Find the magnitude of the torque of a couple | vedclass

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(C) Let the position vectors of the points be $\vec{r}_1 = \hat{i} - \hat{j} + \hat{k}$ and $\vec{r}_2 = 2\hat{i} - 3\hat{j} - \hat{k}$.The force $\ve

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$5\sqrt{5}$

Vedclass · Answer

(C) Let the position vectors of the points be $\vec{r}_1 = \hat{i} - \hat{j} + \hat{k}$ and $\vec{r}_2 = 2\hat{i} - 3\hat{j} - \hat{k}$.The force $\vec{F} = 3\hat{i} + 2\hat{j} - \hat{k}$ acts at $\vec{r}_1$ and $-\vec{F}$ acts at $\vec{r}_2$.The torque $\vec{	au}$ of the couple is given by $\vec{	au} = \vec{r} 	imes \vec{F}$,where $\vec{r} = \vec{r}_1 - \vec{r}_2$.$\vec{r} = (\hat{i} - \hat{j} + \hat{k}) - (2\hat{i} - 3\hat{j} - \hat{k}) = -\hat{i} + 2\hat{j} + 2\hat{k}$.$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & 2 \ 3 & 2 & -1 \end{vmatrix} = \hat{i}(-2 - 4) - \hat{j}(1 - 6) + \hat{k}(-2 - 6) = -6\hat{i} + 5\hat{j} - 8\hat{k}$.The magnitude of the torque is $|\vec{	au}| = \sqrt{(-6)^2 + 5^2 + (-8)^2} = \sqrt{36 + 25 + 64} = \sqrt{125} = 5\sqrt{5}$.

Find the magnitude of the torque of a couple formed by a force $\vec{F} = 3\hat{i} + 2\hat{j} - \hat{k}$ acting at the point $\hat{i} - \hat{j} + \hat{k}$ and the force $-\vec{F}$ acting at the point $2\hat{i} - 3\hat{j} - \hat{k}$.

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