Kelvi

Kelvi

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Kelvi

0 pts

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

medium

System of Particles and Rotational Motion

2018

physics

$-7 \hat{i} - 8 \hat{j} - 4 \hat{k}$

$-4 \hat{i} - \hat{j} - 8 \hat{k}$

Explanation

To find the moment of the force $\vec{F}$ about a point, we use the formula for the moment (torque): $\\ \vec{\tau} = \vec{r} \times \vec{F} \\$

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$-8 \hat{i} - 4 \hat{j} - 7 \hat{k}$

$-7 \hat{i} - 4 \hat{j} - 8 \hat{k}$

where

\vec{r}

is the position vector from the point about which the moment is calculated to the point of application of the force.

\\

Given:

\\ \vec{F} = 4 \hat{i} + 5 \hat{j} - 6 \hat{k} \\

Point of application of the force:

(2, 0, -3) \\

Point about which moment is calculated:

(2, -2, -2) \\

First, calculate the position vector

\vec{r}: \\ \vec{r} = (2 - 2) \hat{i} + (0 + 2) \hat{j} + (-3 + 2) \hat{k} \\ \vec{r} = 0 \hat{i} + 2 \hat{j} - 1 \hat{k} \\

Now, calculate the cross product

\vec{\tau} = \vec{r} \times \vec{F}: \\ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ 4 & 5 & -6 \\ \end{vmatrix} \\

Calculate the determinant:

\\ \vec{\tau} = \hat{i} \left( (2)(-6) - (-1)(5) \right) - \hat{j} \left( (0)(-6) - (-1)(4) \right) + \hat{k} \left( (0)(5) - (2)(4) \right) \\ \vec{\tau} = \hat{i} (-12 + 5) - \hat{j} (0 + 4) + \hat{k} (0 - 8) \\ \vec{\tau} = -7 \hat{i} - 4 \hat{j} - 8 \hat{k} \\

Therefore, the moment of the force about the point is

-7 \hat{i} - 4 \hat{j} - 8 \hat{k}. \\

This corresponds to Option 4.