1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Torque about a point given coordinates in three dimensions

  1. Aug 30, 2014 #1

    squelch

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    Let [itex]\vec{F}=2\hat{i}-3\hat{j}[/itex] act on an object at point (5,1,3). Find the torque about the point (4,1,0)

    2. Relevant equations

    [itex]\tau = \vec F \times \vec r[/itex]

    3. 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 [itex]\vec{r}=-\hat{i}-3\hat{k}[/itex]

    Because [itex]\tau=\vec{F}\times\vec{r}[/itex]:
    [tex]\tau = \left[ {\begin{array}{*{20}{c}}
    2\\
    { - 3}\\
    1
    \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
    { - 1}\\
    0\\
    { - 3}
    \end{array}} \right] = 6\hat i + 5\hat j - \hat k[/tex]
     
  2. jcsd
  3. Aug 30, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I have some notes for you:
    Conceptual niggle: surely the object may be large enough to encompass both points?

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

    $$\vec \tau = \vec r \times \vec F$$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
     
    Last edited: Aug 30, 2014
  4. Aug 30, 2014 #3

    squelch

    User Avatar
    Gold Member

    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 [itex]\vec{r}[/itex] 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 :).
     
  5. Aug 30, 2014 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
  6. Aug 30, 2014 #5

    squelch

    User Avatar
    Gold Member

    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 [itex]\vec{F}=2\hat{i}-3\hat{j}[/itex] acting on an object at point B, (5,1,3). Then [itex]\vec{r}=\vec{AB}=\hat{i}+3\hat{k}[/itex].

    Therefore,
    [tex]\tau = \vec r \times \vec F = \left| {\begin{array}{*{20}{c}}
    {\hat i}&{\hat j}&{\hat k}\\
    1&0&3\\
    2&{ - 3}&0
    \end{array}} \right| = 9\hat i - 5\hat j - 5\hat k[/tex]
     
  7. Aug 30, 2014 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I get different coefficients for j and k.
     
  8. Aug 30, 2014 #7

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    And you are correct. Squelch, you made two small mistakes, probably due to inattention.
     
  9. Aug 30, 2014 #8

    squelch

    User Avatar
    Gold Member

    ImageUploadedByPhysics Forums1409460919.903515.jpg

    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
     
    Last edited: Aug 31, 2014
  10. Aug 31, 2014 #9

    squelch

    User Avatar
    Gold Member

    I tried a different method (taking determinant) and got 9i+6j-3k. Does that match up better?
     
  11. Aug 31, 2014 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

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

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Torque about a point given coordinates in three dimensions
Loading...