1.

The moment of a force 3i-4j+12k acting at a point (1,- 2, 3) will have x-component of moment about a point (2,1,2), Mx = ?​

Answer»

\underline{\textbf{Given:}}

\mathsf{Force,\;\overrightarrow{F}=3\overrightarrow{i}-4\overrightarrow{j}+12\overrightarrow{k}\;acting\;at\;a\;point}

\textsf{(1,-2,3)}

\underline{\textbf{To find:}}

\textsf{Moment of the force the about the point (2,1,2)}

\underline{\textbf{Solution:}}

\mathsf{\overrightarrow{F}=3\overrightarrow{i}-4\overrightarrow{j}+12\overrightarrow{k}}

\mathsf{\overrightarrow{r}=(1\overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k})-(2\overrightarrow{i}+\overrightarrow{j}+2\overrightarrow{k})}

\implies\mathsf{\overrightarrow{r}=-1\overrightarrow{i}-3\overrightarrow{j}+\overrightarrow{k}}

\mathsf{Moment\;of\;force}

\mathsf{\overrightarrow{M}=\overrightarrow{r}{\times}\overrightarrow{F}}

\mathsf{\overrightarrow{M}=\left|\begin{array}{ccc}\overrightarrow{i}&\overrightarrow{j}&\overrightarrow{k}\\-1&-3&1\\3&-4&12\end{array}\right|}

\mathsf{=\overrightarrow{i}(-36+4)-\overrightarrow{j}(-12-3)+\overrightarrow{k}(4+9)}

\mathsf{=-32\overrightarrow{i}-\overrightarrow{j}(-15)+13\overrightarrow{k}}

\mathsf{=-32\overrightarrow{i}+15\overrightarrow{j}+13\overrightarrow{k}}

\mathsf{x\;component\;of\;moment=-32}

\implies\boxed{\mathsf{M_x=-32}}



Discussion

No Comment Found

Related InterviewSolutions