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Cross Product

  1. Feb 17, 2006 #1
    Three forces with magnitudes [tex]F_a, F_b, F_c [/tex] act on a point mass, pulling in unit directions a, b, c, respectively. Thr forces are in 'equilibrium' which means that

    [tex]F_aa + F_bb + F_cc = 0[/tex]​

    By taking the cross product with a, show that

    [tex]F_b(a \times b) = F_c(c \times a)[/tex]​

    and find two similar equations (involving [tex]F_a[/tex] and [tex]F_c[/tex], and [tex]F_a[/tex] and [tex]F_b[/tex], respectively).

    My Work

    I'm quite stuck here, i dont think i understand fully. The only mere working i can muster is:

    [tex] a \times (a \times b) = (a.b)a - (a.a)c [/tex]

    [tex] a \times (c \times a) = (a.a)c - (a.c)a [/tex]

    Can anyone help me out?:uhh:
     
    Last edited: Feb 17, 2006
  2. jcsd
  3. Feb 17, 2006 #2
    That is the wrong approach. It is much simpler. What can you say about this crossproduct: [tex]\hat{u}\times\hat{u}=?[/tex] Also, remember the anticommutation of a crossproduct.
     
  4. Feb 17, 2006 #3
    Is [tex]\hat{u}\times\hat{u}=0?[/tex]

    im not sure what the anticommutation of a crossproduct is:frown: unless its the dot product. Is it something to do with that in each instance the a's will "cross out" and become zero, leaving something more useful? Sorry, i'm really quite lost.
     
  5. Feb 17, 2006 #4
    Ok sorry, should have expressed it differently. By the anticommutation I meant this: [tex]\hat{u}\times\hat{v}=-\hat{v}\times\hat{u}[/tex]. Surely this is something you've had in class, right?
     
  6. Feb 17, 2006 #5

    0rthodontist

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    Yes, u x u = 0.

    When the question says, "simplify by taking the cross product with a," it means just take the cross product with a of each side of your original equation.
     
  7. Feb 17, 2006 #6

    HallsofIvy

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    [tex]F_aa + F_bb + F_cc = 0[/tex]
    so
    [tex] a \times (F_aa+ F_bb+ F_cc)= F_a a \times a+F_b a \times b+ F_c a \times c= 0[/tex]
    [tex]= F_b a \times b+ F_c a \times c= 0[/tex]
    Oh, and remember that a x c= - c x a.
     
  8. Feb 21, 2006 #7
    Damn, i replied to this last week sometime but it seems to not have worked, gutted! Took ages to do, il try remmeber it...

    I see now that you've to take the cross product with the
    [tex]F_aa + F_bb + F_cc[/tex] That helps alot and was all i really needed to be shown. So continuing from HallsofIvy (even tho its prety much done [tanks very mucg btw :smile: ] ) you get

    [tex] F_b(a \times b) - F_c(c \times a) = 0 so F_b(a \times b) = F_c(c \times a) [/tex]

    For the 2 similar equations i got:

    [tex] b \times (F_aa + F_bb + F_cc) = F_a(b \tiems a) + F_b(b \times b) + F_c(b \times c) = 0 \Rightarrow F_a(b \times a) = F_c(c \times b) [/tex]

    and finally,

    [tex] F_a(c \times a) = F_b(b \times c) [/tex]

    All seem right?

    Next the question asks to use one or more of these equations to show that the directions of the three forces must all lie in the same plane.

    Also, explain why the magnitude of each force is proportional to the sine of the angle between the directions of the other two forces.

    For the 2nd part im guessing its using the fact that [tex] a \times b = [|a||b|\sin\theta]n [/tex]

    But for the first part i can't see how to use those equations. Can anyone help me (again :redface: ) ?
     
  9. Feb 21, 2006 #8

    matt grime

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    Assume for now that all of the F_a etc are nonzero, then axb, axc and bxc are all multiples of each other. What does that tell you about a,b, and c? If any one of the F_. are zero then you can do the same kind of analysis, but it just needs you to keep track of things a little more.
     
  10. Feb 21, 2006 #9
    Ok, but i don't see how they are multiples of each other. Assuming they are, i still can;t make the connection between a b and c. Does it mean they are all equal?
     
  11. Feb 21, 2006 #10

    matt grime

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    They are certainly not equal, and if you can't relate what you've done to this then I don't know what else to suggest. You have shown that


    [tex]F_b(a \times b) = F_c(c \times a) [/tex]


    and

    [tex] F_a(c \times a) = F_b(b \times c) [/tex]

    I don't know what else to point out that *you* have written.

    I just cut and pasted these from a post *you* wrote.....
     
  12. Feb 21, 2006 #11
    Hmm, i know your trying to make sure i do the work which i deffinitely agree is a good thing, i learn a lot better plus after all its my work. Could i ask, for a normal set of equations, would i show them all to be in the same plane by proving a common point is present? For here i see that an a x c is equal to both a b x c and an a x b for different F_. is this like a common point?
     
  13. Feb 21, 2006 #12

    matt grime

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    common points? i don't even know what those are nor what they have to do with anything in this question.

    axb is a multiple of axc, and axb is a multiple of bxc, and bxc is a multiple of axc. just work with them...

    if uxv=uxw then ux(v-w)=0

    already the answer has almost been explained before this.
     
  14. Feb 21, 2006 #13
    Ok, got it. Taking the scaler triple product with [tex]F_b(a \times b) = F_c(c \times a) [/tex] gives

    [tex] F_b(a \times b).c = F_c(c \times a).c \rightarrow F_b[c,a,b] = F_c[(c \times c).a] ( noting c \times c = 0) \Rightarrow F_b[a,b,c] = 0 [/tex] Therefore a,b,c = 0 and so a,b,c are coplanar
     
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