# Homework Help: Using Cross Product to Find Torque

1. Nov 7, 2004

Force F = (4.0 N) i + (-2.0 N) k acts on a pebble with position vector r = (2.50 m) j + (-1.3 m) k, relative to the origin

What is the resulting torque acting on the pebble about a point with coordinates (2.0 m, 3.0 m, 2.0 m)?

Ok so I just completely forgot how to work with vectors when they are not acting about the origin so can anyone lead me in the right direction?

2. Nov 7, 2004

### arildno

The force acts on 2.5j-1-3k respective to your origin.
Find the representation of this point when translating your origin to (2,3,2).

3. Nov 7, 2004

### HobieDude16

pretty easy, just take the position crss with the force....
(0i+2.5J-1.3K)(4i+0j-2k)
and then, do the first term of hte first equation times all 3 of the 2nd, then the 2nd times all 3, then the 3rd.... and you have to remember, an i times an i or a j times a j, etc, is zero... and whichever 2 variables you use, the answer of the multiplication is the variable left out, so i times j is k... and it goes in a triangle like this....
i
k j
and you go from whatever the first one is, times the 2nd one, whatever is across is the letter you use, and if it goes clockwise(i.e. k times i), the vector doesnt affect the answer, if its counter clockwise (i.e. j times i) its negative of whatever the answer is.... send me an im at hobiedude16 if you want any help on the other problems, were just about done

4. Nov 7, 2004

### Parth Dave

Relative to (0, 0, 0) the pebbles position is (0, 2.5, -1.3). So what is it's position relative to (2, 3, 2)?

5. Nov 7, 2004

### HobieDude16

haha, my bad, didnt notice you wanted part b of the question
just do vector r minus coordinates and use that instead of vector r, my bad on that one... but same procedure