Moment of a Force-Vector Formulation Learning Goal: To use the vector cross product to calculate the moment produced by a force, or forces, about a specified point on a member. The moment of a force F about the moment axis passing through O and perpendicular to the plane containing O and F can be expressed using the vector cross product, Mo = r x F. In a properly constructed Cartesian coordinate system, the vector cross product can be calculated using a matrix determinant: i j k Ty Tz F₂ Fy F₂ Notice that the order of the elements of the matrix determinant is important; switching rows 2 and 3 of the determinant would change the sign of the moment from positive to negative (or vice versa.) Figure M=rx F = Tz B < 3 of 3 F₂ Part B - Moment due to a force specified as a Cartesian vector Part C - Moment due to two forces As shown, a member is fixed at the origin, point O, and has two applied forces, F₁ and F2, applied at the free end, point B. (Figure 3) IVE ΑΣΦ | | Mo = 311,330,312 The forces are given by F₁ = 115 Ni-110 Nj+45 Nk and F2 has magnitude 165 N and direction angles a = 142.0°, B = 72.0°, and y = 57.8°. The dimensions are x₁ = 1.40 m, y₁ = 1.85 m, and z₁ = 1.10 m. What is the moment about the origin due to the applied forces? Express the individual components of the Cartesian vector to three significant figures, separated by commas. ► View Available Hint(s) Submit Previous Answers Provide Feedback vec 5 X Incorrect; Try Again; 5 attempts remaining B < 2 of 6 ? Review i, j, k] N.m Next >
Moment of a Force-Vector Formulation Learning Goal: To use the vector cross product to calculate the moment produced by a force, or forces, about a specified point on a member. The moment of a force F about the moment axis passing through O and perpendicular to the plane containing O and F can be expressed using the vector cross product, Mo = r x F. In a properly constructed Cartesian coordinate system, the vector cross product can be calculated using a matrix determinant: i j k Ty Tz F₂ Fy F₂ Notice that the order of the elements of the matrix determinant is important; switching rows 2 and 3 of the determinant would change the sign of the moment from positive to negative (or vice versa.) Figure M=rx F = Tz B < 3 of 3 F₂ Part B - Moment due to a force specified as a Cartesian vector Part C - Moment due to two forces As shown, a member is fixed at the origin, point O, and has two applied forces, F₁ and F2, applied at the free end, point B. (Figure 3) IVE ΑΣΦ | | Mo = 311,330,312 The forces are given by F₁ = 115 Ni-110 Nj+45 Nk and F2 has magnitude 165 N and direction angles a = 142.0°, B = 72.0°, and y = 57.8°. The dimensions are x₁ = 1.40 m, y₁ = 1.85 m, and z₁ = 1.10 m. What is the moment about the origin due to the applied forces? Express the individual components of the Cartesian vector to three significant figures, separated by commas. ► View Available Hint(s) Submit Previous Answers Provide Feedback vec 5 X Incorrect; Try Again; 5 attempts remaining B < 2 of 6 ? Review i, j, k] N.m Next >
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
Related questions
Question
Help part A and C PLEASE
![## Moment of a Force—Vector Formulation
### Learning Goal:
To use the vector cross product to calculate the moment produced by a force, or forces, about a specified point on a member.
The moment of a force **F** about the moment axis passing through **O** and perpendicular to the plane containing **O** and **F** can be expressed using the vector cross product, \( M_O = \mathbf{r} \times \mathbf{F} \). In a properly constructed Cartesian coordinate system, the vector cross product can be calculated using a matrix determinant:
\[
\mathbf{M} = \mathbf{r} \times \mathbf{F} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
r_x & r_y & r_z \\
F_x & F_y & F_z \\
\end{vmatrix}
\]
Notice that the order of the elements of the matrix determinant is important; switching rows 2 and 3 of the determinant would change the sign of the moment from positive to negative (or vice versa.)
### Figure:
The diagram shows a three-dimensional coordinate system with axes labeled \( x, y, z \). A member is fixed at the origin, \( O \), with a force \( \mathbf{F} \) applied at the free end, \( B \). The diagram provides a visual representation of \( x_1 = 0.450 \, \text{m} \), \( x_2 = 1.90 \, \text{m} \), \( y_1 = 2.30 \, \text{m} \), and \( z_1 = 1.10 \, \text{m} \).
### Part A: Moment due to a force specified by magnitude and endpoints
As shown, a member is fixed at the origin, point \( O \), and has an applied force **F** (the tension in the rope) applied at the free end, point \( B \). The force has magnitude \( F = 140 \, \text{N} \) and is directed as shown.
What is the moment about the origin due to the applied force **F**?
**Express the individual components of the Cartesian vector to three significant figures, separated by commas.**
### Input Section
- \( M_O = [ \_ , 99.7, -209](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6aa72a1b-7be3-4dcd-bcb5-7886fdb6dafd%2F728a59c4-9bd1-4540-a1be-7ed38af5b1f4%2Fo24c1q_processed.png&w=3840&q=75)
Transcribed Image Text:## Moment of a Force—Vector Formulation
### Learning Goal:
To use the vector cross product to calculate the moment produced by a force, or forces, about a specified point on a member.
