7.5 Q-13

Question Statement

A force $\underline{F} = 2 \hat{i} + \hat{j} - 3 \hat{k}$ is applied at point $A(1, -2, 1)$. Find the moment of the force about the point $B(2, 0, -2)$.

---

Background and Explanation

The moment of a force (or torque) about a point is a vector quantity that measures the rotational effect of the force relative to that point. It is computed using the cross product of the position vector $\overrightarrow{B \vec{A}}$ (from the point where the moment is to be calculated to the point where the force is applied) and the force vector $\overrightarrow{F}$.

The formula for the moment is:

$$\overrightarrow{M} = \overrightarrow{B \vec{A}} \times \overrightarrow{F}$$

Where:

• $\overrightarrow{B \vec{A}}$ is the displacement vector from point $B$ to point $A$,
• $\overrightarrow{F}$ is the force vector,
• The cross product $\times$ gives the moment vector.

---

Solution

Step 1: Define the given vectors

We are given:

• The force vector $\overrightarrow{F} = 2 \hat{i} + \hat{j} - 3 \hat{k}$,
• Point $A(1, -2, 1)$ where the force is applied,
• Point $B(2, 0, -2)$ where the moment is to be calculated.

Step 2: Calculate the displacement vector $\overrightarrow{B \vec{A}}$

The displacement vector $\overrightarrow{B \vec{A}}$ is found by subtracting the coordinates of point $B(2, 0, -2)$ from point $A(1, -2, 1)$:

$$\overrightarrow{B \vec{A}} = A - B = (1, -2, 1) - (2, 0, -2)$$ $$\overrightarrow{B \vec{A}} = (1 - 2, -2 - 0, 1 + 2) = (-1, -2, 3)$$

Thus, the displacement vector is:

$$\overrightarrow{B \vec{A}} = -\hat{i} - 2 \hat{j} + 3 \hat{k}$$

Step 3: Compute the cross product $\overrightarrow{B \vec{A}} \times \overrightarrow{F}$

The moment is calculated by the cross product of $\overrightarrow{B \vec{A}}$ and $\overrightarrow{F}$:

$$\overrightarrow{B \vec{A}} \times \overrightarrow{F} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} -1 & -2 & 3 2 & 1 & -3 \end{array} \right|$$

Now, expand the determinant:

$$\overrightarrow{B \vec{A}} \times \overrightarrow{F} = \hat{i} \left( (-2)(-3) - (3)(1) \right) - \hat{j} \left( (-1)(-3) - (3)(2) \right) + \hat{k} \left( (-1)(1) - (-2)(2) \right)$$

Simplifying each term:

$$= \hat{i} (6 - 3) - \hat{j} (3 - 6) + \hat{k} (-1 + 4)$$ $$= 3 \hat{i} + 3 \hat{j} + 3 \hat{k}$$

Thus, the moment of the force is:

$$\overrightarrow{M} = -3 \hat{i} + 3 \hat{j} + 3 \hat{k}$$

---

Key Formulas or Methods Used

• Moment of Force (Torque):

$$\overrightarrow{M} = \overrightarrow{B \vec{A}} \times \overrightarrow{F}$$

The moment is the cross product of the displacement vector $\overrightarrow{B \vec{A}}$ and the force vector $\overrightarrow{F}$.

• Cross Product: The cross product of two vectors $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$ is calculated as:

$$\vec{A} \times \vec{B} = (A_y B_z - A_z B_y) \hat{i} - (A_x B_z - A_z B_x) \hat{j} + (A_x B_y - A_y B_x) \hat{k}$$

---

Summary of Steps

1. Calculate the displacement vector $\overrightarrow{B \vec{A}}$ by subtracting the coordinates of point $B$ from point $A$.
2. Compute the cross product $\overrightarrow{B \vec{A}} \times \overrightarrow{F}$ to find the moment of the force.
3. Simplify the cross product to obtain the moment vector $\overrightarrow{M} = -3 \hat{i} + 3 \hat{j} + 3 \hat{k}$.