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A force `F=(2hati+3hatj+4hatk)N` is acting at point `P(2m,-3m,6m)` find torque of this force about a point `O` whose position vector is `(2hati-5hatj+3hatk)` m. |
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Answer» `tau=rxxF` Here `r=r_(P)-r_(O)=(2hati+3hatj+6hatk)-(2hati-5hatj+3hatk)=(2hatj+3hatk)m` Now `tau=rxxF=|{:(hati,hatj,hatk),(0,2,3),(2,3,-4):}|=(-17hati+6hatj-4hatk)N-m` |
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