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Rotational Cross Product

  • Thread starter Nanuven
  • Start date
11
0
Force F = (9.0 N) i hat + (-4.0 N) k hat acts on a pebble with position vector r = (3.00 m) j hat + (-1.0 m) k hat, relative to the origin.


(a) What is the resulting torque acting on the pebble about the origin?
( N·m) i hat + ( N·m) j hat + ( N·m) k hat

(b) What is the resulting torque acting on the pebble about a point with coordinates (0.0 m, 0.0 m, 5.0 m)?
( N·m) i hat + ( N·m) j hat + ( N·m) k hat



So all I could come up with was cross product was that torque = r cross F

Therefore T = fRsin[tex]\theta[/tex]

I looked up cross product on Wikipedia (not best site I know but w/e) and it came up with:

i x j = k
j x k = i
k x i = j

I tried those but it doesn't seem to be giving me the right answer. Any help please??



3. The Attempt at a Solution
 

Answers and Replies

Mentz114
Gold Member
5,424
290
[tex]F = \[ \left[ \begin{array}{c} 9 \\\ 0 \\\ -4 \end{array} \right]\][/tex]

[tex]R = \[ \left[ \begin{array}{c} 0 \\\ 3 \\\ -1 \end{array} \right]\][/tex]

[tex]F \times R = \[ \left[ \begin{array}{c} F_2R_3-F_3R_2 \\\ F_3R_1-F_1R_3 \\\ F_1R_2-F_2R_1 \end{array} \right]\][/tex]

[tex]F \times R = \[ \left[ \begin{array}{c} -12 \\\ -9 \\\ -27 \end{array} \right]\][/tex]

I did this with software. I haven't hand checked it.
 
Last edited:
tiny-tim
Science Advisor
Homework Helper
25,789
249
torque = r cross F

i x j = k
j x k = i
k x i = j
Hi Nanuven! :smile:

Also ii = jj = kk = 0.

Treat cross products just the same as any product.

For example:

(ai + bj)(cj + dk)

= acij + adik + bcjj + bdjk

= ack - adj + bdi. :smile:

(btw, use the B tag to do bold … if that doesn't work, type [noparse] before and after.[/noparse])
 

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