2.) please help with problem in photo **Problem Statement:**A force of $ \mathbf{F} = 12\mathbf{i} + 40\mathbf{j} $ [N] is applied to the end of a rod extending from the origin $ (x_0, y_0) = (0, 0) $ meters to point A with coordinates $ (x_A, y_A) = (6.9, 7.3) $ meters. Calculate the moment of force produced by $ \mathbf{F} $. Give the answer to 1 decimal place.**Explanation:**To solve this problem, you need to calculate the moment of force (also known as torque) about the origin. The moment $ \mathbf{M} $ of a force $ \mathbf{F} $ applied at a point with position vector $ \mathbf{r} $ is given by the cross product $ \mathbf{M} = \mathbf{r} \times \mathbf{F} $.- **Position Vector $ \mathbf{r} $:** From the origin to point A: $ \mathbf{r} = 6.9\mathbf{i} + 7.3\mathbf{j} $.- **Force Vector $ \mathbf{F} $:** Given as $ \mathbf{F} = 12\mathbf{i} + 40\mathbf{j} $.- **Moment (Torque) Calculation:** \[ \mathbf{M} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6.9 & 7.3 & 0 \\ 12 & 40 & 0 \end{vmatrix} \] Computing the cross product: \[ \mathbf{M} = (0\mathbf{i} - 0\mathbf{j} + (6.9 \times 40 - 7.3 \times 12)\mathbf{k}) \] \[ \mathbf{M} = (276 - 87.6)\mathbf{k} \] \[ \mathbf{M} = 188.4\mathbf{k} \]The moment of force produced by $ \mathbf{F} $ is \( 188.4 \, \text{Nm} \

A force of F = 12i + 40j [N] is applied to the end of a rod extending from the xo, Yo) = (0, 0) meters to point A with coordinates (XA, YA) = (6.9, 7.3) meters Calculate the moment of force produced by F. Give the answer to 1 decimal

A force of F = 12i + 40j [N] is applied to the end of a rod extending from the xo, Yo) = (0, 0) meters to point A with coordinates (XA, YA) = (6.9, 7.3) meters Calculate the moment of force produced by F. Give the answer to 1 decimal

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

2.) please help with problem in photo

**Problem Statement:**

A force of $ \mathbf{F} = 12\mathbf{i} + 40\mathbf{j} $ [N] is applied to the end of a rod extending from the origin $ (x_0, y_0) = (0, 0) $ meters to point A with coordinates $ (x_A, y_A) = (6.9, 7.3) $ meters.

Calculate the moment of force produced by $ \mathbf{F} $. Give the answer to 1 decimal place.

**Explanation:**

To solve this problem, you need to calculate the moment of force (also known as torque) about the origin. The moment $ \mathbf{M} $ of a force $ \mathbf{F} $ applied at a point with position vector $ \mathbf{r} $ is given by the cross product $ \mathbf{M} = \mathbf{r} \times \mathbf{F} $.

- **Position Vector $ \mathbf{r} $:** From the origin to point A: $ \mathbf{r} = 6.9\mathbf{i} + 7.3\mathbf{j} $.

- **Force Vector $ \mathbf{F} $:** Given as $ \mathbf{F} = 12\mathbf{i} + 40\mathbf{j} $.

- **Moment (Torque) Calculation:**
\[
\mathbf{M} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
6.9 & 7.3 & 0 \\
12 & 40 & 0
\end{vmatrix}
\]

Computing the cross product:
\[
\mathbf{M} = (0\mathbf{i} - 0\mathbf{j} + (6.9 \times 40 - 7.3 \times 12)\mathbf{k})
\]
\[
\mathbf{M} = (276 - 87.6)\mathbf{k}
\]
\[
\mathbf{M} = 188.4\mathbf{k}
\]

The moment of force produced by $ \mathbf{F} $ is \( 188.4 \, \text{Nm} \

Expert Solution

Step 1

$Given F = 12 i + 40 j (x_{o} y_{o}) = (0, 0) (x_{A} y_{B}) = (6.9, 7.3)$

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