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System of two forces

  1. Aug 13, 2012 #1
    1. The problem statement, all variables and given/known data


    The system of two forces {F 1, F 2} is given by
    F 1 = 2 i + 2 j - k acting at (3, 5, 1)
    F 2 = 5 i - j + 3 k acting at (13, 3, 8)
    (i) Find the moment of the system about some point (a, b, c)
    (ii) Hence, or otherwise, show that the system has moment zero about any
    point of the form (a, (a/3) - 2, 2a/3)
    (iii) Determine a third force F 3 and a point on its line of action such that the
    system of forces {F1,F2,F3} is in equilibrium, showing clearly why the
    system is in equilibrium.
    (You may assume that the moment of a system of forces with zero resultant is
    constant.)

    2. Relevant equations



    3. The attempt at a solution

    I did a cross product of the forces and the points and got 7i +j -3k and i -j -2k.

    Then did a cross product with a,b,c and got stuck from there as I couldnt get any answer to work.
     
    Last edited: Aug 13, 2012
  2. jcsd
  3. Aug 13, 2012 #2

    Simon Bridge

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    what was your reasoning behind finding the cross product of the forces? What physical property does that correspond to?

    [itex]\vec{F_1} = 2\hat{\imath} + 2\hat{\jmath}\hat{k}[/itex] makes no sense. [itex] \hat{\jmath} \cdot \hat{k} = 0[/itex] and [itex]\hat{\jmath} \times \hat{k} = \hat{\imath}[/itex] ... similarly for [itex]\vec{F_2}[/itex]
    ... assuming by i.j,k you mean the principle unit vectors.


    What is the definition of "moment" in this context?
     
    Last edited: Aug 13, 2012
  4. Aug 13, 2012 #3
    I just thought it might work because a lot of questions like this require a cross product at some stage. I'm clutching at straws with this question, I'm really lost.
     
  5. Aug 13, 2012 #4

    Simon Bridge

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    No worries - we all get lost sometimes. Please have a go addressing the problem with the forces not making sense and answering the second question.
     
  6. Aug 13, 2012 #5
    it should be F1 = -2i + 2j -k, F2 = 5i -j + 3k

    As for the moment defintion: a measure of the turning effect produced by the system of forces?
     
  7. Aug 13, 2012 #6

    Simon Bridge

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    Cool! Just a typo on the forces.
    Now - what I was after with the definition was the mathematical formula for calculating it - do you have that handy in your notes?

    (Physicists and engineers often use math as a language that we express our ideas in.)
     
  8. Aug 13, 2012 #7
    oh sorry :/... I know there is m = fd where f is force and d is perp distance from pivot. I don't know if this will work here though
     
  9. Aug 13, 2012 #8

    Simon Bridge

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    Good - in physics we use the distance to the pivot times the component of the force perpendicular to the line from the pivot to the force. Moment is a pseudo-vector - so it has a direction - the direction it tries to turn in.

    The distance from the pivot to the force is called the moment arm.
    You know the position of the pivot.
    You know the position of each force.
    The distance between these two points is the length of the vector pointing from the pivot to the force - do you know how to find that?

    Because, if you have that vector (from the pivot, to the force), call it [itex]\vec{r}[/itex] then you can find the moment very easily:[tex]\vec{\tau} = \vec{r}\times\vec{F}[/tex]... here I've used a Greek letter [itex]\tau[/itex] (tau) for the "moment" - which is also called a torque.
    http://en.wikipedia.org/wiki/Torque

    That is where your cross products come in.
     
    Last edited: Aug 13, 2012
  10. Aug 13, 2012 #9
    "The distance between these two points is the length of the vector pointing from the pivot to the force - do you know how to find that?"

    I don't know how to find this no.

    But I do know how to get the rest now thanks, just lacking that last bit.
     
  11. Aug 13, 2012 #10

    Simon Bridge

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    OK - the coordinate of a point is just a vector from the origin to that point.
    So the vector to the pivot would be [itex]\vec{r_0}=(a,b,c)[/itex], the vector pointing to force-1 would be [itex]\vec{r_1} = (3,5,1)[/itex] and the vector pointing to force-2 is [itex]\vec{r_2}=(\cdots )[/itex] - you do that one ;)

    The moment arm is the vector from the pivot to the force ... in general, a vector from point A and point B is given by: [itex]\vec{r_{AB}}=\vec{r_B}-\vec{r_A}[/itex]. Avector from point B to point A (the opposite direction) is [itex]\vec{r_{BA}}=\vec{r_A}-\vec{r_B}[/itex].

    So it follows that the moment arm from the pivot to the first force must be: [itex]\vec{r_{01}}=\vec{r_1}-\vec{r_0} = (3-a,5-b, 1-c)[/itex]

    And the moment arm from the pivot to the second force must be: [itex]\vec{r_{02}}=(\cdots)[/itex] ... you can do that one too ;)

    Give it a go.

    After this homework - you really need to bone up on coordinate geometry.
    Meantime - it's 3:30am here ... I'll be quitting soon. You seem to know how to do cross products so you should be able to finish from here :)
     
    Last edited: Aug 13, 2012
  12. Aug 13, 2012 #11

    Simon Bridge

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    I'm back - how did you get on?
     
  13. Aug 15, 2012 #12
    thanks a million for that helped me out a lot :)
     
  14. Aug 15, 2012 #13

    Simon Bridge

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    Cool - you are probably more used to the coordinate geometry in 2D. Getting to the third dimension just takes a few tweaks ;)
     
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