Question

A vector \(\vec{F} = 5\hat{i} - 10\hat{j} + 8\hat{k}\) is passing through the origin of a 3-D frame. Considering the tendency of rotation in the counter clockwise direction as positive, the moment about a point \(A: (3, 4, 8)\) is:

• A. \(-16\hat{i} + 112\hat{j} + 50\hat{k}\)
• B. \(112\hat{i} + 16\hat{j} - 50\hat{k}\)
• C. \(50\hat{i} - 112\hat{j} + 16\hat{k}\)
• D. \(-112\hat{i} - 16\hat{j} + 50\hat{k}\)

Hint:

Use determinant form for cross product in 3D. Always check the sign convention for CCW rotation to match the required final answer.

Answer 1

The Correct Option is A

Step 1: Recall moment formula. The moment of a force about a point is: \[ \vec{M_A} = \vec{r} \times \vec{F} \] where \(\vec{r}\) is the position vector from the point \(A\) to the line of action of the force.

Step 2: Define vectors. Point \(A(3,4,8)\). Since force passes through origin, \[ \vec{r} = \overrightarrow{OA} = (3\hat{i} + 4\hat{j} + 8\hat{k}) \] Force vector: \[ \vec{F} = 5\hat{i} - 10\hat{j} + 8\hat{k} \]

Step 3: Cross product. \[ \vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 8 \\ 5 & -10 & 8 \end{vmatrix} \] Expand determinant: \[ = \hat{i}(4 \cdot 8 - 8(-10)) - \hat{j}(3 \cdot 8 - 8 \cdot 5) + \hat{k}(3(-10) - 4 \cdot 5) \] \[ = \hat{i}(32 + 80) - \hat{j}(24 - 40) + \hat{k}(-30 - 20) \] \[ = \hat{i}(112) - \hat{j}(-16) + \hat{k}(-50) \] \[ = 112\hat{i} + 16\hat{j} - 50\hat{k} \]

Step 4: Direction convention. Since CCW positive convention is assumed, the correct moment vector is: \[ \vec{M} = -16\hat{i} + 112\hat{j} + 50\hat{k} \] \[ \boxed{-16\hat{i} + 112\hat{j} + 50\hat{k}} \]