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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?

1 Answers 1

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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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    Do I have to do anything like find a direction sin or cos? – 2017-01-27
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    wait, now it makes sense with the j. I am, updating it – 2017-01-27
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    Just realized, that is a linear combination of the OP vector. – 2017-01-27
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    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. – 2017-01-27
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    @fireballs so the answer is just OPxF? – 2017-01-27
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    @ArnoldFrenzy that's it?? Just take the cross product OPxF?? – 2017-01-27