Question

A force F = 5i + 2j – 5k acts on a particle whose
position vector ř = į – 2j + k. What is the torque
about the origin?​

Answer 1

$\mathtt{\huge{\fbox{Solution :)}}}$

Given ,

$\star \: \sf Force \: (f)</p><p></p><p> = 5\hat{ \imath} + 2 \hat{ \jmath} - \hat{ k}</p><p> \\ \\ \sf \star \: Position \: vector \: (r) = \hat{ \imath} - 2 \hat{ \jmath} + \hat{ k}$

We know that , torque is defined as the cross product of position vector of the force and the force

$\large \mathtt{ \fbox{Torque = force \times position \: vector }}$

Thus ,

$\sf \mapsto Torque = (5\hat{ \imath} + 2 \hat{ \jmath} - \hat{ k}) \times (\hat{ \imath} - 2 \hat{ \jmath} + \hat{ k}) \\ \\\sf \mapsto Torque = \hat{ \imath} \bigg((2 \times 1) - (( - 2) \times ( - 1) )\bigg) - \hat{ \jmath} \bigg((5 \times 1) - (1 \times ( - 1)) \bigg) + \hat{ k}\bigg(5 \times ( - 2) - (1 \times 2) \bigg) \\ \\\sf \mapsto Torque = \hat{ \imath} (0) - \hat{ \jmath }(6) + \hat{k} ( - 12) \\ \\ \sf \mapsto Torque = \hat{ \imath} (0) - \hat{ \jmath }(6) - \hat{k} ( 12)$

Hence , the torque is $\fbox{\hat{ \imath} (0) - \hat{ \jmath }(6) - \hat{k} ( 12) }$

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