7.5 Q-11

Question Statement

A force $\underline{F} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k}$ is applied at the point $P(1, -1, 2)$. Find the moment of force about the point $A(2, -1, 3)$.

---

Background and Explanation

The moment of a force about a point (also called the torque) is a vector quantity that describes the rotational effect of a force about that point. It is calculated using the cross product of the position vector $\overrightarrow{A \vec{P}}$ (from the point where the moment is being calculated to the point where the force is applied) and the force vector $\overrightarrow{F}$.

The formula for the moment is:

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

Where:

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

---

Solution

Step 1: Define the given vectors

We are given:

• The force vector $\overrightarrow{F} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k}$,
• Point $P(1, -1, 2)$ where the force is applied,
• Point $A(2, -1, 3)$ where the moment is to be calculated.

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

The displacement vector $\overrightarrow{A \vec{P}}$ is found by subtracting the coordinates of point $A(2, -1, 3)$ from point $P(1, -1, 2)$:

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

Thus, the displacement vector is:

$$\overrightarrow{A \vec{P}} = -\hat{i} + 0 \hat{j} - \hat{k}$$

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

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

$$\overrightarrow{A \vec{P}} \times \overrightarrow{F} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} -1 & 0 & -1 3 & 2 & -4 \end{array} \right|$$

Now, expand the determinant:

$$\overrightarrow{A \vec{P}} \times \overrightarrow{F} = \hat{i} \left( (0)(-4) - (-1)(2) \right) - \hat{j} \left( (-1)(-4) - (-1)(3) \right) + \hat{k} \left( (-1)(2) - (0)(3) \right)$$

Simplifying each term:

$$= \hat{i} (0 + 2) - \hat{j} (4 + 3) + \hat{k} (-2 + 0)$$ $$= 2 \hat{i} - 7 \hat{j} - 2 \hat{k}$$

Thus, the moment of the force is:

$$\overrightarrow{M} = 2 \hat{i} - 7 \hat{j} - 2 \hat{k}$$

---

Key Formulas or Methods Used

• Moment of Force (Torque):

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

The moment (or torque) is the cross product of the displacement vector and the force vector.

• 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 $\overrightarrow{A \vec{P}}$ by subtracting the coordinates of point $A$ from point $P$.
2. Compute the cross product $\overrightarrow{A \vec{P}} \times \overrightarrow{F}$ to find the moment of the force.
3. Simplify the cross product to obtain the moment vector $\overrightarrow{M} = 2 \hat{i} - 7 \hat{j} - 2 \hat{k}$.