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Question: A force \(\overset{\rightarrow}{F =}4\overset{\land}{i} - 5\overset{\land}{j} + 3\overset{\land}{k}\...

A force F=→4i∧−5j∧+3k∧\overset{\rightarrow}{F =}4\overset{\land}{i} - 5\overset{\land}{j} + 3\overset{\land}{k} is acting a pointr1=→i∧+2j∧+3k∧\overset{\rightarrow}{r_{1} =}\overset{\land}{i} + 2\overset{\land}{j} + 3\overset{\land}{k}. The torque acting a but a point is:

A

Zero

B

42i∧−30j∧+6k∧42\overset{\land}{i} - 30\overset{\land}{j} + 6\overset{\land}{k}

C

42i∧+30j∧+6k∧42\overset{\land}{i} + 30\overset{\land}{j} + 6\overset{\land}{k}

D

42i∧+30j∧−6k∧42\overset{\land}{i} + 30\overset{\land}{j} - 6\overset{\land}{k}

Answer

42i∧+30j∧−6k∧42\overset{\land}{i} + 30\overset{\land}{j} - 6\overset{\land}{k}

Explanation

Solution

Position vector of the point at which force is acting r1→=i^+2j^+3k^\overset{\rightarrow}{r_{1}} = \widehat{i} + 2\widehat{j} + 3\widehat{k}

But we have to calculate the torque about another point. So its position vector about that another point

r1′→= 6mur1→−r2→=(i^+2j^+3k^)−(3i^−2j^−3k^)=−2i^+4j^+6k^\overset{\rightarrow}{r'_{1}} = \mspace{6mu}\overset{\rightarrow}{r_{1}} - \overset{\rightarrow}{r_{2}} = (\widehat{i} + 2\widehat{j} + 3\widehat{k}) - (3\widehat{i} - 2\widehat{j} - 3\widehat{k}) = - 2\widehat{i} + 4\widehat{j} + 6\widehat{k}

Now τ→=r1′→×F→=(−2i^+4j^+6k^)×(4i^−5j^+3k^)\overset{\rightarrow}{\tau} = \overset{\rightarrow}{r_{1}^{'}} \times \overset{\rightarrow}{F} = ( - 2\widehat{i} + 4\widehat{j} + 6\widehat{k}) \times (4\widehat{i} - 5\widehat{j} + 3\widehat{k})

\widehat{i} & \widehat{j} & \widehat{k} \\ - 2 & 4 & 6 \\ 4 & - 5 & 3 \end{matrix} \right| = \widehat{i}(12 + 30) - \widehat{j}( - 6 - 24) + \widehat{k}(10 - 16) = (42\widehat{i} + 30\widehat{j} - 6\widehat{k})N ⥂ - ⥂ m$$