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Question: Three forces \(\mathbf{i} + 2\mathbf{j} - 3\mathbf{k},2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\) and ...

Three forces i+2j3k,2i+3j+4k\mathbf{i} + 2\mathbf{j} - 3\mathbf{k},2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k} and ij+k\mathbf{i} - \mathbf{j} + \mathbf{k} are acting on a particle at the point (0, 1, 2). The magnitude of the moment of the forces about the point (1, – 2, 0) is

A

13(p+q+r)\frac{1}{3}(\mathbf{p} + \mathbf{q} + \mathbf{r})

B

6106\sqrt{10}

C

4174\sqrt{17}

D

None of these

Answer

6106\sqrt{10}

Explanation

Solution

Total force

F=(i+2j3k)+(2i+3j+4k)+(ij+k)=4i+4j+2k\overset{\rightarrow}{F} = (\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) + (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) + (\mathbf{i} - \mathbf{j} + \mathbf{k}) = 4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}Moment of the forces about P = r×F=PA×F\mathbf{r} \times \overset{\rightarrow}{F} = \overset{\rightarrow}{PA} \times \overset{\rightarrow}{F}

PA=(01)i+(1+2)j+(20)k=i+3j+2k\overset{\rightarrow}{PA} = (0 - 1)\mathbf{i} + (1 + 2)\mathbf{j} + (2 - 0)\mathbf{k} = - \mathbf{i} + 3\mathbf{j} + 2\mathbf{k}

\therefore Moment about P = (i+3j+2k)×(4i+4j+2k)( - \mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) \times (4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k})

=$\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \

  • 1 & 3 & 2 \ 4 & 4 & 2 \end{matrix} \right| = - 2\mathbf{i} + 10\mathbf{j} - 16\mathbf{k}$

Magnitude of the moment = 2i+10j16k| - 2\mathbf{i} + 10\mathbf{j} - 16\mathbf{k}|

= 212+52+82=290=6102\sqrt{1^{2} + 5^{2} + 8^{2}} = 2\sqrt{90} = 6\sqrt{10}