Moment about the origin

[c4l3b]

Question from my A-level maths book.

Find the moment about the origin O = (0, 0, 0) of a force of 2 units pointing up in the direction of a vertical line and through the point
Q = (10, 5, 30).

\(\displaystyle M = r \times F\)

\(\displaystyle \vec{OQ}=\begin{pmatrix}10 \\ 5 \\ 30 \end{pmatrix}-\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 10 \\ 5 \\ 30 \end{pmatrix}\)
\(\displaystyle r=\begin{pmatrix} 10 \\ 5 \\ 30 \end{pmatrix}\)

Unsure about this part
\(\displaystyle M = \begin{pmatrix} 10 \\ 5 \\ 30 \end{pmatrix} \times 2 [Units] = 20i + 10j + 60k\)

3rd co-ordinate to be the vertical in a 3 co-ordinate system.

So force is 2k?
is that correct?

 

[mmm4444bot]

\(\displaystyle M = r \times F\)

\(\displaystyle M \; = \; <10, \; 5, \; 30> \times \; 2 \; = \; <20, \; 10, \; 60>\)

Hi c4l3b:

M, r, and F are all vectors, so this is a cross-product (not scalar multiplication).

\(\displaystyle \vec{M} = \vec{r} \times \vec{F}\)

Vector \(\displaystyle \vec{r}\) is a position vector, and I'm thinking that it can be drawn from the Origin to ANY point along the direction of \(\displaystyle \vec{F}\), so we can position \(\displaystyle \vec{F}\) with its tail at the point (10,5,30) and then draw the position vector \(\displaystyle \vec{r}\) from the Origin to the tail of \(\displaystyle \vec{F}\).

WARNING: My knowledge with these types of vector exercises is "rusty"; somebody, PLEASE verify my statements.

\(\displaystyle \vec{r} = <10, \; 5, \; 30>\)

We're told that the direction of \(\displaystyle \vec{F}\) is vertical, and its magnitude is 2, so we can write the components of \(\displaystyle \vec{F}\).

\(\displaystyle \vec{F} = <0, \; 0, \; 2>\)

Calculate the components of \(\displaystyle \vec{M}\) by crossing \(\displaystyle \vec{r}\) with \(\displaystyle \vec{F}\)

If you need more help, please show us your work with the cross-product. If I wrote anything that you don't understand, then please ask specific questions.

Cheers,

~ Mark