The moment of a force **F** about the moment axis passing through **O** and perpendicular to the plane containing **O** and **F** can be expressed using the vector cross product, \( M_O = \mathbf{r} \times \mathbf{F} \). In a properly constructed Cartesian coordinate system, the vector cross product can be calculated using a matrix determinant:
\[
\mathbf{M} = \mathbf{r} \times \mathbf{F} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
r_x & r_y & r_z \\
F_x & F_y & F_z \\
\end{vmatrix}
\]
Notice that the order of the elements of the matrix determinant is important; switching rows 2 and 3 of the determinant would change the sign of the moment from positive to negative (or vice versa.)
### Figure:
The diagram shows a three-dimensional coordinate system with axes labeled \( x, y, z \). A member is fixed at the origin, \( O \), with a force \( \mathbf{F} \) applied at the free end, \( B \). The diagram provides a visual representation of \( x_1 = 0.450 \, \text{m} \), \( x_2 = 1.90 \, \text{m} \), \( y_1 = 2.30 \, \text{m} \), and \( z_1 = 1.10 \, \text{m} \).
### Part A: Moment due to a force specified by magnitude and endpoints
As shown, a member is fixed at the origin, point \( O \), and has an applied force **F** (the tension in the rope) applied at the free end, point \( B \). The force has magnitude \( F = 140 \, \text{N} \) and is directed as shown.
What is the moment about the origin due to the applied force **F**?
**Express the individual components of the Cartesian vector to three significant figures, separated by commas.**
### Input Section
- \( M_O = [ \_ , 99.7, -209
![## Moment of a Force—Vector Formulation
### Learning Goal:
To use the vector cross product to calculate the moment produced by a force, or forces, about a specified point on a member.
The moment of a force **F** about the moment axis passing through **O** and perpendicular to the plane containing **O** and **F** can be expressed using the vector cross product:
\[ \mathbf{M}_O = \mathbf{r} \times \mathbf{F} \]
In a properly constructed Cartesian coordinate system, the vector cross product can be calculated using a matrix determinant:
\[
\mathbf{M} = \mathbf{r} \times \mathbf{F} =
\begin{vmatrix}
i & j & k \\
r_x & r_y & r_z \\
F_x & F_y & F_z \\
\end{vmatrix}
\]
Notice that the order of the elements of the matrix determinant is important; switching rows 2 and 3 of the determinant would change the sign of the moment from positive to negative (or vice versa).
### Figure Explanation:
The figure illustrates a 3-dimensional coordinate system. A member is fixed at the origin, point **O**, with two applied forces, **F₁** and **F₂**, acting at the free end, point **B**.
### Problem Statement:
The forces are given by:
\[ \mathbf{F}_1 = 115 \, \text{N} \, \mathbf{i} - 110 \, \text{N} \, \mathbf{j} + 45 \, \text{N} \, \mathbf{k} \]
and **F₂** has magnitude 165 N with direction angles α = 142.0°, β = 72.0°, and γ = 57.8°. The dimensions are x₁ = 1.40 m, y₁ = 1.85 m, and z₁ = 1.10 m.
**Task:** Identify the moment about the origin due to the applied forces.
**Instruction:** Express the individual components of the Cartesian vector to three significant figures, separated by commas.
### Example Attempt:
\[ \mathbf{M}_O = [ 311,330,312 ] \]
- **Response:** Incorrect; Try Again. 5 attempts remaining.
**Note:** Ensure careful calculation and precise input for correct submission.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6aa72a1b-7be3-4dcd-bcb5-7886fdb6dafd%2F728a59c4-9bd1-4540-a1be-7ed38af5b1f4%2Fq8uyz5a_processed.png&w=3840&q=75)
Transcribed Image Text:## Moment of a Force—Vector Formulation
### Learning Goal:
To use the vector cross product to calculate the moment produced by a force, or forces, about a specified point on a member.
The moment of a force **F** about the moment axis passing through **O** and perpendicular to the plane containing **O** and **F** can be expressed using the vector cross product:
\[ \mathbf{M}_O = \mathbf{r} \times \mathbf{F} \]
In a properly constructed Cartesian coordinate system, the vector cross product can be calculated using a matrix determinant:
\[
\mathbf{M} = \mathbf{r} \times \mathbf{F} =
\begin{vmatrix}
i & j & k \\
r_x & r_y & r_z \\
F_x & F_y & F_z \\
\end{vmatrix}
\]
Notice that the order of the elements of the matrix determinant is important; switching rows 2 and 3 of the determinant would change the sign of the moment from positive to negative (or vice versa).
### Figure Explanation:
The figure illustrates a 3-dimensional coordinate system. A member is fixed at the origin, point **O**, with two applied forces, **F₁** and **F₂**, acting at the free end, point **B**.
### Problem Statement:
The forces are given by:
\[ \mathbf{F}_1 = 115 \, \text{N} \, \mathbf{i} - 110 \, \text{N} \, \mathbf{j} + 45 \, \text{N} \, \mathbf{k} \]
and **F₂** has magnitude 165 N with direction angles α = 142.0°, β = 72.0°, and γ = 57.8°. The dimensions are x₁ = 1.40 m, y₁ = 1.85 m, and z₁ = 1.10 m.
**Task:** Identify the moment about the origin due to the applied forces.
**Instruction:** Express the individual components of the Cartesian vector to three significant figures, separated by commas.
### Example Attempt:
\[ \mathbf{M}_O = [ 311,330,312 ] \]
- **Response:** Incorrect; Try Again. 5 attempts remaining.
**Note:** Ensure careful calculation and precise input for correct submission.
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