(D) The torque $\vec{\tau}$ about the origin is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.Given: $\ve

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(D) The torque $\vec{\tau}$ about the origin is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.Given: $\vec{F} = 3 \hat{j} \text{ N}$ and $\vec{r} = 2 \hat{k} \text{ m}$.$\vec{\tau} = \vec{r} \times \vec{F} = (2 \hat{k}) \times (3 \hat{j})$.Using the properties of unit vector cross products: $\hat{k} \times \hat{j} = -\hat{i}$.Therefore,$\vec{\tau} = 6 (\hat{k} \times \hat{j}) = 6(-\hat{i}) = -6 \hat{i} \text{ Nm}$.

Class 11 Physics System of Particles and Rotational Motion Torque and Couple

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Find the torque about the origin when a force of $3 \hat{j} \text{ N}$ acts on a particle whose position vector is $2 \hat{k} \text{ m}$.

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A
$6 \hat{k} \text{ Nm}$
B
$6 \hat{i} \text{ Nm}$
C
$6 \hat{j} \text{ Nm}$
D
$-6 \hat{i} \text{ Nm}$

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System of Particles and Rotational Motion

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Torque and Couple

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Find the torque about the origin when a force of $3 \hat{j} \text{ N}$ acts on a particle whose position vector is $2 \hat{k} \text{ m}$.

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$A$ 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?

If $\vec{F} = (4\hat{i} - 10\hat{j})$ and $\vec{r} = (5\hat{i} - 3\hat{j})$,then calculate the torque $\vec{\tau} = \vec{r} \times \vec{F}$. (in $\hat{k}$)

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