7.5 Q-12

Question Statement

A force $\underline{F} = 4 \hat{i} - 3 \hat{k}$ passes through the point $A(2, -2, 5)$. Find the moment of force about the point $B(1, -3, 1)$.

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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 calculated 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.

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Solution

Step 1: Define the given vectors

We are given:

• The force vector $\overrightarrow{F} = 4 \hat{i} + 0 \hat{j} - 3 \hat{k}$,
• Point $A(2, -2, 5)$ where the force passes through,
• Point $B(1, -3, 1)$ where we need to calculate the moment.

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 $B(1, -3, 1)$ from point $A(2, -2, 5)$:

$$\overrightarrow{B \vec{A}} = A - B = (2, -2, 5) - (1, -3, 1)$$ $$\overrightarrow{B \vec{A}} = (2 - 1, -2 + 3, 5 - 1) = (1, 1, 4)$$

Thus, the displacement vector is:

$$\overrightarrow{B \vec{A}} = \hat{i} + \hat{j} + 4 \hat{k}$$

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 $\overrightarrow{F}$:

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

Now, expand the determinant:

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

Simplifying each term:

$$= \hat{i} (-3) - \hat{j} (-3 - 16) + \hat{k} (0 - 4)$$ $$= -3 \hat{i} - \hat{j} (-19) + (-4) \hat{k}$$ $$= -3 \hat{i} + 19 \hat{j} - 4 \hat{k}$$

Thus, the moment of the force is:

$$\overrightarrow{M} = -3 \hat{i} + 19 \hat{j} - 4 \hat{k}$$

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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:

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

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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 $\overrightarrow{M} = -3 \hat{i} + 19 \hat{j} - 4 \hat{k}$.