A force F = 2.0,N acts on a particle P in the | vedclass

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(B) The torque $\vec{	au}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$,given by $\vec{	au} = \ve

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(B) The torque $\vec{	au}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$,given by $\vec{	au} = \vec{r} 	imes \vec{F}$.The magnitude of the torque is given by $	au = rF \sin 	heta$,where $	heta$ is the angle between the position vector $\vec{r}$ and the force vector $\vec{F}$.Given:$F = 2.0\,N$$r = 3\,m$Since the force $F$ is parallel to the $x-$ axis and the position vector $\vec{r}$ makes an angle of $30^\circ$ with the $x-$ axis,the angle $	heta$ between $\vec{r}$ and $\vec{F}$ is $30^\circ$.Substituting these values into the formula:$	au = (3\,m)(2.0\,N) \sin 30^\circ$$	au = (3\,m)(2.0\,N) \left(\frac{1}{2}ight)$$	au = 3\,N-m$.

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