The force vec{F} = 7widehat{i} + 3widehat{j} | vedclass

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Question

(A) Given position vector $\vec{r} = \hat{i} - \hat{j} + \hat{k}$ and force vector $\vec{F} = 7\hat{i} + 3\hat{j} - 5\hat{k}$.The torque $\vec{	au}$

Vedclass · Accepted Answer

$2\widehat{i} + 12\widehat{j} + 10\widehat{k}$

Vedclass · Answer

(A) Given position vector $\vec{r} = \hat{i} - \hat{j} + \hat{k}$ and force vector $\vec{F} = 7\hat{i} + 3\hat{j} - 5\hat{k}$.The torque $\vec{	au}$ is defined as the cross product of the position vector and the force vector: $\vec{	au} = \vec{r} 	imes \vec{F}$.Calculating the cross product using the determinant method:$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 7 & 3 & -5 \end{vmatrix}$Expanding the determinant:$\vec{	au} = \hat{i}((-1)(-5) - (1)(3)) - \hat{j}((1)(-5) - (1)(7)) + \hat{k}((1)(3) - (-1)(7))$$\vec{	au} = \hat{i}(5 - 3) - \hat{j}(-5 - 7) + \hat{k}(3 + 7)$$\vec{	au} = 2\hat{i} - \hat{j}(-12) + 10\hat{k}$$\vec{	au} = 2\hat{i} + 12\hat{j} + 10\hat{k}$

The force $\vec{F} = 7\widehat{i} + 3\widehat{j} - 5\widehat{k}$ acts on a particle whose position vector is $\vec{r} = \widehat{i} - \widehat{j} + \widehat{k}$. What is the torque of the given force about the origin?

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