# Torque about a point given coordinates in three dimensions

1. Aug 30, 2014

### squelch

1. The problem statement, all variables and given/known data

Let $\vec{F}=2\hat{i}-3\hat{j}$ act on an object at point (5,1,3). Find the torque about the point (4,1,0)

2. Relevant equations

$\tau = \vec F \times \vec r$

3. The attempt at a solution

Please tell me if my procedure is correct.

Let the object occupy point A at (5,1,3) and let (4,1,0) be point B.
Subtracting the distances in each dimension from point A to point B, we see that $\vec{r}=-\hat{i}-3\hat{k}$

Because $\tau=\vec{F}\times\vec{r}$:
$$\tau = \left[ {\begin{array}{*{20}{c}} 2\\ { - 3}\\ 1 \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} { - 1}\\ 0\\ { - 3} \end{array}} \right] = 6\hat i + 5\hat j - \hat k$$

2. Aug 30, 2014

### Simon Bridge

I have some notes for you:
Conceptual niggle: surely the object may be large enough to encompass both points?

If B is the center of rotation, and A is the point the force is applied, then $\vec r = \overrightarrow{BA}$

$$\vec \tau = \vec r \times \vec F$$http://en.wikipedia.org/wiki/Torque
The order is important.

Since your $\vec r$ appears to be $\overrightarrow{AB}$ it may come out in the wash.
However, it is best practice to stick to standard definitions.

It can be useful to make the calculation more explicit too:
$$\vec r \times \vec F = \begin{vmatrix}\hat\imath & \hat\jmath & \hat k\\ r_x & r_y & r_z\\ F_x & F_y & F_z \end{vmatrix}$$
http://en.wikipedia.org/wiki/Cross_product#Matrix_notation

Last edited: Aug 30, 2014
3. Aug 30, 2014

### squelch

The way you described it (calling B the "center of rotation" and pointing out that the two points could be on the same object) really helped.

So the biggest problems you spotted was that I found $\vec{r}$ by subtracting each set of coordinates in the wrong order, and took the cross product in the wrong order? Essentially, that I was seeing things "backwards?" Other than that, was the procedure sound?

Also, the notation you used to make the calculation more explicit is what I had on paper, I just wasn't sure how to express it in latex :).

4. Aug 30, 2014

### Simon Bridge

Find the moment arm vector and cross it with the force vector ... that's the correct procedure all right.
There are other approaches but they are usually messier.

5. Aug 30, 2014

### squelch

Okay, so I corrected it on paper. What I have down is:

Let point A be (4,1,0), the center of rotation for a force $\vec{F}=2\hat{i}-3\hat{j}$ acting on an object at point B, (5,1,3). Then $\vec{r}=\vec{AB}=\hat{i}+3\hat{k}$.

Therefore,
$$\tau = \vec r \times \vec F = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 1&0&3\\ 2&{ - 3}&0 \end{array}} \right| = 9\hat i - 5\hat j - 5\hat k$$

6. Aug 30, 2014

### haruspex

I get different coefficients for j and k.

7. Aug 30, 2014

### nrqed

And you are correct. Squelch, you made two small mistakes, probably due to inattention.

8. Aug 30, 2014

### squelch

I'm not seeing the mistake. I was taught to take the cross product in the way shown in the picture, summing the components from left to right and subtracting the components from right to left.

edit: yeah, wow, i see that mistake now

Last edited: Aug 31, 2014
9. Aug 31, 2014

### squelch

I tried a different method (taking determinant) and got 9i+6j-3k. Does that match up better?

10. Aug 31, 2014

### ehild

It is correct. Previously, you took the product 2*0*k=2k.

ehild