Solving Vector Equation: Obtain m from F=ma

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Discussion Overview

The discussion revolves around the vector equation F = ma and the challenge of isolating the mass m in a vector context. Participants explore various methods to express m while addressing the complexities of vector algebra.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions how to derive a general expression for m from the vector equation F = ma, noting that vector division is not defined.
  • Another participant suggests using the dot product to express m as m = (F · a) / a², assuming that F and a are in the same direction.
  • Several participants propose taking components of the vectors, stating that F_{x_i} = m a_{x_i} can be used to isolate m.
  • Another viewpoint emphasizes that dividing the magnitudes of F and a is valid as long as they are collinear, leading to m = F/a.
  • Clarifications are made regarding the notation used for vector and scalar quantities, with one participant apologizing for potential confusion in their notation.

Areas of Agreement / Disagreement

Participants present multiple methods to express m, indicating a lack of consensus on a single approach. The discussion remains unresolved as different perspectives on handling vector quantities are shared.

Contextual Notes

Participants rely on the assumption that the vectors F and a are collinear, which is not explicitly stated as a given condition in all contributions. The discussion also highlights potential confusion in notation between vector and scalar representations.

Mr Davis 97
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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?
 
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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:
Take any component of F and a and you have ## F_{x_i} = m a_{x_i} ##.
 
MrAnchovy said:
Take any component of F and a and you have ## F_{x_i} = m a_{x_i} ##.
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.
 
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For the directions of ##F## and ##a## are the same, we can get
$$\mathbf{F}=m\mathbf{a}\Rightarrow F\hat{e}=ma\hat{e}\Rightarrow F=ma \Rightarrow m=\frac{F}{a}.$$
 
tommyxu3 said:
For the directions of ##F## and ##a## are the same, we can get
$$\mathbf{F}=m\mathbf{a}\Rightarrow F\hat{e}=ma\hat{e}\Rightarrow F=ma \Rightarrow m=\frac{F}{a}.$$
That last would be better written [itex]\frac{|F|}{|a|}[/itex]
 
Sorry I used ##\mathbf{F}## as the vector force and ##F## as the magnitude of the force.
 
Oh, I see. So you were right all along!
 

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