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Let r equal the line segment OP=i-j+2k. A force F=<10,10,0> is applied at P. Find the torque O produced.

In calculus we learned that torque is equal to the cross product of the force vector and the radius (arm). However I am having the feeling that in this situation, the answer may not be as simple.

Can anyone confirm or refute?

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Yes, by definition $\vec{\tau} = \vec{r} \times \vec{F}$

Assuming $i,j,k$ are the unity vectors specifying the Cartesian coordinates, then:

\begin{equation} \vec{r} \times \vec{F} = \begin{vmatrix} i && j && k \\ 1 && -1 && 2 \\ F_x && F_y && F_z \\ \end{vmatrix} = \begin{vmatrix} i && j && k \\ 1 && -1 && 2 \\ 10 && 10 && 0 \\ \end{vmatrix} = (-20,20,20) \end{equation}

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  • $\begingroup$ Do I have to do anything like find a direction sin or cos? $\endgroup$ – user224997 Jan 27 '17 at 4:17
  • $\begingroup$ wait, now it makes sense with the j. I am, updating it $\endgroup$ – A. Frenzy Jan 27 '17 at 4:18
  • $\begingroup$ Just realized, that is a linear combination of the OP vector. $\endgroup$ – user224997 Jan 27 '17 at 4:18
  • $\begingroup$ You will not need to explicitly find a direction using trig. When you take the cross product, you should get a vector that naturally has a direction. $\endgroup$ – gabe Jan 27 '17 at 4:19
  • $\begingroup$ @fireballs so the answer is just OPxF? $\endgroup$ – user224997 Jan 27 '17 at 4:21

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