Find the torque of a force vec F =( hat i -2 overline j +3 vec k )N about a point O. The position vector of point of application of force about O is vec r =(2 overline i + overline j - overline k )m.
To find the torque of a force about a point, we can use the cross product between the position vector and the force vector. The torque (\(\vec{\tau}\)) is given by:
[tex]\vec{\tau} = \vec{r} \times \vec{F}[/tex]
Given:
[tex]\vec{F} = (i - 2\vec{j} + 3\vec{k}) \, \mathrm{N}[/tex]
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Answer:
To find the torque of a force about a point, we can use the cross product between the position vector and the force vector. The torque (\(\vec{\tau}\)) is given by:
[tex]\vec{\tau} = \vec{r} \times \vec{F}[/tex]
Given:
[tex]\vec{F} = (i - 2\vec{j} + 3\vec{k}) \, \mathrm{N}[/tex]
[tex]\vec{r} = (2\hat{i} + \vec{j} - \vec{k}) \, \mathrm{m}[tex]
Let's calculate the cross product:
[tex]\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & -2 & 3 \end{vmatrix}[/tex]\)
Expanding the determinant, we get:
[tex]\vec{\tau} = (1 \cdot 3 - (-2) \cdot (-1))\hat{i} - (2 \cdot 3 - 1 \cdot (-1))\hat{j} + (2 \cdot (-2) - 1 \cdot 1)\hat{k}[/tex]\)
Simplifying further:
[tex]\vec{\tau} = 5\hat{i} - 5\hat{j} - 5\hat{k}[/tex]\)
Therefore, the torque of the force about point O is [tex]5\hat{i} - 5\hat{j} - 5\hat{k}[/tex]\) Nm.