A force overrightarrow{F} = (hat{i} + 2hat{j} | vedclass

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(B) The torque $\vec{	au}$ about a point $\vec{r}_2$ is given by $\vec{	au} = (\vec{r}_1 - \vec{r}_2) 	imes \vec{F}$.First,calculate the position v

Vedclass · Accepted Answer

$195$

Vedclass · Answer

(B) The torque $\vec{	au}$ about a point $\vec{r}_2$ is given by $\vec{	au} = (\vec{r}_1 - \vec{r}_2) 	imes \vec{F}$.First,calculate the position vector relative to the point of rotation:$\vec{r} = \vec{r}_1 - \vec{r}_2 = (4\hat{i} + 3\hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) = 3\hat{i} + \hat{j} - 2\hat{k}$.Now,calculate the cross product $\vec{	au} = \vec{r} 	imes \vec{F}$:$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & -2 \ 1 & 2 & 3 \end{vmatrix}$$= \hat{i}(1(3) - (-2)(2)) - \hat{j}(3(3) - (-2)(1)) + \hat{k}(3(2) - 1(1))$$= \hat{i}(3 + 4) - \hat{j}(9 + 2) + \hat{k}(6 - 1) = 7\hat{i} - 11\hat{j} + 5\hat{k}$.The magnitude of the torque is $|\vec{	au}| = \sqrt{7^2 + (-11)^2 + 5^2} = \sqrt{49 + 121 + 25} = \sqrt{195}$.Comparing this with $\sqrt{x}$,we get $x = 195$.

$A$ force $\overrightarrow{F} = (\hat{i} + 2\hat{j} + 3\hat{k}) \text{ N}$ acts at a point $\vec{r}_1 = (4\hat{i} + 3\hat{j} - \hat{k}) \text{ m}$. The magnitude of torque about the point $\vec{r}_2 = (\hat{i} + 2\hat{j} + \hat{k}) \text{ m}$ is $\sqrt{x} \text{ N-m}$. The value of $x$ is $........$

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