A force vec{F} = 4hat{i} - 3hat{j} + 4hat{k} | vedclass

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Question

(A) The torque $\vec{	au}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.$\vec{	au} = \vec{r} 	imes

Vedclass · Accepted Answer

$17(\hat{i} - \hat{k})$

Vedclass · Answer

(A) The torque $\vec{	au}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.$\vec{	au} = \vec{r} 	imes \vec{F} = (3\hat{i} + 2\hat{j} + 3\hat{k}) 	imes (4\hat{i} - 3\hat{j} + 4\hat{k})$Using the determinant method for the cross product:$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 3 \ 4 & -3 & 4 \end{vmatrix}$$\vec{	au} = \hat{i} [(2)(4) - (3)(-3)] - \hat{j} [(3)(4) - (3)(4)] + \hat{k} [(3)(-3) - (2)(4)]$$\vec{	au} = \hat{i} [8 + 9] - \hat{j} [12 - 12] + \hat{k} [-9 - 8]$$\vec{	au} = 17\hat{i} - 0\hat{j} - 17\hat{k}$$\vec{	au} = 17(\hat{i} - \hat{k}) \, N \cdot m$

$A$ force $\vec{F} = 4\hat{i} - 3\hat{j} + 4\hat{k} \, N$ acts on a particle at a position $\vec{r} = 3\hat{i} + 2\hat{j} + 3\hat{k}$ from the origin. The torque acting on the particle is:

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