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Dividing vectors?

  1. Aug 16, 2015 #1
    I have a question. Given a vector equation such as F = ma, how can we obtain a general expression for m, the mass? If the equation was scalar, this could easily be done by dividing F by a; however, we are dealing with vectors, and, to my knowledge, a vector divided by another vector is not defined in vector algebra. Therefore, how can we obtain a general expression for m?
  2. jcsd
  3. Aug 16, 2015 #2
    You could always just write [tex]\mathbf{F}=m \mathbf{a} \implies \mathbf{F} \cdot \mathbf{a} = m \mathbf{a} \cdot \mathbf{a} \implies \mathbf{F} \cdot \mathbf{a} = m a^2 \implies m = \frac{\mathbf{F} \cdot \mathbf{a}}{a^2}.[/tex]

    But the acceleration and the net force are always in the same direction. So the dot product is just the scalar product of the magnitudes.
    Last edited: Aug 16, 2015
  4. Aug 16, 2015 #3
    Take any component of F and a and you have ## F_{x_i} = m a_{x_i} ##.
  5. Aug 16, 2015 #4


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    Either pick a non-zero component or, better yet, divide the magnitudes. As long as F and a are collinear (as they must be), this will give the right answer.
  6. Aug 16, 2015 #5
    For the directions of ##F## and ##a## are the same, we can get
    $$\mathbf{F}=m\mathbf{a}\Rightarrow F\hat{e}=ma\hat{e}\Rightarrow F=ma \Rightarrow m=\frac{F}{a}.$$
  7. Aug 18, 2015 #6


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    That last would be better written [itex]\frac{|F|}{|a|}[/itex]
  8. Aug 18, 2015 #7
    Sorry I used ##\mathbf{F}## as the vector force and ##F## as the magnitude of the force.
  9. Aug 18, 2015 #8


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    Oh, I see. So you were right all along!
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