A force vec{F} = 5hat{i} + 2hat{j} - 5hat{k} | vedclass

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(A) Given: $\vec{F} = 5\hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{r} = \hat{i} - 2\hat{j} + \hat{k}$.Torque $\vec{	au}$ is defined as the cross product

Vedclass · Accepted Answer

$8\hat{i} + 10\hat{j} + 12\hat{k}$

Vedclass · Answer

(A) Given: $\vec{F} = 5\hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{r} = \hat{i} - 2\hat{j} + \hat{k}$.Torque $\vec{	au}$ is defined as the cross product of the position vector and the force vector: $\vec{	au} = \vec{r} 	imes \vec{F}$.Using the determinant method:$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 1 \ 5 & 2 & -5 \end{vmatrix}$Expanding the determinant:$\vec{	au} = \hat{i}((-2)(-5) - (1)(2)) - \hat{j}((1)(-5) - (1)(5)) + \hat{k}((1)(2) - (-2)(5))$$\vec{	au} = \hat{i}(10 - 2) - \hat{j}(-5 - 5) + \hat{k}(2 + 10)$$\vec{	au} = 8\hat{i} - (-10)\hat{j} + 12\hat{k}$$\vec{	au} = 8\hat{i} + 10\hat{j} + 12\hat{k}$.

$A$ force $\vec{F} = 5\hat{i} + 2\hat{j} - 5\hat{k}$ acts on a particle whose position vector is $\vec{r} = \hat{i} - 2\hat{j} + \hat{k}$. What is the torque about the origin?

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