Calculating the Moment of a Force Vector

The problem asks us to find the moment of a given force \(\vec F\) applied at a specific point about another point. The moment of a force, often called torque, is a measure of its tendency to cause rotation about a point or axis. It is calculated using the vector cross product.

Understanding the Moment of Force Concept

The moment of a force \(\vec F\) about a point O is given by the vector product:

\[ \vec \tau = \vec r \times \vec F \]

where \(\vec r\) is the position vector from the point O (about which the moment is calculated) to the point of application of the force. In this problem, point O is (2, -1, 3) and the point of application is (1, -1, 2).

Step-by-Step Calculation

1. Identify the Given Force and Points

• Force vector: \(\vec F = 3\hat i + 2\hat j - 4\hat k\)
• Point of application of force (Point A): (1, -1, 2)
• Point about which moment is calculated (Point B): (2, -1, 3)

2. Determine the Position Vector \(\vec r\)

The position vector \(\vec r\) goes from point B to point A. We find this by subtracting the coordinates of B from the coordinates of A.

Position vector of A: \(\vec{r}_A = 1\hat i - 1\hat j + 2\hat k\)

Position vector of B: \(\vec{r}_B = 2\hat i - 1\hat j + 3\hat k\)

The position vector \(\vec r\) from B to A is:

\[ \vec r = \vec{r}_A - \vec{r}_B = (1\hat i - 1\hat j + 2\hat k) - (2\hat i - 1\hat j + 3\hat k) \]

\[ \vec r = (1 - 2)\hat i + (-1 - (-1))\hat j + (2 - 3)\hat k \]

\[ \vec r = -1\hat i + 0\hat j - 1\hat k \]

3. Calculate the Cross Product \(\vec r \times \vec F\)

Now we calculate the moment \(\vec \tau = \vec r \times \vec F\) using the determinant method for the cross product:

\[ \vec \tau = \begin{vmatrix} \hat i & \hat j & \hat k \\ -1 & 0 & -1 \\ 3 & 2 & -4 \end{vmatrix} \]

Expanding the determinant:

\[ \vec \tau = \hat i \begin{vmatrix} 0 & -1 \\ 2 & -4 \end{vmatrix} - \hat j \begin{vmatrix} -1 & -1 \\ 3 & -4 \end{vmatrix} + \hat k \begin{vmatrix} -1 & 0 \\ 3 & 2 \end{vmatrix} \]

Calculate the 2x2 determinants:

• For \(\hat i\): \((0 \times -4) - (-1 \times 2) = 0 - (-2) = 2\)
• For \(\hat j\): \((-1 \times -4) - (-1 \times 3) = 4 - (-3) = 4 + 3 = 7\)
• For \(\hat k\): \((-1 \times 2) - (0 \times 3) = -2 - 0 = -2\)

Substitute these values back:

\[ \vec \tau = \hat i (2) - \hat j (7) + \hat k (-2) \]

\[ \vec \tau = 2\hat i - 7\hat j - 2\hat k \]

Result

The moment of the force \(\vec F\) about the point (2, -1, 3) is \(2\hat i - 7\hat j - 2\hat k\).

Comparison with Options

Let's compare our calculated moment with the given options:

• Option 1: \(\hat i + 4\hat j + 4\hat k\)
• Option 2: \(2\hat i + \hat j + 2\hat k\)
• Option 3: \(2\hat i - 7\hat j - 2\hat k\)
• Option 4: \(2\hat i + 4\hat j - \hat k\)

Our result \(2\hat i - 7\hat j - 2\hat k\) matches Option 3.

Revision Table: Moment of Force

Additional Information: Vector Cross Product Properties

The vector cross product is fundamental to calculating moments. Here are some key properties:

• The result \(\vec A \times \vec B\) is a vector.
• The direction of \(\vec A \times \vec B\) is perpendicular to the plane containing \(\vec A\) and \(\vec B\), determined by the right-hand rule.
• The magnitude is \(|\vec A \times \vec B| = |\vec A| |\vec B| \sin \theta\), where \(\theta\) is the angle between \(\vec A\) and \(\vec B\).
• The cross product is anti-commutative: \(\vec B \times \vec A = -(\vec A \times \vec B)\).
• The cross product of a vector with itself is zero: \(\vec A \times \vec A = 0\).

In the context of moment, the magnitude \(|\vec \tau| = |\vec r| |\vec F| \sin \theta\), where \(\theta\) is the angle between \(\vec r\) and \(\vec F\). The quantity \(|\vec r| \sin \theta\) is the perpendicular distance from the pivot point to the line of action of the force.