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. Part A - Moment due to a force specified by magnitude and endpoints 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: 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. (Figure 1) The force has magnitude F = 180 N and is directed as shown. The dimensions are r, = 0.350 m. r2 = 1.90 m. y1 = 2.30 m. and z1 = 1.20 m. What is the moment about the origin due to the applied force F? ijk Express the individual components of the Cartesian vector to three significant figures, separated by commas. M =rx F =|r. Ty ra. F F, F • View Available Hint(s) 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.) VO AZ 1 vec ? Mo =| i, j, k] N · m Submit Part B - Moment due to a force specified as a Cartesian vector 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. (Figure 2) The force is given by F = 130 N i– 135 Nj+ 70NK The dimensions are a = 1.35 m, y1 = 1.90 m, and z = 1.05 m. . Figure 1 of 3 > 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. • View Available Hint(s) VO AEO It vec + + ? B Mo =[ i, j, k) N - m Submit
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. Part A - Moment due to a force specified by magnitude and endpoints 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: 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. (Figure 1) The force has magnitude F = 180 N and is directed as shown. The dimensions are r, = 0.350 m. r2 = 1.90 m. y1 = 2.30 m. and z1 = 1.20 m. What is the moment about the origin due to the applied force F? ijk Express the individual components of the Cartesian vector to three significant figures, separated by commas. M =rx F =|r. Ty ra. F F, F • View Available Hint(s) 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.) VO AZ 1 vec ? Mo =| i, j, k] N · m Submit Part B - Moment due to a force specified as a Cartesian vector 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. (Figure 2) The force is given by F = 130 N i– 135 Nj+ 70NK The dimensions are a = 1.35 m, y1 = 1.90 m, and z = 1.05 m. . Figure 1 of 3 > 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. • View Available Hint(s) VO AEO It vec + + ? B Mo =[ i, j, k) N - m Submit
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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![### 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:
\[
\mathbf{M} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
r_x & r_y & r_z \\
F_x & F_y & F_z
\end{vmatrix}
\]
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 figure illustrates a 3D coordinate system with a member fixed at the origin \(O\). The member has an applied force **F** at the free end, point \(B\). The following dimensions are given:
- \(x_1, x_2\)
- \(y_1\)
- \(z_1, z_2\)
### 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 \(\mathbf{F}\), the tension in the rope, applied at the free end, point \(B\). (Figure 1)
The force has magnitude \(F = 180 \, N\) and is directed as shown. The dimensions are \(x_1 = 0.350 \, m, x_2 = 1.90 \, m, y_1 = 2.30 \, m,\) and \(z_1 = 1.20 \, m\).
What is the moment about the origin due to the applied force \(\mathbf{F}\)?
*Express the individual components of the Cartesian vector to three significant figures, separated by commas.*
\[ \mathbf{M_O} = \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb1a8ba3-371b-48d7-95b6-c118f740fa83%2F08d3fde9-dd41-4326-8b03-c33c396a43bf%2F6myqjv_processed.png&w=3840&q=75)
Transcribed Image Text:### 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:
\[
\mathbf{M} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
r_x & r_y & r_z \\
F_x & F_y & F_z
\end{vmatrix}
\]
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 figure illustrates a 3D coordinate system with a member fixed at the origin \(O\). The member has an applied force **F** at the free end, point \(B\). The following dimensions are given:
- \(x_1, x_2\)
- \(y_1\)
- \(z_1, z_2\)
### 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 \(\mathbf{F}\), the tension in the rope, applied at the free end, point \(B\). (Figure 1)
The force has magnitude \(F = 180 \, N\) and is directed as shown. The dimensions are \(x_1 = 0.350 \, m, x_2 = 1.90 \, m, y_1 = 2.30 \, m,\) and \(z_1 = 1.20 \, m\).
What is the moment about the origin due to the applied force \(\mathbf{F}\)?
*Express the individual components of the Cartesian vector to three significant figures, separated by commas.*
\[ \mathbf{M_O} = \]
![### Moment due to Two Forces
#### Part C - Moment due to two forces
As shown, a member is fixed at the origin, point \( O \), and has two applied forces, \( \mathbf{F}_1 \) and \( \mathbf{F}_2 \), applied at the free end, point \( B \).
(Figure 3)
The forces are given by \( \mathbf{F}_1 = 85 \, \mathbf{i} - 110 \, \mathbf{j} + 75 \, \mathbf{k} \, \text{N} \) and \( \mathbf{F}_2 \) has magnitude \( 165 \, \text{N} \) and direction angles \( \alpha = 155.0^\circ \), \( \beta = 66.0^\circ \), and \( \gamma = 83.4^\circ \). The dimensions are \( x_1 = 1.40 \, \text{m} \), \( y_1 = 1.85 \, \text{m} \), and \( z_1 = 1.10 \, \text{m} \).
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.
#### Diagram Explanation
The diagram shows a 3D coordinate system with axes labeled \( x \), \( y \), and \( z \). The member is drawn from the origin \( O \) to point \( B \) where the forces \( \mathbf{F}_1 \) and \( \mathbf{F}_2 \) are applied. The dimensions \( x_1 \), \( y_1 \), and \( z_1 \) are marked, indicating the position of point \( B \).
#### Input Field
\[ \mathbf{M}_O = \]
\[ \text{(i, j, k)} \, \text{N} \cdot \text{m} \]
#### Submission Button
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This structure and explanation help in understanding the scenario and provides the necessary steps to calculate the moment about the origin due to the applied forces.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb1a8ba3-371b-48d7-95b6-c118f740fa83%2F08d3fde9-dd41-4326-8b03-c33c396a43bf%2Fio6vpz1_processed.png&w=3840&q=75)
Transcribed Image Text:### Moment due to Two Forces
#### Part C - Moment due to two forces
As shown, a member is fixed at the origin, point \( O \), and has two applied forces, \( \mathbf{F}_1 \) and \( \mathbf{F}_2 \), applied at the free end, point \( B \).
(Figure 3)
The forces are given by \( \mathbf{F}_1 = 85 \, \mathbf{i} - 110 \, \mathbf{j} + 75 \, \mathbf{k} \, \text{N} \) and \( \mathbf{F}_2 \) has magnitude \( 165 \, \text{N} \) and direction angles \( \alpha = 155.0^\circ \), \( \beta = 66.0^\circ \), and \( \gamma = 83.4^\circ \). The dimensions are \( x_1 = 1.40 \, \text{m} \), \( y_1 = 1.85 \, \text{m} \), and \( z_1 = 1.10 \, \text{m} \).
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.
#### Diagram Explanation
The diagram shows a 3D coordinate system with axes labeled \( x \), \( y \), and \( z \). The member is drawn from the origin \( O \) to point \( B \) where the forces \( \mathbf{F}_1 \) and \( \mathbf{F}_2 \) are applied. The dimensions \( x_1 \), \( y_1 \), and \( z_1 \) are marked, indicating the position of point \( B \).
#### Input Field
\[ \mathbf{M}_O = \]
\[ \text{(i, j, k)} \, \text{N} \cdot \text{m} \]
#### Submission Button
Submit
##### Feedback and Navigation
- Provide Feedback
- Next
This structure and explanation help in understanding the scenario and provides the necessary steps to calculate the moment about the origin due to the applied forces.
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