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Question: The torque of the force \(\overset{\rightarrow}{F} = (2\widehat{i} - 3\widehat{j} + 4\widehat{k})N\)...

The torque of the force F→=(2i^−3j^+4k^)N\overset{\rightarrow}{F} = (2\widehat{i} - 3\widehat{j} + 4\widehat{k})N acting at the point r→=(3i^+2j^+3k^)\overset{\rightarrow}{r} = (3\widehat{i} + 2\widehat{j} + 3\widehat{k}) m about the origin be

A

6i^−6j^+12k^6\widehat{i} - 6\widehat{j} + 12\widehat{k}

B

17i^−6j^−13k^17\widehat{i} - 6\widehat{j} - 13\widehat{k}

C

−6i^+6j^−12k^- 6\widehat{i} + 6\widehat{j} - 12\widehat{k}

D

−17i^+6j^+13k^- 17\widehat{i} + 6\widehat{j} + 13\widehat{k}

Answer

17i^−6j^−13k^17\widehat{i} - 6\widehat{j} - 13\widehat{k}

Explanation

Solution

τ→=r→×F→\overset{\rightarrow}{\tau} = \overset{\rightarrow}{r} \times \overset{\rightarrow}{F} =∣i^j^k^3232−34∣= \left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 3 & 2 & 3 \\ 2 & - 3 & 4 \end{matrix} \right|

=[(2×4)−(3×−3)]i^+[(2×3)−(3×4)]j^= \left\lbrack (2 \times 4) - (3 \times - 3) \right\rbrack\widehat{i} + \left\lbrack (2 \times 3) - (3 \times 4) \right\rbrack\widehat{j}

+[(3×−3)−(2×2)]k^=17i^−6j^−13k^+ \left\lbrack (3 \times - 3) - (2 \times 2) \right\rbrack\widehat{k} = 17\widehat{i} - 6\widehat{j} - 13\widehat{k}