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Can you do the Cross Product Backwards?

  1. Aug 12, 2012 #1
    I was doing a question in a test involving the system of moments , you are given 2 force vectors and two points respectively . And you must find a third point and force to make the 3 moments in equilibrium . So I did the cross product with the first force vector and its point and then with the second vector and its point and added the two resultants leaving me with a vector (4i,0j,-6k) (for the 3 vectors to be in equilibrium,when added up must equal to zero) . So i now no i need a vector and point when the cross product rule is applied will equal (-4i,0j,6k) can i do the cross product backwards if so how ?
     
  2. jcsd
  3. Aug 12, 2012 #2
    There's no unique answer, so no, you can't create a 1-to-1 backwards correspondance. However, you can use the right hand rule and some assumptions to at least fine one of the infinitely many solutions that could come together to form this torque.

    Take your right hand and point your thumb in the direction of the torque vector <-4, 0, 6>. You can rotate your hand so that your fingers point in any arbitrary direction as long as your thumb is pointed in the direction of the torque. Choose an easy position for your fingers and claim that direction to be the direction of [itex]\vec{r}[/itex] (I chose [itex]\hat{j}[/itex] so that 'r' was simply <0, 1, 0>..) Then solve from your determinant equation of the cross product what 'F' must be according to your choice of 'r' (<0, 1, 0>). Notice that [itex]F_{j}[/itex] will then be unable to produce a torque and so you won't be able to solve for that value of the force - it can be anything you want.

    Really you can make any assumptions you want as long as your right thumb is pointing in the direction of the torque you want when trying to generate possible 'r' and 'F' vectors that could combine to create that torque. You are solving an 3 equations in 6 unknowns... You need to make a lot of assumptions, there are an infinite number of possible answers you could choose between. Choose one that's mathematically simple like what I did.
     
  4. Aug 12, 2012 #3
    even if there were a way to do the cross products backwards, you still wouldn't get what you want because you are looking for two unknowns r and F with only one equation
    r x F = <-4, 0, 6>

    if you provide either r or F, then you might be able to find a way to do the cross product "backwards" by using [itex] \hat{i} \cdot \left( \vec{r} \times \vec{F} \right) = \hat{i} \left( r_{y}F_{z} - r_{z}F_{y} \right)=-4 [/itex], [itex] \hat{j} \cdot \left( \vec{r} \times \vec{F} \right) = \hat{j} \left( r_{z}F_{x} - r_{x}F_{z} \right)= 0 [/itex], and [itex] \hat{k} \cdot \left( \vec{r} \times \vec{F} \right) = \hat{k} \left( r_{x}F_{y} - r_{y}F_{x} \right)= 6 [/itex]

    you end up with three simultaneous equations with three unknowns if you know either F or r (six unknowns if you know neither)
    [itex]r_{y}F_{z}-r_{z}F{y} = -4[/itex]
    [itex]r_{z}F_{x}-r_{x}F{z} = 0 [/itex]
    [itex]r_{x}F_{y}-r_{y}F{x} = 6[/itex]


    so i suggest you choose values for any 3 of the variables [itex]r_{x},r_{y},r_{z},F_{x},F_{y},F_{z}[/itex] (except r =0 and F =0) and solve for the remaining 3

    By doing that, you are sure that r x F = <-4, 0, 6>
     
  5. Aug 13, 2012 #4
    Well you're not completely free to choose just anything for 3 of those 6 variables. There is the restriction that r and F have to exist in a plane normal to <-4, 0, 6> (right hand rule of the cross product). That's why I was telling him to point his thumb in the direction of <-4, 0, 6> and then rotate his hand in such a way so as to line up his fingers with something easy for the direction of r. I was thinking along the same lines as well until I tried r = <1, 1, 1> and saw that it didn't work. r = <0, 1, 0> does work.

    Really in generalized coordinates you have 3 degrees of freedom in the torque and 4 degrees of freedom in what r and F can be (2 different magnitudes, an angular position for the r in the plane normal to the torque, and then finally the angular difference between r and F still confined to this plane.)
     
  6. Aug 13, 2012 #5
    yes, you're right, thank you :smile:

    I agree with the restriction you are suggesting, and I think it can be mathematically stated as follows:

    [itex]<-4,0,6>\cdot\vec{r} = 0[/itex] (r is perpendicular to <-4,0,6>)
    [itex]-4r_{x}+6r_{z}= 0[/itex]
    [itex]2r_{x}-3r_{z}= 0[/itex]

    OR


    [itex]<-4,0,6>\cdot\vec{F} = 0[/itex] (F is perpendicular to <-4,0,6>)
    [itex]-4F_{x}+6F_{z}= 0[/itex]
    [itex]2F_{x}-3F_{z}= 0[/itex]

    either will do. here is my reasoning:
    >Suppose we use the condition [itex]<-4,0,6>\cdot\vec{r} = 0[/itex]

    >Then, we are sure that <-4,0,6> is perpendicular to r.

    >Because the direction of <-4,0,6> is uniquely determined by r and F, F can only be on the same plane as r, and can only be perpendicular to <-4,0,6>. If not, the right hand rule would give a direction different from the direction of <-4,0,6>.

    now we have 4 equations with 6 unknowns, so it is possible to choose the value of any 2 variables and solve for the rest

    using this method, if you choose [itex]r_{x} =0[/itex] and [itex]r_{y} =1[/itex], you will obtain the solution r = <0,1,0> F = <-6, any number here, -4>

    if we choose [itex]r_{y} =0[/itex] , r = <0,0,0>, but we don't want this because this makes the equations inconsistent
     
  7. Aug 13, 2012 #6
    [itex]\int {rdm} = r_c \int {dm}[/itex]
     
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