate resultant magnitude and direction.
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