1. A cube of side a=3m is acted upon by a force P-20N along the diagonal of a face as shown. a. Determine the moment of force P about the point A. b. Determine the projection of moment of P about the edge AB. c. Determine the projection of moment of P about the diagonal AG of the cube. ( d. Are vector P and diagonal AG parallel or perpendicular to each other? Y D G F E

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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**Title: Analysis of Force on a Cube**

**Problem Statement:**
A cube of side \(a = 3 \, \text{m}\) is acted upon by a force \(P = 20 \, \text{N}\) along the diagonal of a face, as illustrated.

**Tasks:**
a. Determine the moment of force P about the point A.  
b. Determine the projection of the moment of P about the edge AB.  
c. Determine the projection of the moment of P about the diagonal AG of the cube.  
d. Are vector P and diagonal AG parallel or perpendicular to each other?

**Illustration Description:**
The provided illustration is a 3D representation of the cube:

- The cube is labeled with vertices A, B, C, D for the top face, and E, F, G for the bottom face, while O represents the origin.
- The force \(P\) is indicated by an arrow moving from point \(G\) towards vertex \(D\) on the face diagonal.
- Axes are labeled as \(x\), \(y\), and \(z\), with an edge length of \(a = 3 \, \text{m}\).

**Detailed Explanation:**

**1. Determine the Moment of Force P About Point A:**
To determine the moment of a force about a point, we use the vector cross product formula:
\[ \mathbf{M_A} = \mathbf{r_A} \times \mathbf{P} \]

**2. Projection of the Moment of P about Edge AB:**
The projection of the moment about the edge AB can be found by determining the component of the moment vector along the direction of AB.

**3. Projection of the Moment of P about Diagonal AG:**
To find this projection, decompose the moment vector along the diagonal vector AG.

**4. Parallelism or Perpendicularity of Vector P and Diagonal AG:**
Assess the dot product of vector \(P\) and diagonal AG:
\[ \mathbf{P} \cdot \mathbf{AG} \]
If the dot product is zero, the vectors are perpendicular. If not, the angle (θ) between them can be found using:
\[ \cos \theta = \frac{\mathbf{P} \cdot \mathbf{AG}}{|\mathbf{P}| |\mathbf{AG}|} \]

This analysis is applicable for understanding force
Transcribed Image Text:**Title: Analysis of Force on a Cube** **Problem Statement:** A cube of side \(a = 3 \, \text{m}\) is acted upon by a force \(P = 20 \, \text{N}\) along the diagonal of a face, as illustrated. **Tasks:** a. Determine the moment of force P about the point A. b. Determine the projection of the moment of P about the edge AB. c. Determine the projection of the moment of P about the diagonal AG of the cube. d. Are vector P and diagonal AG parallel or perpendicular to each other? **Illustration Description:** The provided illustration is a 3D representation of the cube: - The cube is labeled with vertices A, B, C, D for the top face, and E, F, G for the bottom face, while O represents the origin. - The force \(P\) is indicated by an arrow moving from point \(G\) towards vertex \(D\) on the face diagonal. - Axes are labeled as \(x\), \(y\), and \(z\), with an edge length of \(a = 3 \, \text{m}\). **Detailed Explanation:** **1. Determine the Moment of Force P About Point A:** To determine the moment of a force about a point, we use the vector cross product formula: \[ \mathbf{M_A} = \mathbf{r_A} \times \mathbf{P} \] **2. Projection of the Moment of P about Edge AB:** The projection of the moment about the edge AB can be found by determining the component of the moment vector along the direction of AB. **3. Projection of the Moment of P about Diagonal AG:** To find this projection, decompose the moment vector along the diagonal vector AG. **4. Parallelism or Perpendicularity of Vector P and Diagonal AG:** Assess the dot product of vector \(P\) and diagonal AG: \[ \mathbf{P} \cdot \mathbf{AG} \] If the dot product is zero, the vectors are perpendicular. If not, the angle (θ) between them can be found using: \[ \cos \theta = \frac{\mathbf{P} \cdot \mathbf{AG}}{|\mathbf{P}| |\mathbf{AG}|} \] This analysis is applicable for understanding force
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