Torque about a point given coordinates in three dimensions

[squelch]

Homework Statement

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)

Homework Equations

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

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

 

[Simon Bridge]

I have some notes for you:

Let the object occupy point A at (5,1,3) and let (4,1,0) be point B.

Conceptual niggle: surely the object may be large enough to encompass both points?

Subtracting the distances in each dimension from point A to point B, we see that $\vec{r}=-\hat{i}-3\hat{k}$

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

Because $\tau=\vec{F}\times\vec{r}$:

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

 

[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 :).

 

[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.

 

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

 

[haruspex]

$$9\hat i - 5\hat j - 5\hat k$$

I get different coefficients for j and k.

 

[nrqed]

I get different coefficients for j and k.

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

 

[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

 

[squelch]

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

 

[ehild]

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

ehild