logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  TN XII Math  >>  Vector Algebra
0 votes

Forces $\overrightarrow{2i}+\overrightarrow{7j}, \overrightarrow{2i}-\overrightarrow{5j}+\overrightarrow{6k}, \overrightarrow{-i}+\overrightarrow{2j}-\overrightarrow{k}$ act at a point $P$ Whose position vector is $\overrightarrow{4i}-\overrightarrow{3j}-\overrightarrow{2k}.$ Find the moment of the resultant of three forces acting at $P$ about the point $Q$ whose position vector is $\overrightarrow{6i}+\overrightarrow{j}-\overrightarrow{3k}.$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • 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.
The resultant of the given forces is
$ \overrightarrow F = (2 \overrightarrow i + 7 \overrightarrow j)+(2 \overrightarrow i-5 \overrightarrow j+6 \overrightarrow k) + ( - \overrightarrow i+2 \overrightarrow j- \overrightarrow k)$
$ = 3 \overrightarrow i+4 \overrightarrow j+5 \overrightarrow k$
The moment of $ \overrightarrow F$ acting at $ p( \overrightarrow {OP} = 4 \overrightarrow i-3 \overrightarrow j-2 \overrightarrow k) $ about $ Q( \overrightarrow {OQ} = 6 \overrightarrow i+ \overrightarrow j-3 \overrightarrow k)$ is given by
$ \overrightarrow M = \overrightarrow r \times \overrightarrow F$ where $ \overrightarrow r = \overrightarrow {QP} = ( \overrightarrow {OP}- \overrightarrow {OQ})$
$ = (4 \overrightarrow i-3 \overrightarrow j-2 \overrightarrow k)-(6 \overrightarrow i+ \overrightarrow j-3 \overrightarrow k)$
$ = -2 \overrightarrow i-4 \overrightarrow j+ \overrightarrow k$
$ \overrightarrow M = (-2 \overrightarrow i-4 \overrightarrow j+ \overrightarrow k ) \times (3 \overrightarrow i+4 \overrightarrow j+5 \overrightarrow k)$
$ = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ -2 & -4 & 1 \\ 3 & 4 & 5 \end{vmatrix} = (-20-4) \overrightarrow i-(-10-3) \overrightarrow j+(-8+12) \overrightarrow k$
$ = -24 \overrightarrow i+13 \overrightarrow j+4 \overrightarrow k$
$ | \overrightarrow M| = \sqrt{576+169+16} = \sqrt{761}$
answered Jun 6, 2013 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...