Question

The moment of the force, \(\overrightarrow{F}=4\hat{i}+5\hat{j}-6\hat{k}\) at (2, 0, -3), about the point (2, -2, -2), is given by

• A. \(-8\hat{i}-4\hat{j}-7\hat{k}\)
• B. \(-7\hat{i}-8\hat{j}-4\hat{k}\)
• C. \(-4\hat{i}-\hat{j}-8\hat{k}\)
• D. \(-7\hat{i}-4\hat{j}-8\hat{k}\)

Answer 1

The Correct Option is D

To find the moment of the force \(\overrightarrow{F}=4\hat{i}+5\hat{j}-6\hat{k}\) at the point (2, 0, -3) about the point (2, -2, -2), we use the concept of the moment of a force about a point in three-dimensional space.

The moment of a force \(\overrightarrow{F}\) about a point is given by the cross product of the position vector \(\overrightarrow{r}\) (from the point about which moment is taken to the point of application of the force) and the force vector \(\overrightarrow{F}\):

\(\overrightarrow{M} = \overrightarrow{r} \times \overrightarrow{F}\)

First, we determine the position vector \(\overrightarrow{r}\) from point (2, -2, -2) to point (2, 0, -3):

• The difference in x-coordinates: \(2 - 2 = 0\)
• The difference in y-coordinates: \(0 - (-2) = 2\)
• The difference in z-coordinates: \(-3 - (-2) = -1\)

This gives the position vector:

\(\overrightarrow{r} = 0\hat{i} + 2\hat{j} - 1\hat{k}\)

Now, we compute the cross product \(\overrightarrow{r} \times \overrightarrow{F}\):

\(\overrightarrow{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ 4 & 5 & -6 \end{vmatrix}\)

This determinant expands to:

• Coefficient of \(\hat{i}\): \((2)(-6) - (-1)(5) = -12 + 5 = -7\)
• Coefficient of \(\hat{j}\): \((0)(-6) - (-1)(4) = 0 + 4 = 4\) (change sign to -4)
• Coefficient of \(\hat{k}\): \((0)(5) - (2)(4) = 0 - 8 = -8\)

Therefore, the moment of the force is:

\(\overrightarrow{M} = -7\hat{i} - 4\hat{j} - 8\hat{k}\)

Thus, the correct answer is:

\(-7\hat{i} - 4\hat{j} - 8\hat{k}\)