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F=ma and Vector Division

  1. Oct 12, 2011 #1
    Hi, I have seen a discussion similar to the question I am asking right now on the forum before, but I did not understand it and do not know if a conclusion was reached.

    In F = ma, F, or force net, is a vector and acceleration is a vector. When solving for m, or m = F/a, you are dividing two vectors. Many websites have told me that vector division is "undefined." Can someone please explain thoroughly but simply how this makes sense and how to approach my situation with m = F/a.
     
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  3. Oct 12, 2011 #2

    Doc Al

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    The vectors F and a have the same direction. Just divide their magnitudes.
     
  4. Oct 12, 2011 #3
    True. It gets a little trickier, though, when you consider the analogous law about angular momentum, T = dL/dt = Idω/dt.... ;-)

    BBB
     
  5. Oct 12, 2011 #4
    Thank you both for your responses. So if they are both in the same direction, you can just divide their magnitudes? What about the direction? Is there another way to look at this problem?
     
  6. Oct 12, 2011 #5

    jambaugh

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    As was mentioned, the angular momentum analogue is different in that the torque and angular accelerations will not be in the same direction because the "mass" here is a moment of inertia tensor (think matrix).

    Ultimately one solves for the inertia term by considering multiple cases of force and accelerations. Ultimately one is looking for the inertia function which maps acceleration to force. One must plug in many values to see if it is just multiplication by a constant, or by a constant matrix, or if it depends on other variables such as position and velocity.
     
  7. Oct 12, 2011 #6
    Thanks again. I appreciate the help but I am only in AP Physics B, so I know some physics but I'm not super deep yet. I want to get there at some point in my life, but is there any way to explain this complex idea with a bit more simplicity while still including detail?
     
  8. Oct 12, 2011 #7

    olivermsun

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    F = ma.

    Express the vectors F and a in any form you like. m is a scalar, so it can only scale the vector a. If a is expressed as magnitude and direction [itex](r,\theta)[/itex], then the scalar doesn't change the direction. [itex]F = (R, \theta) = ma = m(r,\theta) = (mr,\theta)[/itex] and you can figure it out from there. If on the other hand you express your vectors in cartesian coordinates, for example, [itex](x, y, z)[/itex], then the rules for multiplication apply there as: [itex]F = (X, Y, Z) = ma = m(x,y,z) = (mx, my, mz)[/itex] and again you can figure how to solve it.
     
  9. Oct 14, 2011 #8

    vela

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    If you have [itex]\vec{F} = m\vec{a}[/itex], it follows that [itex]\|\vec{F}\| = \|m\vec{a}\|= m \|\vec{a}\|[/itex] so that
    [tex]m = \frac{\|\vec{F}\|}{\|\vec{a}\|}[/tex]This is what Doc Al means when he says to just divide the two magnitudes. It wouldn't make sense to say
    [tex]m = \frac{\vec{F}}{\vec{a}}[/tex]because there is no such thing as vector division.
     
  10. Oct 14, 2011 #9

    Andy Resnick

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    I wonder if your underlying problem is conceptual (assuming you can 'turn the crank' as Doc Al mentioned). To be sure, the 'turn the crank' process is not so simple, as others point out in regards to torque- there, torque is not a vector and the moment of inertia is not a scalar.

    Leaving aside whether F = ma, F = d(mv)/dt, F/m = a are equivalent, the 'F' in F = ma and 'ma' in F = ma refer to completely different concepts. It's often said that 'ma' is kinematics (that is, measurable properties of a specific object), while 'F' is dynamics (that is, an underlying 'cause' that cannot be measured).

    Confusion regarding the distinction between acceleration (how an object *responds* to the environment) and force (how the object *interacts* with the environment) leads to statements like 'well, since F = ma, then F/m = a, so if m = 0, then a is infinite!'.

    Does that help?
     
  11. Oct 14, 2011 #10

    jambaugh

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    Forget about the physics for a minute.

    Think about say solving for the line through two points.
    If you look at the equation for a line ax + by = c you'd say "I can't solve for a,b, and c by just knowing x and y!" but you can if you have several x's and y's and know the equation must be satisfied for each case.

    Likewise now for F=mA and its more generalized cousins ( F = I(A) ).
    You apply many forces to the object and measure the accelerations and work out the relationship between these and that's the inertia of the object, the relationship between force and acceleration be it linear, angular, or something more complex.
     
  12. Oct 14, 2011 #11
    It sounds like Metaphysics. Whatever it may be, but it has helped a lot.
     
  13. Oct 14, 2011 #12
    In one sense, vector equations are just handy ways of combing several component equations. Thus, F = ma really means:
    Fx = m ax,
    Fy = m ay,
    Fz = m az.

    Now it should be obvious that m = Fx/ax, etc.
     
  14. Oct 14, 2011 #13
    This is a vector problem. You have to realize that the net force and the acceleration always point in the same direction. Therefore, you are allowed to do arithmetic with the magnitudes. In this case, you can divide the magnitudes of the vectors.

    As mention by others, you can also break up the force and acceleration vectors into components and divide those because the x components of all vectors will point along the x directions thus allowing them to used in arithmetic.
     
  15. Oct 15, 2011 #14
    Let's continue this a bit just to prove that you can get m also by dividing the magnitudes ...
    Square each equation and sum it all up:

    Fx2 + Fy2 + Fz2 = m2(ax2+ax2+ax2)

    Another way of writing this is:

    [tex] |\vec{F}|^2 = m^2 |\vec{a}|^2 [/tex]

    m is positive so from here you see that:

    [tex] m = \frac{|\vec{F}|}{|\vec{a}|} [/tex]
     
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