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Home  >>  TN XII Math  >>  Vector Algebra
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Show that the torque about the point $ A(3, -1, 3 )$ of a force $\overrightarrow{4i}+\overrightarrow{2j}\overrightarrow{k}$ throught the point $\overrightarrow\;B(5, 2, 4)$ is $\overrightarrow{i}+\overrightarrow{2j}-\overrightarrow{8k}$.

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  • If $ \overrightarrow a = a_1\overrightarrow i+a_2\overrightarrow j+a_3\overrightarrow k, \: \overrightarrow b = b_1\overrightarrow i+b_2\overrightarrow j+b_3\overrightarrow k$ then $ \overrightarrow a$ x $ \overrightarrow b = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
  • Moment or torque of force $ \overrightarrow F$ about the point 0 is $ \overrightarrow M = \overrightarrow r \times \overrightarrow f$ where $\overrightarrow r = \overrightarrow {OP}, p$ being the point of application of the force.
$ \overrightarrow F = 4\overrightarrow i+2\overrightarrow j+\overrightarrow k$
To find the moment of $ \overrightarrow F$ acting at $ B ( 5, 2, 4)$
$ [ \overrightarrow {OB} = 5\overrightarrow i+2\overrightarrow j+4\overrightarrow k] $ about the point $ A(3, -1, 3)$
$ [ \overrightarrow {OA} = 3\overrightarrow i-\overrightarrow j+3\overrightarrow k]$
$ \overrightarrow M = \overrightarrow r \times \overrightarrow F = \overrightarrow {AB} \times \overrightarrow F$
$ = [ (5-3)\overrightarrow i + (2+1)\overrightarrow j+(4-3)\overrightarrow k ] \times [ 4\overrightarrow i+2\overrightarrow j+\overrightarrow k]$
$ = (2\overrightarrow i+3\overrightarrow j+\overrightarrow k ) \times (4\overrightarrow i+2\overrightarrow j+\overrightarrow k)$
$ = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 2 & 3 & 1 & \\ 4 & 2 & 1 \end{vmatrix} = (3-2)\overrightarrow i-(2-4)\overrightarrow j+(4-12)\overrightarrow k$
$ = \overrightarrow i+2\overrightarrow j-8\overrightarrow k$
answered Jun 6, 2013 by thanvigandhi_1
 

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