Torque homework help

  • Thread starter gvcalamike
  • Start date
  • #1
Force F = (4.59 N)i - (6.29 N)k acts on a pebble with position vector r = (3.46 m)j - (4.51 m)k, relative to the origin. What is the resulting torque acting on the pebble about (a) the origin and (b) a point with coordinates (3.16 m, 0, -4.97 m)?


*The force F is a vector, as is r, I just don't know how to get the vector symbol above it. i, j, k are "i hat, j hat, k hat, I don't know how to get the symbol above those either. Sorry, only my second post.

Attempt at a solution:

I have no idea where to begin. I think the answer will be the cross product of r x F, but our book doesn't give a good example of cross products. Wouldn't you shift the force vector so that the tail is at the origin O?
 

Answers and Replies

  • #2
jhae2.718
Gold Member
1,170
20


[itex]\mathbf{\tau} = \mathbf{r} \times \mathbf{F}[/tex]
You can compute the cross product in terms of components.

For two vectors [itex]\mathbf{a} = a_x \hat{\imath} + a_y \hat{\jmath}[/itex] and [itex]\mathbf{b} = b_x \hat{\imath} + b_y \hat{\jmath}[/itex], [itex]\mathbf{a} \times \mathbf{b} = (a_xb_y-a_yb_x) \hat{k}[/itex].

You can get this by FOIL-ing the terms or writing the cross product as a 3 x 3 matrix and taking the determinant:
[tex]
\mathbf{a} \times \mathbf{b} = \begin{vmatrix}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
a_x & a_y & 0 \\
b_x & b_y & 0
\end{vmatrix}
[/tex]

For part b, find the new displacement vector from the point to the point defined by r.
 
  • #3


Thanks! I got it.
 

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