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Cross product question

  1. Jan 13, 2019 #1
    1. The problem statement, all variables and given/known data
    Force F⃗ =−11j^N is exerted on a particle at r⃗ = (8i^+5j^)m.
    What is the torque on the particle about the origin? Express your answer using two significant figures. Enter coordinates numerically separated by commas.


    2. Relevant equations


    3. The attempt at a solution

    F: 0i, -11j, 0k
    r: 8i, 5j, 0k

    F X r = (0-0)i - (0-0)k + (0 - -88)j
    = 0, 0, 88

    this is wrong because it should be -88, but I dont see why? It was my understanding that I should plug values into formula i - j + k. Thanks for any help.
     
  2. jcsd
  3. Jan 13, 2019 #2

    PeroK

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    What's the definition of torque?
     
  4. Jan 13, 2019 #3

    Ray Vickson

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    In future, please avoid using symbols like j^N; in typewriter math, this means ##j^N##, which is not what you intend. Just use j N, or perhaps enclose the "N" in parentheses, so write either F = - 11j N or F = -11j (N). Better still, use LaTeX to typeset your formulas and equations, so you could write ##F = - 11\, j## N or ##F = -11\, j## (N). You could even make them look like vectors by using a bold font, or an arrow on top, like this: ##\mathbf{F} = -11 \, \mathbf{j}## N or ##\vec{F} = -11 \vec{j}## N.

    Anyway, back to your question. Is torque equal to ##\mathbf{r \times F}## or is it ##\mathbf{F \times r}##?
     
  5. Jan 13, 2019 #4
    Thanks for the replies . Torque is not communicative so torque = r X F
    Its the cross product of the Force and the distance from pivot.
     
  6. Jan 13, 2019 #5
    OK I think i get it now. it should be in form rXF which when you apply the i-j+k gives 0i, 0j, -88k
    thanks for help
     
  7. Jan 13, 2019 #6

    haruspex

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    I am not familiar with such a formula. I assume it is some way to remember how to get the signs right, but if so I dislike it. It loses the symmetry.
    It is really quite simple. The result is the next one around in cyclic order:
    ixj->k
    jxk->i
    kxi->j
    If you have to switch the inputs to get them in cyclic order, switch the sign too:
    jxi->-k
    kxj->-i
    ixk->-j
     
  8. Jan 13, 2019 #7
    Yup thats exactly what it was .. its not a formula, just way I was remembering it because it was confusing me. thanks for the help
     
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