**Understanding the Magnitude of Resultant Force from Three Given Forces**---**Problem Statement:**What is the magnitude of the resultant force caused by the three forces below? Assume that \( \alpha = 50^\circ \), \( \beta = 35^\circ \), and the magnitude of \( F_1 = 300 \) lb. Note that \( F_3 \) lies in the \( x-y \) plane.---**Diagram Explanation:**In the diagram, there are three forces applied to a system. These forces are labeled as \( F_1 \), \( F_2 \), and \( F_3 \):- \( F_1 \) is a force applied at an angle \( \alpha \) relative to the x-axis.- \( F_2 \) is an applied horizontal force along the negative x-direction with a magnitude of 200 lb.- \( F_3 \) is a force applied at a 30° angle from the y-axis, with a magnitude of 400 lb.There are several angles and directions indicated:- The angle \( \alpha_1 \), which is given as 50° relative to the x-axis.- The angle \( \beta_1 \), which is given as 35° relative to the horizontal plane.- \( F_3 \) makes a 30° angle with the y-axis and lies in the x-y plane.---**Steps to Determine the Resultant Force:**To find the resultant force (R) from the combination of these three forces, you can vectorially sum up the components of each force.1. **Break down each force into its components:** - \( F_1 \): - \( F_{1x} = 300 \times \cos(50^\circ) \) - \( F_{1y} = 300 \times \sin(50^\circ) \) - \( F_{1z} = 300 \times \sin(35^\circ) \) (depending on the context, if this angle is the inclination in the z direction) - \( F_2 \): - This force acts along the -x direction, so: - \( F_{2x} = -200 \) - \( F_{2y} = 0 \) - \( F_{2z} = 0 \) - \( F_3

Vector is a physical quantity that has magnitude and direction both. Scalar is a physical quantity…

What is the magnitude of the resultant force caused by the three forces below? Assume that a, = 50°, B. = 35°, and the magnitude of F. = 300 lb. Note that F. lies in the x-y plane. Fy = 400 lb 30° F = 200 lb a

What is the magnitude of the resultant force caused by the three forces below? Assume that a, = 50°, B. = 35°, and the magnitude of F. = 300 lb. Note that F. lies in the x-y plane. Fy = 400 lb 30° F = 200 lb a

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

**Understanding the Magnitude of Resultant Force from Three Given Forces**

---

**Problem Statement:**

What is the magnitude of the resultant force caused by the three forces below? Assume that \( \alpha = 50^\circ \), \( \beta = 35^\circ \), and the magnitude of \( F_1 = 300 \) lb. Note that \( F_3 \) lies in the \( x-y \) plane.

---

**Diagram Explanation:**

In the diagram, there are three forces applied to a system. These forces are labeled as \( F_1 \), \( F_2 \), and \( F_3 \):

- \( F_1 \) is a force applied at an angle \( \alpha \) relative to the x-axis.
- \( F_2 \) is an applied horizontal force along the negative x-direction with a magnitude of 200 lb.
- \( F_3 \) is a force applied at a 30° angle from the y-axis, with a magnitude of 400 lb.

There are several angles and directions indicated:
- The angle \( \alpha_1 \), which is given as 50° relative to the x-axis.
- The angle \( \beta_1 \), which is given as 35° relative to the horizontal plane.
- \( F_3 \) makes a 30° angle with the y-axis and lies in the x-y plane.

---

**Steps to Determine the Resultant Force:**

To find the resultant force (R) from the combination of these three forces, you can vectorially sum up the components of each force.

1. **Break down each force into its components:**

- \( F_1 \):
- \( F_{1x} = 300 \times \cos(50^\circ) \)
- \( F_{1y} = 300 \times \sin(50^\circ) \)
- \( F_{1z} = 300 \times \sin(35^\circ) \) (depending on the context, if this angle is the inclination in the z direction)

- \( F_2 \):
- This force acts along the -x direction, so:
- \( F_{2x} = -200 \)
- \( F_{2y} = 0 \)
- \( F_{2z} = 0 \)

- \( F_3

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