I don't know how to solve #5 ### Balancing TorquesIn the figure above, a balance is set up with the fulcrum (pivot point) at the 3.0 meter mark.- **Forces applied:** - **3.2 N** at the **0.0 meter mark** - **1.5 N** at the **4.0 meter mark** - **2.5 N** at the **6.0 meter mark**The task is to determine the net torque on the system (sign and direction) about the pivot point.#### Torque Calculation- **Distance from Pivot:** - For the 3.2 N force: 3.0 meters from the pivot. - For the 1.5 N force: 1.0 meter from the pivot. - For the 2.5 N force: 3.0 meters from the pivot.- **Direction:** - The force at 0.0 m creates a **counter-clockwise torque**. - The forces at 4.0 m and 6.0 m create **clockwise torques**.#### Net Torque Options:- A. -18.6 N-m- B. -0.6 N-m- C. 0.6 N-m- D. 18.6 N-mThe diagram visually represents the balance with arrows indicating the direction and magnitude of forces applied at specific points along a beam with the fulcrum at the center. ### #5 Balancing Torques**Diagram Explanation:** The diagram shows a balanced beam with a fulcrum placed at the 3.0-meter mark. The beam is marked at 0.0, 1.0, 2.0, 3.0, 4.0, 5.0, and 6.0 meters. There are arrows indicating the downward forces applied at the following positions:- 0.0 meters: 3.2 N- 4.0 meters: 1.5 N- 6.0 meters: 2.5 N- A force labeled \( F \) needs to be applied at the 5.0-meter mark to balance the system.**Problem Statement:** As explained in problem #4, a balance setup is demonstrated with the fulcrum (pivot point) at the 3.0 meter mark. Forces vertically downward are applied as follows:- At the 0.0 meter mark: 3.2 Newtons- At the 4.0 meter mark: 1.5 Newtons- At the 6.0 meter mark: 2.5 NewtonsThe task is to determine what force (in Newtons) must be applied to the 5.0 meter mark to balance the system (i.e., make the net torque about the pivot point zero).**Options:**- A. 0.3 N- B. 0.6 N- C. 1.2 N- D. 1.8 N

Answered: #4 Balancing Torques 0.0 1.0 2.0 3.0…

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#4 Balancing Torques 0.0 1.0 2.0 3.0 4.0 5.0 6.0 3.2 N 1.5 N 2.5 N in the the fulom (niuot the 2.0

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

I don't know how to solve #5

### Balancing Torques

In the figure above, a balance is set up with the fulcrum (pivot point) at the 3.0 meter mark.

- **Forces applied:**
- **3.2 N** at the **0.0 meter mark**
- **1.5 N** at the **4.0 meter mark**
- **2.5 N** at the **6.0 meter mark**

The task is to determine the net torque on the system (sign and direction) about the pivot point.

#### Torque Calculation

- **Distance from Pivot:**
- For the 3.2 N force: 3.0 meters from the pivot.
- For the 1.5 N force: 1.0 meter from the pivot.
- For the 2.5 N force: 3.0 meters from the pivot.

- **Direction:**
- The force at 0.0 m creates a **counter-clockwise torque**.
- The forces at 4.0 m and 6.0 m create **clockwise torques**.

#### Net Torque Options:
- A. -18.6 N-m
- B. -0.6 N-m
- C. 0.6 N-m
- D. 18.6 N-m

The diagram visually represents the balance with arrows indicating the direction and magnitude of forces applied at specific points along a beam with the fulcrum at the center.

### #5 Balancing Torques

**Diagram Explanation:**
The diagram shows a balanced beam with a fulcrum placed at the 3.0-meter mark. The beam is marked at 0.0, 1.0, 2.0, 3.0, 4.0, 5.0, and 6.0 meters. There are arrows indicating the downward forces applied at the following positions:

- 0.0 meters: 3.2 N
- 4.0 meters: 1.5 N
- 6.0 meters: 2.5 N
- A force labeled \( F \) needs to be applied at the 5.0-meter mark to balance the system.

**Problem Statement:**
As explained in problem #4, a balance setup is demonstrated with the fulcrum (pivot point) at the 3.0 meter mark. Forces vertically downward are applied as follows:

- At the 0.0 meter mark: 3.2 Newtons
- At the 4.0 meter mark: 1.5 Newtons
- At the 6.0 meter mark: 2.5 Newtons

The task is to determine what force (in Newtons) must be applied to the 5.0 meter mark to balance the system (i.e., make the net torque about the pivot point zero).

**Options:**
- A. 0.3 N
- B. 0.6 N
- C. 1.2 N
- D. 1.8 N

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